A308661 Number of ways to write 12*n+5 as (2^a*5^b)^2 + c^2 + d^2, where a,b,c,d are nonnegative integers with a > 0 and c <= d.

1, 2, 3, 3, 2, 3, 3, 5, 5, 4, 5, 3, 5, 5, 5, 6, 3, 6, 4, 3, 5, 4, 7, 6, 6, 6, 2, 8, 8, 5, 5, 5, 6, 5, 6, 10, 6, 6, 8, 4, 6, 8, 8, 7, 3, 10, 5, 7, 9, 6, 7, 3, 9, 7, 2, 7, 6, 9, 8, 6, 8, 6, 8, 9, 5, 4, 7, 6, 4, 5, 7, 8, 5, 8, 7, 6, 4, 8, 10, 6, 10, 3, 6, 9, 6, 11, 5, 9, 4, 4, 8, 8, 10, 9, 7, 4, 5, 11, 7, 9, 10

Offset

0,2

Comments

Conjecture 1: a(n) > 0 for all n = 0,1,2,.... Equivalently, for each nonnegative integer n we can write 3*n+1 as a*(a+1)/2 + b*(b+1)/2 + (2^c*5^d)^2 with a,b,c,d nonnegative integers.

Conjecture 2: For each n = 0,1,2,... we can write 24*n+10 as a^2 + b^2 + (2^c*3^d)^2 with a,b,c,d nonnegative integers and d > 0.

We have verified Conjectures 1 and 2 for n up to 2*10^8 and 10^8 respectively.

By the Gauss-Legendre theorem on sums of three squares, for each n = 0,1,... we can write 4*n+1 (or 4*n+2, or 8*n+3) as the sum of three squares.

Conjecture 1 holds for n < 8.33*10^9. - Giovanni Resta, Jun 19 2019

Links

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000

Example

a(0) = 1 with 12*0 + 5 = (2^1*5^0)^2 + 0^2 + 1^2.
a(4) = 2 with 12*4 + 5 = 53 = (2^1*5^0)^2 + 0^2 + 7^2 = (2^2*5^0)^2 + 1^2 + 6^2.
a(441019) = 2 with 12*441019 + 5 = 5292233 = (2^1*5^2)^2 + 513^2 + 2242^2 = (2^3*5^1)^2 + 757^2 + 2172^2.

Mathematica

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
tab={}; Do[r=0; Do[If[SQ[12n+5-4^a*25^b-x^2], r=r+1], {a, 1, Log[4, 12n+5]}, {b, 0, Log[25, (12n+5)/4^a]}, {x, 0, Sqrt[(12n+5-4^a*25^b)/2]}]; tab=Append[tab, r], {n, 0, 100}]; Print[tab]

Crossrefs

Cf. A000079, A000217, A000244, A000290, A000351, A000378, A001481, A308566, A308584, A308621, A308623, A308640, A308641, A308644, A308656, A308662.

Sequence in context: A289831 A335358 A296611 * A286104 A137779 A191246

Adjacent sequences: A308658 A308659 A308660 * A308662 A308663 A308664

Keyword

nonn

Author

Zhi-Wei Sun, Jun 15 2019

Status

approved