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Number of prime factors p of n such that p^(1+valuation(n,p)) divides sigma(n).
2

%I #20 Feb 17 2022 00:02:29

%S 0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,2,0,0,0,0,

%T 0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,0,1,0,0,1,0,0,0,0,0,1,0,1,0,0,0,2,0,0,

%U 0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,1,0,1,0,0,0,0,0,2,0,0,0,0,0,1

%N Number of prime factors p of n such that p^(1+valuation(n,p)) divides sigma(n).

%H Antti Karttunen, <a href="/A351439/b351439.txt">Table of n, a(n) for n = 1..65537</a>

%H <a href="/index/Si#SIGMAN">Index entries for sequences related to sigma(n)</a>

%F For all n >= 1, a(n) <= A001221(n). [Apparently this is sharp for n > 1].

%F For all n >= 1, a(n) >= A351539(n).

%e For n = 30 = 2*3*5, sigma(30) = 72 = 2^3 * 3^2 and thus for two of the three prime factors of 30, a higher power of the same prime divides sigma(30), and therefore a(30) = 2.

%t {0}~Join~Table[Function[m, Count[FactorInteger[n][[All, 1]], _?(Mod[m, #^(1 + IntegerExponent[n, #])] == 0 &)]][DivisorSigma[1, n]], {n, 2, 108}] (* _Michael De Vlieger_, Feb 16 2022 *)

%o (PARI) A351439(n) = { my(f=factor(n),s=sigma(n)); sum(k=1,#f~,(0==(s%(f[k,1]^(1+f[k,2]))))); };

%Y Cf. A000203, A001221, A007691, A286561, A351539.

%Y Cf. also A336352.

%K nonn

%O 1,30

%A _Antti Karttunen_, Feb 16 2022