A262944 Number of ordered pairs (x,y) with x >= 0 and y > 0 such that n - x^4 - y*(y+1)/2 is a square or a pentagonal number.

1, 2, 2, 2, 2, 3, 4, 3, 1, 3, 5, 3, 2, 2, 5, 5, 3, 3, 5, 5, 3, 6, 6, 3, 3, 8, 6, 5, 5, 3, 7, 5, 5, 3, 4, 4, 8, 9, 3, 5, 7, 6, 3, 5, 5, 7, 5, 3, 4, 5, 6, 6, 9, 4, 5, 7, 7, 5, 4, 4, 7, 6, 1, 5, 5, 7, 7, 7, 1, 6, 10, 8, 6, 3, 4, 3, 6, 4, 6, 9, 5, 7, 9, 3, 5, 8, 9, 8, 3, 3, 11, 10, 6, 6, 8, 12, 5, 6, 4, 7

Offset: 1

Author

Zhi-Wei Sun, Oct 05 2015

Keywords

• nonn

Comments

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 9, 63, 69, 489, 714, 1089.

(ii) For any positive integer n, there are integers x >= 0 and y > 0 such that n - x^4 - y*(y+1)/2 is twice a square or twice a pentagonal number.

(iii) For any positive integer n, there are integers x >= 0 and y > 0 such that n - 2*x^4 - y*(y+1)/2 is a square or a pentagonal number.

See also A262941 and A262945 for similar conjectures.

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(9) = 1 since 9 = 1^4 + 2*3/2 + p_5(2).
a(63) = 1 since 63 = 0^4 + 7*8/2 + p_5(5).
a(69) = 1 since 69 = 2^4 + 7*8/2 + 5^2.
a(489) = 1 since 489 = 3^4 + 12*13/2 + p_5(15).
a(714) = 1 since 714 = 4^4 + 18*19/2 + p_5(14).
a(1089) = 1 since 1089 = 4^4 + 38*39/2 + p_5(8).
		

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, A000290, A000583, A001318, A160325, A254885, A262813, A262815, A262816, A262941, A262945.

Programs

Mathematica

SQ[n_]:=IntegerQ[Sqrt[n]]||(IntegerQ[Sqrt[24n+1]]&&Mod[Sqrt[24n+1]+1,6]==0)
Do[r=0;Do[If[SQ[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}]