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

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, 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, 33, 16, 39, 22, 31, 0, 36, 0, 33, 26, 60, 18, 32, 0, 42, 26, 28, 0, 48, 0, 39

Offset: 1

Author

Thomas Scheuerle, Jul 28 2022

Keywords

• nonn

Comments

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.

Crossrefs

Cf. A000040, A001222, A001414, A003415, A014612, A072195, A328303.

Programs

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]))

Formula

a(n) = A001414(n^(A001222(n) - 1)) = A001414(A072195(n)).

a(n*m) = (A001414(n)+A001414(m))*((A001222(n)+A001222(m)) - 1).

a(n^m) = A001414(n)*m*(A001222(n)*m - 1).

a(A000040(n)) = 0.

a(A001358(n)) = A001414(A001358(n)) = A003415(A001358(n)).

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.