OFFSET
1,2
COMMENTS
Create a triangle with first column T(n,1)=1+4*n for n=0,1,2... The remaining terms T(r,c)=T(r,c-1)+T(r-1,c-1). The sum of the terms in row(n)=a(n+1). - J. M. Bergot, Dec 18 2012
LINKS
Clark Kimberling, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (6,-13,12,-4).
FORMULA
a(n) = -1 + 2^n - 4*n + n*2^(n+1).
a(n) = 6*a(n-1) - 13*a(n-2) + 12*a(n-3) - 4*a(n-4).
G.f.: x*(1 + 5*x - 10*x^2)/(1 - 3*x + 2*x^2 )^2.
MATHEMATICA
(See A213762.)
LinearRecurrence[{6, -13, 12, -4}, {1, 11, 43, 127}, 30] (* Harvey P. Dale, Apr 13 2017 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Jun 20 2012
STATUS
approved