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%I #65 Mar 06 2024 04:50:34
%S 0,5,16,35,64,105,160,231,320,429,560,715,896,1105,1344,1615,1920,
%T 2261,2640,3059,3520,4025,4576,5175,5824,6525,7280,8091,8960,9889,
%U 10880,11935,13056,14245,15504,16835,18240,19721,21280,22919,24640,26445
%N a(n) = n*(n+2)*(n-2)/3.
%C 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.)
%C Number of standard tableaux of shape (n-1,2,1) (n>=3). - _Emeric Deutsch_, May 13 2004
%C Zero followed by partial sums of A028387, starting at n=1. - _Klaus Brockhaus_, Oct 21 2008
%C 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
%C 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
%H G. C. Greubel, <a href="/A077415/b077415.txt">Table of n, a(n) for n = 2..10000</a>
%H Mark Roger Sepanski, <a href="https://doi.org/10.37236/3350">On Divisibility of Convolutions of Central Binomial Coefficients</a>, Electronic Journal of Combinatorics, 21 (1) 2014, #P1.32.
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).
%F a(n) = n*(n+2)*(n-2)/3 = A077414(n) - binomial(n+2,3) = A077414(n) - A000292(n-1).
%F G.f.: x^3*(5 - 4*x + x^2)/(1-x)^4.
%F a(n) = A084990(n-1) - 1. - _Reinhard Zumkeller_, Aug 20 2007
%F 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
%F From _Amiram Eldar_, Jan 06 2021: (Start)
%F Sum_{n>=3} 1/a(n) = 11/32.
%F Sum_{n>=3} (-1)^(n+1)/a(n) = 5/32. (End)
%F E.g.f.: x*(1 + exp(x)*(x^2 + 3*x - 3)/3). - _Stefano Spezia_, Mar 06 2024
%p seq((n^3-4*n)/3, n=2..35); # _Zerinvary Lajos_, Jan 20 2007
%t 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 *)
%t LinearRecurrence[{4, -6, 4, -1}, {0, 5, 16, 35}, 50] (* _Vincenzo Librandi_, Jan 09 2016 *)
%t Table[n*(n + 2)*(n - 2)/3, {n, 2, 50}] (* _G. C. Greubel_, Jan 18 2018 *)
%o (PARI) {a=0; print1(a,","); for(n=1, 42, print1(a=a+n+(n+1)^2, ","))} \\ _Klaus Brockhaus_, Oct 21 2008
%o (PARI) concat(0, Vec(x^3*(5-4*x+x^2)/(1-x)^4 + O(x^100))) \\ _Altug Alkan_, Jan 08 2015
%o (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
%Y Cf. A000292, A028387 (first differences), A033275 (partial sums), A060351, A077414, A084990.
%K nonn,easy
%O 2,2
%A _Wolfdieter Lang_, Nov 29 2002
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