0,2

Equals the triangular numbers convolved with [1, 3, 1, 0, 0, 0,...]. -Gary W. AdamsonandAlexander R. Povolotsky,May 29 2009

FromAnt King,Jun 15 2012: (Start)

a(n) == 1 (mod 5) for all n.

The digital roots of the a(n) form a purely periodic palindromic 9-cycle 1, 6, 7, 4, 6, 4, 7, 6, 1.

The units' digits of the a(n) form a purely periodic palindromic 4-cycle 1, 6, 6, 1.

(End)

Binomial transform of (1, 5, 5, 0, 0, 0,...) and second partial sum of (1, 4, 5, 5, 5,...). -Gary W. Adamson,Sep 09 2015

a(n) = a(-1-n) for all n in Z. -Michael Somos,Jan 25 2019

On the plane start with a single regular pentagon, and repeat the following procedure, "For each edge of any pentagon not already connected to an existing pentagon create a mirror image such that the mirror image does not overlap with an existing pentagon." a(n) is the number of pentagons occupying the plane after n repetitions. -Torlach Rush,Sep 14 2022

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

B. K. Teo and N. J. A. Sloane, Magic numbers in polygonal and polyhedral clusters, Inorgan. Chem. 24 (1985), 4545-4558.

T. D. Noe,Table of n, a(n) for n = 0..1000

Paul Barry,Centered polygon numbers, heptagons and nonagons, and the Robbins numbers,arXiv:2104.01644 [math.CO], 2021.

Simon Plouffe,Approximations de séries génératrices et quelques conjectures,Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.

Simon Plouffe,1031 Generating Functions,Appendix to Thesis, Montreal, 1992

Cliff Reiter,Polygonal Numbers and Fifty One Stars,Lafayette College, Easton, PA (2019).

Eric Weisstein's World of Mathematics,Centered Pentagonal Number.

Index entries for linear recurrences with constant coefficients,signature (3,-3,1).

G.f.: (1 + 3*x + x^2)/(1 - x)^3.Simon Plouffein his 1992 dissertation

Narayana transform (A001263) of [1, 5, 0, 0, 0,...]. -Gary W. Adamson,Dec 29 2007

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3), a(0)=1, a(1)=6, a(2)=16. -Jaume Oliver Lafont,Dec 02 2008

a(n) = 5*A000217(n) + 1 = 5*T(n) + 1, for n = 0, 1, 2, 3,... and where T(n) = n*(n+1)/2 = n-th triangular number. -Thomas M. Green,Nov 25 2009

a(n) = a(n-1) + 5*n, with a(0)=1. -Vincenzo Librandi,Nov 18 2010

a(n) =A028895(n) + 1. -Omar E. Pol,Oct 03 2011

a(n) = 2*a(n-1) - a(n-2) + 5. -Ant King,Jun 12 2012

Sum_{n>=0} 1/a(n) = 2*Pi /sqrt(15) *tanh(Pi/2*sqrt(3/5)) = 1.360613169863... -Ant King,Jun 15 2012

a(n) =A101321(5,n). -R. J. Mathar,Jul 28 2016

E.g.f.: (2 + 10*x + 5*x^2)*exp(x)/2. -Ilya Gutkovskiy,Jul 28 2016

FromAmiram Eldar,Jun 20 2020: (Start)

Sum_{n>=0} a(n)/n! = 17*e/2.

Sum_{n>=0} (-1)^(n+1)*a(n)/n! = 3/(2*e). (End)

a(2)= 5*T(2) + 1 = 5*3 + 1 = 16, a(4) = 5*T(4) + 1 = 5*10 + 1 = 51. -Thomas M. Green,Nov 16 2009

A005891:= proc(n)

1+5*n*(1+n)/2;

end proc:

seq(A005891(n), n=0..40); #R. J. Mathar,Oct 07 2021

FoldList[#1 + #2 &, 1, 5 Range@ 40] (*Robert G. Wilson v,Feb 02 2011 *)

LinearRecurrence[{3, -3, 1}, {1, 6, 16}, 50] (*Harvey P. Dale,Sep 08 2018 *)

Table[ j! Coefficient[Series[Exp[x]*(1 + 5 x^2/2)-1, {x, 0, 20}], x, j], {j, 0, 20}] (*Nikolaos Pantelidis,Feb 07 2023 *)

(PARI) a(n)=5*n*(n+1)/2+1 \\Charles R Greathouse IV,Mar 22 2016

(Magma) [5*n*(n+1)/2 + 1: n in [0..50]]; //G. C. Greubel,Nov 04 2017

Cf.A028895,A001844,A003215,A004068(partial sums),A006322,A001263.

Partial sums ofA008706.

Equals second row ofA167546divided by 2.

Sequence in context:A102214A301679A115007*A108182A244242A092286

Adjacent sequences:A005888A005889A005890*A005892A005893A005894

nonn,easy

Formula corrected and relocated byJohannes W. Meijer,Nov 07 2009

approved