

A349942


Number of ways to write n as a^4 + b^2 + (c^4 + d^2)/25 with a,b,c,d nonnegative integers.


9



1, 4, 6, 4, 3, 5, 4, 1, 1, 4, 8, 7, 2, 4, 6, 2, 4, 12, 13, 6, 7, 9, 4, 1, 2, 11, 19, 11, 2, 10, 10, 2, 6, 12, 12, 9, 11, 9, 8, 4, 3, 16, 18, 7, 1, 13, 10, 1, 4, 7, 17, 15, 11, 11, 10, 2, 4, 12, 11, 9, 4, 13, 12, 5, 3, 15, 25, 10, 10, 12, 8, 3, 4, 9, 17, 17, 4, 14, 16, 3, 5, 20, 20, 14, 13, 12, 14, 4, 3, 12, 30, 22, 3, 12, 13, 4, 4, 16, 24, 20, 11
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OFFSET

0,2


COMMENTS

Conjecture: a(n) > 0 for all n >= 0, and a(n) = 1 only for n = 0, 16^k*m (k = 0,1,2,... and m = 7, 8, 23, 44, 47).
We have verified this for n up to 3*10^5.
As m/n = (m*n^3)/n^4 for any nonnegative integers m and n > 0, the conjecture implies that each nonnegative rational number can be written as x^4 + 25*y^4 + z^2 + w^2 with x,y,z,w rational numbers.
See also A349943 for similar conjectures.


LINKS



EXAMPLE

a(0) = 1 with 0 = 0^4 + 0^2 + (0^4 + 0^2)/25.
a(7) = 1 with 7 = 1^4 + 2^2 + (1^4 + 7^2)/25.
a(8) = 1 with 8 = 0^4 + 2^2 + (0^4 + 10^2)/25.
a(23) = 1 with 23 = 1^4 + 3^2 + (1^4 + 18^2)/25.
a(44) = 1 with 44 = 1^4 + 3^2 + (5^4 + 15^2)/25.
a(47) = 1 with 47 = 1^4 + 6^2 + (3^4 + 13^2)/25.


MATHEMATICA

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
tab={}; Do[r=0; Do[If[SQ[25(nx^4y^2)z^4], r=r+1], {x, 0, n^(1/4)}, {y, 0, Sqrt[nx^4]}, {z, 0, (25(nx^4y^2))^(1/4)}]; tab=Append[tab, r], {n, 0, 100}]; Print[tab]


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



