A111588 Crazy Dice: number of ways to design a pair of n-sided dice with positive integers on their faces, so that the sums when they are tossed occur with the same probabilities as if a pair of standard n-sided dice were tossed.

1, 1, 1, 2, 1, 2, 1, 4, 2, 2, 1, 8, 1, 2, 2, 10, 1, 8, 1, 8, 2, 2, 1, 33, 2, 2, 4, 8, 1, 13, 1, 26, 2, 2, 2, 57, 1, 2, 2, 33, 1, 13, 1, 8, 8, 2, 1, 140, 2, 8, 2, 8, 1, 33, 2, 33, 2, 2, 1, 125, 1, 2, 8, 71, 2, 13, 1, 8, 2, 13, 1, 348, 1, 2, 8, 8, 2, 13, 1, 140, 10, 2, 1, 122, 2, 2, 2, 33, 1, 118, 2

Offset: 1

Author

N. J. A. Sloane, Nov 17 2005

Keywords

• nonn

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

Examples

			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.
		

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.

Crossrefs

Cf. A033833.

Programs

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

Extensions

Edited and extended by Matthew Conroy, Jan 16 2006

Correction to some terms, thanks to Adam Chalcraft. - Matthew Conroy, Apr 04 2010