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 A069126 Centered 13-gonal numbers. 8
 1, 14, 40, 79, 131, 196, 274, 365, 469, 586, 716, 859, 1015, 1184, 1366, 1561, 1769, 1990, 2224, 2471, 2731, 3004, 3290, 3589, 3901, 4226, 4564, 4915, 5279, 5656, 6046, 6449, 6865, 7294, 7736, 8191, 8659, 9140, 9634, 10141, 10661, 11194 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS Centered tridecagonal numbers or centered triskaidecagonal numbers. - Omar E. Pol, Oct 03 2011 LINKS Ivan Panchenko, Table of n, a(n) for n = 1..1000 Eric Weisstein's World of Mathematics, Centered Polygonal Numbers Index entries for linear recurrences with constant coefficients, signature (3,-3,1) FORMULA a(n) = (13n^2 - 13n + 2)/2 Binomial transform of [1, 13, 13, 0, 0, 0,...]; Narayana transform (A001263) of [1, 13, 0, 0, 0,...]. - Gary W. Adamson, Dec 29 2007 a(n) = 13*n+a(n-1)-13 (with a(1)=1). - Vincenzo Librandi, Aug 08 2010 G.f. -x*(1+11*x+x^2) / (x-1)^3 . - R. J. Mathar, Feb 04 2011 a(n) = A152741(n-1) + 1. - Omar E. Pol, Oct 03 2011 From Amiram Eldar, Jun 21 2020: (Start) Sum_{n>=1} 1/a(n) = 2*Pi*tan(sqrt(5/13)*Pi/2)/sqrt(65). Sum_{n>=1} a(n)/n! = 15*e/2 - 1. Sum_{n>=1} (-1)^n * a(n)/n! = 15/(2*e) - 1. (End) EXAMPLE a(5) = 131 because 131 = (13*5^2 - 13*5 + 2)/2 = (325 - 65 + 2)/2 = 262/2 = 131. MATHEMATICA FoldList[#1 + #2 &, 1, 13 Range@ 45] (* Robert G. Wilson v, Feb 02 2011 *) LinearRecurrence[{3, -3, 1}, {1, 14, 40}, 60] (* Harvey P. Dale, Jan 20 2014 *) With[{nn=50}, Total/@Thread[{PolygonalNumber[13, Range[nn]], Range[0, nn-1]^2}]] (* Requires Mathematica version 11 or later *) (* Harvey P. Dale, Aug 29 2016 *) PROG (PARI) a(n)=13*n(n-1)/2+1 \\ Charles R Greathouse IV, Oct 07 2015 CROSSREFS Cf. A005448, A001844, A005891, A003215, A069099, A001263. Sequence in context: A181149 A019063 A101740 * A124707 A126368 A125816 Adjacent sequences:  A069123 A069124 A069125 * A069127 A069128 A069129 KEYWORD easy,nice,nonn AUTHOR Terrel Trotter, Jr., Apr 07 2002 STATUS approved

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Last modified October 24 07:04 EDT 2020. Contains 337975 sequences. (Running on oeis4.)