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

0,5

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.

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.

Example

 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.

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}]

Crossrefs

Cf. A000217, A000290, A005449, A147875.

Sequence in context: A110244 A309020 A278495 * A055188 A164885 A084989

Adjacent sequences: A254620 A254621 A254622 * A254624 A254625 A254626

Keyword

nonn

Author

Zhi-Wei Sun, Feb 03 2015

Status

approved