A254623 Number of ways to write n as x^2 + y*(3*y+1)/2 + z*(5*z+3)/2 with x,y,z nonnegative integers.

1, 1, 1, 1, 2, 1, 2, 2, 2, 1, 1, 3, 1, 2, 1, 4, 4, 1, 1, 3, 4, 1, 2, 2, 3, 1, 1, 4, 3, 5, 3, 5, 2, 1, 2, 3, 4, 1, 4, 2, 5, 1, 3, 5, 4, 3, 3, 2, 3, 4, 2, 5, 2, 6, 4, 5, 3, 5, 2, 1, 2, 3, 8, 1, 6, 4, 3, 2, 3, 5, 6, 5, 2, 4, 2, 3, 5, 6, 7, 5, 1, 6, 3, 4, 3, 4, 8, 2, 5, 5, 4, 3, 3, 6, 4, 4, 3, 7, 1, 2, 6

Offset: 0

Author

Zhi-Wei Sun, Feb 03 2015

Keywords

• nonn

Comments

Conjecture: (i) a(n) > 0 for all n, and a(n) > 1 for all n > 118.

(ii) For each m = 5,7,8, any nonnegative integer n can be written as the sum of two triangular numbers and a second m-gonal number, where the second m-gonal numbers are given by (m-2)*k*(k+1)/2-k (k = 0,1,...).

(iii) For every m = 5,6,7,9,11, any nonnegative integer n can be written as the sum of a triangular number, a square and a second m-gonal number.

Note that k*(3*k+1)/2 (k = 0,1,...) are second pentagonal numbers and k*(5*k+3)/2 (k = 0,1,...) are second heptagonal numbers. The conjecture has been verified for all n = 0.. 2*10^6.

Examples

			a(41) = 1 since 41 = 1^2 + 5*(3*5+1)/2 + 0*(5*0+3)/2.
a(98) = 1 since 98 = 8^2 + 2*(3*2+1)/2 + 3*(5*3+3)/2.
a(118) = 1 since 118 = 2^2 + 3*(3*3+1)/2 + 6*(5*6+3)/2.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
• Zhi-Wei Sun, On universal sums of polygonal numbers, arXiv:1405.0635 [math.NT], 2009-2015.

Crossrefs

Cf. A000217, A000290, A005449, A147875.

Programs

Mathematica

SQ[n_]:=IntegerQ[Sqrt[n]]
Do[r=0;Do[If[SQ[n-y(3y+1)/2-z(5z+3)/2],r=r+1],{y,0,(Sqrt[24n+1]-1)/6},{z,0,(Sqrt[40(n-y(3y+1)/2)+9]-3)/10}];
Print[n," ",r];Continue,{n,0,100}]