%I #56 Jul 04 2022 04:43:07
%S 4,9,13,22,35,57,92,149,241,390,631,1021,1652,2673,4325,6998,11323,
%T 18321,29644,47965,77609,125574,203183,328757,531940,860697,1392637,
%U 2253334,3645971,5899305,9545276,15444581,24989857,40434438,65424295,105858733,171283028
%N Fibonacci sequence beginning 4,9.
%C The associated Pisano series starts as in A001175, but differs for example for modulus 29 where it is 7, not 14. - _R. J. Mathar_, Nov 02 2011
%C The Pisano period also differs for modulus 58, where it is 21 instead of 42. Otherwise, the Pisano periods coincide with those of the Fibonacci numbers. - _Klaus Purath_, Jun 26 2022
%H Vincenzo Librandi, <a href="/A022130/b022130.txt">Table of n, a(n) for n = 0..1000</a>
%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>
%H H. Zhao and X. Li, <a href="http://www.fq.math.ca/Papers1/44-1/quarthzhoa01_2006.pdf">On the Fibonacci numbers of trees</a>, Fib. Quart., 44 (2006), 32-38.
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (1,1).
%F a(n) = 4*Fibonacci(n+2) + Fibonacci(n).
%F G.f.: (4 + 5*x)/(1-x-x^2). - _Philippe Deléham_, Nov 19 2008
%F a(n)= Fibonacci(n-2) + Fibonacci(n+5). - _Gary Detlefs_, Mar 31 2012
%p a:= n-> (<<0|1>, <1|1>>^n.<<4, 9>>)[1,1]:
%p seq(a(n), n=0..40); # _Alois P. Heinz_, Feb 22 2017
%t a={};b=4;c=9;AppendTo[a,b];AppendTo[a,c];Do[b=b+c;AppendTo[a,b];c=b+c;AppendTo[a,c],{n,1,40,1}];a (* _Vladimir Joseph Stephan Orlovsky_, Jul 23 2008 *)
%t LinearRecurrence[{1,1},{4,9},40] (* _Harvey P. Dale_, Dec 15 2011 *)
%o (PARI) a(n)=4*fibonacci(n-1)+9*fibonacci(n) \\ _Charles R Greathouse IV_, Jun 05 2011
%o (Magma) a0:=4; a1:=9; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..35]]; // _Vincenzo Librandi_, Jan 25 2017
%Y Cf. A000032, A001175.
%K nonn,easy
%O 0,1
%A _N. J. A. Sloane_