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A347237
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Möbius transform of A347236.
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3
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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
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1,4
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COMMENTS
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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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LINKS
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FORMULA
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For all n >= 0, a(2^n) = A102901(n).
For all n >= 0, a(3^n) = A120612(n).
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PROG
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(PARI)
A003961(n) = { my(f = factor(n)); for(i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); };
A061019(n) = (((-1)^bigomega(n))*n);
\\ Or alternatively as:
A158523(n) = { my(f=factor(n)); prod(i=1, #f~, my(p=f[i, 1], e=f[i, 2]); ((-1)^e)*(p+1)*(p^(e-1))); };
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CROSSREFS
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Cf. A000040, A001223, A003961, A003972, A008683, A008836, A016825 (positions of zeros), A061019, A102901, A120612, A158523, A347236.
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KEYWORD
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nonn,mult
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AUTHOR
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STATUS
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approved
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