OFFSET
0,3
COMMENTS
For n>=1, row sums of triangle for numbers 10^k*C(m,k) with duplicated diagonals. - Vladimir Shevelev, Apr 13 2012
For n>=1, a(n) equals the numbers of words of length n-1 on alphabet {0,1,...,10} containing no subwords ii, (i=0,1,...,9). - Milan Janjic, Jan 31 2015
a(n) equals the number of sequences over the alphabet {0,1,...,9,10} such that no two consecutive terms have distance 6. - David Nacin, Jun 02 2017
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..1000
M. Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7.
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (10,1).
FORMULA
a(n) = 10*a(n-1) + a(n-2).
a(n) = Sum_{k=0..n} 9^k*A055830(n,k). - Philippe Deléham, Oct 18 2006
G.f.: (1-9*x)/(1-10*x-x^2). - Philippe Deléham, Nov 20 2008
For n>=2, a(n) = F_(n)(10) + F_(n+1)(10), where F_n(x) is Fibonacci polynomial (cf.A049310): F_n(x) = Sum_{i=0,...,floor((n-1)/2)} C(n-i-1,i)*x^(n-2*i-1). - Vladimir Shevelev, Apr 13 2012
MATHEMATICA
LinearRecurrence[{10, 1}, {1, 1}, 30] (* Vincenzo Librandi, Nov 08 2012 *)
CoefficientList[Series[(1-9*x)/(1-10*x-x^2), {x, 0, 50}], x] (* G. C. Greubel, Dec 19 2017 *)
PROG
(Magma) [n le 2 select 1 else 10*Self(n-1) + Self(n-2): n in [1..30]]; // Vincenzo Librandi, Nov 08 2012
(PARI) x='x+O('x^30); Vec((1-9*x)/(1-10*x-x^2)) \\ G. C. Greubel, Dec 19 2017
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
STATUS
approved