A347237 - OEIS

A347237

Möbius transform of A347236.

1, 0, 1, 6, 1, 0, 3, 6, 17, 0, 1, 6, 3, 0, 1, 42, 1, 0, 3, 6, 3, 0, 5, 6, 37, 0, 49, 18, 1, 0, 5, 78, 1, 0, 3, 102, 3, 0, 3, 6, 1, 0, 3, 6, 17, 0, 5, 42, 89, 0, 1, 18, 5, 0, 1, 18, 3, 0, 1, 6, 5, 0, 51, 330, 3, 0, 3, 6, 5, 0, 1, 102, 5, 0, 37, 18, 3, 0, 3, 42, 353, 0, 5, 18, 1, 0, 1, 6, 7, 0, 9, 30, 5, 0, 3, 78, 3, 0, 17

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OFFSET

1,4

COMMENTS

Dirichlet convolution of A003972 (prime shifted phi) with A061019.

Dirichlet convolution of A003961 with A158523.

Multiplicative because A003972 and A061019 are, and also because A347236 is.

From Antti Karttunen, Aug 25 2021: (Start)

All terms are nonnegative because sequence is multiplicative and a(p^k) >= 0 for all primes p and k >= 0.

Proof: For any prime p, sequence a(p^k), k>=0, is obtained as an ordinary convolution of sequences (-p)^k and the first differences of q^k, where q = A151800(p). (E.g., for powers of 2, the sequences convolved are A122803 and A025192, giving A102901.) This convolution is an alternating sum, with the terms 1*(q-1)*q^(k-1), -(p)*(q-1)*q^(k-2), (p^2)*(q-1)*q^(k-3), -(p^3)*(q-1)*q^(k-4), ..., (p^(k-1))*(q-1), -(p^k), for odd k, with sum of each consecutive pair being nonnegative because q >= p+1, while with an even exponent k, the leftover term p^k at the end is also positive, thus the whole sum is nonnegative also in that case.

(End)

LINKS

Antti Karttunen, Table of n, a(n) for n = 1..16383

Index entries for primes, gaps between

Index entries for sequences computed from indices in prime factorization

FORMULA

a(n) = Sum_{d|n} A008683(n/d) * A347236(d).

a(n) = Sum_{d|n} A003972(n/d) * A061019(d).

a(n) = Sum_{d|n} A003961(n/d) * A158523(d).

For all n >= 1, a(A000040(n)) = A001223(n) - 1.

For all n >= 0, a(2^n) = A102901(n).

For all n >= 0, a(3^n) = A120612(n).