A208950 a(4*n) = n*(16*n^2-1)/3, a(2*n+1) = n*(n+1)*(2*n+1)/6, a(4*n+2) = (4*n+1)*(4*n+2)*(4*n+3)/6.

0, 0, 1, 1, 5, 5, 35, 14, 42, 30, 165, 55, 143, 91, 455, 140, 340, 204, 969, 285, 665, 385, 1771, 506, 1150, 650, 2925, 819, 1827, 1015, 4495, 1240, 2728, 1496, 6545, 1785, 3885, 2109, 9139, 2470, 5330, 2870, 12341, 3311, 7095, 3795, 16215, 4324

Offset: 0

Author

Paul Curtz, Mar 03 2012

Keywords

• nonn
• easy

Comments

a(n+2) is divisible by A060819(floor(n/3)).

a(n) is divisible by A176672(floor(n/3)).

Denominator of a(n)/n is of period 24: 1,1,3,4,1,6,1,4,3,1,1,12,1,2,3,4,1,3,1,4,3,2,1,12 (two successive palindromes).

This is the fifth column of the triangle A107711, hence the formula involving gcd(n+2,4) given below follows. - Wolfdieter Lang, Feb 24 2014

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..5000
• Index entries for linear recurrences with constant coefficients, signature (0,0,0,4,0,0,0,-6,0,0,0,4,0,0,0,-1).

Crossrefs

Cf. A107711 (fifth column).

Cf. A000034, A000292, A002415, A051724, A060819, A061037, A107711, A138190, A145979, A176672, A176895.

Programs

Magma

[Binomial(n+1,3)*GCD(n+2,4)/4: n in [0..50]]; // G. C. Greubel, Sep 20 2018

Mathematica

CoefficientList[Series[(x^2 + x^3 + 5 x^4 + 5 x^5 + 31 x^6 + 10 x^7 + 22 x^8 + 10 x^9 + 31 x^10 + 5 x^11 + 5 x^12 + x^13 + x^14)/((1 - x)^4 (1 + x)^4 (1 + 4 x^2 + 6 x^4 + 4 x^6 + x^8)), {x, 0, 47}], x] (* Bruno Berselli, Mar 11 2012 *)

Maxima

A208950(n) := block(
        [a,npr] ,
        if equal(mod(n,4), 0) then (
                a : n/12*(n^2-1)
        ) else if equal(mod(n,2),0) then (
                a : (n-1)*n*(n+1)/6
        ) else (
                npr : (n-1)/2,
                a : npr*(npr+1)*n/6
        ) ,
        return(a)
)$ /* R. J. Mathar, Mar 10 2012 */

PARI

vector(50, n, n--; binomial(n+1,3)*gcd(n+2,4)/4) \\ G. C. Greubel, Sep 20 2018

Formula

a(n) = 4*a(n-4) - 6*a(n-8) + 4*a(n-12) - a(n-16).

a(n+1) = A002415(n+1)/A145979(n-1).

a(n) = A051724(n-1) * A051724(n) * A051724(n+1).

a(n) = A060819(n-1) * A060819(n) * A060819(n+1) / 3.

a(n) * a(n+4) = A061037(n+1) * A061037(n+2) * A061037(n+3) / 9.

a(n) = A138190(n)/A000034(n) for n > 0.

a(n) = A000292(n-1)/A176895(n+2) for n > 0.

a(n)/a(n+4) = n*(n^2-1)/((n+3)*(n+4)*(n+5)).

a(n)/a(n+12) = (n-1)*n*(n+1)/((n+11)*(n+12)*(n+13)).

G.f.: (x^2 + x^3 + 5*x^4 + 5*x^5 + 31*x^6 + 10*x^7 + 22*x^8 + 10*x^9 + 31*x^10 + 5*x^11 + 5*x^12 + x^13 + x^14) / ((1-x)^4*(1+x)^4*(1 + 4*x^2 + 6*x^4 + 4*x^6 + x^8)). - R. J. Mathar, Mar 10 2012

From Wolfdieter Lang, Feb 24 2014: (Start)

G.f.: (1 + x^12 + x*(1+x^10) + 5*x^2*(1+x^8) + 5*x^3*(1+x^7) + 31*x^4*(1+x^4) + 10*x^5*(1+x^2) + 22*x^6)/(1-x^4)^4. This is the preceding g.f. rewritten.

a(n) = binomial(n+1,3)*gcd(n+2,4)/4, n >= 0. From the g.f., see a comment above on A107711. (End)

a(n) = (n*(n-1)*((n+1)*(4+2*(-1)^n + (1+(-1)^n)*(-1)^((2*n+3+(-1)^n)/4))))/48. - Luce ETIENNE, Jan 01 2015

Sum_{n>=2} 1/a(n) = 12 - 27*log(2)/2. - Amiram Eldar, Aug 12 2022