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A069132 Centered 19-gonal numbers. 5
1, 20, 58, 115, 191, 286, 400, 533, 685, 856, 1046, 1255, 1483, 1730, 1996, 2281, 2585, 2908, 3250, 3611, 3991, 4390, 4808, 5245, 5701, 6176, 6670, 7183, 7715, 8266, 8836, 9425, 10033, 10660, 11306, 11971, 12655, 13358, 14080, 14821, 15581, 16360, 17158 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
COMMENTS
Binomial transform of [1, 19, 19, 0, 0, 0, ...] and Narayana transform (A001263) of [1, 19, 0, 0, 0, ...]. - Gary W. Adamson, Jul 28 2011
LINKS
Eric Weisstein's World of Mathematics, Centered Polygonal Numbers
FORMULA
a(n) = (19*n^2 - 19*n + 2)/2.
a(n) = 19*n + a(n-1) - 19 (with a(1)=1). - Vincenzo Librandi, Aug 08 2010
G.f.: x*(1 + 17*x + x^2) / (1-x)^3. - R. J. Mathar, Feb 04 2011
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3); a(0)=1, a(1)=20, a(2)=58. - Harvey P. Dale, Aug 21 2011
From Amiram Eldar, Jun 21 2020: (Start)
Sum_{n>=1} 1/a(n) = 2*Pi*tan(sqrt(11/19)*Pi/2)/sqrt(209).
Sum_{n>=1} a(n)/n! = 21*e/2 - 1.
Sum_{n>=1} (-1)^n * a(n)/n! = 21/(2*e) - 1. (End)
E.g.f.: exp(x)*(1 + 19*x^2/2) - 1. - Nikolaos Pantelidis, Feb 06 2023
EXAMPLE
a(5)= 191 because (19*5^2 - 19*5 + 2)/2 = (475 - 95 + 2)/2 = 382/2 = 191.
MATHEMATICA
FoldList[#1 + #2 &, 1, 19 Range@ 45] (* Robert G. Wilson v, Feb 02 2011 *)
Table[(19n^2-19n+2)/2, {n, 50}] (* or *) LinearRecurrence[{3, -3, 1}, {1, 20, 58}, 50] (* Harvey P. Dale, Aug 21 2011 *)
PROG
(PARI) a(n)=19*binomial(n, 2)+1 \\ Charles R Greathouse IV, Jul 29 2011
CROSSREFS
Cf. centered polygonal numbers listed in A069190.
Sequence in context: A051872 A297397 A104100 * A124713 A126374 A229536
KEYWORD
easy,nice,nonn
AUTHOR
Terrel Trotter, Jr., Apr 07 2002
STATUS
approved

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Last modified March 19 03:33 EDT 2024. Contains 370952 sequences. (Running on oeis4.)