

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.


14



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
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OFFSET

1,2


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.


LINKS

ZhiWei Sun, Table of n, a(n) for n = 1..10000
ZhiWei Sun, On universal sums of polygonal numbers, Sci. China Math. 58(2015), no. 7, 13671396.


EXAMPLE

a(1) = 1 since 1 = 0^4 + 1*2/2 + p_5(0), where p_5(n) denotes the pentagonal number n*(3*n1)/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).


MATHEMATICA

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


CROSSREFS

Cf. A000217, A000290, A000583, A001318, A160325, A254885, A262813, A262815, A262816, A262941, A262945.
Sequence in context: A187758 A171931 A221530 * A322789 A069904 A164978
Adjacent sequences: A262941 A262942 A262943 * A262945 A262946 A262947


KEYWORD

nonn


AUTHOR

ZhiWei Sun, Oct 05 2015


STATUS

approved



