%I #50 Dec 30 2023 23:40:46
%S 1,1,12,133,1475,16358,181413,2011901,22312324,247447465,2744234439,
%T 30434026294,337518523673,3743137786697,41512034177340,
%U 460375513737437,5105642685289147,56622445051918054,627952538256387741,6964100365872183205,77233056562850402996
%N Generalized Fibonacci numbers.
%C For n>=1, row sums of triangle for numbers 11^k*binomial(m,k) with duplicated diagonals. - _Vladimir Shevelev_, Apr 13 2012
%C For n>=1, a(n) equals the numbers of words of length n-1 on alphabet {0,1,...,11} containing no subwords ii, (i=0,1,...,10). - _Milan Janjic_, Jan 31 2015
%H Vincenzo Librandi, <a href="/A015457/b015457.txt">Table of n, a(n) for n = 0..900</a>
%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (11,1).
%F a(n) = 11*a(n-1) + a(n-2).
%F a(n) = Sum_{k=0..n} 10^k*A055830(n,k). - _Philippe Deléham_, Oct 18 2006
%F G.f.: (1-10*x)/(1-11*x-x^2). - _Philippe Deléham_, Nov 21 2008
%F For n>=2, a(n) = F_n(11)+F_(n+1)(11), where F_n(x) is Fibonacci polynomial (cf. A049310): F_n(x) = Sum_{i=0..floor((n-1)/2)} binomial(n-i-1,i)*x^(n-2*i-1). - _Vladimir Shevelev_, Apr 13 2012
%F a(n) = (F(5*n-5) + F(5*n))/5 for F(n) the Fibonacci sequence A000045(n). - _Greg Dresden_, Aug 22 2021
%t LinearRecurrence[{11, 1}, {1, 1}, 30] (* _Vincenzo Librandi_, Nov 08 2012 *)
%t CoefficientList[Series[(1-10*x)/(1-11*x-x^2), {x, 0, 50}], x] (* _G. C. Greubel_, Dec 19 2017 *)
%o (Magma) [n le 2 select 1 else 11*Self(n-1) + Self(n-2): n in [1..30]]; // _Vincenzo Librandi_, Nov 08 2012
%o (PARI) x='x+O('x^30); Vec((1-10*x)/(1-11*x-x^2)) \\ _G. C. Greubel_, Dec 19 2017
%Y Row m=11 of A135597.
%Y Cf. A000045, A049310, A055830.
%K nonn,easy
%O 0,3
%A _Olivier Gérard_