OFFSET
1,4
COMMENTS
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)
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FORMULA
PROG
CROSSREFS
KEYWORD
nonn,mult
AUTHOR
Antti Karttunen, Aug 24 2021
STATUS
approved