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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
Eric Weisstein's World of Mathematics, Centered Polygonal Numbers
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)
E.g.f.: exp(x)*(1 + 13*x^2/2) - 1. - Stefano Spezia, May 15 2022
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
Sequence in context: A181149 A019063 A101740 * A124707 A126368 A125816
KEYWORD
easy,nice,nonn
AUTHOR
Terrel Trotter, Jr., Apr 07 2002
STATUS
approved

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Last modified April 20 07:26 EDT 2024. Contains 371799 sequences. (Running on oeis4.)