OFFSET
1,4
COMMENTS
It is not required that the two dice be identical, it is not required that the entries be bounded by n and we do not ask that the entries be distinct from one another on each cube.
We pretend for the purpose of this sequence that regular n-sided dice exist for all n.
In other words, how many (unordered) pairs of polynomials B(x) = x^b_1 + x^b_2 + ... + x^b_n, C(x) = x^c_1 + x^c_2 + ... + x^c_n, are there with all exponents positive integers, such that B(x)*C(x) = (x+x^2+x^3+...+x^n)^2?
a(n) = 1 means that the only way two n-sided dice can have the same probability distribution as two normal n-sided dice (each side numbered 1 through n) is if they are normal. a(6) = 2 corresponds to normal dice and Sicherman dice (one labeled 1,2,2,3,3,4 and the other 1,3,4,5,6,8). - Charles R Greathouse IV, Jan 19 2017
Records are: 1, 2, 4, 8, 10, 33, 57, 140, 348, 583, 956, 2036, 2393, 3050, ... and they seem to occur at positions given by A033833. - Antti Karttunen, Aug 28 2017
REFERENCES
M. Gardner, "Penrose Tiles to Trapdoor Ciphers", p. 266.
LINKS
Antti Karttunen, Table of n, a(n) for n = 1..239
Donald J. Newman, A Problem Seminar, Springer; see Problem #88.
EXAMPLE
The first nontrivial example is for n=4: {1,2,2,3} and {1,3,3,5} together have the same sum probabilities as a pair of {1,2,3,4}. That is, (x + 2x^2 + x^3)(x + 2x^3 + x^5)=(x + x^2 + x^3 + x^4)^2.
PROG
(PARI) ok(p, e, n)=my(v=Vec(factorback(p, e))); vecmin(v)>=0 && vecsum(v)==n
a(n)=if(n<4, return(1)); my(x='x, f=factor((x^n-1)/(x-1)), p=f[, 1], e=2*f[, 2]~, u=vector(#e, i, [0, e[i]]), s, t); t=vecmax(e); for(i=1, #e, if(e[i]==t, u[i][2]\=2; break)); forvec(v=u, t=e-v; if(cmp(v, t)<=0 && ok(p, v, n) && ok(p, t, n), s++)); s \\ Charles R Greathouse IV, Jan 19 2017
CROSSREFS
KEYWORD
nonn
AUTHOR
N. J. A. Sloane, Nov 17 2005
EXTENSIONS
Edited and extended by Matthew Conroy, Jan 16 2006
Correction to some terms, thanks to Adam Chalcraft. - Matthew Conroy, Apr 04 2010
STATUS
approved