cp's OEIS Frontend

A356174 Replace all prime factors p of n with n/p, then take the integer logarithm (A001414).

%I A356174 #28 Jul 31 2022 13:59:31
%S A356174 0,0,0,4,0,5,0,12,6,7,0,14,0,9,8,24,0,16,0,18,10,13,0,27,10,15,18,22,
%T A356174 0,20,0,40,14,19,12,30,0,21,16,33,0,24,0,30,22,25,0,44,14,24,20,34,0,
%U A356174 33,16,39,22,31,0,36,0,33,26,60,18,32,0,42,26,28,0,48,0,39
%N A356174 Replace all prime factors p of n with n/p, then take the integer logarithm (A001414).
%C A356174 This sequence is closely related to the arithmetic derivative A003415. Let P(x) be a polynomial with zeros in the prime factors p_k of n with multiplicity, defined as P(x) = Product_{p_k|n}(1-p_k) and P'(x) the derivative of P(x), then the product over all roots of P'(x) multiplied by A001222(n) gives us A003415(n). If we do the sum over all roots of P'(x) instead and multiply it by A001222(n), we will get a(n). Note that the roots of P'(x) may have complex values but because complex values will appear in conjugate pairs, the sum will be real.
%F A356174 a(n) = A001414(n^(A001222(n) - 1)) = A001414(A072195(n)).
%F A356174 a(n*m) = (A001414(n)+A001414(m))*((A001222(n)+A001222(m)) - 1).
%F A356174 a(n^m) = A001414(n)*m*(A001222(n)*m - 1).
%F A356174 a(A000040(n)) = 0.
%F A356174 a(A001358(n)) = A001414(A001358(n)) = A003415(A001358(n)).
%F A356174 a(A014612(n))^2 - 12*A003415(A014612(n)) = 0, if A003415(A014612(n)) is not square free (A014612(n) is in A328303). This follows from the discriminant of the quadratic polynomial.
%o A356174 (PARI) a(n)=local(f); if(n<1, 0, f=factor(n^(bigomega(n)-1)); sum(k=1, matsize(f)[1], f[k, 1]*f[k, 2]))
%Y A356174 Cf. A000040, A001222, A001414, A003415, A014612, A072195, A328303.
%K A356174 nonn
%O A356174 1,4
%A A356174 _Thomas Scheuerle_, Jul 28 2022