|
%I #46 Feb 18 2024 01:58:52
%S 1,10,11,21,32,53,85,138,223,361,584,945,1529,2474,4003,6477,10480,
%T 16957,27437,44394,71831,116225,188056,304281,492337,796618,1288955,
%U 2085573,3374528,5460101,8834629,14294730,23129359,37424089,60553448,97977537,158530985
%N Fibonacci sequence beginning 1, 10.
%C a(n-1)=sum(P(10;n-1-k,k),k=0..ceiling((n-1)/2)), n>=1, with a(-1)=9. These are the SW-NE diagonals in P(10;n,k), the (10,1) Pascal triangle A093645. Observation by _Paul Barry_, Apr 29 2004. Proof via recursion relations and comparison of inputs.
%C In general, for b Fibonacci sequence beginning with 1, h, we have:
%C b(n) = (2^(-1-n)*((1 - sqrt(5))^n*(1 + sqrt(5) - 2*h) + (1 + sqrt(5))^n*(-1 + sqrt(5) + 2*h)))/sqrt(5). - _Herbert Kociemba_, Dec 18 2011
%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 (1, 1).
%F a(n)= a(n-1) + a(n-2) for n>=2, a(0)=1, a(1)=10, a(-1):=9.
%F G.f.: (1 + 9*x)/(1 - x - x^2).
%F a(n)=sum{k=0..n, Fib(n-k+1)(9*binomial(1, k)-8*binomial(0, k))}. - _Paul Barry_, May 05 2005
%F a(n) = ((1+sqrt5)^n-(1-sqrt5)^n)/(2^n*sqrt5)+ 4.5*((1+sqrt5)^(n-1)-(1-sqrt5)^(n-1))/(2^(n-2)*sqrt5). Offset 1. a(3)=11. - Al Hakanson (hawkuu(AT)gmail.com), Jan 14 2009
%F From _Bruno Berselli_, Feb 20 2017: (Start)
%F a(n) = 9*A000045(n) + A000045(n+1).
%F a(n) = 11*A000045(n) - A000045(n-2). (End)
%t LinearRecurrence[{1,1},{1,10},40] (* _Harvey P. Dale_, May 17 2017 *)
%o (Magma) a0:=1; a1:=10; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..40]]; // _Bruno Berselli_, Feb 12 2013
%Y Cf. A000045.
%Y a(n) = A109754(9, n+1) = A101220(9, 0, n+1).
%K nonn,easy
%O 0,2
%A _N. J. A. Sloane_
|
