A060384 -id:A060384 - cp's OEIS frontend

A136610 Number of odd digits in Fibonacci numbers.

0, 1, 1, 0, 1, 1, 0, 2, 1, 1, 2, 1, 1, 2, 3, 1, 2, 4, 1, 2, 2, 2, 5, 2, 1, 3, 5, 3, 5, 3, 1, 3, 4, 4, 3, 3, 5, 5, 4, 3, 6, 5, 4, 5, 5, 7, 7, 7, 4, 5, 4, 5, 6, 9, 5, 6, 8, 6, 7, 4, 6, 7, 8, 7, 7, 9, 7, 7, 5, 7, 10, 8, 6, 10, 8, 9, 6, 10, 8, 6, 6

Offset: 0

Parthasarathy Nambi, May 11 2008

			1597 = Fibonacci(17) and has four odd digits, so a(17) = 4.

Ron Knott, The Fibonacci numbers.

Cf. A000045, A138468, A196564.

Cf. A060384, A085855. - R. J. Mathar, Jul 08 2009

nodss := proc(n) local dgs,d; dgs := convert(n,base,10) ; add( d mod 2, d=dgs) ; end: A136610 :=proc(n) nodss(combinat[fibonacci](n)) ; end: seq( A136610(n),n=0..80) ; # R. J. Mathar, Jul 08 2009

a[n_]:=Total[Boole[OddQ/@IntegerDigits[Fibonacci[n]]]] (* James C. McMahon, May 06 2025 *)

a(n) = A196564(A000045(n)). - Michel Marcus, May 06 2025

a(n) = A060384(n) - A138468(n). - James C. McMahon, Jun 07 2025

a(13) corrected and more terms added by R. J. Mathar, Jul 08 2009

A138468 Number of even digits in Fibonacci numbers.

Original entry on oeis.org

1, 0, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 2, 1, 0, 2, 1, 0, 3, 2, 2, 3, 0, 3, 4, 2, 1, 3, 1, 3, 5, 4, 3, 3, 4, 4, 3, 3, 4, 5, 3, 4, 5, 4, 4, 3, 3, 3, 6, 5, 7, 6, 5, 2, 6, 6, 4, 6, 5, 8, 7, 6, 5, 6, 7, 5, 7, 7, 9, 8, 5, 7, 9, 5, 8, 7, 10, 6, 8, 11, 11, 8, 5, 5, 12, 7, 10, 9, 8, 7, 12, 11, 11, 11, 11, 9, 12

Offset: 0

Parthasarathy Nambi, May 18 2008

Zero is counted as an even digit.

			The Fibonacci number 2584 has three even digits in it, so a(18) = 3.

Robert Israel, Table of n, a(n) for n = 0..10000
Ron Knott, The Fibonacci Numbers.

Cf. A000045, A060384, A136610.

Table[Total[Boole[EvenQ/@IntegerDigits[Fibonacci[n]]]],{n,0,96}] (* James C. McMahon, Jun 07 2025 *)

a(n) = A060384(n) - A136610(n). - Robert Israel, May 25 2017

a(0) corrected, and more terms from Robert Israel, May 25 2017

A386758 Number of decimal digits in the n-th Lucas number.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 10, 10, 10, 10, 11, 11, 11, 11, 11, 12, 12, 12, 12, 12, 13, 13, 13, 13, 13, 14, 14, 14, 14, 15, 15, 15, 15, 15, 16, 16, 16, 16, 16, 17, 17, 17, 17, 17

Offset: 0

Hans J. H. Tuenter, Aug 06 2025

As F(n)<=L(n), the number of decimal digits of the Lucas number L(n) is at least as large as the number of decimal digits of the Fibonacci number F(n). Furthermore, the difference is at most one. The indices for which the difference is one is A386760.

			L(0)=2 has one digit, so that a(0)=1; L(5)=11 has two digits, so that a(5)=2.

Hans J. H. Tuenter, Table of n, a(n) for n = 0..10000

Cf. A000032, A001622, A055642, A060384, A097348.

Number of digits of L(p^n): A094057 (p=2), A114469 (p=10).

a:= n-> 1+floor(n*log[10]((1+sqrt(5))/2)):
seq(a(n), n=0..81);

a[n_] := IntegerLength[LucasL[n]]; Array[a, 100, 0] (* Amiram Eldar, Aug 16 2025 *)

a(n) = A055642(A000032(n)).

a(n) = 1 + floor(n*log_10(phi)), where log_10(phi) = A097348, and phi = (1+sqrt(5))/2 = A001622.

A386760 Numbers k such that the number of decimal digits of the Lucas number L(k) is greater than the number of decimal digits of the Fibonacci number F(k).

Original entry on oeis.org

5, 6, 10, 11, 15, 16, 20, 24, 25, 29, 30, 34, 35, 39, 44, 48, 49, 53, 54, 58, 59, 63, 67, 68, 72, 73, 77, 78, 82, 83, 87, 91, 92, 96, 97, 101, 102, 106, 111, 115, 116, 120, 121, 125, 126, 130, 134, 135, 139, 140, 144, 145, 149, 150, 154, 158, 159, 163, 164, 168

Offset: 1

Hans J. H. Tuenter, Aug 13 2025

The difference in the number of decimal digits, A055642(L(k))-A055642(F(k)) = A060384(k)-A386758(k) is either zero or one. In fact, this difference is ceiling(beta-{k*alpha}), with alpha and beta as defined in the Formula section. This implies that, asymptotically, a fraction of beta=0.349485... of the Lucas numbers has one more decimal digit than the corresponding Fibonacci number. This gives the asymptotic behavior of the sequence as a(n)~n/beta. Conjecture: abs(a(n)-n/beta)

			5 is a term since F(5)=5 has length 1 decimal digit, but L(5)=11 has length 2 decimal digits which is greater.

Hans J. H. Tuenter, Table of n, a(n) for n = 1..1000

Cf. A000032, A000045, A001622, A055642, A060384, A097348, A386758.

Select[Range[168],IntegerLength[LucasL[#]]>IntegerLength[Fibonacci[#]]&] (* James C. McMahon, Aug 28 2025 *)

The sequence consists of the integers k>=2, for which {k*alpha}A097348, and phi = (1+sqrt(5))/2 = A001622.

A246554 Concatenation of the n-th Fibonacci number with itself.

Original entry on oeis.org

11, 11, 22, 33, 55, 88, 1313, 2121, 3434, 5555, 8989, 144144, 233233, 377377, 610610, 987987, 15971597, 25842584, 41814181, 67656765, 1094610946, 1771117711, 2865728657, 4636846368, 7502575025, 121393121393, 196418196418, 317811317811, 514229514229

Offset: 1

Indrani Das, Nov 14 2014

a(n) is the n-th Fibonacci number concatenated with itself; concatenation A000045.

Also, the quotient of a(n) divided by the n-th Fibonacci number is 10^d(n)+1, where d(n) is the number of digits in the n-th Fibonacci number (A060384).

			The 7th Fibonacci number, 13, is concatenated with itself to become a(7) = 1313.

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Cf. A247337, A247338, A000045, A060384.

[Seqint(Intseq(Fibonacci(n)) cat Intseq(Fibonacci(n))): n in [1..30]]; // Vincenzo Librandi, Nov 15 2014

A:= proc(n)
local f;
f:= combinat:-fibonacci(n);
(10^length(f)+1)*f;
end proc:
map(A, [$1..100]); # Robert Israel, Nov 16 2014
# second Maple program:
a:= n-> (p-> parse(cat(p$2)))((<<0|1>, <1|1>>^n)[1, 2]):
seq(a(n), n=1..100);  # Alois P. Heinz, Nov 17 2014

Table[FromDigits[Join[Flatten[IntegerDigits[{Fibonacci[n], Fibonacci[n]}]]]], {n, 50}] (* Vincenzo Librandi, Nov 15 2014 *)
#*10^IntegerLength[#]+#&/@Fibonacci[Range[30]] (* Harvey P. Dale, Jul 04 2015 *)

a(n)=(k->eval(Str(k,k)))(fibonacci(n)) \\ Charles R Greathouse IV, Nov 15 2014

a(n) = A000045(n)*(10^A060384(n)+1). - Robert Israel, Nov 16 2014

A362358 Alternating sum of digits of the Fibonacci numbers, with a plus sign for the last digit.

Original entry on oeis.org

0, 1, 1, 2, 3, 5, 8, 2, -1, 1, 0, 1, 1, 2, 3, 5, 8, 2, -1, -10, 0, 12, 1, 2, 3, 5, -3, 13, -1, 1, -11, 1, 12, 2, 3, 5, -3, 13, 10, 1, 0, 1, -10, 13, 3, -17, 19, -9, 10, 1, 0, 1, 12, 13, 3, -6, 8, 2, -1, -10, 0, 1, 12, -9, 3, 5, -3, -9, -23, 1, -22, 34, -10, 2

Offset: 0

Wolfdieter Lang, May 26 2023

a(n) mod 11 = F(n) mod 11 = A105955(n). This is the mod 11 rule applied to F(n) = A000045.

			F(17) = 1597, s(n) = 4 - 1 = 3, a(17) = 7 - 9 + 5 - 1 = 2.

Alois P. Heinz, Table of n, a(n) for n = 0..10000

Cf. A000045, A004090, A007887, A055017, A060384, A105955.

a[n_]:=Sum[(-1)^(IntegerLength[Fibonacci[n]]-i) Part[IntegerDigits[Fibonacci[n]],i],{i,IntegerLength[Fibonacci[n]]}]; Array[a,66,0] (* Stefano Spezia, May 27 2023 *)

a(n) = my(d=Vecrev(digits(fibonacci(n)))); sum(k=1, #d, (-1)^(k+1)*d[k]); \\ Michel Marcus, May 28 2023

Let [f_s(n), f_{s(n)-1}, ..., f_0] be the list of digits of F(n) = A000040(n) with s(n) = A060384(n) - 1, then a(n) = Sum_{j=0..s(n)} (-1)^j*f_j.

a(n) = A055017(A000045(n)), for n >= 0.

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

A329192 Fibonacci numbers with arithmetic mean of digits an integer (sum of digits = a multiple of number of digits).

Original entry on oeis.org

1, 2, 3, 5, 8, 13, 55, 144, 987, 6765, 10946, 9227465, 225851433717, 8944394323791464, 160500643816367088, 83621143489848422977, 59425114757512643212875125, 30960598847965113057878492344, 127127879743834334146972278486287885163

Offset: 1

José de Jesús Camacho Medina, Nov 07 2019

			55 is a term as the arithmetic mean of digits is an integer: (5+5)/2 = 5.
144 is a term as the arithmetic mean of digits is an integer: (1+4+4)/3 = 3.
6765 is a term as the arithmetic mean of digits is an integer: (6+7+6+5)/4 = 6.

Cf. A000045, A004090, A060384, A069709, A164947.

Mathematica
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a(n) = A000045(A164947(n+1)).

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A136610 Number of odd digits in Fibonacci numbers.

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A138468 Number of even digits in Fibonacci numbers.

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A386758 Number of decimal digits in the n-th Lucas number.

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A386760 Numbers k such that the number of decimal digits of the Lucas number L(k) is greater than the number of decimal digits of the Fibonacci number F(k).

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A246554 Concatenation of the n-th Fibonacci number with itself.

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A362358 Alternating sum of digits of the Fibonacci numbers, with a plus sign for the last digit.

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

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