OFFSET
1,2
COMMENTS
Centered hendecagonal (11-gonal) numbers. - Omar E. Pol, Oct 03 2011
Numbers of the form (2*m+1)^2 + k*m*(m+1)/2: in this case is k=3. See also A254963. - Bruno Berselli, Feb 11 2015
LINKS
T. D. Noe, Table of n, a(n) for n = 1..1000
X. Acloque, Polynexus Numbers and other mathematical wonders. [broken link]
Leo Tavares, Illustration: Clipped Stars.
Eric Weisstein's World of Mathematics, Centered Polygonal Numbers.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
FORMULA
a(n) = 1 + Sum_{j=0..n-1} (11*j). - Xavier Acloque, Oct 22 2003
Binomial transform of [1, 11, 11, 0, 0, 0, ...]; Narayana transform (A001263) of [1, 11, 0, 0, 0, ...]. - Gary W. Adamson, Dec 29 2007
a(n) = 11*n + a(n-1) - 11 with n > 1, a(1)=1. - Vincenzo Librandi, Aug 08 2010
G.f.: -x*(1+9*x+x^2)/(x-1)^3. - R. J. Mathar, Jun 05 2011
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3); a(0)=1, a(1)=12, a(2)=34. - Harvey P. Dale, Jun 25 2011
a(n) = A152740(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(3/11)*Pi/2)/sqrt(33).
Sum_{n>=1} a(n)/n! = 13*e/2 - 1.
Sum_{n>=1} (-1)^n * a(n)/n! = 13/(2*e) - 1. (End)
E.g.f.: exp(x)*(1 + 11*x^2/2) - 1. - Elmo R. Oliveira, Oct 18 2024
EXAMPLE
a(5)=111 because 111 = (11*5^2 - 11*5 + 2)/2 = (275 - 55 + 2)/2 = 222/2.
MATHEMATICA
FoldList[#1 + #2 &, 1, 11 Range@ 45] (* Robert G. Wilson v *)
Table[(11n^2-11n+2)/2, {n, 60}] (* or *) LinearRecurrence[{3, -3, 1}, {1, 12, 34}, 60] (* Harvey P. Dale, Jun 25 2011 *)
PROG
(PARI) a(n)=(11*n^2-11*n+2)/2 \\ Charles R Greathouse IV, Sep 24 2015
CROSSREFS
KEYWORD
nonn,easy,nice
AUTHOR
Terrel Trotter, Jr., Apr 07 2002
EXTENSIONS
More terms from Harvey P. Dale, Jun 25 2011
Name rewritten by Bruno Berselli, Feb 11 2015
STATUS
approved