A008963 -id:A008963 - cp's OEIS frontend

A261607 Initial digit of Fibonacci number F(n) in base 60.

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 1, 2, 3, 6, 10, 16, 26, 43, 1, 1, 3, 4, 7, 12, 20, 33, 54, 1, 2, 3, 6, 10, 16, 26, 42, 1, 1, 3, 4, 7, 12, 20, 33, 54, 1, 2, 3, 6, 10, 16, 26, 42, 1, 1, 2, 4, 7, 12, 20, 33, 53, 1, 2, 3, 6, 9, 16, 25, 42, 1, 1, 2, 4, 7, 12

Offset: 0

Reinhard Zumkeller, Sep 09 2015

Alois P. Heinz, Table of n, a(n) for n = 0..10000 (first 1001 terms from Reinhard Zumkeller)
Eric Weisstein's World of Mathematics, Sexagesimal.
Wikipedia, Sexagesimal.

Cf. A000045, A261575, A261585, A261606, A008963, A001622.

Haskell
```
a261607 = last . a261575_row
```

IntegerDigits[Fibonacci[Range[0, 75]], 60][[All, 1]] (* Michael De Vlieger, Jan 22 2022 *)

a(n) = if (n, digits(fibonacci(n), 60)[1], 0); \\ Michel Marcus, Jan 22 2022

a(n) = A261575(n, A261585(n)-1).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Sum_{d=1..59} d*log(1+1/d)/log(60) = 13.92958... . - Amiram Eldar, Jan 12 2023
For n>9, a(n) = floor(60^{alpha*n-beta}), where alpha=log_60(phi), beta=log_60(5)/2, {x}=x-floor(x) denotes the fractional part of x, and phi = (1+sqrt(5))/2 = A001622. - Hans J. H. Tuenter, Aug 26 2025

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

A153726 Initial digit of Catalan number A000108(n).

Original entry on oeis.org

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

Offset: 0

Omar E. Pol, Dec 30 2008

Amiram Eldar, Table of n, a(n) for n = 0..10000

Cf. A000030, A008963, A077648, A152669.

First[IntegerDigits[#]]&/@CatalanNumber[Range[0,120]] (* Harvey P. Dale, Jan 20 2014 *)

a(n)=A000030(A000108(n)).

More terms from Harvey P. Dale, Jan 20 2014

A367556 Comma transform of the Fibonacci numbers.

Original entry on oeis.org

1, 11, 12, 23, 35, 58, 81, 32, 13, 45, 58, 91, 42, 33, 76, 9, 71, 72, 44, 16, 51, 61, 12, 74, 87, 51, 31, 83, 15, 98, 1, 92, 93, 85, 79, 51, 22, 73, 96, 61, 51, 12, 64, 77, 31, 1, 32, 34, 67, 91, 52, 43, 95, 38, 21, 52, 73, 25, 99, 11, 2, 14, 16, 21, 31, 52, 84

Offset: 0

Alois P. Heinz, Nov 22 2023

See A367360 for further information.

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

Cf. A000045, A003893, A008963, A138844, A367360.

F:= combinat[fibonacci]:
a:= n-> parse(cat(""||(F(n))[-1], ""||(F(n+1))[1])):
seq(a(n), n=0..92);

With[{nmax=100},Map[10Mod[#[[1]],10]+IntegerDigits[#[[2]]][[1]]&,Partition[Fibonacci[Range[0,nmax+1]],2,1]]] (* Paolo Xausa, Nov 24 2023 *)

from sympy import fibonacci
from itertools import islice, pairwise, count
def S(): yield from (fibonacci(i) for i in count(0))
def C(g): # generator of comma transform of sequence passed as a generator
    yield from (10*(t%10) + int(str(u)[0]) for t, u in pairwise(g))
def agen(): return C(S())
print(list(islice(agen(), 67))) # Michael S. Branicky, Jan 05 2024

a(n) = 10 * A003893(n) + A008963(n+1).

A105150 Approximation to leading digit of n-th Fibonacci number.

Original entry on oeis.org

0, 1, 1, 2, 3, 5, 8, 1, 2, 3, 5, 8, 1, 2, 3, 5, 8, 1, 2, 3, 5, 8, 1, 2, 3, 5, 8, 1, 2, 3, 5, 8, 1, 2, 3, 5, 8, 1, 2, 3, 5, 8, 1, 2, 3, 5, 8, 1, 2, 3, 5, 8, 1, 2, 3, 5, 8, 1, 2, 3, 5, 8, 1, 2, 3, 5, 8, 1, 2, 3, 5, 8, 1, 2, 3, 5, 8, 1, 2, 3, 5, 8, 1, 2, 3, 5, 8, 1, 2, 3, 5, 8, 1, 2, 3, 5, 8, 1, 2, 3, 5, 8, 1, 2, 3

Offset: 0

Reinhard Zumkeller, Apr 10 2005

a(0) = 1, a(1) = a(2) = 1 and for n > 2:
a(n) = floor(w/10) + (w mod 10)*0^floor(w/10) where
w = (x+y)*0^(z-1) + y + z*0^floor((11-z)/10),
x = a(n-3), y = a(n-2) and z = a(n-1);
a(n) = A008963(n) = A000030(A000045(n)) for n<=14.

			n=11, x=2, y=3, z=5: w = (2+3) * 0^(5-1)+3+5*0^[(11-5)/10] = 5*0^4+3+5*0^0 = 0+3+5*1 = 8, a(11) = [8/10] + (8 mod 10) * 0^[8/10] = 0 + 8*0^0 = 8;
n=12, x=3, y=5, z=8: w = (3+5) * 0^(8-1)+5+8*0^[(11-8)/10] = 8*0^7+5+8*0^0 = 0+5+8*1 = 13, a(12) = [13/10] + (13 mod 10) * 0^[13/10] = 1 + 3*0^1 = 1;
n=13, x=5, y=8, z=1: w = (5+8) * 0^(1-1)+8+1*0^[(11-1)/10] = 13*0^0+8+1*0^1 = 13*1+8+1*0 = 21, a(13) = [21/10] + (21 mod 10) * 0^[21/10] = 2 + 1*0^2 = 2.

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A261607 Initial digit of Fibonacci number F(n) in base 60.

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

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A153726 Initial digit of Catalan number A000108(n).

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A367556 Comma transform of the Fibonacci numbers.

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A105150 Approximation to leading digit of n-th Fibonacci number.

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