|
|
A069125
|
|
a(n) = (11*n^2 - 11*n + 2)/2.
|
|
13
|
|
|
1, 12, 34, 67, 111, 166, 232, 309, 397, 496, 606, 727, 859, 1002, 1156, 1321, 1497, 1684, 1882, 2091, 2311, 2542, 2784, 3037, 3301, 3576, 3862, 4159, 4467, 4786, 5116, 5457, 5809, 6172, 6546, 6931, 7327, 7734, 8152, 8581, 9021, 9472, 9934, 10407, 10891
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
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
|
|
|
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) = 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
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)
|
|
EXAMPLE
|
a(5)=111 because 111 = (11*5^2 - 11*5 + 2)/2 = (275 - 55 + 2)/2 = 222/2.
|
|
MATHEMATICA
|
Table[(11n^2-11n+2)/2, {n, 60}] (* or *) LinearRecurrence[{3, -3, 1}, {1, 12, 34}, 60] (* Harvey P. Dale, Jun 25 2011 *)
|
|
PROG
|
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,easy,nice
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|