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Fibonacci sequence beginning 1, 19.
3

%I #30 Feb 22 2024 10:38:43

%S 1,19,20,39,59,98,157,255,412,667,1079,1746,2825,4571,7396,11967,

%T 19363,31330,50693,82023,132716,214739,347455,562194,909649,1471843,

%U 2381492,3853335,6234827,10088162,16322989,26411151,42734140,69145291,111879431,181024722

%N Fibonacci sequence beginning 1, 19.

%C a(n-1) = Sum(P(19;n-1-k,k),k=0..ceiling((n-1)/2)), n>=1, with a(-1)=18. These are the SW-NE diagonals in P(19;n,k), the (19,1) Pascal triangle. Cf. A093645 for the (10,1) Pascal triangle. Observation by _Paul Barry_, Apr 29 2004. Proof via recursion relations and comparison of inputs.

%H Paolo Xausa, <a href="/A022109/b022109.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 <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), n >= 2, a(0) = 1, a(1) = 19.

%F G.f.: (1+18*x)/(1-x-x^2).

%t LinearRecurrence[{1, 1}, {1, 19}, 35] (* _Paolo Xausa_, Feb 22 2024 *)

%o (Magma) a0:=1; a1:=19; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..30]]; // _Bruno Berselli_, Feb 12 2013

%Y a(n) = A109754(18, n+1) = A101220(18, 0, n+1).

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_