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Numerators of the continued fraction n+1/(n+1/...) [n times].
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%I #40 Jul 05 2022 11:36:21

%S 1,5,33,305,3640,53353,927843,18674305,426938895,10928351501,

%T 309601751184,9616792908241,324971855514293,11868363584907985,

%U 465823816409224245,19553538801258341377,874091571490181406680

%N Numerators of the continued fraction n+1/(n+1/...) [n times].

%C The n-th term of the Lucas sequence U(n,-1). The denominator is the (n-1)-th term. Adjacent terms of the sequence U(n,-1) are relatively prime. - _T. D. Noe_, Aug 19 2004

%H Alois P. Heinz, <a href="/A084845/b084845.txt">Table of n, a(n) for n = 1..386</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/LucasSequence.html">Lucas Sequence</a>

%F a(n) = Sum_{k=0..floor(n/2)}* binomial(n-k, k)*n^(n-2k). - _Michel Lagneau_

%F a(n) = [x^n] 1/(1 - n*x - x^2). - _Paul D. Hanna_, Dec 27 2012

%F a(n) = (s^(n+1) - (-s)^(-n-1))/(2*s - n), where s = (n + sqrt(n^2 + 4))/2. - _Vladimir Reshetnikov_, May 07 2016

%F a(n) = A117715(n+1,n). - _Alois P. Heinz_, Aug 12 2017

%e a(4) = 305 since 4+1/(4+1/(4+1/4)) = 305/72.

%p A084845 := proc(n)

%p fibonacci(n+1,n) ;

%p end proc:

%p seq(A084845(n),n=1..20) ; # _Zerinvary Lajos_, Dec 01 2006

%t myList[n_] := Module[{ex = {n}}, Do[ex = {ex, n}, {n - 1}]; Flatten[ex]] Table[Numerator[FromContinuedFraction[myList[n]]], {n, 1, 20}]

%t Table[s=n; Do[s=n+1/s, {n-1}]; Numerator[s], {n, 20}] (* _T. D. Noe_, Aug 19 2004 *)

%o (PARI) {a(n)=polcoeff(1/(1-n*x-x^2+x*O(x^n)),n)} \\ _Paul D. Hanna_, Dec 27 2012

%o (Python)

%o from sympy import fibonacci

%o def a117715(n, m): return 0 if n==0 else fibonacci(n, m)

%o def a(n): return a117715(n + 1, n)

%o print([a(n) for n in range(1, 31)]) # _Indranil Ghosh_, Aug 12 2017

%Y Cf. A084844 (denominators).

%Y Cf. A097690, A097691, A117715.

%K frac,nonn

%O 1,2

%A _Hollie L. Buchanan II_, Jun 08 2003