A028412 -id:A028412 - cp's OEIS frontend

A049667 a(n) = Fibonacci(7*n)/13.

0, 1, 29, 842, 24447, 709805, 20608792, 598364773, 17373187209, 504420793834, 14645576208395, 425226130837289, 12346203370489776, 358465123875040793, 10407834795746672773, 302185674200528551210, 8773792386611074657863

Offset: 0

Clark Kimberling

G. C. Greubel, Table of n, a(n) for n = 0..650
Tanya Khovanova, Recursive Sequences
Shaoxiong Yuan, Generalized Identities of Certain Continued Fractions, arXiv:1907.12459 [math.NT], 2019.
Index entries for linear recurrences with constant coefficients, signature (29, 1).

A column of array A028412.

Cf. A000045, A134498.

[Fibonacci(7*n)/13: n in [0..30]]; // G. C. Greubel, Dec 02 2017

a:= n-> (<<0|1>, <1|29>>^n)[1, 2]:
seq(a(n), n=0..20);  # Alois P. Heinz, Sep 20 2017

Fibonacci[(7*Range[0,20])]/13 (* or *) LinearRecurrence[{29,1},{0,1},20] (* Harvey P. Dale, Sep 17 2017 *)

numlib::fibonacci(7*n)/13 $ n = 0..25; // Zerinvary Lajos, May 09 2008

a(n)=fibonacci(7*n)/13 \\ Charles R Greathouse IV, Oct 07 2016

[fibonacci(7*n)/13 for n in range(0, 17)] # Zerinvary Lajos, May 15 2009

G.f.: x/(1 - 29*x - x^2).
a(n) = A134498(n)/13.
a(n) = F(n, 29), the n-th Fibonacci polynomial evaluated at x=29. - T. D. Noe, Jan 19 2006
a(n) = 29*a(n-1) + a(n-2), n > 1; a(0)=0, a(1)=1. - Philippe Deléham, Nov 22 2008
For n >= 1, a(n) equals the denominator of the continued fraction [29, 29, ..., 29] (with n copies of 29). The numerator of that continued fraction is a(n+1). - Greg Dresden and Shaoxiong Yuan, Jul 26 2019
a(n) = ((-1)^n*7*F(n) + 14*5*F(n)^3 + (-1)^n*7*5^2*F(n)^5 + 5^3*F(n)^7)/13, n >= 0. See the general D. Jennings formula given in comment on triangle A111125, where also the reference is given. Here the fourth row (k=3) applies. - Wolfdieter Lang, Sep 01 2012
G.f.: G(0)*x/(2-29*x), where G(k)= 1 + 1/(1 - (x*(845*k-841))/((x*(845*k+4)) - 58/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 15 2013
O.g.f.: x*exp(Sum_{n >= 1} Lucas(7*n)*x^n/n) = x + 29*x^2 + 842*x^3 + .... - Peter Bala, Oct 11 2019

A051294 a(n) = F(n^2)/F(n), where F(n) = A000045(n) is the n-th Fibonacci number.

Original entry on oeis.org

1, 3, 17, 329, 15005, 1866294, 598364773, 505248088463, 1114384187445409, 6440451785077489365, 97415813466381445596089, 3858093084890921488916776332, 400009475456580321242184872389193, 108580324845164033411588204172311746011, 77162132055850720265395402022419687035969985

Offset: 1

Leroy Quet

Clark Kimberling, Table of n, a(n) for n = 1..69 (all terms with <= 1000 digits)

Main diagonal of array A028412.

a:= n-> (f->f(n^2)/f(n))(k->(<<0|1>, <1|1>>^k)[1, 2]):
seq(a(n), n=1..15);  # Alois P. Heinz, May 08 2025

Table[Fibonacci[n^2]/Fibonacci[n],{n,15}]  (* Harvey P. Dale, Apr 12 2011 *)

PARI
```
a(n)=fibonacci(n^2)/fibonacci(n)
```

{Lucas(n)=fibonacci(n-1)+fibonacci(n+1)}
{a(n)=polcoeff(1/(1-Lucas(n)*x+(-1)^n*x^2+x*O(x^n)),n-1)} /* Paul D. Hanna, Jan 28 2012 */

a(n) = [x^(n-1)] 1/(1 - Lucas(n)*x + (-1)^n*x^2), where Lucas(n) = A000204(n). - Paul D. Hanna, Jan 28 2012

More terms from Benoit Cloitre, Jan 05 2003

A047946 a(n) = 5F(n)^2 + 3(-1)^n where F(n) are the Fibonacci numbers A000045.

Original entry on oeis.org

3, 2, 8, 17, 48, 122, 323, 842, 2208, 5777, 15128, 39602, 103683, 271442, 710648, 1860497, 4870848, 12752042, 33385283, 87403802, 228826128, 599074577, 1568397608, 4106118242, 10749957123, 28143753122, 73681302248, 192900153617, 505019158608, 1322157322202

Offset: 0

John W. Layman, May 21 1999

Form the matrix A=[1,1,1;2,1,0;1,0,0]. a(n)=trace(A^n). - Paul Barry, Sep 22 2004
The set of prime divisors of elements of this sequence with the exception of 3 is the set of primes that do not divide odd Fibonacci numbers. - Tanya Khovanova, May 19 2008
If a(n) is prime then n is a power of 3 (Boase, 1998). The only values of k not exceeding 12 for which a(3^k) is prime are 0 and 1. - Amiram Eldar, Jun 19 2022

Mansur Boase, Problem 1558, Mathematics Magazine, Vol. 71, No. 4 (1998), p. 316; Primes in a Recursively Defined Sequence, Solution to Problem 1558 by TAMUK Problem Solvers, ibid., Vol. 72, No. 4 (1999), pp. 330-331.
Tanya Khovanova, Divisibility of Odd Fibonaccis, 2008.
Claudio de Jesús Pita Ruiz Velasco, On s-Fibonomials, Journal of Integer Sequences, Vol. 14 (2011), Article 11.3.7.
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Cf. A000045, A000032, A005248.
Second row of array A028412.
Cf. A133247 (prime numbers p such that no odd Fibonacci number is divisible by p).

Table[LucasL[n]^2 - (-1)^n, {n, 0, 30}] (* Amiram Eldar, Feb 02 2022 *)

PARI
```
a(n)=5*fibonacci(n)^2+3*(-1)^n
```

from sympy import fibonacci
def A047946(n): return 5*fibonacci(n)**2+(-3 if n&1 else 3) # Chai Wah Wu, Jul 29 2022

a(n) = F(3n)/F(n), n>0.
a(n) = 2*a(n-1)+2*a(n-2)-a(n-3).
a(n) = 3a(n-1)-a(n-2)+5(-1)^n.
a(n) = A005248(n) + (-1)^n.
G.f.: ( 3-4*x-2*x^2 ) / ( (1+x)*(x^2-3*x+1) ).
for n>0 a(n) = A000045(3n)/A000045(n) - Benoit Cloitre, Aug 30 2003
For n>0, the linear recurrence for the sequence F(n*k)^2 has signature (a(n),a(n),-1) for n odd, and (a(n),-a(n), 1) for n even. For example, the linear recurrence for the sequence F(3*k)^2 has signature (17, 17, -1) (conjectured). - Greg Dresden, Aug 30 2021
a(n) = Lucas(n)^2 - (-1)^n. - Amiram Eldar, Feb 02 2022

Entry improved by comments from Michael Somos.

A049669 a(n) = Fibonacci(9*n)/34.

Original entry on oeis.org

0, 1, 76, 5777, 439128, 33379505, 2537281508, 192866774113, 14660412114096, 1114384187445409, 84707858657965180, 6438911642192799089, 489441992665310695944, 37204030354205805690833

Offset: 0

Clark Kimberling

nonn

G. C. Greubel, Table of n, a(n) for n = 0..530
S. Falcon, Generalized Fibonacci Sequences Generated from a k-Fibonacci Sequence, Journal of Mathematics Research Vol. 4, No. 2; April 2012. - From _N. J. A. Sloane_, Sep 22 2012
Tanya Khovanova, Recursive Sequences
Shaoxiong Yuan, Generalized Identities of Certain Continued Fractions, arXiv:1907.12459 [math.NT], 2019.
Index entries for linear recurrences with constant coefficients, signature (76, 1).

A column of array A028412.

Cf. A000045.

[Fibonacci(9*n)/(34): n in [0..30]]; // G. C. Greubel, Dec 02 2017

with (combinat):seq(fibonacci(3*n,4)/17, n=0..13); # Zerinvary Lajos, Apr 20 2008

Fibonacci[9Range[0,20]]/34 (* or *) LinearRecurrence[{76,1},{0,1},20] (* Harvey P. Dale, Jan 20 2013 *)

numlib::fibonacci(9*n)/34 $ n = 0..25; // Zerinvary Lajos, May 09 2008

for(n=0,30, print1(fibonacci(9*n)/34, ", ")) \\ G. C. Greubel, Dec 02 2017

G.f.: x/(1-76*x-x^2), 76=L(9)=A000032(9) (Lucas).
a(n) = 76*a(n-1) + a(n-2), n>1, a(0)=0, a(1)=1. - Philippe Deléham, Nov 23 2008
a(n) = (9*F(n) + (-1)^n*30*5*F(n)^3 + 27*5^2*F(n)^5 + (-1)^n*9*5^3*F(n)^7 + 5^4*F(n)^9)/34, n >= 0. See the general D. Jennings formula given in a comment on the triangle A111125, where also the reference is given. Here the fifth row (k=4) applies. - Wolfdieter Lang, Sep 01 2012
For n >= 1, a(n) equals the denominator of the continued fraction [76, 76, ..., 76] (with n copies of 76). The numerator of that continued fraction is a(n+1). - Greg Dresden and Shaoxiong Yuan, Jul 26 2019
E.g.f.: exp(38*x)*sinh(17*sqrt(5)*x)/(17*sqrt(5)). - Stefano Spezia, Aug 05 2019

More terms from James Sellers, Jan 20 2000

A383742 Square array A(n,k), n>=0, k>=0, read by antidiagonals downwards, where column k is the expansion of g.f. x/(1 - A002203(k)x + (-1)^kx^2).

Original entry on oeis.org

0, 0, 1, 0, 1, 2, 0, 1, 2, 3, 0, 1, 6, 5, 4, 0, 1, 14, 35, 12, 5, 0, 1, 34, 197, 204, 29, 6, 0, 1, 82, 1155, 2772, 1189, 70, 7, 0, 1, 198, 6725, 39236, 39005, 6930, 169, 8, 0, 1, 478, 39203, 551532, 1332869, 548842, 40391, 408, 9, 0, 1, 1154, 228485, 7761996, 45232349, 45278310, 7722793, 235416, 985, 10

Offset: 0

Seiichi Manyama, May 07 2025

			Square array begins:
  0,  0,    0,     0,       0,        0, ...
  1,  1,    1,     1,       1,        1, ...
  2,  2,    6,    14,      34,       82, ...
  3,  5,   35,   197,    1155,     6725, ...
  4, 12,  204,  2772,   39236,   551532, ...
  5, 29, 1189, 39005, 1332869, 45232349, ...

Columns k=0..6 give A001477, A000129, A001109, A041085(n-1), A091761, A292423, A097731(n-1).
Rows n=0..5 give A000004, A000012, A002203, A383720, A383740, A383741.
Main diagonal gives A380083.
Cf. A028412.

A[n_, k_] := Fibonacci[k*n, 2]/Fibonacci[k, 2]; A[n_, 0] := n; Table[A[k, n - k], {n, 0, 10}, {k, 0, n}] // Flatten (* Amiram Eldar, May 08 2025 *)

pell(n) = ([2, 1; 1, 0]^n)[2, 1];
a(n, k) = if(k==0, n, pell(k*n)/pell(k));

A(0,k) = 0, A(1,k) = 1; A(n,k) = A002203(k) * A(n-1,k) - (-1)^k * A(n-2,k) for n > 1.

A(n,k) = Pell(k*n)/Pell(k) for k > 0.

A083564 a(n) = L(n)*L(2n), where L(n) are the Lucas numbers (A000204).

Original entry on oeis.org

3, 21, 72, 329, 1353, 5796, 24447, 103729, 439128, 1860621, 7880997, 33385604, 141421803, 599075421, 2537719272, 10749959329, 45537545553, 192900159396, 817138154247, 3461452823129, 14662949371128, 62113250430021

Offset: 1

Gary W. Adamson, Jun 12 2003

a(n+1)/a(n) -> (phi)^3 = ((1 + sqrt(5))/2)^3 = 4.236067...

			a(4) = Lucas(4)*Lucas(8) = 7*47 = 329.

Vincenzo Librandi, Table of n, a(n) for n = 1..160
C. Pita, On s-Fibonomials, J. Int. Seq. 14 (2011) # 11.3.7
Index entries for linear recurrences with constant coefficients, signature (3, 6, -3, -1).

Third row of array A028412.

[Lucas(n)*Lucas(2*n): n in [1..25]]; // Vincenzo Librandi, May 03 2011

Table[Fibonacci[n*4]/Fibonacci[n],{n,50}] (* Vladimir Joseph Stephan Orlovsky, May 02 2011 *)

From Benoit Cloitre, Aug 30 2003: (Start)
a(n) = 3*a(n-1) + 6*a(n-2) - 3*a(n-3) - a(n-4);
a(n) = Fibonacci(4*n)/Fibonacci(n) = A000045(4*n)/A000045(n). (End)
a(n) = Lucas(3*n) + (-1)^n*Lucas(n).
From R. J. Mathar, Oct 27 2008: (Start)
G.f.: x*(3+12*x-9*x^2-4*x^3)/((1+x-x^2)*(1-4*x-x^2)).
a(n) = A061084(n+1) + 2*A001077(n). (End)
a(n) = (1+phi)^n + (-phi)^n + (2*phi+1)^n + (3-2*phi)^n, phi = (1+sqrt(5))/2. - Gary Detlefs, Dec 09 2012

A103326 a(n) = Fibonacci(5n)/Fibonacci(n).

Original entry on oeis.org

5, 55, 305, 2255, 15005, 104005, 709805, 4873055, 33379505, 228841255, 1568358005, 10750060805, 73681030805, 505019869255, 3461450947505, 23725155368255, 162614587921805, 1114577087604805, 7639424691459005, 52361396626646255, 358890349406803505

Offset: 1

Ralf Stephan, Feb 03 2005

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (5,15,-15,-5,1).

Fourth row of array A028412.

Cf. A000032, A000045, A001656, A088545.

[Fibonacci(5*n)/Fibonacci(n): n in [1..50]]; // Vincenzo Librandi, Apr 20 2011

p:= (1+5^(1/2))/2: q:=(1-5^(1/2))/2:
seq(simplify(q^(4*n)+(p-2)^n+(q-2)^n+(3*p+2)^n+(-1)^(2*n)/4+3/4),n=1..19);

Vec(-5*x*(x^4-4*x^3-9*x^2+6*x+1)/((x-1)*(x^2-7*x+1)*(x^2+3*x+1)) + O(x^30)) \\ Colin Barker, Jun 03 2016

a(n) = L(4n) + (-1)^n*L(2n) + 1, where L(n) = A000032, the Lucas numbers.
a(n) = 1 + L(n)*L(3n). - Neven Juric, Jan 05 2009
a(n) = 25*(Fibonacci(n)^4 + (-1)^n*Fibonacci(n)^2) + 5. - Gary Detlefs, Dec 22 2012
G.f.: -5*x*(x^4 - 4*x^3 - 9*x^2 + 6*x + 1) /((x - 1)*(x^2 - 7*x + 1)*(x^2 + 3*x + 1)). - Colin Barker, Jul 16 2013
a(n) = 5*A088545(n). - Joerg Arndt, Jul 16 2013
exp(Sum_{n >= 1} a(n)*x^n/n) = Sum_{n >= 0} A001656(n)*x^n. - Peter Bala, Mar 30 2015
a(n) = 1 + (1/2*(7 - 3*sqrt(5)))^n + (1/2*(-3 - sqrt(5)))^n + (1/2*(-3 + sqrt(5)))^n + (1/2*(7 + 3*sqrt(5)))^n. - Colin Barker, Jun 03 2016

More terms from Colin Barker, Jul 16 2013

A103624 Quotient F(n(n+1))/{F(n)*F(n+1)}, where F(n) is the n-th Fibonacci number A000045(n).

Original entry on oeis.org

1, 4, 24, 451, 20801, 2576099, 827294629, 698114862576, 1540142884690276, 8900260727038783399, 134626641626040936157699, 5331733869372412024840703621, 552800277142057127306392295957801

Offset: 1

Lekraj Beedassy, Mar 25 2005

Robert Israel, Table of n, a(n) for n = 1..69

Cf. A000045, A205505, A051294, A028412, A037943.

seq(combinat:-fibonacci(n*(n+1))/combinat:-fibonacci(n)/combinat:-fibonacci(n+1),n=1..25); # Robert Israel, Jan 11 2018

Table[Fibonacci[n(n+1)]/(Fibonacci[n]Fibonacci[n+1]),{n,20}] (* Harvey P. Dale, Aug 04 2012 *)

{a(n)=fibonacci(n*(n+1))/(fibonacci(n)*fibonacci(n+1))}  \\ Paul D. Hanna, Jan 28 2012

a(n) = phi^(n^2 - n - 1) + O(phi^(n^2 - 3n)). [Charles R Greathouse IV, Feb 01 2012]

a(2) corrected by Paul D. Hanna, Jan 28 2012

A214982 a(n) = (Fibonacci(5n)/Fibonacci(n) - 5)/50.

Original entry on oeis.org

0, 1, 6, 45, 300, 2080, 14196, 97461, 667590, 4576825, 31367160, 215001216, 1473620616, 10100397385, 69229018950, 474503107365, 3252291758436, 22291541752096, 152788493829180, 1047227932532925, 7177806988136070

Offset: 1

Clark Kimberling, Oct 28 2012

nonn

See the comments at A028412.

Clark Kimberling, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (5, 15, -15, -5, 1).

Cf. A000045, A028412.

(See A028412.)
Table[(Fibonacci[5n]/Fibonacci[n]-5)/50,{n,25}] (* or *) LinearRecurrence[ {5,15,-15,-5,1},{0,1,6,45,300},30] (* Harvey P. Dale, Nov 03 2013 *)

a(n) = (Fibonacci(5n)/Fibonacci(n) - 5)/50.
Empirical G.f.: -x^2*(x+1)/((x-1)*(x^2-7*x+1)*(x^2+3*x+1)). - Colin Barker, Nov 22 2012
a(n) = 5*a(n-1) + 15*a(n-2) - 15*a(n-3) - 5*a(n-4) + a(n-5), with a(1)=0, a(2)=1, a(3)=6, a(4)=45, a(5)=300. - Harvey P. Dale, Nov 03 2013
a(n) = (1/2)*Fibonacci(n)^2*(Fibonacci(n)^2 + (-1)^n) shows that a(n) is always an integer. - Peter Bala, Nov 29 2013

A205505 Fibonacci(n*(n+1)) / Fibonacci(n).

Original entry on oeis.org

1, 8, 72, 2255, 166408, 33489287, 17373187209, 23735905327584, 84707858657965180, 792123204706451722511, 19386236394149894806708656, 1242293991563772001787883943693, 208405704482555536994509895576090977, 91533085042008706066658193727853843719640

Offset: 1

Paul D. Hanna, Jan 28 2012

Cf. A103624, A051294, A028412.

Table[Fibonacci[n(n+1)]/Fibonacci[n],{n,20}] (* Harvey P. Dale, Mar 30 2012 *)

PARI
```
{a(n)=fibonacci(n*(n+1))/fibonacci(n)}
```

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

a(n) = [x^n] 1/(1 - Lucas(n)*x + (-1)^n*x^2), where Lucas(n) = A000204(n).

Forms a diagonal in table A028412.

cp's OEIS Frontend

A049667 a(n) = Fibonacci(7*n)/13.

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A051294 a(n) = F(n^2)/F(n), where F(n) = A000045(n) is the n-th Fibonacci number.

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A047946 a(n) = 5*F(n)^2 + 3*(-1)^n where F(n) are the Fibonacci numbers A000045.

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A049669 a(n) = Fibonacci(9*n)/34.

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A383742 Square array A(n,k), n>=0, k>=0, read by antidiagonals downwards, where column k is the expansion of g.f. x/(1 - A002203(k)*x + (-1)^k*x^2).

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A083564 a(n) = L(n)*L(2n), where L(n) are the Lucas numbers (A000204).

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A103326 a(n) = Fibonacci(5n)/Fibonacci(n).

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A047946 a(n) = 5F(n)^2 + 3(-1)^n where F(n) are the Fibonacci numbers A000045.

A383742 Square array A(n,k), n>=0, k>=0, read by antidiagonals downwards, where column k is the expansion of g.f. x/(1 - A002203(k)x + (-1)^kx^2).