OFFSET
0,3
COMMENTS
Conjecture: (i) a(n) > 0 for all n = 0,1,2,...., and a(n) = 1 only for n = 0, 1, 4, 6, 8, 9, 18, 19, 21, 33, 40, 43.
(ii) For every m = 5,6,7,... and each integer d > 1 not divisible by 4, any sufficiently large integer can be written as x + y + z + d*w (or x + y + 2*z + d*w) with x,y,z,w m-gonal numbers.
Part (i) of the conjecture is stronger than the classical result that any natural number is the sum of five pentagonal numbers.
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, A result similar to Lagrange's theorem, J. Number Theory 162(2016), 190-211.
EXAMPLE
a(4) = 1 since 4 = 0*(3*0-1)/2 + 1*(3*1-1)/2 + 1*(3*1-1)/2 + 1*(3*1-1).
a(18) = 1 since 18 = 1*(3*1-1)/2 + 2*(3*2-1)/2 + 3*(3*3-1)/2 + 0*(3*0-1).
a(19) = 1 since 19 = 0*(3*0-1)/2 + 2*(3*2-1)/2 + 3*(3*3-1)/2 + 1*(3*1-1).
a(21) = 1 since 21 = 1*(3*1-1)/2 + 2*(3*2-1)/2 + 2*(3*2-1)/2 + 2*(3*2-1).
a(33) = 1 since 33 = 0*(3*0-1)/2 + 1*(3*1-1)/2 + 4*(3*4-1)/2 + 2*(3*2-1).
a(40) = 1 since 40 = 0*(3*0-1)/2 + 2*(3*2-1)/2 + 5*(3*5-1)/2 + 0*(3*0-1).
a(43) = 1 since n = 1*(3*1-1)/2 + 2*(3*2-1)/2 + 5*(3*5-1)/2 + 1*(3*1-1).
MATHEMATICA
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
pQ[n_]:=pQ[n]=SQ[24n+1]&&(n==0||Mod[Sqrt[24n+1]+1, 6]==0)
Do[r=0; Do[If[pQ[n-x*(3x-1)-y(3y-1)/2-z(3z-1)/2], r=r+1], {x, 0, (Sqrt[12n+1]+1)/6}, {y, 0, (Sqrt[8(n-x*(3x-1))+1]+1)/6}, {z, y, (Sqrt[12(n-x*(3x-1)-y(3y-1)/2)+1]+1)/6}]; Print[n, " ", r]; Continue, {n, 0, 80}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Aug 10 2016
STATUS
approved