A154323 Central coefficients of number triangle A113582.
1, 2, 10, 37, 101, 226, 442, 785, 1297, 2026, 3026, 4357, 6085, 8282, 11026, 14401, 18497, 23410, 29242, 36101, 44101, 53362, 64010, 76177, 90001, 105626, 123202, 142885, 164837, 189226, 216226, 246017, 278785, 314722, 354026, 396901, 443557, 494210, 549082, 608401, 672401, 741322, 815410, 894917, 980101
Offset: 0
Links
- G. C. Greubel, Table of n, a(n) for n = 0..1000
- Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).
Programs
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Magma
[(n^4 + 2*n^3 + n^2 + 4)/4: n in [0..40]]; // Vincenzo Librandi, Feb 13 2015
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Mathematica
s = 1; lst = {s}; Do[s += n^3; AppendTo[lst, s], {n, 1, 42, 1}]; lst (* Zerinvary Lajos, Jul 12 2009 *) Table[n!^3*Det[Array[KroneckerDelta[#1,#2](((#1^3+1)/(#1^3))-1)+1&,{n,n}]],{n,1,30}] (* John M. Campbell, May 20 2011 *) LinearRecurrence[{5, -10, 10, -5, 1}, {1, 2, 10, 37, 101}, 25] (* or *) Table[(n^4 + 2*n^3 + n^2 + 4)/4, {n,0,25}] (* G. C. Greubel, Sep 11 2016 *)
Formula
a(n) = (n^4 + 2*n^3 + n^2 + 4)/4.
G.f.: (1 - 3*x + 10*x^2 - 3*x^3 + x^4)/(1-x)^5.
a(n) = 1 + C(n+1,2)^2 = 1 + A000537(n).
From G. C. Greubel, Sep 11 2016: (Start)
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
E.g.f.: (1/4)*(4 + 4*x + 14*x^2 + 8*x^3 + x^4)*exp(x). (End)
a(n) = a(n-1)+n^3. - Charles U. Lonappan, Jun 09 2021
Comments