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

Author

Antti Karttunen, Jul 24 2017

Keywords

• nonn

Examples

			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.
		

Links

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

Crossrefs

Cf. A008437, A063725.

Bisections: A000004, A008442.

Programs

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

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))))))))))

Formula

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