A004187 -id:A004187 - cp's OEIS frontend

A094565 Triangle read by rows: binary products of Fibonacci numbers.

1, 2, 3, 5, 6, 8, 13, 15, 16, 21, 34, 39, 40, 42, 55, 89, 102, 104, 105, 110, 144, 233, 267, 272, 273, 275, 288, 377, 610, 699, 712, 714, 715, 720, 754, 987, 1597, 1830, 1864, 1869, 1870, 1872, 1885, 1974, 2584, 4181, 4791, 4880, 4893, 4895, 4896, 4901, 4935, 5168, 6765

Offset: 1

Clark Kimberling, May 12 2004

Row n consists of n numbers, first F(2n-1) and last F(2n).
Central numbers: (1,6,40,273,...) = A081016.
Row sums: A001870.
Alternating row sums: 1,1,7,7,48,48,329,329; the sequence b=(1,7,48,329,...) is A004187, given by b(n)=F(4n+2)-b(n-1) for n>=2, with b(1)=1.
In each row, the difference between neighboring terms is a Fibonacci number.

			Triangle begins:
   1;
   2,   3;
   5,   6    8;
  13,  15,  16,  21;
  34,  39,  40,  42,  55;
  89, 102, 104, 105, 110, 144; ...

G. C. Greubel, Rows n = 1..100 of triangle, flattened
Clark Kimberling, Orderings of products of Fibonacci numbers, Fibonacci Quarterly 42:1 (2004), pp. 28-35.

Cf. A000045, A094566, A094568.

Flat(List([1..12], n-> List([1..n], k-> Fibonacci(2*k)*Fibonacci(2*n-2*k+1) ))); # G. C. Greubel, Jul 15 2019

[Fibonacci(2*k)*Fibonacci(2*n-2*k+1): k in [1..n], n in [1..12]]; // G. C. Greubel, Jul 15 2019

Table[Fibonacci[2*k]*Fibonacci[2*n-2*k+1], {n,12}, {k,n}]//Flatten (* G. C. Greubel, Jul 15 2019 *)

row(n) = vector(n, k, fibonacci(2*k)*fibonacci(2*n-2*k+1));
tabl(nn) = for(n=1, nn, print(row(n))); \\ Michel Marcus, May 03 2016

[[fibonacci(2*k)*fibonacci(2*n-2*k+1) for k in (1..n)] for n in (1..12)] # G. C. Greubel, Jul 15 2019

Row n: F(2)F(2n-1), F(4)F(2n-3), ..., F(2n)F(1).

A179943 Triangle read by rows, antidiagonals of an array (r,k), r=(0,1,2,...), generated from 2 X 2 matrices of the form [1,r; 1,(r+1)].

Original entry on oeis.org

1, 1, 2, 1, 3, 3, 1, 4, 8, 4, 1, 5, 15, 21, 5, 1, 6, 24, 56, 55, 6, 1, 7, 35, 115, 209, 144, 7, 1, 8, 48, 204, 551, 780, 377, 8, 1, 9, 63, 329, 1189, 2640, 2911, 987, 9, 1, 10, 80, 496, 2255, 6930, 12649, 10864, 2584, 10, 1, 11, 99, 711, 3905, 15456, 40391, 60605, 40545, 6765, 11

Offset: 0

Gary W. Adamson, Aug 07 2010

Row sums = A179944: (1, 3, 7, 17, 47, 148, 518,...)
Row 1 = A001906, row 2 = A001353, row 3 = A004254, row 4 = A001109, row 5 = A004187, row 6 = A001090, row 7 = A018913, row 9 = A004189.
Let S_m(x) be the m-th Chebyshev S-polynomial, described by Wolfdieter Lang in his draft [Lang], defined by S_0(x)=1, S_1(x)=x and S_m(x)=x*S_{m-1}(x)-S_{m-2}(x) (m>1). Let A = (A(r,c)) denote the rectangular array (not the triangle). Then A(r,c) = S_c(r+2), r,c=0,1,2,.... - L. Edson Jeffery, Aug 14 2011
As to the array, (n+1)-th row is the INVERT transform of n-th row. - Gary W. Adamson, Jun 30 2013
If the array sequences are labeled (2,3,4,...) for the n-th sequence, convergence tends to (n + sqrt(n^2 - 4))/2. - Gary W. Adamson, Aug 20 2013

			First few rows of the array:
  1, 2,  3,   4,    5,    6,     7,...
  1, 3,  8,  21,   55,  144,   377,...
  1, 4, 15,  56,  209,  780,  2911,...
  1, 5, 24, 115,  551, 2640, 12649,...
  1, 6, 35, 204, 1189, 6930, 40391,...
Taking antidiagonals, we obtain triangle A179943:
  1;
  1, 2;
  1, 3, 3;
  1, 4, 8, 4;
  1, 5, 15, 21, 5;
  1, 6, 24, 56, 55, 6;
  1, 7, 35, 115, 209, 144, 7;
  1, 8, 48, 204, 551, 780, 377, 8;
  1, 9, 63, 329, 1189, 2640, 2911, 987, 9;
  1, 10, 80, 496, 2255, 6930, 12649, 10864, 2584, 10;
  1, 11, 99, 711, 3905, 15456, 40391, 60605, 40545, 6765, 11;
  1, 12, 120, 980, 6319, 30744, 105937, 235416, 290376, 151316, 17711, 12;
  ...
Examples: Row 1 of the array: (1, 3, 8, 21, 55, 144,...); 144 = term (1,5) of the array = term (2,1) of M^6; where M = the 2 X 2 matrix [1,1; 1,2] and M^6 = [89,144; 144,233].
Term (1,5) of the array = 144 = (r+2)*(term (1,4)) - (term (1,3)) = 3*55 - 21.

Alois P. Heinz, Rows n = 0..140, flattened
Robert G. Donnelly, Molly W. Dunkum, Sasha V. Malone, and Alexandra Nance, Symmetric Fibonaccian distributive lattices and representations of the special linear Lie algebras, arXiv:2012.14991 [math.CO], 2020.
Wolfdieter Lang, Chebyshev S-polynomials: ten applications.

Cf. A001906, A001353, A004254, A001109, A004187, A001090, A018913, A004189.

invtr:= proc(b) local a;
          a:= proc(n) option remember; local i;
          `if`(n<1, 1, add(a(n-i) *b(i-1), i=1..n+1)) end
        end:
A:= proc(n) A(n):= `if`(n=0, k->k+1, invtr(A(n-1))) end:
seq(seq(A(d-k)(k), k=0..d), d=0..10);  # Alois P. Heinz, Jul 17 2013
# using observation by Gary W. Adamson

a[, 0] = 0; a[, 1] = 1; a[r_, k_] := a[r, k] = (r+1)*a[r, k-1] - a[r, k-2]; Table[a[r-k+2, k], {r, 0, 10}, {k, 1, r+1}] // Flatten (* Jean-François Alcover, Feb 23 2015 *)

Antidiagonals of an array, (r,k), a(k) = (r+2)*a(k-1) - a*(k-2), r=0,1,2,... where (r,k) = term (2,1) in the 2 X 2 matrix [1,r; 1,r+1]^(k+1).

G.f. for row r of array: 1/(1 - (r+2)*x + x^2). - L. Edson Jeffery, Oct 26 2012

A203319 a(n) = nFibonacci(n) Sum_{d|n} 1/(d*Fibonacci(d)).

Original entry on oeis.org

1, 3, 7, 19, 26, 81, 92, 267, 358, 848, 980, 3061, 3030, 7976, 11042, 25099, 27150, 78642, 79440, 219884, 270704, 584862, 659112, 1977909, 1950651, 4735370, 6204499, 14189096, 14912642, 43168586, 41734340, 110786987, 135815060, 290854380, 339428752, 953889058, 893839230

Offset: 1

Paul D. Hanna, Jan 01 2012

nonn

			L.g.f.: L(x) = x + 3*x^2/2 + 7*x^3/3 + 19*x^4/4 + 26*x^5/5 + 81*x^6/6 +...
where
L(x) = x*(1 + x + 2*x^2 + 3*x^3 + 5*x^4 + 8*x^5 +...+ F(n+1)*x^n +...) +
x^2/2*(1 + 3*x^2 + 8*x^4 + 21*x^6 + 55*x^8 +...+ F(2*n+2)*x^(2*n) +...) +
x^3/3*(1 + 4*x^3 + 17*x^6 + 72*x^9 +...+ F(3*n+3)/2*x^(3*n) +...) +
x^4/4*(1 + 7*x^4 + 48*x^8 + 329*x^12 +...+ F(4*n+4)/3*x^(4*n) +...) +
x^5/5*(1 + 11*x^5 + 122*x^10 + 1353*x^15 +...+ F(5*n+5)/5*x^(5*n) +...) +
x^6/6*(1 + 18*x^6 + 323*x^12 + 5796*x^18 +...+ F(6*n+6)/8*x^(6*n) +...) +...
here F(n) = Fibonacci(n) = A000045(n).
Equivalently,
L(x) = x/(1-x-x^2) + (x^2/2)/(1-3*x^2+x^4) + (x^3/3)/(1-4*x^3-x^6) + (x^4/4)/(1-7*x^4+x^8) +...+ (x^n/n)/(1 - Lucas(n)*x^n + (-1)^n*x^(2*n)) +...
here Lucas(n) = A000032(n).
Exponentiation of the l.g.f. equals the g.f. of A203318:
exp(L(x)) = 1 + x + 2*x^2 + 4*x^3 + 9*x^4 + 16*x^5 + 36*x^6 + 64*x^7 +...+ A203318(n)*x^n +...

Paul D. Hanna, Table of n, a(n) for n = 1..200

Cf. A203318, A203321; A203414 (Pell variant).

Cf. A000032 (Lucas), A000045 (Fibonacci), A001906, A001076, A004187, A049666, A049660, A049667.

a[n_] := n Fibonacci[n] DivisorSum[n, 1/(# Fibonacci[#]) &]; Array[a, 40] (* Jean-François Alcover, Dec 23 2015 *)

{a(n)=if(n<1,0, n*fibonacci(n)*sumdiv(n,d,1/(d*fibonacci(d))) )}

{Lucas(n)=fibonacci(n-1)+fibonacci(n+1)}
{a(n)=n*polcoeff(sum(m=1,n+1,(x^m/m)/(1-Lucas(m)*x^m+(-1)^m*x^(2*m)+x*O(x^n))),n)}

{Lucas(n)=fibonacci(n-1)+fibonacci(n+1)}
{a(n)=local(L=x); L=sum(m=1, n, x^m/m*exp(sum(k=1, floor((n+1)/m), Lucas(m*k)*x^(m*k)/k)+x*O(x^n))); n*polcoeff(L,n)}

{a(n)=local(A=1+x+x*O(x^n),F=1/(1-x-x^2+x*O(x^n))); A=exp(sum(m=1, n+1, x^m/m*round(prod(k=0, m-1, subst(F, x, exp(2*Pi*I*k/m)*x+x*O(x^n)))))); n*polcoeff(log(A), n)}

Equals the logarithmic derivative of A203318.
L.g.f.: L(x) = Sum_{n>=1} a(n)*x^n/n satisfies:
(1) L(x) = Sum_{n>=1} x^n/n * Sum_{k>=0} F(n*k+n)/F(n) * x^(n*k) where F(n) = Fibonacci(n).
(2) L(x) = Sum_{n>=1} x^n/n * exp( Sum_{k>=1} Lucas(n*k)*x^(n*k)/k ) where Lucas(n) = A000032(n).
(3) L(x) = Sum_{n>=1} x^n/n * 1/(1 - Lucas(n)*x^n + (-1)^n*x^(2*n)) where Lucas(n) = A000032(n).
(4) L(x) = Sum_{n>=1} x^n/n * G_n(x^n) where G_n(x^n) = Product_{k=0..n-1} G(u^k*x) where G(x) = 1/(1-x-x^2) and u is an n-th root of unity.

A290903 p-INVERT of the positive integers, where p(S) = 1 - 5*S.

Original entry on oeis.org

5, 35, 240, 1645, 11275, 77280, 529685, 3630515, 24883920, 170556925, 1169014555, 8012544960, 54918800165, 376419056195, 2580014593200, 17683683096205, 121205767080235, 830756686465440, 5694091038177845, 39027880580779475, 267501073027278480

Offset: 0

Clark Kimberling, Aug 17 2017

Suppose s = (c(0), c(1), c(2), ...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x). Taking p(S) = 1 - S gives the "INVERT" transform of s, so that p-INVERT is a generalization of the "INVERT" transform (e.g., A033453).

See A290890 for a guide to related sequences.

			(See the example at A290902.)

Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (7, -1)

Cf. A000027, A004187, A290890.

z = 60; s = x/(1 - x)^2; p = 1 - 5 s;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290903 *)
u/5 (* A004187 shifted *)

G.f.: 5/(1 - 7 x + x^2).
a(n) = 7*a(n-1) - a(n-2).
a(n) = 5*A004187(n+1) for n >= 0.

A372817 Table read by antidiagonals: T(m,n) = number of 1-metered (m,n)-parking functions.

Original entry on oeis.org

1, 0, 2, 0, 3, 3, 0, 4, 8, 4, 0, 6, 21, 15, 5, 0, 8, 55, 56, 24, 6, 0, 12, 145, 209, 115, 35, 7, 0, 16, 380, 780, 551, 204, 48, 8, 0, 24, 1000, 2912, 2640, 1189, 329, 63, 9, 0, 32, 2625, 10868, 12649, 6930, 2255, 496, 80, 10, 0, 48, 6900, 40569, 60606, 40391, 15456, 3905, 711, 99, 11

Offset: 1

Spencer Daugherty, May 13 2024

			For T(3,2) the 1-metered (3,2)-parking functions are 111, 121, 211, 212.
Table begins:
  1,  2,    3,     4,     5,      6,      7, ...
  0,  3,    8,    15,    24,     35,     48, ...
  0,  4,   21,    56,   115,    204,    329, ...
  0,  6,   55,   209,   551,   1189,   2255, ...
  0,  8,  145,   780,  2640,   6930,  15456, ...
  0, 12,  380,  2912, 12649,  40391, 105937, ...
  0, 16, 1000, 10868, 60606, 235416, 726103, ...
  ...

Spencer Daugherty, Pamela E. Harris, Ian Klein, and Matt McClinton, Metered Parking Functions, arXiv:2406.12941 [math.CO], 2024.

Main diagonal is A097690 and first row of A372816.
First, second, and third diagonals above main are A097691, A342167, A342168.
Second column A029744. Second row A005563. Third row A242135.
Cf. A001353, A004254, A001109, A004187, A372818, A372821.

T(m,n) = (n*(n+sqrt(n^2 - 4))-2)/(n*(n+sqrt(n^2 - 4))-4)*((n+sqrt(n^2-4))/2)^m + (n*(n-sqrt(n^2 - 4))-2)/(n*(n-sqrt(n^2 - 4))-4)*((n-sqrt(n^2-4))/2)^m.

T(m,n) = n*T(m-1,n) - T(m-2,n) with T(0,n) = 1.

A049686 a(n) = Fibonacci(8n)/3.

Original entry on oeis.org

0, 7, 329, 15456, 726103, 34111385, 1602508992, 75283811239, 3536736619241, 166151337293088, 7805576116155895, 366695926122033977, 17226902951619441024, 809297742799991694151, 38019767008647990184073, 1786119751663655546957280, 83909608561183162716808087, 3941965482623944992143022809

Offset: 0

Clark Kimberling

a(n) = (t(i+4n) - t(i))/(t(i+2n+1) - t(i+2n-1)), where (t) is any sequence of the form t(n+2) = 8t(n+1) - 8t(n) + t(n-1) or of the form t(n+1) = 7t(n) - t(n-1) without regard to initial values as long as t(i+2n+1) - t(i+2n-1) != 0. - Klaus Purath, Jun 23 2024

			a(2) = F(8 * 2) / 3 = F(16) / 3 = 987 / 3 = 329. - _Indranil Ghosh_, Feb 05 2017

Indranil Ghosh, Table of n, a(n) for n = 0..597
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (47,-1).

Cf. A000045, A004187, A049668.

List([0..20], n-> Fibonacci(8*n)/3 ); # G. C. Greubel, Dec 14 2019

[Fibonacci(8*n)/3: n in [0..20]]; // G. C. Greubel, Dec 14 2019

with(combinat); seq( fibonacci(8*n)/3, n=0..20); # G. C. Greubel, Dec 14 2019

Fibonacci[8(Range[20]-1)]/3 (* G. C. Greubel, Dec 14 2019 *)
LinearRecurrence[{47,-1},{0,7},20] (* Harvey P. Dale, Dec 27 2019 *)

a(n) = fibonacci(8*n)/3; \\ Michel Marcus, Feb 05 2017

[fibonacci(8*n)/3 for n in (0..20)] # G. C. Greubel, Dec 14 2019

a(n) = 47*a(n-1) - a(n-2), n>1. a(0)=0, a(1)=7.
G.f.: 7*x/(1-47*x+x^2).
a(n) = A004187(2n).
a(n) = 7*A049668(n). - R. J. Mathar, Oct 26 2015
E.g.f.: 2*exp(47*x/2)*sinh(21*sqrt(5)*x/2)/(3*sqrt(5)). - Stefano Spezia, Dec 14 2019

Better description and more terms from Michael Somos

A073134 Table by antidiagonals of T(n,k)=n*T(n,k-1)-T(n,k-2) starting with T(n,1)=1.

Original entry on oeis.org

1, 1, 1, 0, 2, 1, -1, 3, 3, 1, -1, 4, 8, 4, 1, 0, 5, 21, 15, 5, 1, 1, 6, 55, 56, 24, 6, 1, 1, 7, 144, 209, 115, 35, 7, 1, 0, 8, 377, 780, 551, 204, 48, 8, 1, -1, 9, 987, 2911, 2640, 1189, 329, 63, 9, 1, -1, 10, 2584, 10864, 12649, 6930, 2255, 496, 80, 10, 1, 0, 11, 6765, 40545, 60605, 40391, 15456, 3905, 711, 99, 11, 1, 1, 12

Offset: 1

Henry Bottomley, Jul 16 2002

			Rows start:
  1, 1,  0, -1,  -1,   0,    1, ...;
  1, 2,  3,  4,   5,   6,    7, ...;
  1, 3,  8, 21,  55, 144,  377, ...;
  1, 4, 15, 56, 209, 780, 2911, ...;
  ...

Shmuel T. Klein, Combinatorial Representation of Generalized Fibonacci Numbers, Fib. Quarterly 29 (2) (1991) 124-131, variable U_n^m. [From _R. J. Mathar_, Feb 19 2010]

Rows include A010892, A000027, A001906, A001353, A004254, A001109, A004187, A001090, A018913, A004189, A004190. Columns include (with some gaps) A000012, A000027, A005563, A057722.

Cf. A094954.

T(n,k) = sum(j=0,k-1,A049310(k-1,j)*n^j) \\ Jason Yuen, Aug 20 2024

T(n, k) = A073133(n, k)-2*A073135(n, k-2).

T(n, k) = Sum_{j=0..k-1} A049310(k-1, j)*n^j.

A092499 Chebyshev polynomials S(n-1,21) with Diophantine property.

Original entry on oeis.org

0, 1, 21, 440, 9219, 193159, 4047120, 84796361, 1776676461, 37225409320, 779956919259, 16341869895119, 342399310878240, 7174043658547921, 150312517518628101, 3149388824232642200, 65986852791366858099

Offset: 0

Rainer Rosenthal, Apr 05 2004

Sequence R_21: Starts with 0,1,21 and satisfies A*C=B^2-1 for successive A,B,C.
The natural numbers a(n)=n satisfy the recurrence a(n-1)*a(n+1)=a(n)^2-1. Let R_r denote the sequence starting with 0,1,r and with this recurrence. We see that R_2 = "the natural numbers" and we find R_3 = A001906. These R_r form a "family" of sequences, which coincides with the m-family (r=m-2, n -> n+1) provided by Wolfdieter Lang (see A078368). This sequence R_21 is strongly related to A041833, which gives the denominators in the continued fraction of sqrt(437).
All positive integer solutions of Pell equation b(n)^2 - 437*a(n)^2 = +4 together with b(n)=A097777(n), n>=0.
For n>=2, a(n) equals the permanent of the (n-1)X(n-1) tridiagonal matrix with 21's along the main diagonal, and i's along the superdiagonal and the subdiagonal (i is the imaginary unit). - John M. Campbell, Jul 08 2011
For n>=1, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,20}. - Milan Janjic, Jan 25 2015

			a(3)=440 because a(1)*440 = a(2)^2-1.

Indranil Ghosh, Table of n, a(n) for n = 0..756
Tanya Khovanova, Recursive Sequences
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (21,-1).

Cf. R_3=A001906, R_4=A001353, R_5=A004254, R_6=A001109, R_7=A004187, R_8=A001090, R_9=A018913, R_10=A004189, R_11=A004190, R_12=A004191, R_13=A078362, R_14=A007655, R_15=A078364, R_16=A077412, R_17=A078366, R_18=A049660, R_19=A078368, R_20=A075843, R_21=this, sequence, R_22=A077421. See also A041219 and A041917.

LinearRecurrence[{21,-1},{0,1},30] (* Harvey P. Dale, Apr 23 2015 *)

[lucas_number1(n,21,1) for n in range(0,20)] # Zerinvary Lajos, Jun 25 2008

a(0)=0, a(1)=1, a(2)=21 and a(n-1)*a(n+1) = a(n)^2-1
a(n) = S(n-1, 21)=U(n-1, 21/2) with S(n, x)=U(n, x/2) Chebyshev's polynomials of the 2nd kind, A049310. S(-1, x)= 0 = U(-1, x).
a(n) = S(2*n-1, sqrt(23))/sqrt(23), n>=1.
a(n) = 21*a(n-1)-a(n-2), n >= 1; a(0)=0, a(1)=1.
a(n) = (ap^n-am^n)/(ap-am) with ap := (21+sqrt(437))/2 and am := (21-sqrt(437))/2.
G.f.: x/(1-21*x+x^2).
a(n+1) = Sum_{k, 0<=k<=n} A101950(n,k)*20^k. - Philippe Deléham, Feb 10 2012
Product {n >= 1} (1 + 1/a(n)) = 1/19*(19 + sqrt(437)). - Peter Bala, Dec 23 2012
Product {n >= 2} (1 - 1/a(n)) = 1/42*(19 + sqrt(437)). - Peter Bala, Dec 23 2012

Extension, Chebyshev and Pell comments from Wolfdieter Lang, Aug 31 2004

Corrected by T. D. Noe, Nov 07 2006

A221366 The simple continued fraction expansion of F(x) := Product_{n >= 0} (1 - x^(4n+3))/(1 - x^(4n+1)) when x = (1/2)(7 - 3sqrt(5)).

Original entry on oeis.org

1, 5, 1, 45, 1, 320, 1, 2205, 1, 15125, 1, 103680, 1, 710645, 1, 4870845, 1, 33385280, 1, 228826125, 1, 1568397605, 1, 10749957120, 1, 73681302245, 1, 505019158605, 1, 3461452808000, 1, 23725150497405, 1

Offset: 0

Peter Bala, Jan 15 2013

The function F(x) := Product_{n >= 0} (1 - x^(4*n+3))/(1 - x^(4*n+1)) is analytic for |x| < 1. When x is a quadratic irrational of the form x = 1/2*(N - sqrt(N^2 - 4)), N an integer greater than 2, the real number F(x) has a predictable simple continued fraction expansion. The first examples of these expansions, for N = 2, 4, 6 and 8, are due to Hanna. See A174500 through A175503. The present sequence is the case N = 7. See also A221364 (N = 3), A221365 (N = 5) and A221367 (N = 9).

If we denote the present sequence by [1, c(1), 1, c(2), 1, c(3), ...] then for k = 1, 2, ..., the simple continued fraction expansion of F((1/2*(7 - sqrt(45)))^k) is given by the sequence [1; c(k), 1, c(2*k), 1, c(3*k), 1, ...]. Examples are given below.

			F(1/2*(7 - sqrt(45))) = 1.16725 98258 10214 95210 ... = 1 + 1/(5 + 1/(1 + 1/(45 + 1/(1 + 1/(320 + 1/(1 + 1/(2205 + ...))))))).
F((1/2*(7 - sqrt(45)))^2) = 1.02173 93445 69104 86504 ... = 1 + 1/(45 + 1/(1 + 1/(2205 + 1/(1 + 1/(103680 + 1/(1 + 1/(4870845 + ...))))))).
F((1/2*(7 - sqrt(45)))^3) = 1.00311 52648 91110 10148 ... = 1 + 1/(320 + 1/(1 + 1/(103680 + 1/(1 + 1/(33385280 + 1/(1 + 1/(10749957120 + ...))))))).

Peter Bala, Some simple continued fraction expansions for an infinite product, Part 1
Index entries for linear recurrences with constant coefficients, signature (0,8,0,-8,0,1).

Cf. A004187, A033890, A049682, A081070.

Cf. A174500 (N = 4), A221364 (N = 3), A221365 (N = 5), A221369 (N = 9).

LinearRecurrence[{0,8,0,-8,0,1},{1,5,1,45,1,320},40] (* or *) Riffle[ LinearRecurrence[{8,-8,1},{5,45,320},20],1,{1,-1,2}] (* Harvey P. Dale, Jan 04 2018 *)

a(2*n-1) = (1/2*(7 + sqrt(45)))^n + (1/2*(7 - sqrt(45)))^n - 2 = A081070(n); a(2*n) = 1.
a(4*n-1) = 45*A049682(n) = 45*(A004187(n))^2;
a(4*n+1) = 5*(A033890(n))^2.
a(n) = 8*a(n-2)-8*a(n-4)+a(n-6). G.f.: -(x^4+5*x^3-7*x^2+5*x+1) / ((x-1)*(x+1)*(x^2-3*x+1)*(x^2+3*x+1)). - Colin Barker, Jan 20 2013

A049678 a(n) = F(8*n+4)/3, where F=A000045 (the Fibonacci sequence).

Original entry on oeis.org

1, 48, 2255, 105937, 4976784, 233802911, 10983760033, 516002918640, 24241153416047, 1138818207635569, 53500214605455696, 2513371268248782143, 118074949393087305025, 5547009250206854554032, 260591359810329076734479, 12242246901835259751966481

Offset: 0

Clark Kimberling

			a(2) = F(8 * 2 + 4) / 3 = F(20) / 3 = 6765 / 3 = 2255. - _Indranil Ghosh_, Feb 04 2017

Indranil Ghosh, Table of n, a(n) for n = 0..596
Tanya Khovanova, Recursive Sequences
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences Integers, Volume 12A (2012) The John Selfridge Memorial Volume
Index entries for linear recurrences with constant coefficients, signature (47,-1).

Cf. A004187, A099483, A100047.

[Fibonacci(8*n+4)/3: n in [0..30]]; // G. C. Greubel, Dec 02 2017

CoefficientList[Series[(1+x)/(1-47x+x^2),{x,0,20}],x]  (* Harvey P. Dale, Feb 18 2011 *)
Table[Fibonacci[8*n+4]/3, {n,0,30}] (* G. C. Greubel, Dec 02 2017 *)

for(n=0,30, print1(fibonacci(8*n+4)/3, ", ")) \\ G. C. Greubel, Dec 02 2017

a(n) = 47*a(n-1) - a(n-2), n>1. a(0)=1, a(1)=48.
G.f.: (1+x)/(1-47*x+x^2).
From Peter Bala, Mar 23 2015: (Start)
a(n) = A004187(2*n + 1); a(n) = A099483(4*n + 1).
a(n) = ( Fibonacci(8*n + 8 - 2*k) + Fibonacci(8*n + 2*k) )/( Fibonacci(8 - 2*k) + Fibonacci(2*k) ), for k an arbitrary integer.
a(n) = ( Fibonacci(8*n + 8 - 2*k - 1) - Fibonacci(8*n + 2*k + 1) )/( Fibonacci(8 - 2*k - 1) - Fibonacci(2*k + 1) ), for k an arbitrary integer.
The aerated sequence (b(n))n>=1 = [1, 0, 48, 0, 2255, 0, 105937, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -45, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047 for the connection with Chebyshev polynomials. (End)

Better description and more terms from Michael Somos

2 more terms from Indranil Ghosh, Feb 04 2017

cp's OEIS Frontend

A094565 Triangle read by rows: binary products of Fibonacci numbers.

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A179943 Triangle read by rows, antidiagonals of an array (r,k), r=(0,1,2,...), generated from 2 X 2 matrices of the form [1,r; 1,(r+1)].

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A203319 a(n) = n*Fibonacci(n) * Sum_{d|n} 1/(d*Fibonacci(d)).

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A290903 p-INVERT of the positive integers, where p(S) = 1 - 5*S.

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A372817 Table read by antidiagonals: T(m,n) = number of 1-metered (m,n)-parking functions.

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A049686 a(n) = Fibonacci(8n)/3.

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A073134 Table by antidiagonals of T(n,k)=n*T(n,k-1)-T(n,k-2) starting with T(n,1)=1.

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A203319 a(n) = nFibonacci(n) Sum_{d|n} 1/(d*Fibonacci(d)).

A221366 The simple continued fraction expansion of F(x) := Product_{n >= 0} (1 - x^(4n+3))/(1 - x^(4n+1)) when x = (1/2)(7 - 3sqrt(5)).