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A125140
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SEPSigma(n) = (-1)^(Sum_i r_i)*Sum_{d|n} (-1)^(Sum_j Max(r_j))*d = Product_i (Sum_{s_i=1..r_i} p_i^s_i)+(-1)^r_i where n = Product_i p_i^r_i, d = Product_j p_j^r_j, p_j^max(r_j) is the largest power of p_j dividing n.
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2
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1, 1, 2, 7, 4, 2, 6, 13, 13, 4, 10, 14, 12, 6, 8, 31, 16, 13, 18, 28, 12, 10, 22, 26, 31, 12, 38, 42, 28, 8, 30, 61, 20, 16, 24, 91, 36, 18, 24, 52, 40, 12, 42, 70, 52, 22, 46, 62, 57, 31, 32, 84, 52, 38, 40, 78, 36, 28, 58, 56, 60, 30, 78, 127, 48, 20, 66, 112, 44, 24, 70, 169, 72
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OFFSET
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1,3
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COMMENTS
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SEP stands for Signed by Exponents of Prime factors.
By "Max(r_j)" I mean the following: If d|m, d=p^e*q^f, m=p^x*q^y*r^z then Max(e)=x, Max(f)=y.
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LINKS
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FORMULA
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a(n) = Product_i (-1)^r_i + ((p_i^(r_i+1)-p_i)/(p_i-1)), where p_i and r_i range over the primes and their exponents in the prime factorization of n.
a(n) = Product_{p^e || n} (-1)^e + ((p^(1+e)-p)/(p-1)), where p and e range over the primes and their exponents in the prime factorization of n.
Dirichlet g.f.: zeta(s-1) * zeta(2*s) * Product_{p prime} (1 - 1/p^s + 2/p^(2*s-1)).
Sum_{k=1..n} a(k) ~ c * n^2, where c = (1/2) * Product_{p prime} (1 - (p^2 - 2*p - 1)/(p^4 - 1)) = 0.48777088716109463306... . (End)
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EXAMPLE
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If n = 240, d = 12 then 2^max(r_j) = 2^max(2) = 2^4, 3^max(r_j) = 3^max(1) = 3^1, SEPSigma(240) = (1+2+4+8+16)*(-1+3)*(-1+5) = 248.
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MAPLE
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A125140 := proc(n) local ifs, i, a, r, p ; ifs := ifactors(n)[2] ; a := 1 ; for i from 1 to nops(ifs) do r := op(2, op(i, ifs)) ; p := op(1, op(i, ifs)) ; a := a*(p*(1-p^r)/(1-p)+(-1)^r) ; od ; RETURN(a) ; end: for n from 1 to 80 do printf("%d, ", A125140(n)) ; od ; # R. J. Mathar, Jun 07 2007
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MATHEMATICA
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f[p_, e_] := (p^(e+1) - p)/(p - 1) + (-1)^e; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 18 2023 *)
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PROG
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(PARI) A125140(n) = { my(f=factor(n), p, e); prod(k=1, #f~, p = f[k, 1]; e = f[k, 2]; ((-1)^e) + (((p^(e+1))-p) / (p-1))); }; \\ Antti Karttunen, Feb 21 2022
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CROSSREFS
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KEYWORD
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nonn,easy,mult
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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