July 2005 - mathfun@mailman.xmission.com

[math-fun] Applied graph theory
by Michael Kleber 15 Jul '05

15 Jul '05

This is more frustrating than it rationally ought to be: http://home.cwru.edu/~jnt5/Planarity/ --Michael Kleber -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.

2 1

[math-fun] FW: divisibility conjectures
by Schroeppel, Richard 09 Jul '05

09 Jul '05

This seems like it might appeal to some folks here. Rich rcs(a)cs.arizona.edu -----Original Message----- From: Number Theory List on behalf of Don Davis Sent: Fri 7/8/2005 8:38 PM To: NMBRTHRY(a)LISTSERV.NODAK.EDU Subject: divisibility conjectures Some conjectures about divisibility by 2 have arisen in my work in algebraic topology. The first one involves Stirling numbers of the second kind. It would give a nice form for homotopy groups of some important topological spaces. We write nu(-) for the exponent of 2 in an integer. Conj 1: If j \ge n and L is sufficiently large (e.g. L>3n/2), then nu(j! S(2^L + n -1, j)) \ge n-1 + nu([n/2]!). It is not difficult to remove the 2^L and the n-1 on the RHS, obtaining the following conjecture, which implies Conj 1. Conj 2: If j>m, then nu(sum ((j choose 2k) k^m)) \ge nu([(m+1)/2]!). Instead of dealing with expressions k^m, one might prefer to work entirely with binomial coefficients. We have the following conjecture, which implies Conj 2. Our sums are taken over all sensible values of k. Conj 3: nu( sum ((j choose (4k+2)) (k choose i))) \ge nu([j/2]!) - i - nu(i!). This can be interpreted as a coefficient in a power series, of which it appears that not only the coefficient in question is highly divisible, but also all subsequent coefficients. We obtain the following conjecture, which implies Conj 3. Conj 4: nu( coeff( (1+x)^j / (1-x^4)^(i+1), x^k) ) \ge nu([j/2]!) - i - nu(i!) for k > j-4i-4. It is quite dramatic that the coefficient of x^(j-4i-4) will be odd, but then apparently all the rest are highly divisible. It is enough to prove this when j = 4a, and probably when i+1 is a 2-power. Experimenting with related power series, by removing the 4's, we are led to the following conjecture, which is closely related to Conj 4. My feeling is that this last conjecture is one which others might have seen, and that a proof of it would probably lead to a proof of Conj 4. Conj 5: If 2^t > 2d, then nu( coeff( ((1+x)/(1-x))^(2^t) (1+x)^(2d), x^k) ) \ge t + d + nu(d!) + 1 if k > 2d. Just to clarify...the power series in Conj 5 equals (1+x)^(2^t + 2d) / (1-x)^(2^t). If anyone on this listserve can point to any literature related to these conjectures, or offer suggestions toward a proof, I would be most appreciative. If you can prove one, we can discuss an appropriate way to publish it. Don Davis dmd1(a)lehigh.edu http://www.lehigh.edu/~dmd1

1 0

Re: [math-fun] Lattice / finite group question
by dasimov＠earthlink.net 09 Jul '05

09 Jul '05

Gene writes: << No. Inversion through the origin is always a lattice isometry, so the isometry group must have even order. >> Aha -- good point. Are there further restrictions? --Dan

3 2

[math-fun] Self-erasure surviving numbers
by Eric Angelini 08 Jul '05

08 Jul '05

[Eric A.] much too affirmative, wrote a few days ago: > No SESN > 10 begins with "1" -- see why? > No SESN > 299 begins with "2", etc. > > The sequence is finite, thus. > > Last term? ... [Michael A. Cleverly] objects that 114 *is* a SESN: 11(4)1(1)4(1)14(1)1(4)114(1) ^^ ^ ^ ^^ !!! So I might be wrong about the finitude of the sequence, sorry. And I'll have to add a few more terms to the seq. in the OEIS... Best, É.

1 0

[math-fun] Lattice / finite group question
by dasimov＠earthlink.net 07 Jul '05

07 Jul '05

Let G be any finite group. Questions: 1) Is there always some n such that there is some n-dimensional lattice L whose isometry group (keeping the origin fixed) is isomorphic to G ? Call such an isometry group I_0(L). 2) If so, is there always some n and some lattice L that is a (not nec. proper) subgroup of the integer lattice Z^n, such that I_0(L) is isomorphic to G ? ------------------------------------------ Note: It's easy to see that if n = #(G), then the symmetric group S_n is a *subgroup* of I_0(Z^n), and hence, so is G. Close, but not quite what we are looking for. ------------------------------------------ (We define a n-dimensional lattice to be any discrete subgroup -- isomorphic to the group Z^n -- of the abelian group and Euclidean space R^m, necessarily for some m >= n.) --Dan

2 1

[math-fun] "Self-erasure surviving numbers"
by Eric Angelini 07 Jul '05

07 Jul '05

Hello Seqfan and math-fun, Take an integer like 36, for example. Concatenate an infinite amount of copies. You get: 363636363636363636363636363636... - read the leftmost digit -"3"-, - jump *over* 3 digits (to the right), land on (3) and erase it: 3636(3)6363636363636363636363636... ^ - read the leftmost unread digit, jump only visible digits, erase: 3636(3)6363(6)36363636363636363636... ^^ - repeat until you see a substring like this [...(a)36(b)...] [(a) and (b) are erased digits - "36" is the integer you are, testing]: bingo, you have found a "Self-erasure survi- ving number" (SESN): "36" is such a number: 3636(3)63(6)3(6)36(3)(6)3(6)36363636363636... ^^^^ ^^ .. <-- hit This erasing technique gives sometimes quite complicated pat- terns. "16", for instance, is not a SESN -- but it takes a while to see: 16(1)616(1)61(6)1(6)(1)6(1)(6)1(6)1(6)1(6)1(6)(1)6(1)616(1)61(6)1(6) ^^ ^^^ ^^ ^ ^ ^ ^ ^ ^ ^ ^^^ ^^ ^ |_______________________________________________| recurrent pattern The first SESN I have found by hand are: 0 1 2 3 4 5 6 7 8 9 10 20 23 24 25 26 27 28 29 30 32 36 37 38 39 40 42 ... [BTW, reading "0" means erasing the closest visible digit immediately to the right] No SESN > 10 begins with "1" -- see why? No SESN > 29 begins with "2", etc. The sequence is finite, thus. Last term? And what about recurrent patterns: do all integers behave like that? Could some strings be definitely "chaotic"? I guess no... Best, É.

5 4

[math-fun] "Self-erasure surviving numbers" (2)
by Eric Angelini 07 Jul '05

07 Jul '05

> No SESN > 29 begins with "2", etc. ... Oooops, mean: No SESN > 299 begins with "2", etc. Best, É.

1 0

[math-fun] nonstandard soccer ball
by R. William Gosper 06 Jul '05

06 Jul '05

Small rhombicosidodecahedron spotted at Wal-Mart. I couldn't resist firing up the pyramid solid angle formula, 1 - cos(v) 1 - --------------------------------- 2 %pi sqrt(cos(v) - cos(-----)) n 2 (------------------------- + 1) %pi sqrt(2) sin(---) n 2 n acos(-------------------------------------) v cos(-) 2 (all n vertex angles = v) to confirm Weisstein's radius/edge ratio: r sqrt(4 sqrt(5) + 11) - = -------------------- ~ 2.23295050941569. e 2 Much smaller than I expected, given 62 faces. Confession: although very easy to set up, the formula was too big to clear of radicals, and I resorted to Rich's numeric lattice reducer to get the quartic. The solid angle of a base corner is surprisingly nicer: v sin(-) 2 %pi -------- + sin(---) %pi n sin(---) n 2 acos(-------------------). v sin(-) + 1 2 Adding two squares, a triangle, and a pentagon gives the solid angle of the rhombi...dron vertex: 4 sqrt(5) - 5 acos(-------------) + %pi ~ 4.44630893315853 15 Which (finite) semiregular polyhedron have vertex solid angles a rational multiple of pi? So far, I have found only the regular polygonal prisms, since solid_angle(pi/2,pi/2,a) = a. There seems to be no 3D analog of the polygon sum-of-vertex-angles formula. Except for the prisms and the five regular solids, even the central angles subtended by the faces are rarely nice. The solid angle of the vertex of an ngonal antiprism is %pi %pi sqrt(3) tan(---) (4 cos(---) + 5) 2 n n 2 acos(---------------------------------), 9 a derivation which took inexcusably long. E.g., n=3 sez the vertex solid angle of an octahedron is 2 acos 7/9. --rwg

1 1

[math-fun] pretty series for log
by Joerg Arndt 05 Jul '05

05 Jul '05

Recently I stepped on a series similar to 1 / 1+x \ 1 3 1 5 1 2k+1 - log | --- |=x+- x +- x +...+---- x +... 2 \ 1-x / 3 5 2k+1 The relation is / 2 \ 2 3 5 6 1 | 1+3x+3x | 3 5 3 7 3 11 3 13 - log | -------- |=x - -- x - -- x + -- x + -- x +...= 6 | 2 | 5 7 11 13 - \ 1-3x+3x / / 6k+0 6k+2 6k+3 6k+5 \ \~~ oo| 3 12k+1 3 12k+5 3 12k+7 3 12k+11 | = > | + ----- x - ----- x - ----- x + ------ x | /__ k=0\ 12k+1 12k+5 12k+7 12k+11 / Alternative form: / +-+ 2 \ 5 7 11 13 1 | 1+\|3 x+x | x x x x ------ log | ----------- |=x - -- - -- + --- + --- +...= +-+ | +-+ 2 | 5 7 11 13 - 2 \|3 \ 1-\|3 x+x / / 12k+1 12k+5 12k+7 12k+11 \ \~~ oo| x x x x | = > | + ------ - ------ - ------ + ------- | /__ k=0\ 12k+1 12k+5 12k+7 12k+11 / Seeking for relations of that type, I spotted the following, given in D.H.Bailey, R.E.Crandall: "On the Random Character of Fundamental Constant Expansions": / 2 \ 2 4 5 / 3k+1 3k+2 \ 1 | 1+x+x | x x x \~~ oo| x x | - log | ------- |=x + -- + -- + -- +... = > | + ----- + ----- | 3 | 2 | 2 4 5 - /__ k=0\ 3k+1 3k+2 / \ 1-2x+x / I could not any further relation of that type. -- p=2^q-1 prime <== q>2, cosh(2^(q-2)*log(2+sqrt(3)))%p=0 Life is hard and then you die.

1 0

[math-fun] Self-powers numbers (SPN)
by Eric Angelini 05 Jul '05

05 Jul '05

Hello Math-fun and Seqfan, N=325648 is interesting if read so: "3" means "a cube is visible in N" (yes, it is "8" -- 8=2*2*2) "2" means "a square is visible in N" (yes, it is "25" -- 25=5*5) (4 is ok too, being 2*2) "5" means "a power 5 is visible in N" (yes, it is "32" = 2*2*2*2*2) "6" means "a power 6 is visible in N" (yes, it is "64" = 2*2*2*2*2*2) "4" means "a power 4 is visible in N" (yes, it is "256" = 4*4*4*4*) "8" means "a power 8 is visible in N" (yes, it is "256" =2*2*2*2*2*2*2*2) N=832564 is a SPN too, of course. ["visible" means "as a whole": "25" is NOT visible in 235] Question: Can someone compute all such SPN _which don't include any 0's or 1's_ ? This restriction applies because 0^a=0 and 1^b=1, which brings a lot of unwan- ted SPN like 117 or 308: "1" means "a power 1 is visible in N" (yes, it is "1" -- 1^1=1) "7" means "a power 7 is visible in N" (yes, it is "1" again -- 1^7=1) or "3" means "a cube is visible in N" (yes, it is "0" -- 0*0*0=0) "0" means "a power 0 is visible in N" (yes, it is "0" again -- any a^0=0) "8" means "a power 8 is visible in N" (yes, it is "0" again and again -- 0^8=0) Best, É.

1 0