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 A154323 Central coefficients of number triangle A113582. 5
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS a(n) equals n!^3 times the determinant of the n X n matrix whose (i,j)-entry is KroneckerDelta[i, j] (((i^3 + 1)/(i^3)) - 1) + 1. - John M. Campbell, May 20 2011 Let b(0)=b(1)=1; b(n)=max(b(n-1)+(n-1)^3, b(n-2)+(n-2)^3); then a(n)=b(n+1). - Yalcin Aktar, Jul 28 2011 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). 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 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 *) PROG (Magma) [(n^4 + 2*n^3 + n^2 + 4)/4: n in [0..40]]; // Vincenzo Librandi, Feb 13 2015 CROSSREFS Sequence in context: A151022 A144895 A236767 * A191349 A073110 A034547 Adjacent sequences: A154320 A154321 A154322 * A154324 A154325 A154326 KEYWORD easy,nonn AUTHOR Paul Barry, Jan 07 2009 STATUS approved

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Last modified December 2 16:43 EST 2022. Contains 358510 sequences. (Running on oeis4.)