A054783 -id:A054783 - cp's OEIS frontend

A058635 a(n) = Fibonacci(2^n).

1, 1, 3, 21, 987, 2178309, 10610209857723, 251728825683549488150424261, 141693817714056513234709965875411919657707794958199867

Offset: 0

Robert G. Wilson v, Jan 16 2001

nonn

The next term has 107 digits.
From Peter Bala, Oct 30 2013: (Start)
Apart from giving the numerators in the Engel series representation of (1/2)*(7 - sqrt(5)), as noted below by Cloitre, this sequence (excluding the initial term) is also a generalized Pierce expansion defined as follows. Let x and b be positive real numbers. We define a Pierce expansion of x to the base b to be a nondecreasing sequence [a(1), a(2), a(3), ...] of positive integers such that we have an alternating series representation x = b/a(1) - b^2/(a(1)*a(2)) + b^3/(a(1)*a(2)*a(3)) - ....
The present sequence, apart from the initial term, is a Pierce expansion of the real number x := (1/2)*(3 - sqrt(5)) to the base b := 1/sqrt(5). The associated series representation begins (1/2)*(3 - sqrt(5)) = b/1 - b^2/(1*3) + b^3/(1*3*21) - b^4/(1*3*21*987) + .... Cf. A071579 and A230338.
More generally, for n >= 0, the sequence [a(n+1), a(n+2), a(n+3), ...] gives a Pierce expansion of ( (1/2)*(3 - sqrt(5)) )^(2^n) to the base b = 1/sqrt(5). Some examples are given below. (End)
a(n) is the denominator of the n-th iterate when Newton's method is applied to the function x^2 - x - 1 with initial guess x = 1. The n-th iterate is A192222(n)/a(n). - Jason Zimba, Jan 20 2023

			Let b = 1/sqrt(5) and x = (1/2)*(3 - sqrt(5)). We have the following Pierce expansions to base b:
x = b/1 - b^2/(1*3) + b^3/(1*3*21) - b^4/(1*3*21*987) + ....
x^2 = b/3 - b^2/(3*21) + b^3/(3*21*987) - b^4/(3*21*987*2178309) + ....
x^4 = b/21 - b^2/(21*987) + b^3/(21*987*2178309) - ....
x^8 = b/987 - b^2/(987*2178309) + .... - _Peter Bala_, Oct 30 2013

Jay Kappraff, Beyond Measure, A Guided Tour Through Nature, Myth and Number, World Scientific, 2002, p. 446.

Vincenzo Librandi, Table of n, a(n) for n = 0..12
Mohammad K. Azarian, Fibonacci Identities as Binomial Sums, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 38, 2012, pp. 1871-1876. Mathematical Reviews, MR2959001. Zentralblatt MATH, Zbl 1255.05003.
Mohammad K. Azarian, Fibonacci Identities as Binomial Sums II, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 42, 2012, pp. 2053-2059. Mathematical Reviews, MR2980853. Zentralblatt MATH, Zbl 1255.05004.
S. B. Ekhad and D. Zeilberger, How To Generate As Many Somos-Like Miracles as You Wish, arXiv preprint arXiv:1303.5306 [math.CO], 2013.
John Gill and Matthew Miller, Newton's Method and Ratios of Fibonacci Numbers, Fibonacci Quarterly, 19(1):1-3, February 1981.
H. Hu, Z.-W. Sun and J.-X. Liu, Reciprocal sums of second order recurrent sequences, Fib. Quart. 39(2001), no. 3, 214-220.
Hideyuki Ohtsuka, Problem H-767, Advanced Problems and Solutions, The Fibonacci Quarterly, Vol. 53, No. 1 (2015), p. 88; Nested Radicals and Fibonacci Numbers, Solution to Problem H-767 by the proposer, ibid., Vol. 55, No. 1 (2017), pp. 90-91.

Cf. A000032, A000045, A000079, A054783, A001566, A071579, A230338.

[Fibonacci(2^n): n in [0..10]]; // Vincenzo Librandi, Mar 25 2014

a:= n-> (<<0|1>, <1|1>>^(2^n))[1,2]:
seq(a(n), n=0..10);  # Alois P. Heinz, Nov 21 2014

Table[ Fibonacci[ 2^n ], {n, 0, 9} ]
G = (1 + Sqrt[5])/2; Table[Expand[(G^(2^n) - (1 - G)^(2^n))/Sqrt[5]], {n, 1, 7}] (* Artur Jasinski, Oct 05 2008 *)
Table[Round[(4/5)^(1/2)*Cosh[2^n*ArcCosh[((5/4)^(1/2))]]], {n, 1, 10}] (* Artur Jasinski, Oct 05 2008 *)

a(n)=fibonacci(2^n) \\ Charles R Greathouse IV, Oct 03 2016

a(n) = a(n-1)*A001566(n-2). - Joe Keane (jgk(AT)jgk.org), May 31 2002
Sum_{n>=0} 1/a(n) = (1/2)*(7-sqrt(5)). - Benoit Cloitre, Jan 26 2003
1/phi^2 = (0.6180339...)^2 = 2/(3+sqrt(5)) = Sum_{n>=2} 1/a(n) = 1/3 + 1/21 + 1/987 + 1/2178309 + ... - Gary W. Adamson, Jun 12 2003
From Artur Jasinski, Oct 05 2008: (Start)
a(n) = (G^(2^n) - (1 - G)^(2^n))/sqrt(5) where G = GoldenRatio = (1 + sqrt(5))/2.
a(n) = sqrt(4/5)*cosh((2^n)*arccosh(sqrt(5/4))). (End)
a(n) = (a(n-1)^3 / a(n-2)^2 + 5 * a(n-1) * a(n-2)^2) / 2, for n > 1. - Lee A. Newberg, Jul 20 2010
Recurrence equations from Peter Bala, Oct 30 2013: (Start)
a(n)/a(n-1) = (a(n-1)/a(n-2))^2 - 2 for n >= 3.
a(n)/a(n-1) = 5*a(n-2)^2 + 2 for n >= 3.
a(n) = a(n-1)*sqrt(5*a(n-1)^2 + 4) for n >= 2. (End)
0 = a(n)^2 * ( a(n+3) - 2*a(n+2) ) - a(n+1)*a(n+2) * ( a(n+2) - 2*a(n+1)) if n > 0. - Michael Somos, Mar 24 2014
From Amiram Eldar, Dec 02 2021: (Start)
a(n) = A000045(A000079(n)).
Limit_{n->oo} sqrt(a(1)^2 + sqrt(a(2)^2 + sqrt(a(3)^2 + ... + sqrt(a(n))))) = 3 (Ohtsuka, 2015). (End)
a(n) = Product_{k=0..n-1} L(2^k), for n >= 1, where L(k) is the k-th Lucas number (A000032). - Amiram Eldar, Mar 30 2023

A250486 A(n,k) is the n^k-th Fibonacci number; square array A(n,k), n>=0, k>=0, read by antidiagonals.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 3, 2, 1, 0, 1, 21, 34, 3, 1, 0, 1, 987, 196418, 987, 5, 1, 0, 1, 2178309, 37889062373143906, 10610209857723, 75025, 8, 1

Offset: 0

Alois P. Heinz, Nov 24 2014

			Square array A(n,k) begins:
  1,  0,  0,      0,       0,    0,  0,  0,   ...
  1,  1,  1,      1,       1,    1,  1,  1,   ...
  1,  1,  3,      21,      987,  2178309,     ...
  1,  2,  34,     196418,  37889062373143906, ...
  1,  3,  987,    10610209857723,             ...
  1,  5,  75025,  59425114757512643212875125, ...
  1,  8,  14930352,                           ...
  1,  13, 7778742049,                         ...

Alois P. Heinz, Antidiagonals n = 0..10, flattened
Wikipedia, Fibonacci number

Columns k=0-8 give: A000012, A000045, A054783, A182149, A250490, A250491, A250492, A250493, A250494.
Rows n=0-10 give: A000007, A000012, A058635, A045529, A145231, A145232, A145233, A145234, A250487, A250488, A250489.
Main diagonal gives A250495.
Cf. A000045.

A:= (n, k)-> (<<0|1>, <1|1>>^(n^k))[1, 2]:
seq(seq(A(n, d-n), n=0..d), d=0..8);

A[n_, k_] := MatrixPower[{{0, 1}, {1, 1}}, n^k][[1, 2]]; A[0, 0] = 1;
Table[A[n, d-n], {d, 0, 8}, {n, 0, d}] // Flatten (* Jean-François Alcover, May 28 2019, from Maple *)

A(n,k) = [0, 1; 1, 1]^(n^k)[1,2].

A081667 a(n) = Fibonacci(binomial(n+2,2)).

Original entry on oeis.org

1, 2, 8, 55, 610, 10946, 317811, 14930352, 1134903170, 139583862445, 27777890035288, 8944394323791464, 4660046610375530309, 3928413764606871165730, 5358359254990966640871840, 11825896447871834976429068427, 42230279526998466217810220532898

Offset: 0

Paul Barry, Mar 26 2003

Diagonal of Fibonacci-Pascal triangle A045995.

Alois P. Heinz, Table of n, a(n) for n = 0..96
Peter M. Chema, Illustration of first 12 terms on a square spiral
T. Kotek, J. A. Makowsky, Recurrence Relations for Graph Polynomials on Bi-iterative Families of Graphs, arXiv preprint arXiv:1309.4020 [math.CO], 2013.

Cf. A000045, A000217, A033192, A045995, A054783.

with(combinat): seq(fibonacci((n^2-n)/2),n=2..16); # Zerinvary Lajos, May 18 2008
# second Maple program:
a:= n-> (<<0|1>, <1|1>>^((n+1)*(n+2)/2))[1, 2]:
seq(a(n), n=0..20);  # Alois P. Heinz, Jan 20 2017

Table[Fibonacci[Binomial[n+2,2]],{n,0,20}] (* Harvey P. Dale, Dec 03 2014 *)

[fibonacci(binomial(n,2)) for n in range(2, 17)] # Zerinvary Lajos, Nov 30 2009

a(n) = sqrt(5)2^(-n(n+3)/2)(sqrt(5)+1)^((n^2+3n+2)/2)/10 + sqrt(5)2^(-n(n + 3)/2)(sqrt(5)-1)^((n^2+3n+ 2)/2)(-1)^(n(n+3)/2)/10.
a(n) = A045995(n+2,2).
a(n) = A000045(A000217(n+1)). - Peter M. Chema, Sep 18 2016. See the name.

Name edited by Michel Marcus, Sep 25 2016

A102310 Square array read by antidiagonals: Fibonacci(k*n).

Original entry on oeis.org

1, 1, 1, 2, 3, 2, 3, 8, 8, 3, 5, 21, 34, 21, 5, 8, 55, 144, 144, 55, 8, 13, 144, 610, 987, 610, 144, 13, 21, 377, 2584, 6765, 6765, 2584, 377, 21, 34, 987, 10946, 46368, 75025, 46368, 10946, 987, 34, 55, 2584, 46368, 317811, 832040, 832040, 317811, 46368, 2584, 55

Offset: 1

Ralf Stephan, Jan 06 2005

			1,  1,   2,    3,     5, ...
1,  3,   8,   21,    55, ...
2,  8,  34,  144,   610, ...
3, 21, 144,  987,  6765, ...
5, 55, 610, 6765, 75025, ...

R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. 2nd Edition. Addison-Wesley, Reading, MA, 1994, p. 294.

Freddy Barrera, Antidiagonals n = 1..50, flattened

Equals A000045(A003991(k, n)).
Columns include A000045, A001906, A014445, A033888, A102312.
Main diagonal is in A054783. Antidiagonal sums are in A102311.

/* As triangle */ [[Fibonacci(k*(n-k+1)): k in [1..n]]: n in [1.. 15]]; // Vincenzo Librandi, Jul 04 2019

Table[Fibonacci[k*(n-k+1)], {n, 1, 10}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jun 10 2017 *)

F = fibonacci # A000045
def A(n, k):
    return F((n-1)*k)*F(k+1) + F((n-1)*k - 1)*F(k)
[A(n, k) for d in (1..10) for n, k in zip((d..1, step=-1), (1..d))] # Freddy Barrera, Jun 24 2019

For prime p, the formula holds: Fibonacci(k*p) = Fibonacci(p) * Sum_{i=0..floor((k-1)/2)} C(k-i-1, i)*(-1)^(i*p+i)*Lucas(p)^(k-2i-1).

A(n, k) = F((n-1)*k)*F(k+1) + F((n-1)*k-1)*F(k), where F(n) = A000045(n). - Freddy Barrera, Jun 24 2019

A182149 a(n) = Fibonacci(n^3).

Original entry on oeis.org

0, 1, 21, 196418, 10610209857723, 59425114757512643212875125, 619220451666590135228675387863297874269396512, 215414832505658809004682396169711233230800418578767753330908886771798637

Offset: 0

Vincenzo Librandi, Jul 07 2012

Vincenzo Librandi, Table of n, a(n) for n = 0..16

Cf. A054783, A250490.

Magma
```
[Fibonacci(n^3): n in [0..10]];
```

a:= n-> (<<0|1>, <1|1>>^(n^3))[1,2]:
seq(a(n), n=0..8);  # Alois P. Heinz, Aug 09 2018

Mathematica
```
Table[Fibonacci[n^3], {n, 20}]
```

a(n) = A000045(A000578(n)).

A207972 Expansion of g.f.: exp( Sum_{n>=1} 5Fibonacci(n^2) x^n/n ).

Original entry on oeis.org

1, 5, 20, 115, 1665, 82650, 12847310, 5620114060, 6659421195205, 21082748688390045, 177217804775828062850, 3941798437750184226876305, 231505293200405380457355524620, 35848160499603817968830380832049915, 14619744406297572472084577939841875791890

Offset: 0

Paul D. Hanna, Feb 22 2012

Moss and Ward prove that this is an integral sequence. - Peter Bala, Nov 28 2022

Let A(x) be the g.f. for this sequence. Note that the expansion of A(x)^(1/5) = exp( Sum_{n>=1} Fibonacci(n^2) * x^n/n ) does not have integer coefficients.

			G.f.: A(x) = 1 + 5*x + 20*x^2 + 115*x^3 + 1665*x^4 + 82650*x^5 + ...
such that
log(A(x))/5 = x + 3*x^2/2 + 34*x^3/3 + 987*x^4/4 + 75025*x^5/5 + 14930352*x^6/6 + 7778742049*x^7/7 + ... + Fibonacci(n^2)*x^n/n + ...

Patrick Moss and Tom Ward, Fibonacci along even powers is (almost) realizable, arXiv:2011.13068 [math.NT], 2020; Fibonacci Quart. 60 (2022), no. 1, 40-47.

Cf. A054888, A207969, A207970, A207971, A054783, A207834, A211892.

{a(n)=polcoeff(exp(sum(k=1,n,5*fibonacci(k^2)*x^k/k)+x*O(x^n)),n)}
for(n=0,16,print1(a(n),", "))

A204060 G.f.: Sum_{n>=1} Fibonacci(n^2)*x^(n^2).

Original entry on oeis.org

1, 0, 0, 3, 0, 0, 0, 0, 34, 0, 0, 0, 0, 0, 0, 987, 0, 0, 0, 0, 0, 0, 0, 0, 75025, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 14930352, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 7778742049, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 10610209857723, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 37889062373143906

Offset: 1

Paul D. Hanna, Jan 12 2012

nonn

Compare g.f. to the Lambert series identity: Sum_{n>=1} lambda(n)*x^n/(1-x^n) = Sum_{n>=1} x^(n^2).

Liouville's function lambda(n) = (-1)^k, where k is number of primes dividing n (counted with multiplicity).

			G.f.: A(x) = x + 3*x^4 + 34*x^9 + 987*x^16 + 75025*x^25 + 14930352*x^36 +...
where A(x) = x/(1-x-x^2) + (-1)*1*x^2/(1-3*x^2+x^4) + (-1)*2*x^3/(1-4*x^3-x^6) + (+1)*3*x^4/(1-7*x^4+x^8) + (-1)*5*x^5/(1-11*x^5-x^10) + (+1)*8*x^6/(1-18*x^6+x^12) +...+ lambda(n)*fibonacci(n)*x^n/(1 - Lucas(n)*x^n + (-1)^n*x^(2*n)) +...

Cf. A203847, A054783, A008836 (lambda), A000204 (Lucas), A000045.

Cf. A209614 (variant).

PARI
```
{a(n)=issquare(n)*fibonacci(n)}
```

{lambda(n)=local(F=factor(n));(-1)^sum(i=1,matsize(F)[1],F[i,2])}
{Lucas(n)=fibonacci(n-1)+fibonacci(n+1)}
{a(n)=polcoeff(sum(m=1,n,lambda(m)*fibonacci(m)*x^m/(1-Lucas(m)*x^m+(-1)^m*x^(2*m)+x*O(x^n))),n)}

G.f.: Sum_{n>=1} lambda(n)*Fibonacci(n)*x^n/(1 - Lucas(n)*x^n + (-1)^n*x^(2*n)), where lambda(n) = A008836(n) and Lucas(n) = A000204(n).

A204327 a(n) = Pell(n^2).

Original entry on oeis.org

1, 12, 985, 470832, 1311738121, 21300003689580, 2015874949414289041, 1111984844349868137938112, 3575077977948634627394046618865, 66992092050551637663438906713182313772, 7316660981177400006023755031791634132229378601

Offset: 1

Paul D. Hanna, Jan 14 2012

nonn

			G.f.: A(x) = x + 12*x^2 + 985*x^3 + 470832*x^4 + 1311738121*x^5 +...

Seiichi Manyama, Table of n, a(n) for n = 1..51

Cf. A000129 (Pell), A000290 (n^2), A204274, A165938, A165937, A166879, A054783.

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

a(n) = ( (1+sqrt(2))^(n^2) - (1-sqrt(2))^(n^2) ) / (2*sqrt(2)).

A211488 a(n) = Fibonacci(n^2) - Fibonacci(n).

Original entry on oeis.org

0, 2, 32, 984, 75020, 14930344, 7778742036, 10610209857702, 37889062373143872, 354224848179261915020, 8670007398507948658051832, 555565404224292694404015791664, 93202207781383214849429075266681736, 40934782466626840596168752972961528245770

Offset: 1

Vincenzo Librandi, Jul 09 2012

Vincenzo Librandi, Table of n, a(n) for n = 1..69

Cf. A000045, A054783.

[Fibonacci(n^2) - Fibonacci(n): n in [1..15]];

Table[Fibonacci[n^2]-Fibonacci[n],{n,20}]

a(n) = A054783(n) - A000045(n). - Michel Marcus, May 02 2025

A225686 a(n) = Fibonacci(2*n^2), a "Somos-like" sequence.

Original entry on oeis.org

1, 21, 2584, 2178309, 12586269025, 498454011879264, 135301852344706746049, 251728825683549488150424261, 3210056809456107725247980776292056, 280571172992510140037611932413038677189525

Offset: 1

N. J. A. Sloane, May 23 2013

nonn

Seiichi Manyama, Table of n, a(n) for n = 1..48
S. B. Ekhad and D. Zeilberger, How To Generate As Many Somos-Like Miracles as You Wish, arXiv:1303.5306 [math.CO], 2013.

Cf. A000045, A006720, A054783.

[Fibonacci(2*n^2): n in [1..10]]; // G. C. Greubel, Aug 09 2018

A225686 := proc(n)
    if n <= 5 then
        op(n,[1, 21, 2584, 2178309, 12586269025]) ;
    else
        ( 2303*procname(n - 4)*procname(n - 3)*procname(n - 1)
        + 2255*procname(n - 3)^2*procname(n - 2)
        + 329*procname(n - 4)*procname(n - 1)^2
        - 15792*procname(n - 4)*procname(n - 2)^2
        + 329*procname(n - 4)*procname(n - 3)^2
        - 2303*procname(n - 4)^2*procname(n - 2)
        + 441*procname(n - 2)
        - procname(n-2)^3
        -2961*procname(n-4)
        - procname(n-5)*procname(n-2)*procname(n-1)
        + 329*procname(n-5)*procname(n-3)*procname(n-2) )
        / 48/procname(n-4)/procname(n-2) ;
    end if;
end proc: # R. J. Mathar, Jul 09 2013
# second Maple program:
a:= n-> (<<0|1>, <1|1>>^(2*n^2))[1,2]:
seq(a(n), n=1..12);  # Alois P. Heinz, Aug 09 2018

a[ n_] := Fibonacci[2 n^2]; (* Michael Somos, Dec 05 2016 *)

{a(n) = fibonacci(2 * n^2)}; /* Michael Somos, Dec 05 2016 */

a(1) = 1, a(2) = 21, a(3) = 2584, a(4) = 2178309, a(5) = 12586269025, and for n>=6, a(n) = ( 2303a(n - 4)a(n - 3)a(n - 1) + 2255a(n - 3)^2 a(n - 2) + 329a(n - 4)a(n - 1)^2 - 15792a(n - 4)a(n - 2)^2 + 329a(n - 4)a(n - 3)^2 - 2303a(n - 4)^2 a(n - 2) + 441a(n - 2) - a(n-2)^3-2961a(n-4) - a(n-5)a(n-2)a(n-1) + 329a(n-5)a(n-3)a(n-2) )/( 48a(n-4)a(n-2) ).

0 = a(n)*(+233805165*a(n+4) - 726110*a(n+6)) + a(n+1)*(-76921899285*a(n+3) + 75284537349*a(n+7)) + a(n+2)*(+11222647920*a(n+2) + 3613692630240*a(n+4) - 1138829425306704*a(n+6) - 34837488*a(n+8)) + a(n+3)*(-527230649330*a(n+3) + 526991761443*a(n+7) + 329*a(n+9)) + a(n+4)*(+516007999155*a(n+4) - 1636636155*a(n+6) - 4976784*a(n+8)) + a(n+6)*(2255*a(n+6)). for all n in Z. - Michael Somos, Dec 05 2016

cp's OEIS Frontend

A058635 a(n) = Fibonacci(2^n).

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A250486 A(n,k) is the n^k-th Fibonacci number; square array A(n,k), n>=0, k>=0, read by antidiagonals.

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A081667 a(n) = Fibonacci(binomial(n+2,2)).

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A102310 Square array read by antidiagonals: Fibonacci(k*n).

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A182149 a(n) = Fibonacci(n^3).

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A207972 Expansion of g.f.: exp( Sum_{n>=1} 5*Fibonacci(n^2) * x^n/n ).

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A204060 G.f.: Sum_{n>=1} Fibonacci(n^2)*x^(n^2).

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A207972 Expansion of g.f.: exp( Sum_{n>=1} 5Fibonacci(n^2) x^n/n ).