A309337 a(n) = n^3 if n odd, 3*n^3/4 if n even.

0, 1, 6, 27, 48, 125, 162, 343, 384, 729, 750, 1331, 1296, 2197, 2058, 3375, 3072, 4913, 4374, 6859, 6000, 9261, 7986, 12167, 10368, 15625, 13182, 19683, 16464, 24389, 20250, 29791, 24576, 35937, 29478, 42875, 34992, 50653, 41154, 59319, 48000, 68921, 55566, 79507, 63888, 91125

Offset

0,3

Comments

Moebius transform of A078307.

Links

Amiram Eldar, Table of n, a(n) for n = 0..10000

Index entries for linear recurrences with constant coefficients, signature (0,4,0,-6,0,4,0,-1).

Formula

G.f.: x * (1 + 6*x + 23*x^2 + 24*x^3 + 23*x^4 + 6*x^5 + x^6)/(1 - x^2)^4.

G.f.: Sum_{k>=1} J_3(k) * x^k/(1 + x^k), where J_3() is the Jordan function (A059376).

Dirichlet g.f.: zeta(s-3) * (1 - 2^(1-s)).

a(n) = n^3 * (7 - (-1)^n)/8.

a(n) = Sum_{d|n} (-1)^(n/d + 1) * J_3(d).

Sum_{n>=1} 1/a(n) = 25*zeta(3)/24 = 1.252142607457910713958...

Multiplicative with a(2^e) = 3*2^(3*e-2), and a(p^e) = p^(3*e) for odd primes p. - Amiram Eldar, Oct 26 2020

a(n) = Sum_{1 <= i, j, k <= n} (-1)^(1 + gcd(i,j,k,n)) = Sum_{d | n} (-1)^(d+1) * J_3(n/d). Cf. A129194. - Peter Bala, Jan 16 2024

Mathematica

a[n_] := If[OddQ[n], n^3, 3 n^3/4]; Table[a[n], {n, 0, 45}]
nmax = 45; CoefficientList[Series[x (1 + 6 x + 23 x^2 + 24 x^3 + 23 x^4 + 6 x^5 + x^6)/(1 - x^2)^4, {x, 0, nmax}], x]
LinearRecurrence[{0, 4, 0, -6, 0, 4, 0, -1}, {0, 1, 6, 27, 48, 125, 162, 343}, 46]
Table[n^3 (7 - (-1)^n)/8, {n, 0, 45}]

Crossrefs

Cf. A000578, A016755, A059376, A078307, A129194, A193356, A309338.

Sequence in context: A292384 A183603 A067876 * A174226 A100784 A174974

Adjacent sequences: A309334 A309335 A309336 * A309338 A309339 A309340

Keyword

nonn,easy,mult

Author

Ilya Gutkovskiy, Jul 24 2019

Status

approved