

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.


LINKS



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.


KEYWORD

nonn


AUTHOR



STATUS

approved



