A134490 -id:A134490 - cp's OEIS frontend

A103134 a(n) = Fibonacci(6n+4).

3, 55, 987, 17711, 317811, 5702887, 102334155, 1836311903, 32951280099, 591286729879, 10610209857723, 190392490709135, 3416454622906707, 61305790721611591, 1100087778366101931, 19740274219868223167, 354224848179261915075, 6356306993006846248183

Offset: 0

Creighton Dement, Jan 24 2005

Gives those numbers which are Fibonacci numbers in A103135.
Generally, for any sequence where a(0)= Fibonacci(p), a(1) = F(p+q) and Lucas(q)*a(1) +- a(0) = F(p+2q), then a(n) = L(q)*a(n-1) +- a(n-2) generates the following Fibonacci sequence: a(n) = F(q(n)+p). So for this sequence, a(n) = 18*a(n-1) - a(n-2) = F(6n+4): q=6, because 18 is the 6th Lucas number (L(0) = 2, L(1)=1); F(4)=3, F(10)=55 and F(16)=987 (F(0)=0 and F(1)=1). See Lucas sequence A000032. This is a special case where a(0) and a(1) are increasing Fibonacci numbers and Lucas(m)*a(1) +- a(0) is another Fibonacci. - Bob Selcoe, Jul 08 2013
a(n) = x + y where x and y are solutions to x^2 = 5*y^2 - 1. (See related sequences with formula below.) - Richard R. Forberg, Sep 05 2013

Colin Barker, Table of n, a(n) for n = 0..750
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (18,-1).
Index entries for sequences related to Chebyshev polynomials.

Subsequence of A033887.
Cf. A000032, A000045, A001906, A001519, A015448, A014445, A033888, A033889, A033890, A033891, A049310, A049660, A102312, A099100, A134490, A134491, A134492, A134493, A134494, A134495, A103134, A134497, A134498, A134499, A134500, A134501, A134502, A134503, A134504.
Cf. A103135.

[Fibonacci(6*n +4): n in [0..100]]; // Vincenzo Librandi, Apr 17 2011

Table[Fibonacci[6n+4], {n, 0, 30}]
LinearRecurrence[{18,-1},{3,55},20] (* Harvey P. Dale, Mar 29 2023 *)
Table[ChebyshevU[3*n+1, 3/2], {n, 0, 20}] (* Vaclav Kotesovec, May 27 2023 *)

a(n)=fibonacci(6*n+4) \\ Charles R Greathouse IV, Feb 05 2013

G.f.: (x+3)/(x^2-18*x+1).
a(n) = 18*a(n-1) - a(n-2) for n>1; a(0)=3, a(1)=55. - Philippe Deléham, Nov 17 2008
a(n) = A007805(n) + A075796(n), as follows from comment above. - Richard R. Forberg, Sep 05 2013
a(n) = ((15-7*sqrt(5)+(9+4*sqrt(5))^(2*n)*(15+7*sqrt(5))))/(10*(9+4*sqrt(5))^n). - Colin Barker, Jan 24 2016
a(n) = S(3*n+1, 3) = 3*S(n,18) + S(n-1,18), with the Chebyshev S polynomials (A049310), S(-1, x) = 0, and S(n, 18) = A049660(n+1). - Wolfdieter Lang, May 08 2023

Edited by N. J. A. Sloane, Aug 10 2010

A134504 a(n) = Fibonacci(7n + 6).

Original entry on oeis.org

8, 233, 6765, 196418, 5702887, 165580141, 4807526976, 139583862445, 4052739537881, 117669030460994, 3416454622906707, 99194853094755497, 2880067194370816120, 83621143489848422977, 2427893228399975082453

Offset: 0

Artur Jasinski, Oct 28 2007

Index entries for linear recurrences with constant coefficients, signature (29,1).

Cf. A000045, A001906, A001519, A033887, A015448, A014445, A033888, A033889, A033890, A033891, A102312, A099100, A134490, A134491, A134492, A134493, A134494, A134495, A103134, A134497, A134498, A134499, A134500, A134501, A134502, A134503, A134504.

[Fibonacci(7*n +6): n in [0..100]]; // Vincenzo Librandi, Apr 17 2011

Table[Fibonacci[7n+6], {n, 0, 30}]
LinearRecurrence[{29,1},{8,233},20] (* Harvey P. Dale, Jul 21 2021 *)

a(n)=fibonacci(7*n+6) \\ Charles R Greathouse IV, Jun 11 2015

G.f.: (-8-x) / (-1 + 29*x + x^2). - R. J. Mathar, Jul 04 2011
a(n) = A000045(A017053(n)). - Michel Marcus, Nov 08 2013
a(n) = 29*a(n-1) + a(n-2). - Wesley Ivan Hurt, Mar 15 2023

Offset changed from 1 to 0 by Vincenzo Librandi, Apr 17 2011

A100334 An inverse Catalan transform of Fibonacci(2n).

Original entry on oeis.org

0, 1, 2, 2, 0, -5, -13, -21, -21, 0, 55, 144, 233, 233, 0, -610, -1597, -2584, -2584, 0, 6765, 17711, 28657, 28657, 0, -75025, -196418, -317811, -317811, 0, 832040, 2178309, 3524578, 3524578, 0, -9227465, -24157817, -39088169, -39088169, 0, 102334155, 267914296, 433494437, 433494437, 0, -1134903170

Offset: 0

Paul Barry, Nov 17 2004

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-4,2,-1).

Cf. A001906, A109466.

Cf. A102312 (F(5n)), A134489 (F(5n+2)), A134490 (F(5n+3)).

I:=[0,1,2,2]; [n le 4 select I[n] else 3*Self(n-1) -4*Self(n-2) +2*Self(n-3) -Self(n-4): n in [1..41]]; // G. C. Greubel, Jan 30 2023

Table[FullSimplify[GoldenRatio^n*Sqrt[2/5 + 2*Sqrt[5]/25]*Sin[Pi*n/5 + Pi/5] - (1/GoldenRatio)^n*Sqrt[2/5 - 2*Sqrt[5]/25]*Sin[2*Pi*n/5 + 2*Pi/5]], {n, 0, 41}] (* Arkadiusz Wesolowski, Oct 26 2012 *)
LinearRecurrence[{3,-4,2,-1}, {0,1,2,2}, 41] (* G. C. Greubel, Jan 30 2023 *)

def A100334(n): return sum((-1)^k*binomial(n-k,k)*fibonacci(2*n-2*k) for k in range(1+(n//2)))
[A100334(n) for n in range(41)] # G. C. Greubel, Jan 30 2023

G.f.: x*(1-x)/(1-3*x+4*x^2-2*x^3+x^4).
a(n) = (phi)^n*sqrt(2/5+2*sqrt(5)/25)*sin(Pi*(n+1)/5) -(1/phi)^n*sqrt(2/5-2*sqrt(5)/25)*sin(2*Pi*(n+1)/5), where phi=(1+sqrt(5))/2;
a(n) = Sum_{k=0..floor(n/2)} (C(n-k, k)*(-1)^k*Sum_{j=0..n-k} C(n-k, j)*F(j));
a(n) = Sum_{k=0..floor(n/2)} C(n-k, k)*(-1)^k*Fibonacci(2n-2k).
a(n) = 3*a(n-1)-4*a(n-2)+2*a(n-3)-a(n-4). - Paul Curtz, May 13 2008
a(n) = Sum_{k=0..n} A109466(n,k)*A001906(k). - Philippe Deléham, Oct 30 2008
a(5*n) = -F(-5*n), a(5*n+1) = -F(-5*n-2), a(5*n+2) = a(5*n+3) = F(-5*n-3), a(5*n+4) = 0. - Ehren Metcalfe, Apr 04 2019

A134494 a(n) = Fibonacci(6n+2).

Original entry on oeis.org

1, 21, 377, 6765, 121393, 2178309, 39088169, 701408733, 12586269025, 225851433717, 4052739537881, 72723460248141, 1304969544928657, 23416728348467685, 420196140727489673, 7540113804746346429, 135301852344706746049, 2427893228399975082453

Offset: 0

Artur Jasinski, Oct 28 2007

Colin Barker, Table of n, a(n) for n = 0..750
Index entries for linear recurrences with constant coefficients, signature (18,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A000045, A001906, A001519, A015448, A014445, A033887-A033891, A049310, A049660, A099100, A102312, A103134, A134490 - A134504.

[Fibonacci(6*n +2): n in [0..100]]; // Vincenzo Librandi, Apr 17 2011

seq( combinat[fibonacci](6*n+2),n=0..10) ; # R. J. Mathar, Apr 17 2011

Table[Fibonacci[6n+2], {n, 0, 30}]
Table[ChebyshevU[3*n, 3/2], {n, 0, 20}] (* Vaclav Kotesovec, May 27 2023 *)

a(n)=fibonacci(6*n+2) \\ Charles R Greathouse IV, Jun 11 2015

Vec((1+3*x)/(1-18*x+x^2) + O(x^100)) \\ Altug Alkan, Jan 24 2016

From R. J. Mathar, Jul 04 2011: (Start)
G.f.: ( 1+3*x ) / ( 1-18*x+x^2 ).
a(n) = 3*A049660(n)+A049660(n+1). (End)
a(n) = A000045(A016933(n)). - Michel Marcus, Nov 07 2013
a(n) = ((5-3*sqrt(5)+(5+3*sqrt(5))*(9+4*sqrt(5))^(2*n)))/(10*(9+4*sqrt(5))^n). - Colin Barker, Jan 24 2016
a(n) = S(3*n, 3) = S(n,18) + 3*S(n-1,18), with the Chebyshev S polynomials (A049310), S(-1, x) = 0, and S(n, 18) = A049660(n+1). - Wolfdieter Lang, May 08 2023

Index in definition corrected by T. D. Noe, Joerg Arndt, Apr 17 2011

A134501 a(n) = Fibonacci(7n + 3).

Original entry on oeis.org

2, 55, 1597, 46368, 1346269, 39088169, 1134903170, 32951280099, 956722026041, 27777890035288, 806515533049393, 23416728348467685, 679891637638612258, 19740274219868223167, 573147844013817084101, 16641027750620563662096

Offset: 0

Artur Jasinski, Oct 28 2007

Index entries for linear recurrences with constant coefficients, signature (29,1).

Cf. A000045, A001906, A001519, A033887, A015448, A014445, A033888, A033889, A033890, A033891, A102312, A099100, A134490-A134495, A103134, A134497 - A134504.

[Fibonacci(7*n+3): n in [0..100]]; // Vincenzo Librandi, Apr 16 2011

Mathematica
```
Table[Fibonacci[7n+3], {n, 0, 30}]
```

a(n)=fibonacci(7*n+3) \\ Charles R Greathouse IV, Jun 11 2015

From R. J. Mathar, Jul 04 2011: (Start)
G.f.: (-2+3*x) / (-1 + 29*x + x^2).
a(n) = 2*A049667(n+1) - 3*A049667(n). (End)
a(n) = A000045(A017017(n)). - Michel Marcus, Nov 07 2013

Offset changed to 0 by Vincenzo Librandi, Apr 16 2011

A134502 a(n) = Fibonacci(7n + 4).

Original entry on oeis.org

3, 89, 2584, 75025, 2178309, 63245986, 1836311903, 53316291173, 1548008755920, 44945570212853, 1304969544928657, 37889062373143906, 1100087778366101931, 31940434634990099905, 927372692193078999176, 26925748508234281076009

Offset: 0

Artur Jasinski, Oct 28 2007

Index entries for linear recurrences with constant coefficients, signature (29,1)

Cf. A000045, A001906, A001519, A033887, A015448, A014445, A033888, A033889, A033890, A033891, A102312, A099100, A134490-A134495, A103134, A134497-A134504.

[Fibonacci(7*n +4): n in [0..100]]; // Vincenzo Librandi, Apr 17 2011

Mathematica
```
Table[Fibonacci[7n+4], {n, 0, 30}]
```

a(n)=fibonacci(7*n+4) \\ Charles R Greathouse IV, Jun 11 2015

From R. J. Mathar, Jul 04 2011: (Start)
G.f.: (-3-2*x) / (-1 + 29*x + x^2).
a(n) = 2*A049667(n) + 3*A049667(n+1). (End)
a(n) = A000045(A017029(n)). - Michel Marcus, Nov 07 2013

Offset changed from 1 to 0 by Vincenzo Librandi, Apr 17 2011

A134491 a(n) = Fibonacci(5n+4).

Original entry on oeis.org

3, 34, 377, 4181, 46368, 514229, 5702887, 63245986, 701408733, 7778742049, 86267571272, 956722026041, 10610209857723, 117669030460994, 1304969544928657, 14472334024676221, 160500643816367088

Offset: 0

Artur Jasinski, Oct 28 2007

Index entries for linear recurrences with constant coefficients, signature (11,1)

Cf. A000045, A016897, A102312, A099100, A134490.

[Fibonacci(5*n +4): n in [0..100]]; // Vincenzo Librandi, Apr 17 2011

Mathematica
```
Table[Fibonacci[5n + 4], {n, 0, 30}]
```

a(n)=fibonacci(5*n+4) \\ Charles R Greathouse IV, Oct 16 2015

G.f.: ( -3-x ) / ( -1+11*x+x^2 ). - R. J. Mathar, Apr 17 2011

a(n) = A000045(A016897(n)). - Michel Marcus, Nov 07 2013

A134489 a(n) = Fibonacci(5*n + 2).

Original entry on oeis.org

1, 13, 144, 1597, 17711, 196418, 2178309, 24157817, 267914296, 2971215073, 32951280099, 365435296162, 4052739537881, 44945570212853, 498454011879264, 5527939700884757, 61305790721611591, 679891637638612258

Offset: 0

Artur Jasinski, Oct 28 2007

The o.g.f. of {F(m*n + 2)}_{n>=0}, for m = 1, 2, ..., is

G(m,x) = (1 + F(m - 2)*x) / (1 - L(m)*x + (-1)^m*x^2), with F = A000045 and F(-1) = 1, and L = A000032. - Wolfdieter Lang, Feb 06 2023

Index entries for linear recurrences with constant coefficients, signature (11,1).

Cf. A000045, A001906, A001519, A033887, A015448, A014445, A033888-A033891, A102312, A099100, A134490-A134495, A103134, A134497-A134504.

[Fibonacci(5*n+2): n in [0..50]]; // Vincenzo Librandi, Apr 20 2011

Table[Fibonacci[5n + 2], {n, 0, 30}]
LinearRecurrence[{11,1},{1,13},20] (* Harvey P. Dale, May 05 2022 *)

From R. J. Mathar, Jul 04 2011: (Start)
G.f.: (-1-2*x) / (-1 + 11*x + x^2).
a(n) = 2*A049666(n) + A049666(n+1). (End)
a(n) = A000045(A016873(n)). - Michel Marcus, Nov 05 2013

A138110 Table T(d,n) read column by column: the n-th term in the sequence of the d-th differences of A138112, d=0..4.

Original entry on oeis.org

0, 0, 0, 1, -1, 0, 0, 1, 0, -1, 0, 1, 1, -1, -1, 1, 2, 0, -2, -1, 3, 2, -2, -3, 0, 5, 0, -5, -3, 3, 5, -5, -8, 0, 8, 0, -13, -8, 8, 13, -13, -21, 0, 21, 13, -34, -21, 21, 34, 0, -55, 0, 55, 34, -34, -55, 55, 89, 0, -89, 0, 144, 89, -89, -144, 144, 233, 0, -233, -144, 377, 233, -233, -377, 0, 610, 0, -610, -377, 377

Offset: 0

Paul Curtz, May 04 2008

Ignoring signs, the sequence contains A000045(2)=1 ten times and each of the following Fibonacci numbers A000045(i>2) four times.

			All 5 rows of the table T(d,n) are:
.0,.0,.0,.1,.3,.5,.5,..0,-13,-34,-55,-55,...0,.144,...
.0,.0,.1,.2,.2,.0,-5,-13,-21,-21,..0,.55,.144,.233,...
.0,.1,.1,.0,-2,-5,-8,.-8,..0,.21,.55,.89,..89,...0,...
.1,.0,-1,-2,-3,-3,.0,..8,.21,.34,.34,..0,.-89,-233,...
-1,-1,-1,-1,.0,.3,.8,.13,.13,..0,-34,-89,-144,-144,...

Cf. A102312 (A049666), A099100, A134489, A134490, A134491.

T(0,n)=A138112(n). T(d,n)= T(d-1,n+1)-T(d-1,n), d=1..4.
T(1,n)=A100334(n-1). T(2,n)=A103311(n). T(3,n) = -A138003(n-2). T(4,n)= -A105371(n).
sum_(d=0..4) T(d,n)=0 (columns sum to zero).

Edited by R. J. Mathar, Jul 04 2008

A141053 Most-significant decimal digit of Fibonacci(5n+3).

Original entry on oeis.org

2, 2, 2, 2, 2, 3, 3, 3, 4, 4, 5, 5, 6, 7, 8, 8, 9, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 4, 4, 5, 5, 6, 7, 7, 8, 9, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 5, 5, 6, 6, 7, 8, 9, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2

Offset: 0

Paul Curtz, Aug 01 2008

Leading digit of A134490(n).
From Johannes W. Meijer, Jul 06 2011: (Start)
The leading digit d, 1 <= d <= 9, of A141053 follows Benford’s Law. This law states that the probability for the leading digit is p(d) = log_10(1+1/d), see the examples.
We observe that the last digit of A134490(n), i.e. F(5*n+3) mod 10, leads to the Lucas sequence A000032(n) (mod 10), i.e. a repetitive sequence of 12 digits [2, 1, 3, 4, 7, 1, 8, 9, 7, 6, 3, 9] with p(0) = p(5) = 0, p(1) = p(3) = p(7) = p(9) = 1/6 and p(2) = p(4) = p(6) = p(8) = 1/12. This does not obey Benford’s Law, which would predict that the last digit would satisfy p(d) = 1/10, see the links. (End)

			From _Johannes W. Meijer_, Jul 06 2011: (Start)
d     p(N=2000) p(N=4000) p(N=6000) p(Benford)
1      0.29900   0.29950   0.30033   0.30103
2      0.17700   0.17675   0.17650   0.17609
3      0.12550   0.12525   0.12517   0.12494
4      0.09650   0.09675   0.09700   0.09691
5      0.07950   0.07950   0.07933   0.07918
6      0.06700   0.06675   0.06700   0.06695
7      0.05800   0.05825   0.05800   0.05799
8      0.05150   0.05125   0.05100   0.05115
9      0.04600   0.04600   0.04567   0.04576
Total  1.00000   1.00000   1.00000   1.00000 (End)

Hans J. H. Tuenter, Table of n, a(n) for n = 0..10000
Kevin Brown, Benford's Law.
Eric Weisstein's World of Mathematics, Benford's Law.
Wikipedia, Benford's Law.
Index entries for sequences related to Benford's law

Cf. A000045 (F(n)), A008963 (Initial digit F(n)), A105511-A105519, A003893 (F(n) mod 10), A130893, A186190 (First digit tribonacci), A008952 (Leading digit 2^n), A008905 (Leading digit n!), A045510, A112420 (Leading digit Collatz 3*n+1 starting with 1117065), A007524 (log_10(2)), A104140 (1-log_10(9)). - Johannes W. Meijer, Jul 06 2011

Cf. A001622, A097348.

A134490 := proc(n) combinat[fibonacci](5*n+3) ; end proc:
A141053 := proc(n) convert(A134490(n),base,10) ; op(-1,%) ; end proc:
seq(A141053(n),n=0..70) ; # R. J. Mathar, Jul 04 2011

Table[IntegerDigits[Fibonacci[5n+3]][[1]],{n,0,70}] (* Harvey P. Dale, Jun 22 2025 *)

a(n) = floor(F(5*n+3)/10^(floor(log(F(5*n+3))/log(10)))). - Johannes W. Meijer, Jul 06 2011

For n>0, a(n) = floor(10^{alpha*n+beta}), where alpha=5*log_10(phi)-1, beta=log_10(1+2/sqrt(5)), {x}=x-floor(x) denotes the fractional part of x, log_10(phi) = A097348, and phi = (1+sqrt(5))/2 = A001622. - Hans J. H. Tuenter, Aug 27 2025

Edited by Johannes W. Meijer, Jul 06 2011

cp's OEIS Frontend

A103134 a(n) = Fibonacci(6n+4).

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A134504 a(n) = Fibonacci(7n + 6).

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A100334 An inverse Catalan transform of Fibonacci(2n).

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A134494 a(n) = Fibonacci(6n+2).

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A134501 a(n) = Fibonacci(7n + 3).

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A134502 a(n) = Fibonacci(7n + 4).

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A134491 a(n) = Fibonacci(5n+4).

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