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A067856
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Sum_{n >= 1} a(n)/n^s = 1/(Sum_{n >= 1} (-1)^(n + 1)/n^s).
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30
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1, 1, -1, 2, -1, -1, -1, 4, 0, -1, -1, -2, -1, -1, 1, 8, -1, 0, -1, -2, 1, -1, -1, -4, 0, -1, 0, -2, -1, 1, -1, 16, 1, -1, 1, 0, -1, -1, 1, -4, -1, 1, -1, -2, 0, -1, -1, -8, 0, 0, 1, -2, -1, 0, 1, -4, 1, -1, -1, 2, -1, -1, 0, 32, 1, 1, -1, -2, 1, 1, -1, 0, -1, -1, 0, -2, 1, 1, -1, -8, 0, -1, -1, 2, 1, -1, 1, -4, -1, 0, 1, -2, 1, -1, 1, -16, -1, 0, 0, 0
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OFFSET
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1,4
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COMMENTS
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The first 31 terms equal the values of the Ramanujan sum c_n(8) -- see for example A085906 -- but a(32) <> c_{32}(8). - R. J. Mathar, Apr 02 2011
Let Mu(n) = (-1)^(n+1)*a(n), an analog of the Möbius function mu(n). Then for arithmetic functions f(n) and g(n) we have the following analog of the Möbius inversion formula: f(n) = Sum_{d divides n} (-1)^((1+d)*(1+n/d))*g(d) iff g(n) = Sum_{d divides n} (-1)^((1+d)*(1+n/d))*Mu(n/d)*f(d).
Each of the following two equations implies the other: F(x) = Sum_{n >= 1} (-1)^(n+1)*G(n*x); G(x) = Sum_{n >= 1} a(n)*F(n*x). See G. Pólya and G. Szegő, Part V111, Chap. 1, Nos. 66-68.2. (End)
Let D(n) denote the set of partitions of n into distinct parts. Then Sum_{parts k in D(n)} a(k) = |D(n-1)| = A000009(n-1). For example, D(6) = {6, 1 + 5, 2 + 4, 1 + 2 + 3} and the sum a(6) + (a(1) + a(5)) + (a(2) + a(4)) + (a(1) + a(2) + a(3)) = 3 = |D(5)|. - Peter Bala, Mar 14 2019
For p prime >= 2, Petrogradsky (2003) defined the multiplicative functions 1_p and mu_p in the following way:
1_p(n) = 1 when gcd(n,p) = 1 and 1_p(n) = 1 - p when gcd(n,p) = p;
mu_p(n) = mu(n) when gcd(n,p) = 1 and mu_p(n) = mu(m)*(p^s - p^(s-1)) when n = m*p^s with gcd(m,p) = 1 and s >= 1.
We have 1_2(n) = A062157(n), 1_3(n) = A061347(n) (with offset 1), a(n) = mu_2(n), and A117997(n) = mu_3(n) for n >= 1.
Some of the results by other contributors can be generalized:
(i) Rogel's (1897) formula becomes Sum_{d | n} 1_p(d) * mu_p(n/d) = 0 for n > 1. Thus, 1_p is the Dirichlet inverse of mu_p.
(ii) R. J. Mathar's Dirichlet g.f. for mu_p becomes 1/(zeta(s) * (1 - p^(1-s))). The Dirichlet g.f. for 1_p is zeta(s) * (1 - p^(1-s)).
(iii) Benoit Cloitre's formula becomes 1 = Sum_{k=1..n} mu_p(k)*g_p(n/k), where g_p(x) = floor(x) - p*floor(x/p) = floor(x) mod p.
(iv) Paul D. Hanna's formula becomes Sum_{n >= 1} (mu_p(n)/n)*log((1 - x^(n*p))/(1 - x^n)) = x.
(v) The definition of A117997 generalizes to Sum_{d | n} mu_p(d) = n, if n = p^s for s >= 0, and = 0, otherwise.
(vi) By differentiating both sides of (iv) w.r.t. x and multiplying both sides by x, we get Sum_{n >= 1} mu_p(n)*(x^n + 2*x^(2*n) + ... + (p-1)*x^(n*(p-1)))/(1 + x^n + x^(2*n) + ... + x^(n*(p-1))) = x, which generalizes one of Peter Bala's formulas. It can be thought as a "generalized Lambert series".
(vii) Obviously, f(n) = Sum_{d | n} 1_p(n/d)*g(d) if and only if g(n) = Sum_{d | n} mu_p(n/d)*f(d). (End)
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REFERENCES
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G. Pólya and G. Szegő, Problems and Theorems in Analysis Volume II. Springer_Verlag 1976.
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LINKS
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Franz Rogel, Transformationen arithmetischer Reihen, S.-B. Kgl. Bohmischen Ges. Wiss. Article LI (1897), Prague (31 pages); see pp. 10-11 and especially Eqs. (21) - (24). [There are some obvious typos there; especially Eq. (24) should become Sum_{t | v} (-1)^(v/t) * c(t) = 0 for v > 1, which is the equation a(n) = Sum{k | n, 1 < k} (-1)^k a(n/k), for n >= 2, in the Formula section below. - Petros Hadjicostas, Jul 21 2019]
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FORMULA
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a(1) = 1 and a(n) = Sum{k | n, 1 < k} (-1)^k a(n/k) for n >= 2; the sum is over the divisors, k, of n, where k > 1. If n is odd, a(n) = mu(n), where mu(.) is the Moebius function. If n is even, a(n) = mu(m)* 2^(k-1), where n = m*2^k, m is odd integer, and k is a positive integer.
Sum_{n > 0} a(n)*x^n/(1 + x^n) = x. Moebius transform of A048298. Multiplicative with a(2^e) = 2^(e - 1), a(p) = -1 for p > 2, a(p^e) = 0 for p > 2 and e > 1. - Vladeta Jovovic, Jan 02 2003
Sum_{n > 0} a(n)*log(1 + x^n)/n = x. - Paul D. Hanna, May 06 2003
a(n) = 0 if and only if n is divisible by the square of an odd prime (A038838). - Michael Somos, Aug 22 2006
1 = Sum_{k=1..n} a(k)*g(n/k), where g(x) = floor(x) - 2*floor(x/2). - Benoit Cloitre, Nov 11 2010
Dirichlet g.f.: 1/( zeta(s) * (1 - 2^(1-s)) ). - R. J. Mathar, Apr 02 2011
Sum_{n >= 1} a(n)*x^n/(1 + x^n) = x
Sum_{n >= 1} a(n)*x^n/(1 - x^n) = x + 2*x^2 + 4*x^4 + 8*x^8 + 16*x^16 + ...
Sum_{n >= 1} a(n)*x^n/(1 + (-x)^n) = x + 2*(x^2 + x^4 + x^8 + x^16 + ...)
Sum_{n >= 1} a(n)*x^n/(1 - (-x)^n) = x + 2*(x^4 + 3*x^8 + 7*x^16 + 15*x^32 + ...). (End)
G.f. A(x) satisfies: A(x) = x + A(x^2) - A(x^3) + A(x^4) - A(x^5) + ... - Ilya Gutkovskiy, May 11 2019
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MATHEMATICA
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a[n_] := If[n == 1, 1, Product[{p, e} = pe; If[2 == p, e--, If[e > 1, p = 0, p = -1]]; p^e, {pe, FactorInteger[n]}]];
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PROG
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(PARI) {a(n)=local(k); if(n<1, 0, k=valuation(n, 2); moebius(n/2^k)*2^max(0, k-1))} /* Michael Somos, Aug 22 2006 */
(PARI) A067856(n) = { my(f=factor(n)); for(i=1, #f~, if(2==f[i, 1], f[i, 2]--, if(f[i, 2]>1, f[i, 1]=0, f[i, 1]=-1))); factorback(f); }; \\ Antti Karttunen, Sep 27 2018, after Vladeta Jovovic_'s multiplicative formula.
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CROSSREFS
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KEYWORD
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sign,mult
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AUTHOR
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STATUS
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approved
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