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%I #22 Oct 04 2022 12:36:26
%S 1,1,3,4,7,6,12,8,15,13,18,12,28,14,24,24,31,18,39,20,42,32,36,24,60,
%T 6,42,40,56,30,72,32,63,48,54,48,91,38,60,56,90,42,96,44,84,78,72,48,
%U 124,57,18,72,98,54,120,72,120,80,90,60,168,62,96,104,127,84,144,68,126,96,144,72,195,74,114,24,140
%N Sum of divisors of n that are not divisible by 25. a(0) = 1.
%H Antti Karttunen, <a href="/A227131/b227131.txt">Table of n, a(n) for n = 0..16384</a>
%H <a href="/index/Su#sums_of_divisors">Index entries for sequences related to sums of divisors</a>.
%F a(n) is multiplicative with a(0) = 1, a(5^e) = 6 if e>0, a(p^e) = (p^(e+1) - 1) / (p - 1) otherwise.
%F G.f. is a period 1 Fourier series which satisfies f(-1 / (25 t)) = 25 (t/i)^2 f(t) where q = exp(2 Pi i t).
%F G.f.: 1 + Sum_{k>0} k * x^k / (1 - x^k) - Sum_{k>0} 25 * k * x^(25*k) / (1 - x^(25*k)).
%F Sum_{k=1..n} a(k) ~ (2*Pi^2/25) * n^2. - _Amiram Eldar_, Oct 04 2022
%e G.f. = 1 + q + 3*q^2 + 4*q^3 + 7*q^4 + 6*q^5 + 12*q^6 + 8*q^7 + 15*q^8 + 13*q^9 + ...
%e 75 has six divisors: 1, 3, 5, 15, 25, 75, but both 25 and 75 are divisible by 25, thus not counted, and we have a(75) = 1+3+5+15 = 24. - _Antti Karttunen_, Nov 23 2017
%t a[ n_] := If[ n < 1, Boole[ n == 0], Sum[ If[ Mod[ d, 25] > 0, d, 0], {d, Divisors @ n}]];
%o (PARI) {a(n) = if( n<1, n==0, sumdiv( n, d, if( d%25, d)))};
%o (PARI) {a(n) = if( n<1, n==0, 1 * (sigma(n) - if( n%25==0, 25 * sigma( n / 25))))};
%o (Sage) A = ModularForms( Gamma0(25), 2, prec=66) . basis(); A[0] + A[1] + 3*A[2] + 4*A[3] + 7*A[4];
%o (Magma) A := Basis( ModularForms( Gamma0(25), 2), 66); A[1] + A[2] + 3*A[3] + 4*A[4] + 7*A[5]; /* _Michael Somos_, Jun 12 2014 */
%Y Cf. A000118, A004011, A008653, A028594, A028887, A096726, A107505.
%K nonn,mult
%O 0,3
%A _Michael Somos_, Jul 02 2013
%E More terms from _Antti Karttunen_, Nov 23 2017
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