A262945 Number of ordered pairs (x,y) with x >= 0 and y >= 0 such that n - x^4 - 2*y^2 is a triangular number or a pentagonal number.

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

Offset: 0

Author

Zhi-Wei Sun, Oct 05 2015

Keywords

• nonn

Comments

Conjecture: a(n) > 0 for every n = 0,1,2,..., and a(n) = 1 only for the following 55 values of n: 0, 26, 27, 32, 41, 42, 50, 65, 80, 97, 112, 122, 130, 160, 196, 227, 239, 272, 322, 371, 612, 647, 736, 967, 995, 1007, 1106, 1127, 1205, 1237, 1240, 1262, 1637, 1657, 1757, 2912, 2987, 3062, 3107, 3524, 3647, 3902, 5387, 5587, 5657, 6047, 6107, 11462, 13427, 14717, 15002, 17132, 20462, 30082, 35750.

See also A262941, A262944, A262954 and A262955 for similar conjectures.

Examples

			a(26) = 1 since 26 = 2^4 + 2*0^2 + 4*5/2.
a(32) = 1 since 32 = 0^4 + 2*4^2 + 0*1/2.
a(41) = 1 since 41 = 1^4 + 2*3^2 + p_5(4), where p_5(n) denotes the pentagonal number n*(3*n-1)/2.
a(196) = 1 since 196 = 1^4 + 2*5^2 + p_5(10).
a(3524) = 1 since 3524 = 0^4 + 2*22^2 + 71*72/2.
a(3647) = 1 since 3647 = 0^4 + 2*34^2 + p_5(30).
a(6047) = 1 since 6047 = 5^4 + 2*39^2 + p_5(40).
a(6107) = 1 since 6107 = 0^4 + 2*1^2 + 110*111/2.
a(11462) = 1 since 11462 = 9^4 + 2*5^2 + 98*99/2.
a(13427) = 1 since 13427 = 7^4 + 2*0^2 + 148*149/2.
a(14717) = 1 since 14717 = 8^4 + 2*72^2 + 22*23/2.
a(15002) = 1 since 15002 = 0^4 + 2*86^2 + 20*21/2.
a(17132) = 1 since 17132 = 3^4 + 2*30^2 + p_5(101).
a(20462) = 1 since 20462 = 0^4 + 2*26^2 + 195*196/2.
a(30082) = 1 since 30082 = 11^4 + 2*63^2 + 122*123/2.
a(35750) = 1 since 35750 = 0^4 + 2*44^2 + 252*253/2.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
• Zhi-Wei Sun, On universal sums of polygonal numbers, Sci. China Math. 58(2015), no. 7, 1367-1396.

Crossrefs

Cf. A000217, A000290, A000326, A000583, A262941, A262944, A262954, A262955.

Programs

Mathematica

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