

A262941


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


19



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

1,3


COMMENTS

Conjecture: a(n) > 0 for all n > 0. In other words, any positive integer n can be written as x^4 + 2^k*y^2 + z*(z+1)/2, where k is 1 or 2, and x,y,z are integers with z > 0.
This has been verified for n up to 2*10^6. We also guess that a(n) = 1 only for n = 1, 2, 13, 16, 50, 59, 239, 493, 1156, 1492, 1984, 3332.


LINKS



EXAMPLE

a(1) = 1 since 1 = 0^4 + 0^2 + 1*2/2 with 0 even.
a(2) = 1 since 2 = 1^4 + 0^2 + 1*2/2 with 0 even.
a(13) = 1 since 13 = 1^4 + 2* 1^2 + 4*5/2.
a(16) = 1 since 16 = 1^4 + 0^2 + 5*6/2 with 0 even.
a(50) = 1 since 50 = 1^4 + 2^2 + 9*10/2 with 2 even.
a(59) = 1 since 59 = 0^4 + 2^2 + 10*11/2 with 2 even.
a(239) = 1 since 239 = 0^4 + 2* 2^2 + 21*22/2 with 2 even.
a(493) = 1 since 493 = 2^4 + 18^2 + 17*18/2 with 18 even.
a(1156) = 1 since 1156 = 1^4 + 2*24^2 + 2*3/2 with 24 even.
a(1492) = 1 since 1492 = 2^4 + 2* 7^2 + 52*53/2.
a(1984) = 1 since 1984 = 5^4 + 18^2 + 45*46/2 with 18 even.
a(3332) = 1 since 3332 = 5^4 + 52^2 + 2*3/2 with 52 even.


MATHEMATICA

SQ[n_]:=IntegerQ[Sqrt[n/2]]IntegerQ[Sqrt[n/4]]
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, A254885, A262813, A262827, A262944, A262945, A262954, A262955, A262956.


KEYWORD

nonn


AUTHOR



STATUS

approved



