|
%I #6 Mar 16 2015 00:41:23
%S 1,1,2,2,1,2,1,3,2,2,2,1,3,3,3,3,2,4,3,2,3,2,4,1,5,4,4,4,3,6,3,4,4,2,
%T 3,3,5,6,4,6,4,5,5,6,4,3,4,6,5,4,6,7,6,5,6,5,4,4,7,7,6,5,7,8,8,4,5,5,
%U 6,4,6,9,8,6,6,9,6,9,8,8,6,6,10,6,7,9,6,8,5,9,6,5,10,8,11,6,7,10,7,9,8
%N Number of ways to write n as w*(3w+1)/2 + x*(3x-1)/2 + y*(3y-1)/2 + z*(3z-1)/2, where w,x,y,z are nonnegative integers with x <= y <= z.
%C Conjecture: a(n) > 0 for all n. In other words, any nonnegative integer n can be expressed as the sum of three pentagonal numbers and a second pentagonal number.
%C See also A255350 for a similar conjecture.
%H Zhi-Wei Sun, <a href="/A256132/b256132.txt">Table of n, a(n) for n = 0..6000</a>
%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1503.03743">A result similar to Lagrange's theorem</a>, arXiv:1503.03743 [math.NT], 2015.
%e a(4) = 1 since 4 = 1*(3*1+1)/2 + 0*(3*0-1)/2 + 1*(3*1-1)/2 + 1*(3*1-1)/2.
%e a(11) = 1 since 11 = 0*(3*0+1)/2 + 1*(3*1-1)/2 + 2*(3*2-1)/2 + 2*(3*2-1)/2.
%e a(23) = 1 since 23 = 0*(3*0+1)/2 + 0*(3*0-1)/2 + 1*(3*1-1)/2 + 4*(3*4-1)/2.
%t GenPen[n_]:=IntegerQ[Sqrt[24n+1]]&&Mod[Sqrt[24n+1],6]==1
%t Do[r=0;Do[If[GenPen[n-x(3x-1)/2-y(3y-1)/2-z(3z-1)/2],r=r+1],{x,0,(Sqrt[8n+1]+1)/6},{y,x,(Sqrt[12(n-x(3x-1)/2)+1]+1)/6},
%t {z,y,(Sqrt[24(n-x(3x-1)/2-y(3y-1)/2)+1]+1)/6}];Print[n," ",r];Continue,{n,0,100}]
%Y Cf. A000326, A005449, A255350.
%K nonn
%O 0,3
%A _Zhi-Wei Sun_, Mar 15 2015
|
