A077415 a(n) = n*(n+2)*(n-2)/3.

0, 5, 16, 35, 64, 105, 160, 231, 320, 429, 560, 715, 896, 1105, 1344, 1615, 1920, 2261, 2640, 3059, 3520, 4025, 4576, 5175, 5824, 6525, 7280, 8091, 8960, 9889, 10880, 11935, 13056, 14245, 15504, 16835, 18240, 19721, 21280, 22919, 24640, 26445

Offset: 2

Author

Wolfdieter Lang, Nov 29 2002

Keywords

• nonn
• easy

Comments

a(n) is the number of independent components of a 3-tensor t(a,b,c) which satisfies t(a,b,c)=t(b,a,c) and sum(t(a,a,c),a=1..n)=0 for all c and t(a,b,c)+t(b,c,a)+t(c,a,b)=0, with a,b,c range 1..n. (3-tensor in n-dimensional space which is symmetric and traceless in one pair of its indices and satisfies the cyclic identity.)

Number of standard tableaux of shape (n-1,2,1) (n>=3). - Emeric Deutsch, May 13 2004

Zero followed by partial sums of A028387, starting at n=1. - Klaus Brockhaus, Oct 21 2008

For n>=4, a(n-1) is the number of permutations of 1,2...,n with the distribution of up (1) - down (0) elements 0...0101 (the first n-4 zeros), or, the same, a(n-1) is up-down coefficient {n,5} (see comment in A060351). - Vladimir Shevelev, Feb 14 2014

For n>=3, a(n) equals the second immanant of the (n-1) X (n-1) tridiagonal matrix with 2's along the main diagonal, and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jan 08 2016

Links

• G. C. Greubel, Table of n, a(n) for n = 2..10000
• Mark Roger Sepanski, On Divisibility of Convolutions of Central Binomial Coefficients, Electronic Journal of Combinatorics, 21 (1) 2014, #P1.32.
• Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Crossrefs

Cf. A000292, A028387 (first differences), A033275 (partial sums), A060351, A077414, A084990.

Programs

Magma

[n*(n+2)*(n-2)/3: n in [2..50]]; /* or */ I:=[0,5,16,35]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..50]]; // Vincenzo Librandi, Jan 09 2016

Maple

seq((n^3-4*n)/3, n=2..35); # Zerinvary Lajos, Jan 20 2007

Mathematica

Print[Table[Sum[(-1)^i*2^(n-2*i-1)*Binomial[n-i-1, i]*(n-2*i-2), {i, 0, Floor[(n-1)/2]}], {n, 2, 100}]] ;  (* John M. Campbell, Jan 08 2016 *)
LinearRecurrence[{4, -6, 4, -1}, {0, 5, 16, 35}, 50] (* Vincenzo Librandi, Jan 09 2016 *)
Table[n*(n + 2)*(n - 2)/3, {n, 2, 50}] (* G. C. Greubel, Jan 18 2018 *)

PARI

{a=0; print1(a,","); for(n=1, 42, print1(a=a+n+(n+1)^2, ","))} \\ Klaus Brockhaus, Oct 21 2008

PARI

concat(0, Vec(x^3*(5-4*x+x^2)/(1-x)^4 + O(x^100))) \\ Altug Alkan, Jan 08 2015

Formula

a(n) = n*(n+2)*(n-2)/3 = A077414(n) - binomial(n+2,3) = A077414(n) - A000292(n-1).

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

a(n) = A084990(n-1) - 1. - Reinhard Zumkeller, Aug 20 2007

a(n) = Sum_{i=0..floor((n-1)/2)} (-1)^i * 2^(n-2*i-1) * binomial(n-i-1, i) * (n-2*i-2). - John M. Campbell, Jan 08 2016

From Amiram Eldar, Jan 06 2021: (Start)

Sum_{n>=3} 1/a(n) = 11/32.

Sum_{n>=3} (-1)^(n+1)/a(n) = 5/32. (End)

E.g.f.: x*(1 + exp(x)*(x^2 + 3*x - 3)/3). - Stefano Spezia, Mar 06 2024