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A067856 Sum_{n >= 1} a(n)/n^s = 1/(Sum_{n >= 1} (-1)^(n + 1)/n^s). 30

%I #93 Jan 14 2021 06:37:43

%S 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,

%T 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,

%U -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

%N Sum_{n >= 1} a(n)/n^s = 1/(Sum_{n >= 1} (-1)^(n + 1)/n^s).

%C Dirichlet inverse of A062157. - _R. J. Mathar_, Jul 15 2010

%C 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

%C From _Peter Bala_, Mar 12 2019: (Start)

%C 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).

%C 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)

%C 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

%C From _Petros Hadjicostas_, Jul 25 2020: (Start)

%C For p prime >= 2, Petrogradsky (2003) defined the multiplicative functions 1_p and mu_p in the following way:

%C 1_p(n) = 1 when gcd(n,p) = 1 and 1_p(n) = 1 - p when gcd(n,p) = p;

%C 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.

%C 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.

%C Some of the results by other contributors can be generalized:

%C (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.

%C (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)).

%C (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.

%C (iv) _Paul D. Hanna_'s formula becomes Sum_{n >= 1} (mu_p(n)/n)*log((1 - x^(n*p))/(1 - x^n)) = x.

%C (v) The definition of A117997 generalizes to Sum_{d | n} mu_p(d) = n, if n = p^s for s >= 0, and = 0, otherwise.

%C (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".

%C (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)

%D G. Pólya and G. Szegő, Problems and Theorems in Analysis Volume II. Springer_Verlag 1976.

%H Antti Karttunen, <a href="/A067856/b067856.txt">Table of n, a(n) for n = 1..65537</a>

%H Peter Bala, <a href="/A067856/a067856_1.pdf">A signed Dirichlet product of arithmetical functions</a>

%H V. M. Petrogradsky, <a href="https://core.ac.uk/download/pdf/82129961.pdf">Witt's formula for restricted Lie algebras</a>, Advances in Applied Mathematics, 30 (2003), 219-227.

%H V. M. Petrogradsky, <a href="https://doi.org/10.1016/S0196-8858(02)00533-X">Witt's formula for restricted Lie algebras</a>, Advances in Applied Mathematics, 30 (2003), 219-227.

%H Franz Rogel, <a href="https://www.zobodat.at/pdf/SB-Ges-Wiss-Prag_1896_2_0001-0687.pdf">Transformationen arithmetischer Reihen</a>, 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]

%F 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.

%F 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

%F Sum_{n > 0} a(n)*log(1 + x^n)/n = x. - _Paul D. Hanna_, May 06 2003

%F a(n) = 0 if and only if n is divisible by the square of an odd prime (A038838). - _Michael Somos_, Aug 22 2006

%F 1 = Sum_{k=1..n} a(k)*g(n/k), where g(x) = floor(x) - 2*floor(x/2). - _Benoit Cloitre_, Nov 11 2010

%F Dirichlet g.f.: 1/( zeta(s) * (1 - 2^(1-s)) ). - _R. J. Mathar_, Apr 02 2011

%F From _Peter Bala_, Mar 13 2019: (Start)

%F Sum_{n >= 1} a(n)*x^n/(1 + x^n) = x

%F Sum_{n >= 1} a(n)*x^n/(1 - x^n) = x + 2*x^2 + 4*x^4 + 8*x^8 + 16*x^16 + ...

%F Sum_{n >= 1} a(n)*x^n/(1 + (-x)^n) = x + 2*(x^2 + x^4 + x^8 + x^16 + ...)

%F Sum_{n >= 1} a(n)*x^n/(1 - (-x)^n) = x + 2*(x^4 + 3*x^8 + 7*x^16 + 15*x^32 + ...). (End)

%F 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

%t 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]}]];

%t a /@ Range[1, 100] (* _Jean-François Alcover_, Oct 01 2019 *)

%o (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 */

%o (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.

%Y Cf. A000009, A038712, A038838, A048298 (inverse Mobius transform), A061347, A062157, A085906, A117997, A321088 (Euler transform), A321558.

%K sign,mult

%O 1,4

%A _Leroy Quet_, Feb 15 2002

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