A352632 Number of ways to write n as (a+b)*(a+2*b) + c^4 + 2*d^4, where a is a positive integer, and b,c,d are nonnegative integers.

1, 1, 1, 2, 1, 2, 2, 1, 2, 1, 1, 2, 1, 1, 2, 2, 3, 2, 2, 2, 1, 3, 1, 2, 3, 2, 3, 3, 1, 3, 3, 2, 3, 2, 1, 4, 3, 4, 2, 2, 3, 4, 3, 3, 3, 3, 2, 5, 4, 2, 3, 4, 2, 3, 1, 4, 5, 4, 1, 4, 3, 2, 5, 4, 3, 4, 4, 4, 2, 2, 1, 6, 4, 3, 2, 2, 2, 4, 2, 4, 5, 6, 3, 5, 2, 3, 4, 5, 3, 5, 3, 4, 6, 1, 3, 6, 5, 4, 5, 3

Offset

1,4

Comments

Conjecture: a(n) > 0 for all n > 0. In other words, any positive integer n can be written as u*v + x^4 + 2*y^4, where u,v,x,y are nonnegative integers with 0 < u <= v < 2*u.

It seems that a(n) = 1 only for n = 1, 2, 3, 5, 8, 10, 11, 13, 14, 21, 23, 29, 35, 55, 59, 71, 94, 127, 134, 245, 329, 395, 404.

See also A352627, A352628 and A352629 for similar conjectures.

Links

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Example

a(11) = 1 with 11 = (3+0)*(3+2*0) + 0^4 + 2*1^4.
a(23)= 1 with 23 = (3+1)*(3+2*1) + 1^4 + 2*1^4.
a(134) = 1 with 134 = (10+1)*(10+2*1) + 0^4 + 2*1^4.
a(245) = 1 with 245 = (10+1)*(10+2*1) + 3^4 + 2*2^4.
a(329) = 1 with 329 = (6+6)*(6+2*6) + 3^4 + 2*2^4.
a(395) = 1 with 395 = (2+11)*(2+2*11) + 3^4 + 2*1^4.
a(404) = 1 with 404 = (15+2)*(15+2*2) + 3^4 + 2*0^4.

Mathematica

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
tab={}; Do[r=0; Do[If[SQ[4(n-2d^4-c^4)+b^2], r=r+1], {d, 0, ((n-1)/2)^(1/4)}, {c, 0, (n-1-2d^4)^(1/4)}, {b, 0, Sqrt[(n-1-2d^4-c^4)/2]}]; tab=Append[tab, r], {n, 1, 100}]; Print[tab]

Crossrefs

Cf. A000290, A000583, A352627, A352628, A352629.

Sequence in context: A336532 A100429 A049710 * A025143 A174314 A237253

Adjacent sequences: A352629 A352630 A352631 * A352633 A352634 A352635

Keyword

nonn

Author

Zhi-Wei Sun, Mar 25 2022

Status

approved