This site is supported by donations to The OEIS Foundation.

 Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
 A264853 a(n) = n*(n + 1)*(5*n^2 + 5*n - 4)/12. 2
 0, 1, 13, 56, 160, 365, 721, 1288, 2136, 3345, 5005, 7216, 10088, 13741, 18305, 23920, 30736, 38913, 48621, 60040, 73360, 88781, 106513, 126776, 149800, 175825, 205101, 237888, 274456, 315085, 360065, 409696, 464288, 524161, 589645, 661080, 738816, 823213, 914641 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,3 COMMENTS Partial sums of centered 10-gonal (or decagonal) pyramidal numbers. Subsequence of A204221. In fact, a(n) is of the form (k^2-1)/15 for k = 5*n*(n+1)/2-1. - Bruno Berselli, Nov 27 2015 LINKS OEIS Wiki, Figurate numbers Eric Weisstein's World of Mathematics, Pyramidal Number Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1). FORMULA G.f.: x*(1 + 8*x + x^2)/(1 - x)^5. a(n) = Sum_{k = 0..n} A004466(k). a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). - Vincenzo Librandi, Nov 27 2015 MATHEMATICA Table[n (n + 1) (5 n^2 + 5 n - 4)/12, {n, 0, 50}] LinearRecurrence[{5, -10, 10, -5, 1}, {0, 1, 13, 56, 160}, 40] (* Harvey P. Dale, Aug 14 2017 *) PROG (MAGMA) [n*(n+1)*(5*n^2+5*n-4)/12: n in [0..50]]; // Vincenzo Librandi, Nov 27 2015 (PARI) a(n)=n*(n+1)*(5*n^2+5*n-4)/12 \\ Charles R Greathouse IV, Jul 26 2016 CROSSREFS Cf. A004466, A204221. Cf. similar sequences listed in A264854. Sequence in context: A290396 A061161 A212053 * A210290 A007202 A222161 Adjacent sequences:  A264850 A264851 A264852 * A264854 A264855 A264856 KEYWORD nonn,easy AUTHOR Ilya Gutkovskiy, Nov 26 2015 STATUS approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent | More pages
The OEIS Community | Maintained by The OEIS Foundation Inc.

Last modified September 20 07:11 EDT 2018. Contains 315226 sequences. (Running on oeis4.)