OFFSET
1,1
COMMENTS
Conjecture: satisfies a linear recurrence having signature (23,-143,121). - Harvey P. Dale, May 12 2019
This conjecture is true since a(n) - a(n-1) yields the recurrence 1 + 10*n + 11*n*a(n-1) - (n-1)*a(n) = 0 with polynomial coefficients in n. - Georg Fischer, Feb 19 2021
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..900
Paul Leyland, Factors of Cullen and Woodall numbers.
Paul Leyland, Generalized Cullen and Woodall numbers.
Index entries for linear recurrences with constant coefficients, signature (23,-143,121).
FORMULA
From Elmo R. Oliveira, Sep 07 2024: (Start)
G.f.: x*(121*x^2 - 11*x - 10)/((x - 1)*(11*x - 1)^2).
E.g.f.: 1 + exp(x)*(11*x*exp(10*x) - 1).
a(n) = 23*a(n-1) - 143*a(n-2) + 121*a(n-3) for n > 3.
a(n) = A064749(n) - 2. (End)
MAPLE
k:= 11; f:= gfun:-rectoproc({1 + (k-1)*n + k*n*a(n-1) - (n-1)*a(n) = 0, a(1) = k-1}, a(n), remember): map(f, [$1..20]); # Georg Fischer, Feb 19 2021
MATHEMATICA
Table[n*11^n-1, {n, 20}] (* Harvey P. Dale, May 12 2019 *)
PROG
(Magma) [n*11^n - 1: n in [1..20]]; // Vincenzo Librandi, Sep 16 2011
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
N. J. A. Sloane, Oct 19 2001
STATUS
approved