Reminder: The OEIS is hiring a new managing editor, and the application deadline is January 26.
%I #8 Mar 31 2012 14:40:21
%S 1,1,2,3,5,7,9,11,14,16,18,19,20,20,19,18,16,14,11,9,7,5,3,2,1,1,0,0,
%T 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
%U 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0
%N Number of partitions of n with k <= 5 parts and each part p <= 5.
%C There are only 26 nonzero terms.
%C Contribution from _Toby Gottfried_, Feb 19 2009: (Start)
%C a(n) is the number of partitions of n+5 into exactly 5 parts with each part p: 1 <= p <= 6
%C i.e. the number of different ways to get a total of n+5 with 5 (normal, 6-sided) dice in any order (End)
%F G.f.: = 1+z+2*z^2+3*z^3+5*z^4+7*z^5+9*z^6+11*z^7+14*z^8+16*z^9+18*z^10+19*z^11+20*z^12+20*z^13+19*z^14+18*z^15+16*z^16+14*z^17+11*z^18+9*z^19 +7*z^20+5*z^21+3*z^22+2*z^23+z^24+z^25.
%e a(7)=11 because we can write 7=1+2+2+2 or 5+2 or 1+2+4 or 3+4 or 1+3+3 or 1+1+1+1+3 or 1+1+2+3 or 2+2+3 or 1+1+1+2+2 1+1+1+4 or 1+1+5.
%e A total of 8 comes from 1+1+1+1+4, 1+1+1+2+3, 1+1+2+2+2 and a(3) = 3 [8 = 3+5] [From _Toby Gottfried_, Feb 19 2009]
%Y See A102420 for k=5 and p<=5.
%Y Cf. A000041, A102420, A063746.
%Y Contribution from _Toby Gottfried_, Feb 19 2009: (Start)
%Y A102420 has the numbers for 4 dice
%Y A063260 gives the number of permuted rolls of each possible total for any number of dice. (End)
%K easy,nonn
%O 0,3
%A _Thomas Wieder_, Jan 09 2005