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 A168566 a(n) = (n-1)*(n+2)*(n^2 + n + 2)/4. 2
 0, 8, 35, 99, 224, 440, 783, 1295, 2024, 3024, 4355, 6083, 8280, 11024, 14399, 18495, 23408, 29240, 36099, 44099, 53360, 64008, 76175, 89999, 105624, 123200, 142883, 164835, 189224, 216224, 246015, 278783, 314720, 354024, 396899, 443555, 494208 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS The products of two consecutive numbers in this sequence may be evaluated in terms of the Frobenius numbers for 5 consecutive integers, A138985(n) = F(n): for n>0, a(2n-1)*a(2n) = F(4n^2-2)^2 - (2n)^2; a(2n)*a(2n+1) = F(4n^2+4n)^2 - (2n+1)^2. - Charlie Marion, Jan 23 2012 LINKS Vincenzo Librandi, Table of n, a(n) for n = 1..10000 Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1). FORMULA G.f.: x^2*(-8 + 5*x - 4*x^2 + x^3)/(x-1)^5. - R. J. Mathar, Jan 04 2011 a(n) = A000217(n)^2 - 1 = A000537(n)-1. - Charlie Marion, Sep 27 2011 From G. C. Greubel, Jul 26 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 - 4*x + 14*x^2 + 8*x^3)*exp(x)). (End) MATHEMATICA s=0; lst={s}; Do[s+=n^3; AppendTo[lst, s], {n, 2, 5!}]; lst (* Vladimir Joseph Stephan Orlovsky, Apr 27 2010 *) Table[(n-1)*(n+2)*(n^2 + n + 2)/4, {n, 1, 25}] (* G. C. Greubel, Jul 26 2016 *) LinearRecurrence[{5, -10, 10, -5, 1}, {0, 8, 35, 99, 224}, 40] (* Harvey P. Dale, Jan 21 2019 *) PROG (MAGMA) [(n-1)*(n+2)*(n^2+n+2)/4: n in [1..50]] (PARI) a(n)=(n-1)*(n+2)*(n^2+n+2)/4 \\ Charles R Greathouse IV, Jul 26 2016 CROSSREFS Cf. A000578 (first differences). Sequence in context: A303383 A265840 A212903 * A058102 A212674 A279743 Adjacent sequences:  A168563 A168564 A168565 * A168567 A168568 A168569 KEYWORD nonn,easy AUTHOR Vincenzo Librandi, Nov 30 2009 STATUS approved

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Last modified June 27 12:48 EDT 2022. Contains 354896 sequences. (Running on oeis4.)