OFFSET
0,3
COMMENTS
Conjecture: (i) a(n) > 0 for all n.
(ii) Any nonnegative integer n can be written as w + b*x + c*y + d*z with w,x,y,z pentagonal numbers, provided that (b,c,d) is one of the following 15 triples: (1,1,2), (1,2,2), (1,2,3), (1,2,4), (1,2,5), (1,2,6), (1,3,6), (2,2,4), (2,2,6), (2,3,4), (2,3,5), (2,3,7), (2,4,6), (2,4,7), (2,4,8).
I have shown the following related result: For m > 4 and 0 < a <= b <= c <= d, if every nonnegative integer can be written as a*w + b*x + c*y + d*z with w,x,y,z m-gonal numbers, then either m = 6 and (a,b,c,d) = (1,1,2,4), or m = 5 and a = 1 and (b,c,d) is among the 15 triples listed in part (ii) of the conjecture.
In the preprint arXiv:1608.02022, Xiang-Zi Meng and Zhi-Wei Sun confirmed part (i) of the conjecture, and they also proved that for each triple (b,c,d) = (1,2,2),(1,2,4) any natural number can be written as w + b*x + c*y + d*z with w,x,y,z pentagonal numbers. Zhi-Wei Sun, Aug 09 2016
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Xiang-Zi Meng and Zhi-Wei Sun, Sums of four polygonal numbers with coefficients, arXiv:1608.02022 [math.NT], 2016.
Zhi-Wei Sun, On universal sums of polygonal numbers, arXiv:0905.0635 [math.NT], 2009-2015.
Zhi-Wei Sun, A result similar to Lagrange's theorem, J. Number Theory 162(2016), 190-211.
EXAMPLE
a(65) = 1 since 65 = 1*(2*1-1) + 4*(2*4-1) + 2*2*(2*2-1) + 4*2*(2*2-1) = 1 + 28 + 2*6 + 4*6 with 1,28,6,6 hexagonal numbers.
a(104) = 1 since 104 = 1*(2*1-1) + 7*(2*7-1) + 2*2*(2*2-1) + 4*0*(2*0-1) = 1 + 91 + 2*6 + 4*0 with 1,91,6,0 hexagonal numbers.
MATHEMATICA
H[n_]:=IntegerQ[Sqrt[8n+1]]&&(n==0||Mod[Sqrt[8n+1]+1, 4]==0)
Do[r=0; Do[If[Mod[n-x(2x-1)-y(2y-1)-2z(2z-1), 4]==0&&H[(n-x(2x-1)-y(2y-1)-2z(2z-1))/4], r=r+1], {x, 0, (Sqrt[4n+1]+1)/4}, {y, x, (Sqrt[8(n-x(2x-1))+1]+1)/4},
{z, 0, (Sqrt[4(n-x(2x-1)-y(2y-1))+1]+1)/4}]; Print[n, " ", r]; Continue, {n, 0, 100}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Mar 14 2015
STATUS
approved