A254661 Number of ways to write n as the sum of a triangular number, an even square and a second pentagonal number.

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

Offset

0,4

Comments

Conjecture: (i) a(n) > 0 for all n. Also, a(n) = 1 only for n = 0, 1, 2, 4, 9, 11, 13, 20, 35, 69, 75, 188.

(ii) For each a = 2,3, any nonnegative integer n can be written as x(x+1)/2 + a*y^2 + z*(3*z+1)/2 with x,y,z nonnegative integers.

Compare part (i) of this conjecture with the conjecture in A160325.

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.

Example

a(20) = 1 since 20 = 1*2/2 + 2^2 + 3*(3*3+1)/2.
a(35) = 1 since 35 = 7*8/2 + 0^2 + 2*(3*2+1)/2.
a(69) = 1 since 69 = 2*3/2 + 8^2 + 1*(3*1+1)/2.
a(75) = 1 since 75 = 9*10/2 + 2^2 + 4*(3*4+1)/2.
a(188) = 1 since 188 = 1*2/2 + 0^2 + 11*(3*11+1)/2.

Mathematica

TQ[n_]:=IntegerQ[Sqrt[8n+1]]
Do[r=0; Do[If[TQ[n-4y^2-z(3z+1)/2], r=r+1], {y, 0, Sqrt[n/4]}, {z, 0, (Sqrt[24(n-4y^2)+1]-1)/6}];
Print[n, " ", r]; Label[aa]; Continue, {n, 0, 100}]

Crossrefs

Cf. A000217, A000290, A016742, A005449, A160325.

Sequence in context: A284044 A126305 A088431 * A052304 A049874 A060501

Adjacent sequences: A254658 A254659 A254660 * A254662 A254663 A254664

Keyword

nonn

Author

Zhi-Wei Sun, Feb 04 2015

Status

approved