A262955 Number of ordered pairs (x,y) with x >= 0 and y > 0 such that n - x^4 - y*(y+1)/2 is a pentagonal number (A000326) or twice a pentagonal number.

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

Offset: 1

Author

Zhi-Wei Sun, Oct 05 2015

Keywords

• nonn

Comments

Conjecture: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 10, 35, 75, 134, 415, 515, 1465, 2365, 3515, 4140.

Examples

			a(1) = 1 since 1 = 0^4 + 1*2/2 + p_5(0), where p_5(n) denotes the pentagonal number n*(3*n-1)/2.
a(10) = 1 since 10 = 0^4 + 4*5/2 + p_5(0).
a(35) = 1 since 35 = 1^4 + 4*5/2 + 2*p_5(3).
a(75) = 1 since 75 = 2^4 + 5*6/2 + 2*p_5(4).
a(134) = 1 since 134 = 2^4 + 1*2/2 + p_5(9).
a(415) = 1 since 415 = 0^4 + 21*22/2 + 2*p_5(8).
a(515) = 1 since 515 = 0^4 + 6*7/2 + 2*p_5(13).
a(1465) = 1 since 1465 = 5^4 + 35*36/2 + p_5(12).
a(2365) = 1 since 2365 = 5^4 + 8*9/2 + 2*p_5(24).
a(3515) = 1 since 3515 = 5^4 + 51*52/2 + 2*p_5(23).
a(4140) = 1 since 4140 = 1^4 + 90*91/2 + 2*p_5(4).
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
• Zhi-Wei Sun, On universal sums of polygonal numbers, Sci. China Math. 58(2015), no. 7, 1367-1396.

Crossrefs

Cf. A000217, A000326, A000583, A262941, A262944, A262945, A262954.

Programs

Mathematica

PenQ[n_]:=IntegerQ[Sqrt[24n+1]]&&(n==0||Mod[Sqrt[24n+1]+1,6]==0)
PQ[n_]:=PenQ[n]||PenQ[n/2]
Do[r=0;Do[If[PQ[n-x^4-y(y+1)/2],r=r+1],{x,0,n^(1/4)},{y,1,(Sqrt[8(n-x^4)+1]-1)/2}];Print[n," ",r];Continue,{n,1,100}]