A117715 -id:A117715 - cp's OEIS frontend

A084844 Denominators of the continued fraction n + 1/(n + 1/...) [n times].

1, 2, 10, 72, 701, 8658, 129949, 2298912, 46866034, 1082120050, 27916772489, 795910114440, 24851643870041, 843458630403298, 30918112619119426, 1217359297034666112, 51240457936070359069, 2296067756927144738850, 109127748348241605689981

Offset: 1

Hollie L. Buchanan II, Jun 08 2003

The (n-1)-th term of the Lucas sequence U(n,-1). The numerator is the n-th term. Adjacent terms of the sequence U(n,-1) are relatively prime. - T. D. Noe, Aug 19 2004
From Flávio V. Fernandes, Mar 05 2021: (Start)
Also, the n-th term of the n-th metallic sequence (the diagonal through the array A073133, and its equivalents, which is rows formed by sequences beginning with A000045, A000129, A006190, A001076, A052918) as shown below (for n>=1):
   0    1    0    1     0      1  ... A000035
   0   [1]   1    2     3      5  ... A000045
   0    1   [2]   5    12     29  ... A000129
   0    1    3  [10]   33    109  ... A006190
   0    1    4   17   [72]   305  ... A001076
   0    1    5   26   135   [701] ... A052918. (End)

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

Alois P. Heinz, Table of n, a(n) for n = 1..387
Eric Weisstein's World of Mathematics, Lucas Sequence

Cf. A084845 (numerators).

Cf. A000045, A097690, A097691, A117715, A290864 (primes in this sequence).

A084844 :=proc(n) combinat[fibonacci](n, n) end:
seq(A084844(n), n=1..30); # Zerinvary Lajos, Jan 03 2007

myList[n_] := Module[{ex = {n}}, Do[ex = {ex, n}, {n - 1}]; Flatten[ex]] Table[Denominator[FromContinuedFraction[myList[n]]], {n, 1, 20}]
Table[s=n; Do[s=n+1/s, {n-1}]; Denominator[s], {n, 20}] (* T. D. Noe, Aug 19 2004 *)
Table[Fibonacci[n, n], {n, 1, 20}] (* Vladimir Reshetnikov, May 07 2016 *)
Table[DifferenceRoot[Function[{y,m},{y[2+m]==n*y[1+m]+y[m],y[0]==0,y[1]==1}]][n],{n,1,20}] (* Benedict W. J. Irwin, Nov 03 2016 *)

from sympy import fibonacci
def a(n):
    return fibonacci(n, n)
print([a(n) for n in range(1, 31)]) # Indranil Ghosh, Aug 12 2017

a(n) = (s^n - (-s)^(-n))/(2*s - n), where s = (n + sqrt(n^2 + 4))/2. - Vladimir Reshetnikov, May 07 2016
a(n) = y(n,n), where y(m+2,n) = n*y(m+1,n) + y(m,n), with y(0,n)=0, y(1,n)=1 for all n. - Benedict W. J. Irwin, Nov 03 2016
a(n) ~ n^(n-1). - Vaclav Kotesovec, Jun 03 2017
a(n) = A117715(n,n). - Bobby Jacobs, Aug 12 2017
a(n) = [x^n] x/(1 - n*x - x^2). - Ilya Gutkovskiy, Oct 10 2017
a(n) == 0 (mod n) for even n and 1 (mod n) for odd n. - Flávio V. Fernandes, Dec 08 2020
a(n) == 0 (mod n) for even n and 1 (mod n^2) for odd n; see A065599. - Flávio V. Fernandes, Dec 25 2020
a(n) == 0 (mod 2*(n/2)^2) for even n and 1 (mod n^2) for odd n; see A129194. - Flávio V. Fernandes, Feb 06 2021

A084845 Numerators of the continued fraction n+1/(n+1/...) [n times].

Original entry on oeis.org

1, 5, 33, 305, 3640, 53353, 927843, 18674305, 426938895, 10928351501, 309601751184, 9616792908241, 324971855514293, 11868363584907985, 465823816409224245, 19553538801258341377, 874091571490181406680

Offset: 1

Hollie L. Buchanan II, Jun 08 2003

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

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

Alois P. Heinz, Table of n, a(n) for n = 1..386
Eric Weisstein's World of Mathematics, Lucas Sequence

Cf. A084844 (denominators).

Cf. A097690, A097691, A117715.

A084845 := proc(n)
    fibonacci(n+1,n) ;
end proc:
seq(A084845(n),n=1..20) ; # Zerinvary Lajos, Dec 01 2006

myList[n_] := Module[{ex = {n}}, Do[ex = {ex, n}, {n - 1}]; Flatten[ex]] Table[Numerator[FromContinuedFraction[myList[n]]], {n, 1, 20}]
Table[s=n; Do[s=n+1/s, {n-1}]; Numerator[s], {n, 20}] (* T. D. Noe, Aug 19 2004 *)

{a(n)=polcoeff(1/(1-n*x-x^2+x*O(x^n)),n)} \\ Paul D. Hanna, Dec 27 2012

from sympy import fibonacci
def a117715(n, m): return 0 if n==0 else fibonacci(n, m)
def a(n): return a117715(n + 1, n)
print([a(n) for n in range(1, 31)]) # Indranil Ghosh, Aug 12 2017

a(n) = Sum_{k=0..floor(n/2)}* binomial(n-k, k)*n^(n-2k). - Michel Lagneau
a(n) = [x^n] 1/(1 - n*x - x^2). - Paul D. Hanna, Dec 27 2012
a(n) = (s^(n+1) - (-s)^(-n-1))/(2*s - n), where s = (n + sqrt(n^2 + 4))/2. - Vladimir Reshetnikov, May 07 2016
a(n) = A117715(n+1,n). - Alois P. Heinz, Aug 12 2017

A124152 a(n) = Fibonacci(6, n).

Original entry on oeis.org

0, 8, 70, 360, 1292, 3640, 8658, 18200, 34840, 61992, 104030, 166408, 255780, 380120, 548842, 772920, 1065008, 1439560, 1912950, 2503592, 3232060, 4121208, 5196290, 6485080, 8017992, 9828200, 11951758, 14427720, 17298260, 20608792, 24408090, 28748408

Offset: 0

Zerinvary Lajos, Dec 01 2006

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A117715 formatted as a triangular array: row 7.

Cf. A000045.

with(combinat, fibonacci):seq(fibonacci(6, i), i=0..35);

LinearRecurrence[{6,-15,20,-15,6,-1},{0,8,70,360,1292,3640},40] (* Harvey P. Dale, Apr 18 2019 *)

concat(0, Vec(2*x*(4 + 11*x + 30*x^2 + 11*x^3 + 4*x^4) / (1 - x)^6 + O(x^30))) \\ Colin Barker, Apr 06 2017

[lucas_number1(6,n,-1) for n in range(0, 30)] # Zerinvary Lajos, May 16 2009

From Colin Barker, Apr 06 2017: (Start)
G.f.: 2*x*(4 + 11*x + 30*x^2 + 11*x^3 + 4*x^4) / (1 - x)^6.
a(n) = n*(3 + 4*n^2 + n^4).
a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6) for n>5.
(End)

A117716 Triangle T(n,k) read by rows: the coefficient [x^n] of x^2/(1-(k+1)*x-x^3) in row n, columns 0 <= k <= n.

Original entry on oeis.org

0, 0, 0, 1, 1, 1, 1, 2, 3, 4, 1, 4, 9, 16, 25, 2, 9, 28, 65, 126, 217, 3, 20, 87, 264, 635, 1308, 2415, 4, 44, 270, 1072, 3200, 7884, 16954, 32960, 6, 97, 838, 4353, 16126, 47521, 119022, 264193, 534358, 9, 214, 2601, 17676, 81265, 286434, 835569, 2117656, 4815801, 10050030

Offset: 0

Roger L. Bagula, Apr 13 2006, corrected Apr 15 2006

			Triangle begins as:
  0;
  0,  0;
  1,  1,   1;
  1,  2,   3,    4;
  1,  4,   9,   16,   25;
  2,  9,  28,   65,  126,  217;
  3, 20,  87,  264,  635, 1308,  2415;
  4, 44, 270, 1072, 3200, 7884, 16954, 32960;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A000930 (column 0), A008998 (column 1), A052541 (column 2), A052927 (column 3), A001093 (row 5), A185065 (row 6), A117715, A117724.

m:=12;
R:=PowerSeriesRing(Integers(), m+2);
A117716:= func< n,k | Coefficient(R!( x^2/(1-(k+1)*x-x^3) ), n) >;
[[A117716(n,k): k in [0..n]]: n in [0..m]]; // G. C. Greubel, Jul 23 2023

A117716 := proc(n,m)
        x^2/(1-(m+1)*x-x^3) ;
        if n < 0 then
                0;
        else
                coeftayl(%,x=0,n) ;
        end if;
end proc: # R. J. Mathar, May 14 2013

T[n_, k_]:= T[n, k]= Coefficient[Series[x^2/(1-(k+1)*x-x^3), {x,0,n+ 2}], x, n];
Table[T[n,k], {n,0,12}, {k,0,n}]//Flatten

def A117716(n,k):
    P. = PowerSeriesRing(QQ)
    return P( x^2/(1-(k+1)*x-x^3) ).list()[n]
flatten([[A117716(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Jul 23 2023

Edited by G. C. Greubel, Jul 23 2023

A290864 Numbers k such that the k-th Fibonacci polynomial evaluated at k is prime.

Original entry on oeis.org

2, 5, 71, 8419

Offset: 1

Bobby Jacobs, Aug 12 2017

Numbers k such that A084844(k) = A117715(k,k) is prime.
a(5) > 9200. - Giovanni Resta, Aug 13 2017
Except for a(1), all terms == 1 or 5 (mod 6). - Robert Israel, Aug 13 2017

			5 is in the sequence because A117715(5,5) = 701 is prime.

Eric Weisstein's World of Mathematics, Fibonacci Polynomial
Wikipedia, Fibonacci polynomials

Cf. A000045, A011973, A084844, A117715, A162515, A168561.

select(t -> isprime(combinat:-fibonacci(t,t)), [2,seq(seq(6*i+j,j=[1,5]),i=0..100)]); # Robert Israel, Aug 13 2017

Select[Range[100], PrimeQ@ Fibonacci[#, #] &] (* Giovanni Resta, Aug 13 2017 *)

a(4) from Giovanni Resta, Aug 13 2017

cp's OEIS Frontend

A084844 Denominators of the continued fraction n + 1/(n + 1/...) [n times].

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A084845 Numerators of the continued fraction n+1/(n+1/...) [n times].

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A124152 a(n) = Fibonacci(6, n).

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A117716 Triangle T(n,k) read by rows: the coefficient [x^n] of x^2/(1-(k+1)*x-x^3) in row n, columns 0 <= k <= n.

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A290864 Numbers k such that the k-th Fibonacci polynomial evaluated at k is prime.

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