OFFSET
1,25
COMMENTS
Or number of ways n^2 can be written as the sum of two positive squares: a(5) = 1: 3^2 + 4^2 = 5^2; a(25) = 2: 7^2 + 24^2 = 15^2 + 20^2 = 25^2. - Alois P. Heinz, Aug 01 2019
REFERENCES
A. H. Beiler, Recreations in the Theory of Numbers, New York: Dover, pp. 116-117, 1966.
LINKS
Stanislav Sykora, Table of n, a(n) for n = 1..20000
Ron Knott, Pythagorean Triples and Online Calculators
F. Richman, Pythagorean Triples
A. Tripathi, On Pythagorean triples containing a fixed integer, Fib. Q., 46/47 (2008/2009), 331-340. See Theorem 7.
Eric Weisstein's World of Mathematics, Pythagorean Triple
FORMULA
Let n = 2^e_2 * product_i p_i^f_i * product_j q_j^g_j where p_i == 1 mod 4, q_j == 3 mod 4; then a(n) = (1/2)*(product_i (2*f_i + 1) - 1). - Beiler, corrected
8*a(n) + 4 = A046109(n) for n > 0. - Ralf Stephan, Mar 14 2004
a(n) = 0 for n in A004144. - Lekraj Beedassy, May 14 2004
a(A084645(k)) = 1. - Ruediger Jehn, Jan 14 2022
a(A084646(k)) = 2. - Ruediger Jehn, Jan 14 2022
a(A084647(k)) = 3. - Jean-Christophe Hervé, Dec 01 2013
a(A084648(k)) = 4. - Jean-Christophe Hervé, Dec 01 2013
a(A084649(k)) = 5. - Jean-Christophe Hervé, Dec 01 2013
a(n) = A063725(n^2) / 2. - Michael Somos, Mar 29 2015
a(n) = Sum_{k=1..n} Sum_{i=1..k} [i^2 + k^2 = n^2], where [ ] is the Iverson bracket. - Wesley Ivan Hurt, Dec 10 2021
a(A002144(k)^n) = n. - Ruediger Jehn, Jan 14 2022
MAPLE
f:= proc(n) local F, t;
F:= select(t -> t[1] mod 4 = 1, ifactors(n)[2]);
1/2*(mul(2*t[2]+1, t=F)-1)
end proc:
map(f, [$1..100]); # Robert Israel, Jul 18 2016
MATHEMATICA
a[1] = 0; a[n_] := With[{fi = Select[ FactorInteger[n], Mod[#[[1]], 4] == 1 & ][[All, 2]]}, (Times @@ (2*fi+1)-1)/2]; Table[a[n], {n, 1, 99}] (* Jean-François Alcover, Feb 06 2012, after first formula *)
PROG
(PARI) a(n)={my(m=0, k=n, n2=n*n, k2, l2);
while(1, k=k-1; k2=k*k; l2=n2-k2; if(l2>k2, break); if(issquare(l2), m++)); return(m)} \\ brute force, Stanislav Sykora, Mar 18 2015
(PARI) {a(n) = if( n<1, 0, sum(k=1, sqrtint(n^2 \ 2), issquare(n^2 - k^2)))}; /* Michael Somos, Mar 29 2015 */
(PARI) a(n) = {my(f = factor(n/(2^valuation(n, 2)))); (prod(k=1, #f~, if ((f[k, 1] % 4) == 1, 2*f[k, 2] + 1, 1)) - 1)/2; } \\ Michel Marcus, Mar 08 2016
(Python)
from math import prod
from sympy import factorint
def A046080(n): return prod((e<<1)+1 for p, e in factorint(n).items() if p&3==1)>>1 # Chai Wah Wu, Sep 06 2022
KEYWORD
nonn
AUTHOR
STATUS
approved