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A077415
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a(n) = n*(n+2)*(n-2)/3.
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19
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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
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OFFSET
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2,2
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COMMENTS
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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
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
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LINKS
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FORMULA
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G.f.: x^3*(5 - 4*x + x^2)/(1-x)^4.
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
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
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MAPLE
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MATHEMATICA
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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 *)
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PROG
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(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
(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
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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