A253187 Number of ordered ways to write n as the sum of a pentagonal number, a second pentagonal number and a generalized decagonal number.

1, 2, 2, 2, 1, 1, 1, 3, 4, 2, 2, 1, 4, 3, 3, 4, 2, 3, 1, 3, 2, 2, 5, 3, 3, 3, 3, 6, 3, 6, 4, 2, 3, 1, 7, 2, 4, 5, 5, 4, 1, 5, 5, 2, 3, 4, 4, 5, 5, 5, 3, 5, 7, 6, 4, 3, 1, 6, 6, 8, 5, 3, 6, 4, 7, 4, 2, 6, 5, 5, 3, 4, 8, 3, 3, 3, 6, 6, 7, 9, 6, 2, 5, 6, 7, 7, 4, 6, 6, 7, 5, 3, 10, 6, 3, 4, 5, 7, 3, 10, 7

Offset

0,2

Comments

Conjecture: a(n) > 0 for all n. Also, for any ordered pair (k,m) among (5,7), (5,9), (5,13), (6,5), (6,7), (7,5), each nonnegative integer n can be written as the sum of a k-gonal number, a second k-gonal number and a generalized m-gonal number.

See also the author's similar conjectures in A254574, A254631, A255916 and the two linked papers.

Links

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000

Zhi-Wei Sun, On universal sums of polygonal numbers, arXiv:0905.0635 [math.NT], 2009-2015.

Zhi-Wei Sun, On universal sums a*x^2+b*y^2+f(z), a*T_x+b*T_y+f(z) and a*T_x+b*y^2+f(z), arXiv:1502.03056 [math.NT], 2015.

Example

a(33) = 1 since 33 = 0*(3*0-1)/2 + 4*(3*4+1)/2 + 1*(4*1+3).
a(56) = 1 since 56 = 4*(3*4-1)/2 + 2*(3*2+1)/2 + 3*(4*3+3).

Mathematica

DQ[n_]:=IntegerQ[Sqrt[16n+9]]
Do[r=0; Do[If[DQ[n-x(3x-1)/2-y(3y+1)/2], r=r+1], {x, 0, (Sqrt[24n+1]+1)/6}, {y, 0, (Sqrt[24(n-x(3x-1)/2)+1]-1)/6}];
Print[n, " ", r]; Continue, {n, 0, 100}]

Crossrefs

Cf. A000326, A000384, A000566, A005449, A014105, A074377, A085787, A147875, A254574, A254631.

Sequence in context: A320278 A302111 A124278 * A139755 A213366 A212648

Adjacent sequences: A253184 A253185 A253186 * A253188 A253189 A253190

Keyword

nonn

Author

Zhi-Wei Sun, Apr 07 2015

Status

approved