OFFSET
0,2
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.
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.
EXAMPLE
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.
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}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Oct 05 2015
STATUS
approved