Conference rusure::math

    Every binomial in Z[x] has an irreducible factor with at most 3 terms.
    	-- Davenport, Factorisation of Sparse Polynomials, in Proceedings
    	of Eurocal '83.  Springer-Verlag, 1983.  pp. 214-224.
    
    (I believe a binomial in Z[x] is something of the form ax^b+cx^d, where
    a, b, c, and d are integers and b and d are non-negative and unequal.)
    
    A strict limit is not known for trinomials, but it is at least 8; in
    1981, Brenner showed this with:
    
    	x^14+Ax^2-B = P(x)Q(x),
    
    where
    
    	A = 10307342028165274266525889239823042363346833922876
    	    075828731741952918006405132760000000,
    	B = 12841208948559362541055001023860063131661452126861
    	    6325666077934457967306716740794924967132900000000
    
    and where
    
    	P(x) = x^7+2pax^6+2pbx^5+2pcx^4+2pdx^3+2pex^2+2pfx+pg
    and
    	Q(x) = x^7-2pax^6+2pbx^5-2pcx^4+2pdx^3-2pex^2+2pfx-pg
    
    are irreducible with
    
    	p = 17
    	a = 470820
    	b = 3768415030800
    	c = 44978423066488340100
    	d = 478593017683468165593108000
    	e = 937523138375225928321188658341000
    	f = 123149310992981992534109963118406585650000
    	g = 666582695222076790155887095950281112938435190000
    
    (Somebody want to check this?)

    Let L be the perimeter of some convex set in the plane.  Let A be the
    area contained within the set.  Let r be the radius of the largest
    circle that can be inscribed in the set.  Denote pi with p.

    It was previously known that
    
    	L^2 - 4pA >= 0.
    
    Bonnesen showed:

    	L^2-4pA >= (L-2pr)^2,
    	L^2-4pA >= (A/r-pr)^2, and
    	L^2-4pA >= (L-2A/r)^2.

    Bonnesen derived these independently; they each have different proofs. 
    Observe that the left sides are identical and the rights sides are
    different.  The right sides don't even have the same sets of variables.

    Osserman showed the above three inequalities are algebraically
    equivalent.  Without any use of geometry, just high-school algebra,
    they can be transformed into each other!

    I've entered the derivations below.


    				-- edp
    
    L^2 - 4pA >= (L - 2pr)^2.
    L^2 - 4pA >= L^2 - 4Lpr + 4*p^2*r^2.
    -4pA >= -4Lpr + 4*p^2*r^2.
    -A >= -Lr + pr^2.
    Lr - pr^2 - A >= 0.

    L^2 - 4pA >= (A/r - pr)^2.
    L^2 - 4pA >= A^2/r^2 - 2pA + p^2*r^2.
    L^2 >= A^2/r^2 + 2pA + p^2*r^2.
    L^2*r^2 >= A^2 + 2pAr^2 + p^2*r^4.
    (Lr)^2 >= (A + pr^2)^2.
    Lr >= A + pr^2.
    Lr - pr^2 - A >= 0.

    L^2 - 4pA >= (L - 2A/r)^2.
    L^2 - 4pA >= L^2 - 4LA/r + 4A^2/r^2.
    -4pA >= -4LA/r + 4A^2/r^2.
    -p >= -L/r + A/r^2.
    -pr^2 >= -Lr + A.
    Lr - pr^2 - A >= 0.

    The following inequality is similar to the Bonneson inequality for
    convex sets, but this one applies to triangles and is due to Stan
    Rabinowitz.

    A is the area of the triangle, L is its perimeter, r is the radius of
    the largest circle that can be inscribed in the triangle, and R is the
    radius of the smallest circle that can be circumscribed around the
    triangle.

    Then L^2 - 12 sqrt(3) A >= ~35.098 r (R - 2r)

    where ~35.098 is a root of x^3 - 280 x^2 + 10368 x - 62208 = 0.

    There is equality iff the triangle is equilateral or similar to an
    isosceles triangle with sides of 1, 1, and lambda, where lambda is the
    largest root of 31 x^3 - 28 x^2 - 16 x + 4 = 0.

    Stan wonders if
    
    	1 + 1 / (2 + (1 / (3 + . . . 1/n) ) )
    
    can be expressed in terms of
    
    	1/1 + 1/2 + 1/3 + 1/4 + . . . + 1/n.

    Let F = (1 + 2x - 2x^2 + 4x^3 + 4x4) (1 + 2x4 - 2x8 + 4x^12 - 10x^16 +
    28x20 - 84x^24).
    
    Then F^2 has _fewer_ terms than F.
    
    Reference:  Davenport and Siret and Tournier, _Computer Algebra_,
    Academic Press, London:  1988, page 68.

.5> Then F^2 has _fewer_ terms than F.

    F has 29 terms (it's a 28-degree polynomial), and F^2 has 28 terms.
    Is there such an F with smaller degree?  Or with fewer terms?

.6>>	.5> Then F^2 has _fewer_ terms than F.
>>
>>	    F has 29 terms (it's a 28-degree polynomial), and F^2 has 28 terms.
>>	    Is there such an F with smaller degree?  Or with fewer terms?

	My thought was, for each postive integer m is there a
	polynomial F such that F^2 has at least m fewer terms
	than F does?  That is, instead of looking for a simpler
	F, look for one that produces a larger gap.

	Dan

    .4 sounds suspiciously similar to a problem where Stan wondered
    about the asymptotic limit of a continued fraction...

   If P/Q = 1/(1 + 1/(2 + 1/(3 + ...

   then the convergents satisfy

	P[n] = n P[n-1] + P[n-2]
	Q[n] = n Q[n-1] + Q[n-2]

   while if S/T = 1/1 + 1/2 + 1/3 + ...

   then

	S[n] = n S[n-1] + T[n-1]
	T[n] = n T[n-1] = n!

   Can these recurrences be related to each other in closed form?

    - Jim

    Stan showed a photocopy of a page from a Graph Theory book while
    discussing a problem having to do with constructing minimal graphs of
    identical length straight lines and all nodes having N connections.
    
    Note 900.* in the ROBTOB::BRAIN_BOGGLERS conference discusses this
    topic at some length.
    
    Also, could anyone tell me what the name & author of the book was?  It
    is not visible in the photocopy.
    
    --  Barry

    Re .9:
    
    In other photocopies apparently from the same book, the entire title is
    visible:  _Pearls in Graph Theory_.
    
    
    				-- edp

In case there's another fool like me who wants to verify all this, note the 
following typo:

>    15a+15b, 14a+21b, 21a+23b, 17a+226b, 14a+17b, 13a+23b, 16a+16b,
                                    ***

- perhaps should be 26, not 226.

And if that supposition is correct, there appear to be at least *two* more
typos, although I can't identify them: there is a discrepancy in the sum of 
squares that cannot be accounted for by a single error.

Lynn Yarbrough 

    (Here is a corrected version.  Besides "226b" -> "22b", there was a
    mistake in the first term and another place that had "vb" instead of
    "b".  I don't see a third error in the coefficients.)
    
    Below are two lists of terms.  In each list, take each term, raise it
    to the k-th power, and add the powers.  The results for the two lists
    will be equal when k is an integer from 1 to 13, inclusive.  Stan did
    not say how these were arrived at!
    
    3a+b, 3a+3b, a+4b, 4a+8b, 10a+5b, 6a+6b, 7a+12b, 7a+2b, 10a+6b,
    12a+5b, 6a+7b, 11a+5b, 9a+14b, 15a+8b, 5a+13b, 13a+10b, 9a+16b,
    12a+17b, 9a+11b, 2a+9b, 4a+14b, 8a+17b, 13a+15b, 5a+14b, 12a+20b,
    10a+7b, 14a+10b, 10a+11b, 9a+17b, 16a+19b, 7a+18b, 7a+14b, 9a+13b,
    8a+19b, 12a+21b, 19a+23b, 20a+20b, 6a+9b, 7a+6b, 14a+8b, 17a+15b,
    14a+12b, 18a+10b, 17a+16b, 19a+15b, 19a+11b, 10a+10b, 17a+12b, 16a+18b,
    12a+19b, 14a+9b, 21a+15b, 13a+14b, 18a+12b, 22a+15b, 24a+20b, 16a+22b,
    17a+13b, 13a+19b, 21a+16b, 11a+21b, 15a+24b, 20a+22b, 14a+24b, 16a+23b,
    19a+27b, 20a+23b, 17a+18b, 16a+24b, 19a+17b, 22a+21b, 23a+28b, 25a+25b,
    23a+26b
    
    and
    
    a+5b, 4a+b, 2a+2b, 3a+10b, 7a+5b, 9a+4b, 12a+11b, 12a+7b, 9a+8b,
    9a+3b, 7a+9b, 10a+13b, 13a+6b, 12a+12b, 9a+7b, 5a+6b, 12a+8b, 11a+14b,
    5a+10b, 7a+15b, 16a+11b, 8a+10b, 13a+8b, 19a+16b, 4a+5b, 3a+12b,
    10a+15b, 7a+10b, 6a+16b, 11a+18b, 8a+15b, 16a+12b, 6a+15b, 6a+17b,
    10a+20b, 9a+20b, 14a+23b, 16a+14b, 12a+6b, 17a+9b, 16a+9b, 14a+18b,
    20a+12b, 20a+14b, 10a+17b, 18a+14b, 17a+21b, 20a+13b, 19a+19b, 23a+17b,
    22a+24b, 7a+13b, 13a+21b, 18a+19b, 10a+18b, 15a+11b, 19a+14b, 21a+19b,
    15a+15b, 14a+21b, 21a+23b, 17a+22b, 14a+17b, 13a+23b, 16a+16b,
    19a+20b, 17a+26b, 14a+22b, 17a+25b, 19a+24b, 23a+19b, 24a+27b, 22a+28b,
    25a+24b.

Aha! I assumed that the error in 226 was a repeated 2, not the 6.

    "226b" -> "22b"

So indeed there were still two errors, in the first term and in the '26'
I used.

Here's the result that I forgot to give at my talk
last week:
 
2682440^4 + 15365639^4 + 18796760^4 = 20615673^4
 
discovered by Noam Elkies in 1988 is a disproof
of Euler's conjecture.
 
Later, Frye, found the smaller example:
 
95800^4 + 217519^4 + 414560^4 = 422481^4.

    Here is another item like that of .13.  The equation is true for any
    integer k from 1 to 36, inclusive.
    
    2(3c+4)^k +  34(3c+44)^k +  5952(3c+42)^k +  1152(3c+6)^k + 
    273356(3c+40)^k +  5185920(3c+38)^k +  96460(3c+8)^k + 
    49304794(3c+36)^k +  259570688(3c+34)^k +  787135632(3c+32)^k + 
    1302845184(3c+30)^k +  2777664(3c+10)^k +  688544100(3c+28)^k + 
    36120954(3c+12)^k +  505930230(3c+21)^k +  239120640(3c+14)^k + 
    837053072(3c+16)^k +  2754176400(3c+23)^k +  1447605760(3c+18)^k + 
    2524661700(3c+25)^k +  631065636(3c+20)^k +  271426080(3c+27)^k =
    1(3c+45)^k +  69(3c+5)^k +  560(3c+43)^k +  45882(3c+41)^k + 
    1309800(3c+39)^k +  12392(3c+7)^k +  17298129(3c+37)^k + 
    121331584(3c+35)^k +  578458(3c+9)^k +  484355160(3c+33)^k + 
    1103799392(3c+31)^k +  1211238882(3c+29)^k +  10956144(3c+11)^k + 
    484823136(3c+15)^k +  100706045(3c+13)^k +  1788218880(3c+22)^k + 
    1216893592(3c+17)^k +  3029594040(3c+24)^k +  1302845184(3c+19)^k + 
    1468894080(3c+26)^k.

    Brackets denote subscripts.  E.g., x[1] is x with a "1" subscript.
    
    Euler conjectured:
    
    	x[1]^n + x[2]^n + . . . + x[k]^n = y^n
    
    has no (nontrivial) solutions if k < n.
    
    Lander and Parkin showed in 1966:
    
    	27^5 + 84^5 + 110^5 + 133^5 = 144^5.
    
    (Compare to note .15.)

    Is there a formula for
    
    	sum(k=1 to n) of floor(k*phi)
    
    where phi = (1+sqrt(5))/2 and floor(x) is the greatest integer not
    greater than x?
    
    (Rabinowitz and Zeitlin)
    
    How about
    
    	sum(k=1 to n) of floor(k*sqrt(5))	?

    u[n] = u[n-1] + u[n-2]
    
    u[1] <= u[2], integers
    
    [Definition of a general Fibonacci sequence?  -- edp]
    
    If two Fibonacci sequences intersect in 3 points, then they agree from
    some point on.
    
    S. Stein, the intersection of Fibonacci sequences, 1962, pp. 399-402.

    Given an integer n, find the next larger Fibonacci number.  Express
    your answer as a formula in terms of n.
    
    (Rabinowitz)

    I think this answers the previous problem.
    
    Let a = 1/sqrt(5), b = (1+sqrt(5)/2), d = (1-sqrt(5)/2).
    
    Then the least Fibonacci number greater than n is:
    
    a*b^(floor(ln((n+.8)/a)/ln(b))+1)-a*d^(floor(ln((n+.8)/a)/ln(b))+1)
    
    or
    
    a*(b^m-d^m), where m = 1 + floor( ln((n+.8)/a) / ln(b)).
    
    
    				-- edp

    L[0] = 2.  L[1] = 1.
    
    L[n+2] = L[n+1]+L[n].
    
    n L[n]
    0	 2	 2+2 = 4 = 2^2.
    1	 1
    2	 3	 3-2 = 1 = 1^2.
    3    4
    4	 7	 7+2 = 9 = 3^2.
    5   11
    6   18	18-2 = 16 = 4^2.
    7	29
    8	47	47+2 = 49 = 7^2.
    
    L[2n] + 2*(-1)^n = L[n]^2
    
    or L[2n] is near a square.

    For n>3,
    
    (13 sqrt(5) - 19)/10 * L[2n+1] + 4.4*(-1)^n is near a square.
    
    (Stan)

    G[n+2] = G[n+1] + G[n].
    
    G[1] = 1786 772701 928802 632268 715130 455793.
    G[2] = 1059 683225 053915 111058 165141 686995.
    
    Then { G[i] } contains no primes.
    
    (Ron Graham)
    
    Reference:  Donald Knuth, A Fibonacci-Like Sequence of Composite
    Numbers, Math. Magazine 63 (1990) 21-25.

    Consider
    
    	X[0] = 1, X[1] = 2, X[2] = 3, X[3] = 5,
    
    where X[n+1] = X[n]*X[n+n]/(n+1), giving
    
    	1, 2, 3, 5, 10, 28, 154, 3520, 1551880, 267593772160, . . .
    
    Is the sequence always integral?
    
    No!  X[43] is not an integer.
    
    Richard Guy, The Strong Law of Small Numbers.  AMM 95 (1988) 697-712.

    G[1] and G[2] in note .24 are relatively prime.
    
    
    				-- edp

re .25:
>    Consider
>    	X[0] = 1, X[1] = 2, X[2] = 3, X[3] = 5,
>    where X[n+1] = X[n]*X[n+n]/(n+1), giving
>    	1, 2, 3, 5, 10, 28, 154, 3520, 1551880, 267593772160, . . .

Eh?  I get X[4] = 5.  X[4] can occur in three possible places in the
recurrence, either:

	X[3] = X[2]*X[4]/3	(n=2)
	X[4] = X[3]*X[6]/4	(n=3)
	X[5] = X[4]*X[8]/5	(n=4)

I'll assume the first defines X[4] (and the others define X[6], X[8], etc),
so that 5 = 3*X[4]/3, or X[4] = 5.

In fact, the described sequence continues:

	1, 2, 3, 5, 5, X[5], 4, X[7], X[5], X[9], 24 X[11]/X[5], 7 X[7]/4,
	X[13], 8 X[5]/X[7], X[15], 9 X[9]/X[5], X[17], 240/(X[5] X[9]), X[19],
	...

where the subscripted X[n] are actually free variables.

    If you make the definition:
    
    ... where X[n+1] = X[n]*(X[n]+n)/(n+1), ...
    
    then the values given are correct.
    
    Not that I can now prove that x[43] is not an integer, (although there
    is obviously a straightforward proof :-)
    
    dick
    

>    G[1] and G[2] in note .24 are relatively prime.

That's generally true of Fibonacci-like sequences. Working backwards from any 
point in the sequence, you find that the least member of the sequence is 
the GCD of any adjacent pair of them. Thus if the sequence starts with 1, 
every such pair is relatively prime.

MAPLE factors G(1) readily. G(2) appears to contain one or two large 
primes; I couldn't wait for the factorization to complete.

>>	>    G[1] and G[2] in note .24 are relatively prime.

	I think the point was, that if they weren't relatively
	prime, then it would be "obvious" that no members of the
	sequence were prime (all being multiples of the gcd of
	the first two)

	Dan

    Re .29, .30:
    
    Always, we should expect that G[1]-G[2] is prime, otherwise the
    composite sequence could start earlier.
    
    
    				-- edp

.28>    Not that I can now prove that x[43] is not an integer, (although there
.28>    is obviously a straightforward proof :-)

Yes, there is.  Letting Y[n] be X[n] mod 43, we have:

	Y[0] = 1	Y[1] = 2	Y[2] = 3	Y[3] = 5
	Y[4] = 10	Y[5] = 28	Y[6] = 25 = mod(154,43)

Now Y[6]*(Y[6]+6)/7 = 775/7.  Either X[7] is not an integer (in which case
we now know that the sequence does not produce only integers), or
Y[7] = mod( (775+k*43)/7, 43), where k is an integer that makes 775+k*43 a
multiple of 7 (there is really only one; i.e., (k = 2) mod 7).  So let's take
Y[7] = mod(123,43) = 37, and continue.

							Y[7] = 37
	Y[8] = 10	Y[9] = 20	Y[10] = 15	Y[11] = 38
	...
	Y[40] = 8	Y[41] = 23	Y[42] = 33	Y[43] = ?

For Y[43], we have {33*(33+42)+k*43}/43, which is *never* an integer.
Since Y[43] is not an integer, X[43] is not an integer.  QED

.28>    there is obviously a straightforward proof :-)

The other straightforward proof is to calculate x[43].  This is tedious,
since the number of digits nearly doubles in each step of the recurrence.
I figure X[43] is roughly 5.409162 * 10**178485291567, which is well beyond
a googol; fwiw, X[135] (or so) exceeds a googolplex.

    a[n] = ((34n^3-51n^2+27n-5)*a[n-1] - (m-1)^3*a[n-2])/n^3.
    
    a[0] = 1.  a[1] = 5.
    
    The sequence always yields integers.
    
    Ap�ry showed:
    
    	a[n] = sum(k = 0 to n) C(n,k)^2 * C(n+k,k)^2.
    
    C(m,n) = number of ways to select n things from m without order.
    
    Reference:  Chang, Recurrence Relations and Integer Sequences. 
    Utilitas Math. 32 (1987) 95-99.

    The Somos sequence:
    
    a[0] = a[1] = a[2] = a[3] = a[4] = a[5] = 1.
    
    a[n] = ( a[n-1]*a[n-5] + a[n-2]*a[n-4] + a[n-3]^2 ) / a[n-6].
    
    Is the sequence always integral?
    
    Michael Somos, Problem 1470, Crux Mathematicorum 15 (1989) 208.

>>    The Somos sequence:
>>[...]
>>    Is the sequence always integral?

	I have a truly marvelous proof that every term of the
	sequence is rational...

	Dan

    The Seven Circles Theorem says that if circle is surrounded by six
    circles each tangential to the center circle and each tangential to its
    two neighbors, then the three lines drawn between the three opposite
    pairs of exterior circles intersect in a point (not necessarily the
    center of the center circle).
    
    I'm not sure about the conditions for this.  It applies if no circle is
    contained in any other.  It also seems to apply if three of the
    surrounding circles (every other one) are actually large and oriented
    to contain their neighbors.  I am not sure which other combinations are
    possible.
    
    There is a degenerate case of three surrounding circles containing
    their neighbors -- they degenerate the lines, and a triangle is formed
    around the center circle.  Thus, if a circle is inscribed in a triangle
    and three circles are drawn in the corners of the triangle, tangential
    to the two lines of their corners and to the center circle, the three
    lines formed by drawing lines from a tangent of a corner circle with
    the center circle to the opposing tangent of the triangle with the
    center circle will meet in a point.
    
    Stan raised some questions about where this point was in relation to
    the center of the circle.

    I(r) is a circle with radius r inscribed in a triangle ABC, and the
    circles O1(r1), O2(r2), and O3(r3) respectively touch AB and AC, BA and
    BC, and CA and CB, and all touch I(r) externally.  Show that
    
    	r = sqrt(r1*r2) + sqrt(r2*r3) + sqrt(r3*r1).
    
    O(r) and O'(r') are two non-concentric circles.  Inscribed in the space
    between them is a loop of 14 circles Oi(ri) (i = 1, ..., 14).  Each
    circle touches the two circles O and O', internally and externally,
    respectively, and also the circles on either side of it.  Show that
    
    	1/r1 + 1/r8 = 1/r4 + 1/r11.
    
    The situation is as in the preceding problem, but the loop of contact
    circles contains 6 circles.  Find r5 in terms of r1, r2, and r3. 
    (Answer is below.)
    
    In the ellipse O(a,b) there is inscribed a chain of 10 contact circles
    Ci(ri) (i = 1, 2, ..., 10) each of which touches the ellipse at two
    distinct points.  Show that
    
    	r7(r1+r7) = r4(r4+r10).
    
    r5 = r1*r2*r3 / (r1*r3 + 2*r1*r2 + 2*r2*r3).

    In regard to the Seven Circles Theorem, is there a formula connecting
    the radii of the various circles?

    Gaussian binomial coefficients:
    
     n     (1-q^n)(1-q^(n-1))...(1-q^(n-k+1)
    ( )  = ---------------------------------
     k q      (1-q^k)(1-q^(k-1)...(1-q^1)
    
    The expression on the left-hand side of the equation is an n above a k
    in parentheses with a subscript q.
    
    Give an elementary proof that these polynomials are unimodal
    (coefficients increase then decrease).

    I think I'll skip the patterns from Pascal's triangle.  And Conway's
    prime-producing machine can be found in topic 10.
    
    
    				-- edp

    Here is a 14-line program that computes pi to 1000 decimal places.  It
    runs in 10 seconds on a Mac-IIci.
                                     
    
	integer vect(3350), buffer(201)
	data vect/3350*2/, more/0/
	do 2 n=1,201
		karry = 0
		do 3 l = 3350,1,-1
			num=100000*vect(l)+karry*l
			karry=num/(2*l-1)
 3			vect(l)=num-karry*(2*l-1)
		k=karry/100000
		buffer(n)=more+k
 2		more=karry-k*100000
	write(*,100) buffer
 100	format(1xi1,'.'/(1x,10i5.5))
	end

Can you please explain that format statement ?

Thanks.

/Eric

    Re .42:
    
    It's been a whlie since I did FORTRAN, but I think it says:
    
    	(1xi1,'.'/(1x,10i5.5))
    
    Skip 1 space (1x), print a one-digit integer (i1), a period ('.'), and
    a new-line (/).  Then the last item is a space (1x) followed by ten
    (10) repetitions of five-digit integers (i5) with at least five digits
    printed (.5, to force leading zeroes).
    
    Something in that must force the last specification to be used
    repeatedly -- perhaps the fact that it is surrounded by parentheses and
    is the last item?
    
    
    				-- edp

    Yes, if you reach the end of a FORMAT statement and still have data to
    output, you (a) skip to a new line and (b) go back to the last open
    paren and scan rightwards again.
    
    Why, back in the old days we had nothing BUT FORTRAN and assembler, so
    everybody learned FORTRAN.  We had to write FORMAT statements by
    candlelight, and burned punched cards to keep out the cold...
    
      John

>    Here is a 14-line program that computes pi to 1000 decimal places.  It
>    runs in 10 seconds on a Mac-IIci.

The program runs in 8 sec on my VS3100; MAPLE "evalf(Pi,1001);" takes 1.8 sec.
CPU. Interesting.

Re FORMAT statements: I recall seeing a program several years ago, 
consisting of a WRITE, a FORMAT, and an END, that would reproduce its 
source. That trick is possible in most languages.

Lynn 

If you print a one-digit number, what happens to the other digits ?  Do
they get thrown away, or do they get printed, using the next item in
the format statement ?

(I'm asking, because I thought I could "trivially" translate that program
into C but as you can see I'm getting hung up...)

/Eric

> If you print a one-digit number, what happens to the other digits ?
    
    SS$_NOPARSE, I can't figure out that question
    
    Anyway, you should be able to write a quick program that does what
    you want.  Just remember to give each FORMAT statement a different
    statement number (1-99999).
    
      John

        John, I think he meant "What happens if the number has
        more digits than you allow for in the FORMAT statement?".
        It's been a while since I've read a FORTRAN manual, but
        as I recall it takes factorials and integer square roots
        until the number fits, then prints it.
        
        Dan

  Re .46

  To force printing of leading zeroes in C use a format like "%08d"
  which has a field width of 8 with leading zeroes.  "%8d" pads with
  leading blanks instead.

  Fortran will show stars if the number won't fit.  I forget the way
  to obtain leading zeroes but there is one. (RTFM)  Don't have a clue
  what Dan means, unless the smiley was missing...

  - Jim

        re .-1 unless the smiley was missing...
        
        Oops, I went back and added it to the title. :-)
        
        Dan

what I meant was, suppose you say

	j(1) = 123
	j(2) = 456789222
	write 1, j(1),j(2)
1	format (i1,i6)

I don't know if that format statement is right, but pretend it's whatever
format statement means "one-digit integer", "six-digit integer".

Will our output look like this:

	1   234567

or will it look like this:

	3   789222

Do you understand my question now ?  It's "what happens to the unused digits"?

Thanks.

/Eric

    .51:
    
>Will our output look like this:
>
>	1   234567
>
>or will it look like this:
>
>	3   789222
    
    
    As Jim says, you will see:
    
    	*   ******
    
    For a good example of this, try using the VMS "MONITOR POOL" command
    on a system with a lot of non-paged pool.  (Actually it's in PL/I but
    the effect is the same).
    
    --  Barry

    Gregory's series for pi:
    
    	pi/4 = 1 - 1/3 + 1/5 - 1/7 + 1/9 - 1/11 + 1/13 - ...
    
    converges very slowly.  500,000 terms gives pi correct to only 5
    decimal places.
    
    That is,
    
    	4 * sum(k=1 to 500,000 of (-1)^(k-1)/(2k-1)) =
    		3.141590653589793240462643383269502884197...
    
    The 0 in the sixth decimal place should be a two.  Looking at the
    number again:
    
    		3.141590653589793240462643383269502884197...
                       -          --          -
    
    The underlined digits are wrong.  The others are correct!
    
    Reference:  Borwein et al.  American Mathematical Monthly 96 (1989)
    681-687.

      <<< Note 1315.53 by JARETH::EDP "Always mount a scratch monkey." >>>

>>    		3.141590653589793240462643383269502884197...
>>                     -          --          -

                3.14159265358979323846?????????????

(if memory serves)

    sum(k=1 to n of cos^2r(k*pi/n)) = C(2r,r)*n/4^r if n > r.
    
    Rabinowitz, problem 1463, _Crux Mathematicorum_ 15 (1989) 207.
    
    [I'm not sure if Stan asserted this is correct or presented it as a
    challenge to prove or disprove.  -- edp]
    
    [The "^2r" is an exponent of the cosine.  C(m,n) is the number of
    combinations of m things taken n at a time.  -- edp]

    sum(k=1 to n, restricted to (k,n)=1, of sin^2m(k*pi/n)) =
    C(2m,m)*(phi(n)-mu(n))/4^m.
    
    phi(n) = Euler phi function.
    mu(n) = Moebius function.
    
    All proper divisors of n being > m.
    
    (Rabinowitz)

    tan(3pi/11) + 4 sin(2pi/11) = sqrt(11).
    
    cos(pi/13) + cos(3pi/13) + cos(9pi/13) = (1+sqrt(13))/4.

    If n is odd, then
    
    	product(k=1 to (n-1)/2 of tan(k*pi/n)) = sqrt(n).

    [The next item is a program from a math package; I believe it is an aid
    to testing identities of the sort in .57.  -- edp]
    
    TrigNumPatterns = {
    	Degree   :> Pi/180,
    	x_ == y_ :> x-y,
    	3^(-1/2) :> Sqrt[3]/3,
    	2^(-1/2) :> Sqrt[2]/2,
    	Sqrt[2]  :> w^(1/8) + w^(-1/8),
    	Sqrt[3]  :> w^(1/12) + w^(-1/12),
    	Sqrt[n_Integer?EvenQ]	:> Sqrt[2] Sqrt[n/2],
    	Sqrt[n_Integer?OddQ]	:> Product[Tan[k Pi/n], {k, (n-1)/2 },
    	Tan[x_]  :> Sin[x]/Cos[x],
    	Sin[x_]  :> Cos[Pi/2 - x],
    	Cos[x_]  :> (1/2) (w^(x/(2Pi)) + w^(-x/(2Pi)))
    };
    
    Simp[arg_] :=
    	Block[{eq,n,wform,w,cw,k},
    	eq=arg/.x_==y_->x-y;
    	eq=eq//.TrigNumPatterns;
    	eq=Expand[eq];
    	wform=Numerator[Together[eq]];
    	wform=Expand[wform];;
    	If[wform == 0,Return[True]];
    	n=Apply[LCM,Map[Denominator,Exponent[wform,w,List]]];
    	wform = wform/.w->w^n;
    	cw = Cyclotomic[n,w];
    	wform = PolynomialRemainder[wform, cw, w];
    	If[wform == 0,Return[True]];
    	Return[{wform,w==E^(2I Pi/n)}]
    	]

    The next page of Stan's handout contains one example each of a planar
    regular graph of degree three and degree four.  (I hope I have the
    criteria right, can anybody confirm this?)
    
    A regular graph is one for which each vertex has the same number of
    edges.  A planar graph is one that can be embedded (drawn) in a plane.
    
    The graph shown for degree three is not the smallest.  Finding the
    smallest is left as an exercise for the reader.
    
    It is not known if the graph shown for degree four, Harborth's graph,
    is the smallest.
    
    For degrees greater than four, such graphs do not exist.  The proof is
    left as an exercise.
    
    
    				-- edp

    The remaining pages deal with a peculiar mapping situation.  In the
    famous four-color problem, one attempted to color countries on a map,
    where the countries were adjacent if they shared a common boundary (not
    a single point).
    
    Countries are usually connected.  However, these pages, from _Pearls in
    Graph Theory_, consider collections of disjoint countries.  The
    collections are called pires.  In particular, a collection of m
    disjoint countries is an m-pire.  Two empires are adjacent if they
    share a common edge.
    
    Heawood found a non-symmetric map of twelve mutually adjacent 2-pires. 
    80 years later, Scott Kim found a symmetric map of twelve mutually
    adjacent 2-pires.
    
    Theorem:  Every m-pire map can be colored by 6m colors.
    
    Herbert Taylor's map of 18 mutually adjacent 3-pires is shown. 
    (Almost.  The map shown actually contains 17 3-pires and one 2-pire,
    but the 2-pire touches all other 2-pires, so a third country can be
    added to the map anywhere [not touching the other two countries of the
    2-pire] without destroying any adjacencies.)
    
    Theorem:  For every m > 1, there exists a map in the plane of 6m
    mutually adjacent m-pires.
    
    A symmetric map of 25 mutually adjacent 5-pires is shown.
    
    A map of 30 mutually adjacent 5-pires is shown.
    
    Finally, there is some discussion of multiple-planet maps, in which the
    countries of an empire are on different spheres (equivalent to planes
    for mapping purposes).  The handouts had only one page of this, with an
    illustration of Sulanke's earth-moon map, which contains 11 earth-moon
    2-pires.  It seems each 2-pire has one country on the "earth" and one
    on the "moon".  They are not all mutually adjacent.  I'm not sure of
    all the criteria here.
    
    
    				-- edp

        re .61,
        
>>    Theorem:  Every m-pire map can be colored by 6m colors.
        
>>    Theorem:  For every m > 1, there exists a map in the plane of 6m
>>    mutually adjacent m-pires.
        
        The second explicitly states "in the plane".  The first
        leaves that implicit; it too is talking about planar
        maps.
        
        Dan

    Let a convex polygon, inscribed in a circle, be divided into
    triangles by diagonals from one vertex.  Then the sum of the
    radii of the circles inscribed in these triangles is the same,
    whichever vertex is chosen.

>    The next page of Stan's handout contains one example each of a planar
>    regular graph of degree three and degree four.  (I hope I have the
>    criteria right, can anybody confirm this?)
>    
>    A regular graph is one for which each vertex has the same number of
>    edges.  A planar graph is one that can be embedded (drawn) in a plane.
>    
>    The graph shown for degree three is not the smallest.  Finding the
>    smallest is left as an exercise for the reader.
>    
>    It is not known if the graph shown for degree four, Harborth's graph,
>    is the smallest.
>    
>    For degrees greater than four, such graphs do not exist.  The proof is
>    left as an exercise.
>    
>    
>    				-- edp

	Re: .60.   I don't understand this.   If by "degree" you mean the
"number of edges incident to each vertex", then surely smallest graphs of
degree 3 & 4 are the tetrahedron and octahredron.   There is some criterion
that you're missing, I suspect...

Andrew.

PS, for degree 5, the icosohedron would work.

    It is discussing the existence and if so the smallest size
    only of *planar* regular graphs.
    
    Dan

>    It is discussing the existence and if so the smallest size
>    only of *planar* regular graphs.

	Yes, and all the examples I gave are planar, Dan.   Draw one on a
sphere, puncture a hole somewhere, and str-e-e-e-tch into a plane.

	I'm with edp when he says that the sun-moon graph for 11 2-pires 
doesn't make sense.   (Is that .62 or somewhere around there?   What a messy
note.   We need some discplined moderatorship to untease all the conversations
which have gone on here...)

Andrew.

    re .-1,
    
>	Yes, and all the examples I gave are planar, Dan.   Draw one on a
>sphere, puncture a hole somewhere, and str-e-e-e-tch into a plane.
    
    Oops.
    
    Maybe all of the edges have to be straight? :-)
    
    Dan

RE: .67 (Dan)

>    Maybe all of the edges have to be straight? :-)

    They would be Dan.  You'll have to settle for "all the same length." :-)

				Topher

    Re: .60 as corrected in .68.
    
    	The smallest cubic plane graph with every edge the same length has
    8 vertices. Take two copies of the following graph:
    
    	*
       / \
      /   \
     *-----*
      \   /
       \ /
        *
    
    Then we can position the two shapes so that they do not overlap, and so
    two edges can be drawn to link them. No smaller graph will work.
    
    Regards,
    Andy.

    re .69
    
    I am having difficulty grokking.
    
    How can you position two of the "diamonds" so that they do not overlap
    and yet still join two pairs of corresponding vertices with edges of
    the common edge length? Can you provide a diagram?

Start with the two diamonds side-by-side, so that the point in the middle
is touching.  The connecting lines are horizontal, and the proper length.
Now slide the left diamond up, and the right diamond down a little bit.
(They'll also move closer together a bit, but the sides of the diamonds
slope away enough that this is no problem.)    

    	*--__
       / \   ~*
      /   \  / \
     *-----*/   \
      \   /*-----*
       \ /  \   /
        *--__\ /
             ~*

    Re: .60 as corrected in .68.
    
        Consider the following graph fragment (called the "curve"):
    
    *-----*-----.
     \   / \   /
      \ /   \ /
       *-----*
       |     |
       |     |
       *-----.
    
	Copies of this can be stuck together in all kinds of ways. In
    particular, 12 copies suffice to make a dodecagonal shape which is
    regular of degree 4, and in which every edge has unit length.
    
    	I have no idea whether there is a smaller graph with the property.
    
    Regards,
    Andy.

	As the original posting suggested, there is no quintic plane
'isometric' graph. I'll sketch the proof here.

	Suppose such a graph exists. Fiddling around with Euler's formula, we 
get:

	F_3 = 20 + 2*F_4 + 5*F_5 + 8*F_6 + ...

where F_i is the number of faces of degree i in the graph (i-gons).

	Each triangle is equilateral and so contains 3 angles of 60 degrees.
Call this angle the "critical angle". Five angles meet at each vertex, so they
can't *all* be critical angles. Yet analyzing the formula above, more than 60%
of the angles in the graph are critical ones. So there is a vertex with four
critical angles, and a fifth which is 120 degrees from some other polygon, P.

	P "soaks up" 4 critical angles. But by the formula, its existence will
imply the existence of more triangles, and hence more critical angles. Any
k-gon (k>=7) will imply more critical angles than it can soak up. We need to
study the smaller k-gons more carefully.

	k=4 (two opposed angles of 120 degrees) and k=6 (regular) will imply
exactly as many critical angles as they soak up. Any other configuration, or
k=5, cannot even do as well as these. So there is no solution.

	Note that the results for k=4 & 6 do leave the door open for some
decorative *infinite* quintic plane 'isometric' graphs...

Regards,
Andrew. 

Here's one infinite quintic plane isometric graph:

	    *   *---*---*   *---*---*
	   / \ / \   \ / \ / \   \ /
	  *---*---*---*---*---*---*
	 /     \ / \ /     \ / \ /
	*       *---*       *---*
	 \     / \ / \     / \ / \
	  *---*---*---*---*---*---*
	   \ / \ /   / \ / \ /   / \
	    *   *---*---*   *---*---*
	   / \ / \   \ / \ / \   \ /
	  *---*---*---*---*---*---*
	 /     \ / \ /     \ / \ /
	*       *---*       *---*
	 \     / \ / \     / \ / \
	  *---*---*---*---*---*---*
	   \ / \ /   / \ / \ /   / \
	    *   *---*---*   *---*---*

    I don't think the graph in the previous reply works. There a vertex in
    the middle which only has 4 edges. I'd draw it, but I'm only coming
    from a very slow link.
    
    A

Oops, you're right.  It's this one:

	  *---*---*---*---*---*---*
	   \ / \ /   / \ / \ /   / \
	    *   *---O---*   *---O---*
	   / \ / \   \ / \ / \   \ /
	  *---*---*---*---*---*---*

	So what *infinite* quintic plane isometric graphs are there?

	Let me focus on identifying the repeating patterns, although the
approach that I take will pick up some non-repeating ones as well.

	The repeated pattern can be viewed as a tiling of a torus. The
Euler's formula which addresses this case is similar to the previous one, but
the constant term disappears:

	F_3 = 2*F_4 + 5*F_5 + 8*F_6 ....

	Examining this formula, we can see that at least 60% of the angles are
critical ones. As before, F_k = 0 for k = 5, or k >= 7. Any hexagon present
must be regular, but now any rhombus is legal. At each vertex, exactly three
critical angles must be present.

	Suppose that we have a rhombus which does *not* have angles 60 and 120.
Then at any vertex incident with this rhombus, we must (apart from the
obligatory three critical angles) see another rhombus, so that the degrees
add up to 180. So viewing the tiling as an infinite tiling of the plane, the
rhombus will be part of an infinite chain stretching straight across the plane,
bounded between two chains of equilateral triangles. So all such chains of
rhombi must be parallel.

	Thus we are left with the problem of tiling
		(a) the plane
		(b) the half-plane
		(c) an infinite strip
with the following shapes:
		(i) regular hexagon
		(ii) the exceptional rhombus
		(iii) equilateral triangle.

	This can be regarded as taking a tiling of equilateral triangles, and
then removing certain edges, such that the degree of each edge is 0 or 5.

	A hexagon will set one vertex at degree 0, and establish the degree of
each neighbouring vertex at 5. An exceptional rhombus will establish the 
degree of two vertices at 5. Each vertex must be adjacent to exactly one
hexagon or exceptional rhombus, but not both.

	So our goal is to tile the lattice of the vertices of the equilateral
triangles with the following two shapes:

			* *



                        * *
                       * * *
			* *

	This can be done in lots of ways. Even to tile an infinite strip with
the domino can be done in an infinite number of ways, including an infinite
number of repeating ways. [Except for the case when the infinite strip has
width 1!]

	This characterizes all of the solutions which do not create an excess
of critical angles. A question I am currently unclear about is whether for an
infinite graph, I can create a finite or infinite excess of critical angles. My
intuition says no.

Cheers,
Andrew.

    According to note 900.l in the Brain Bogglers notesfile, there's a
    known example of a quartic graph with 32 vertices (as compared to mine, 
    which had 60 vertices) Does anyone have a picture of this graph, to
    save me the hassle of finding it?
    
    Cheers,
    Andrew.