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%I #19 Jun 21 2022 05:09:37
%S 0,0,0,0,25,0,0,0,0,25,0,0,169,0,25,0,289,0,0,25,0,0,0,0,25,169,0,0,
%T 841,25,0,0,0,289,25,0,1369,0,169,25,1681,0,0,0,25,0,0,0,0,25,289,169,
%U 2809,0,25,0,0,841,0,25,3721,0,0,0,194,0,0,289,0,25,0,0,5329,1369,25,0,0,169,0,25,0,1681,0,0,314
%N Sum of squares of primes = 1 mod 4 dividing n.
%H Antti Karttunen, <a href="/A005079/b005079.txt">Table of n, a(n) for n = 1..10000</a>
%F Additive with a(p^e) = p^2 if p = 1 (mod 4), 0 otherwise.
%F a(n) = A005063(n) - A005083(n) - 4*A059841(n). - _Antti Karttunen_, Jul 11 2017
%t Array[DivisorSum[#, #^2 &, And[PrimeQ@ #, Mod[#, 4] == 1] &] &, 85] (* _Michael De Vlieger_, Jul 11 2017 *)
%t f[p_, e_] := If[Mod[p, 4] == 1, p^2, 0]; a[n_] := Plus @@ f @@@ FactorInteger[n]; a[1] = 0; Array[a, 100] (* _Amiram Eldar_, Jun 21 2022 *)
%o (Scheme) (define (A005079 n) (if (= 1 n) 0 (+ (if (= 1 (modulo (A020639 n) 4)) (A000290 (A020639 n)) 0) (A005079 (A028234 n))))) ;; _Antti Karttunen_, Jul 11 2017
%o (PARI) a(n) = my(f=factor(n)); sum(k=1, #f~, if (((p=f[k,1])%4) == 1, p^2)); \\ _Michel Marcus_, Jul 11 2017
%Y Cf. A000290, A005063, A005078, A005080, A005081, A005083, A059841.
%K nonn
%O 1,5
%A _N. J. A. Sloane_
%E More terms from _Antti Karttunen_, Jul 11 2017
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