A290081 a(n) = number of ways of writing n as the sum of two odd positive squares.

0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 2

Offset

0,11

Links

Antti Karttunen, Table of n, a(n) for n = 0..11050

Formula

a(2n) = A008442(n), a(2n+1) = 0.

Example

a(2) = 1 as 2 = 1 + 1.
a(10) = 2 as 10 = 1 + 9 = 9 + 1.
a(50) = 3 as 50 = 1 + 49 = 49 + 1 = 25 + 25.

Prog

(Scheme) (define (A290081 n) (cond ((< n 2) 0) ((odd? n) 0) (else (let loop ((k (- (A000196 n) (modulo (+ 1 (A000196 n)) 2))) (s 0)) (if (< k 1) s (loop (- k 2) (+ s (A010052 (- n (* k k))))))))))
(PARI) upto(n) = {my(m, v, res = vector(n)); m = (sqrtint(n)+1)\2; v = vector(m, i, (2*i-1)^2); forvec(x = vector(2, i, [1, #v]), s = v[x[1]] + v[x[2]]; if(s <= n, res[s]+=(1+(x[1]!=x[2]))), 1); concat(0, res)}  \\ David A. Corneth, Jul 24 2017
(PARI)
A008442(n) = if( n<1 || n%4!=1, 0, sumdiv(n, d, (d%4==1) - (d%4==3))); \\ This function from Michael Somos, Apr 24 2004
A290081(n) = if(n%2, 0, A008442(n/2));
(Python)
from sympy import divisors
def A290081(n): return 0 if n&1 else 0 if (m:=n>>1)&3!=1 else sum(((a:=d&3)==1)-(a==3) for d in divisors(m, generator=True)) # Chai Wah Wu, May 17 2023

Crossrefs

Cf. A008437, A063725.

Bisections: A000004, A008442.

Sequence in context: A318655 A056626 A366123 * A347706 A348381 A010103

Adjacent sequences: A290078 A290079 A290080 * A290082 A290083 A290084

Keyword

nonn

Author

Antti Karttunen, Jul 24 2017

Status

approved