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

1,4

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.

Links

Table of n, a(n) for n=1..74.

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.

Prog

(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]))

Crossrefs

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

Sequence in context: A352167 A089389 A046779 * A339436 A255369 A292177

Adjacent sequences: A356171 A356172 A356173 * A356175 A356176 A356177

Keyword

nonn

Author

Thomas Scheuerle, Jul 28 2022

Status

approved