A129194 a(n) = (n/2)^2*(3 - (-1)^n).

0, 1, 2, 9, 8, 25, 18, 49, 32, 81, 50, 121, 72, 169, 98, 225, 128, 289, 162, 361, 200, 441, 242, 529, 288, 625, 338, 729, 392, 841, 450, 961, 512, 1089, 578, 1225, 648, 1369, 722, 1521, 800, 1681, 882, 1849, 968, 2025, 1058, 2209, 1152, 2401, 1250, 2601, 1352

Offset

0,3

Comments

The numerator of the integral is 2,1,2,1,2,1,...; the moments of the integral are 2/(n+1)^2. See 2nd formula.

The sequence alternates between twice a square and an odd square, A001105(n) and A016754(n).

Partial sums of the positive elements give the absolute values of A122576. - Omar E. Pol, Aug 22 2011

Partial sums of the positive elements give A212760. - Omar E. Pol, Dec 28 2013

Conjecture: denominator of 4/n - 2/n^2. - Wesley Ivan Hurt, Jul 11 2016

Multiplicative because both A000290 and A040001 are. - Andrew Howroyd, Jul 25 2018

References

G. Pólya and G. Szegő, Problems and Theorems in Analysis II (Springer 1924, reprinted 1976), Part Eight, Chap. 1, Sect. 7, Problem 73.

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..960

Olivier Bordelles, A Multidimensional Cesaro Type Identity and Applications, J.

Int. Seq. 18 (2015) # 15.3.7.

John M. Campbell, An Integral Representation of Kekulé Numbers, and Double Integrals Related to Smarandache Sequences, arXiv preprint arXiv:1105.3399 [math.GM], 2011.

Index entries for linear recurrences with constant coefficients, signature (0,3,0,-3,0,1).

Formula

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

a(n+1) = denominator((1/(2*Pi))*Integral_{t=0..2*Pi} exp(i*n*t)(-((Pi-t)/i)^2)), i=sqrt(-1).

a(n) = 3*a(n-2) - 3*a(n-4) + a(n-6) for n > 5. - Paul Curtz, Mar 07 2011

a(n) is the numerator of the coefficient of x^4 in the Maclaurin expansion of exp(-n*x^2). - Francesco Daddi, Aug 04 2011

O.g.f. as a Lambert series: x*Sum_{n >= 1} J_2(n)*x^n/(1 + x^n), where J_2(n) denotes the Jordan totient function A007434(n). See Pólya and Szegő. - Peter Bala, Dec 28 2013

From Ilya Gutkovskiy, Jul 11 2016: (Start)

E.g.f.: x*((2*x + 1)*sinh(x) + (x + 2)*cosh(x))/2.

Sum_{n>=1} 1/a(n) = 5*Pi^2/24. [corrected by Amiram Eldar, Sep 11 2022] (End)

a(n) = A000290(n) / A040001(n). - Andrew Howroyd, Jul 25 2018

Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/24 (A222171). - Amiram Eldar, Sep 11 2022

From Peter Bala, Jan 16 2024: (Start)

a(n) = Sum_{1 <= i, j <= n} (-1)^(1 + gcd(i,j,n)) = Sum_{d | n} (-1)^(d+1) * J_2(n/d), that is, the Dirichlet convolution of the pair of multiplicative functions f(n) = (-1)^(n+1) and the Jordan totient function J_2(n) = A007434(n). Hence this sequence is multiplicative. Cf. A193356 and A309337.

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

a(n) = Sum_{1 <= i, j <= n} (-1)^(n + gcd(i, n)*gcd(j, n)) = Sum_{d|n, e|n} (-1)^(n+e*d) * phi(n/d)*phi(n/e). - Peter Bala, Jan 22 2024

Maple

A129194:=n->n^2*(3-(-1)^n)/4: seq(A129194(n), n=0..80); # Wesley Ivan Hurt, Jul 11 2016

Mathematica

Table[n^2*(3-(-1)^n)/4, {n, 0, 60}] (* Wesley Ivan Hurt, Jul 11 2016 *)
LinearRecurrence[{0, 3, 0, -3, 0, 1}, {0, 1, 2, 9, 8, 25}, 60] (* Harvey P. Dale, Dec 27 2023 *)

Prog

(Magma) [n^2*(3-(-1)^n)/4: n in [0..60]]; // Vincenzo Librandi, Apr 26 2011
(PARI) a(n) = lcm(2, n^2)/2; \\ Andrew Howroyd, Jul 25 2018
(SageMath) [n^2*(1+(n%2))/2 for n in range(61)] # G. C. Greubel, Apr 04 2023

Crossrefs

Cf. A000290, A001105, A007434, A010713, A016742, A016754, A040001.

Cf. A061038, A061040, A061050, A105398, A129204, A152020, A222171, A309337.

Sequence in context: A215025 A162954 A357993 * A300780 A272347 A371255

Adjacent sequences: A129191 A129192 A129193 * A129195 A129196 A129197

Keyword

easy,frac,nonn,mult

Author

Paul Barry, Apr 02 2007

Extensions

More terms from Michel Marcus, Dec 28 2013

Status

approved