A058635 -id:A058635 - cp's OEIS frontend

A319749 a(n) is the numerator of the Heron sequence with h(0)=3.

3, 11, 119, 14159, 200477279, 40191139395243839, 1615327685887921300502934267457919, 2609283532796026943395592527806764363779539144932833602430435810559

Offset: 0

Paul Weisenhorn, Sep 27 2018

The denominator of the Heron sequence is in A319750.
The following relationship holds between the numerator of the Heron sequence and the numerator of the continued fraction A041018(n)/A041019(n) convergent to sqrt(13).
n even: a(n)=A041018((5*2^n-5)/3).
n  odd: a(n)=A041018((5*2^n-1)/3).
More generally, all numbers c(n)=A078370(n)=(2n+1)^2+4 have the same relationship between the numerator of the Heron sequence and the numerator of the continued fraction convergent to 2n+1.
sqrt(c(n)) has the continued fraction 2n+1; n,1,1,n,4n+2.
hn(n)^2-c(n)*hd(n)^2=4 for n>1.
From Peter Bala, Mar 29 2022: (Start)
Applying Heron's method (sometimes called the Babylonian method) to approximate the square root of the function x^2 + 4, starting with a guess equal to x, produces the sequence of rational functions [x, 2*T(1,(x^2+2)/2)/x, 2*T(2,(x^2+2)/2)/( 2*x*T(1,(x^2+2)/2) ), 2*T(4,(x^2+2)/2)/( 4*x*T(1,(x^2+2)/2)*T(2,(x^2+2)/2) ), 2*T(8,(x^2+2)/2)/( 8*x*T(1,(x^2+2)/2)*T(2,(x^2+2)/2)*T(4,(x^2+2)/2) ), ...], where T(n,x) denotes the n-th Chebyshev polynomial of the first kind. The present sequence is the case x = 3. Cf. A001566 and A058635 (case x = 1), A081459 and A081460 (essentially the case x = 4). (End)

			A078370(2)=29.
hn(0)=A041046(0)=5; hn(1)=A041046(3)=27; hn(2)=A041046(5)=727;
hn(3)=A041046(13)=528527.

P. Liardet and P. Stambul, Séries d'Engel et fractions continuées, Journal de Théorie des Nombres de Bordeaux 12 (2000), 37-68.
Wikipedia, Engel Expansion
Wikipedia, Methods of computing square roots
Index entries for sequences of form a(n+1)=a(n)^2 + ...

2*T(2^n,x/2) modulo differences of offset: A001566 (x = 3 and x = 7), A003010 (x = 4), A003487 (x = 5), A003423 (x = 6), A346625 (x = 8), A135927 (x = 10), A228933 (x = 18).

Cf. A041018, A041019, A057076, A078370, A081459, A010122, A319750, A041046.

hn[0]:=3:  hd[0]:=1:
for n from 1 to 6 do
hn[n]:=(hn[n-1]^2+13*hd[n-1]^2)/2:
hd[n]:=hn[n-1]*hd[n-1]:
   printf("%5d%40d%40d\n", n, hn[n], hd[n]):
end do:
#alternative program
a := n -> if n = 0 then 3 else simplify( 2*ChebyshevT(2^(n-1), 11/2) ) end if:
seq(a(n), n = 0..7); # Peter Bala, Mar 16 2022

def aupton(nn):
    hn, hd, alst = 3, 1, [3]
    for n in range(nn):
        hn, hd = (hn**2 + 13*hd**2)//2, hn*hd
        alst.append(hn)
    return alst
print(aupton(7)) # Michael S. Branicky, Mar 16 2022

h(n) = hn(n)/hd(n); hn(0)=3; hd(0)=1.
hn(n+1) = (hn(n)^2+13*hd(n)^2)/2.
hd(n+1) = hn(n)*hd(n).
A041018(n) = A010122(n)*A041018(n-1) + A041018(n-2).
A041019(n) = A010122(n)*A041019(n-1) + A041019(n-2).
From Peter Bala, Mar 16 2022: (Start)
a(n) = 2*T(2^(n-1),11/2) for n >= 1, where T(n,x) denotes the n-th Chebyshev polynomial of the first kind.
a(n) = 2*T(2^n, 3*sqrt(-1)/2) for n >= 2.
a(n) = ((11 + 3*sqrt(13))/2)^(2^(n-1)) + ((11 - 3*sqrt(13))/2)^(2^(n-1)) for n >= 1.
a(n+1) = a(n)^2 - 2 for n >= 1.
a(n) = A057076(2^(n-1)) for n >= 1.
Engel expansion of (1/6)*(13 - 3*sqrt(13)); that is, (1/6)*(13 - 3*sqrt(13)) = 1/3 + 1/(3*11) + 1/(3*11*119) + .... (Define L(n) = (1/2)*(n - sqrt(n^2 - 4)) for n >= 2 and show L(n) = 1/n + L(n^2-2)/n. Iterate this relation with n = 11. See also Liardet and Stambul, Section 4.)
sqrt(13) = 6*Product_{n >= 0} (1 - 1/a(n)).
sqrt(13) = (9/5)*Product_{n >= 0} (1 + 2/a(n)). See A001566. (End)

a(6) and a(7) added by Peter Bala, Mar 16 2022

A385248 Number of digits in the decimal expansion of Fibonacci(2^n).

Original entry on oeis.org

1, 1, 1, 2, 3, 7, 14, 27, 54, 107, 214, 428, 856, 1712, 3424, 6848, 13696, 27393, 54785, 109570, 219140, 438279, 876558, 1753116, 3506231, 7012462, 14024923, 28049846, 56099693, 112199385, 224398770, 448797540, 897595080, 1795190160, 3590380321, 7180760641

Offset: 0

Juande Santander-Vela, Jul 28 2025

Binet's formula is Fibonacci(k) = (phi^k - psi^k)/sqrt(5), with phi being the golden ratio (1 + sqrt(5))/2, and psi = (1 - sqrt(5))/2. For even values of k, Fibonacci(k) = floor((phi^k)/sqrt(5)) since psi^(2*k)/sqrt(5) < 0.17^k for all k > 0, and from which the formula below.

Paolo Xausa, Table of n, a(n) for n = 0..3000

Cf. A000079, A000045, A001622, A055642, A058635, A060384, A068070.

a:= n-> `if`(n=0, 1, floor(2^n*log[10]((1+sqrt(5))/2)-log[10](5)/2)+1):
seq(a(n), n=0..35);  # Alois P. Heinz, Jul 30 2025

a[n_] := IntegerLength[Fibonacci[2^n]]; Array[a, 30, 0] (* Amiram Eldar, Jul 30 2025 *)
A385248[n_] := If[n == 0, 1, Floor[2^n*Log10[GoldenRatio] - Log10[5]/2] + 1];
Array[A385248, 50, 0] (* Paolo Xausa, Aug 07 2025 *)

a(n) = #Str(fibonacci(2^n)); \\ Michel Marcus, Jul 30 2025

from sympy import Rational, log, sqrt # uses symbolic computation
phi = (1+sqrt(5))/2
def a(n): return 1 if n==0 or n==1 else int(2**n *log(phi)/log(10)-Rational(1,2)*log(5)/log(10))+1

a(n) = A055642(A058635(n)).
a(n) = A060384(A000079(n)).
a(n) = floor(2^n * log10(phi) - (1/2) * log10(5)) + 1, for n >= 1.
Limit_{n->oo} a(n+1)/a(n) = 2.

More terms from Michel Marcus, Jul 30 2025

a(29)-a(35) from Amiram Eldar, Jul 30 2025

A290498 Numbers m such that the set of distinct prime divisors of the number of divisors of Fibonacci(m) is equal to the set of distinct prime divisors of m.

Original entry on oeis.org

1, 4, 8, 16, 24, 32, 60, 64, 72, 96, 128, 192, 256, 300, 336, 512, 576, 648, 900, 1008, 1024, 1080, 1250

Offset: 1

Altug Alkan, Aug 04 2017

Thanks to squarefree terms of A058635, numbers of the form 2^k appear in this sequence for k > 1. However it is not proven yet whether it is always true.
From Jon E. Schoenfield, Aug 05 2017: (Start)
The difficulty in extending this sequence is that it becomes hard to obtain the complete prime factorization of Fibonacci(m) as m increases. However, since every number having an odd number of divisors is a square, and the largest Fibonacci number that is also a square is Fibonacci(12) = 144, we can confine the search for terms > 12 to even numbers only.
Even for values of m for which we are unable to completely factorize Fibonacci(m), we can determine with a high degree of confidence whether m is in the sequence by considering only the multiplicities of the smaller primes in those factorizations, because multiplicities greater than 1 in the prime factorizations of Fibonacci numbers rarely occur among the larger prime factors. If, in place of the actual complete factorization of Fibonacci(m) for each examined value of m, we were to use only the multiplicities of the prime factors of Fibonacci(m) that are less than 10000 (which are quickly and easily counted using trial division), the terms we would obtain for this sequence would begin with 1, 4, 8, 16, 24, 32, 60, 64, 72, 96, 128, 192, 256, 300, 336, 512, 576, 648, 900, 1008, 1024, 1080, 1250, 1500, 1536, 1620, 1920, 2048, 2352, 2500, 2592, 2700, 4096, 4608, 5000, 5184, 5400, 5832, 7500, 8100, 8192, 8448, 8640, 9072, 9600, 10000, 13608, 15000, ...
Perhaps surprisingly, we would get the same terms (up through at least a(141) = 960000) if, instead of the multiplicities of prime factors <= 10000, we were to use the multiplicities of just the prime factors <= 13. (End)

			72 is a term because d(Fibonacci(2^3*3^2)) = 2^9*3.
300 is a term because d(Fibonacci(2^2*3*5^2)) = 2^15*3^2*5.

Blair Kelly, Fibonacci and Lucas Factorizations

Cf. A000005, A000045, A022307, A038575, A058635, A063375, A081381, A086597, A135939.

Select[Range[12]~Join~Range[14, 300, 2], Apply[SameQ, Map[FactorInteger[#][[All, 1]] &, {#, DivisorSigma[0, Fibonacci@ #]}]] &] (* Michael De Vlieger, Aug 07 2017 *)

is(n) = factor(numdiv(fibonacci(n)))[,1]==factor(n)[,1] \\ David A. Corneth, Aug 04 2017

a(16)-a(19) from David A. Corneth, Aug 04 2017
a(20)-a(22) from Jon E. Schoenfield, Aug 05 2017
a(23) from Amiram Eldar, Oct 14 2019

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A319749 a(n) is the numerator of the Heron sequence with h(0)=3.

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A385248 Number of digits in the decimal expansion of Fibonacci(2^n).

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A290498 Numbers m such that the set of distinct prime divisors of the number of divisors of Fibonacci(m) is equal to the set of distinct prime divisors of m.

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