A023172 -id:A023172 - cp's OEIS frontend

A168041 Fibonacci numbers that are divisible by their index.

1, 5, 144, 46368, 75025, 14930352, 4807526976, 1548008755920, 498454011879264, 51680708854858323072, 16641027750620563662096, 5358359254990966640871840, 59425114757512643212875125, 555565404224292694404015791808

Offset: 1

Vladimir Joseph Stephan Orlovsky, Nov 17 2009

nonn

G. C. Greubel, Table of n, a(n) for n = 1..100

Cf. A000045, A023172, A167745.

f[n_]:=Fibonacci[n]/n; lst={};Do[If[IntegerQ[f[n]],AppendTo[lst,Fibonacci[n]]],{n,4*5!}];lst

a(n) = A000045(A023172(n)).

A220168 Numbers k that divide Fibonacci(k+2).

Original entry on oeis.org

1, 4, 34, 46, 88, 94, 106, 166, 214, 226, 274, 334, 346, 394, 454, 466, 514, 526, 586, 634, 646, 694, 706, 754, 766, 886, 934, 1006, 1114, 1126, 1174, 1186, 1234, 1294, 1306, 1354, 1366, 1486, 1546, 1594, 1654, 1714, 1726, 1774, 1894, 1906, 1954, 1966, 2026

Offset: 1

Alex Ratushnyak, May 03 2013

nonn

Giovanni Resta, Table of n, a(n) for n = 1..1000

Cf. A000045.
Cf. A023172 (numbers k that divide Fibonacci(k)).
Cf. A069104 (numbers k that divide Fibonacci(k+1)).
Cf. A123976 (numbers k that divide Fibonacci(k-1)).
Cf. A159051 (numbers k that divide Fibonacci(k-2)).

Select[Range[2000], Mod[Fibonacci[#+2], #] == 0 &] (* T. D. Noe, Feb 05 2014 *)

is(n)=((Mod([1,1;1,0],n))^(n+2))[1,2]==0 \\ Charles R Greathouse IV, Feb 03 2014

prpr = prev = 1
for i in range(3, 3000):
    prpr, prev = prev, prpr+prev
    if prev % (i-2) == 0:  print(i-2, end=', ')

A254141 The average of a(n) consecutive Fibonacci numbers is never an integer.

Original entry on oeis.org

8, 16, 21, 28, 32, 40, 52, 55, 56, 64, 65, 68, 69, 80, 84, 85, 87, 88, 92, 93, 99, 104, 105, 112, 117, 119, 128, 132, 133, 136, 140, 141, 145, 148, 152, 153, 155, 156, 160, 161, 164, 165, 171, 172, 176, 184, 187, 188, 196, 200, 203, 204, 205, 207, 208, 209, 212

Offset: 1

Paolo P. Lava, Jan 26 2015

nonn

Subset of A033949 and A175594 (essentially the same sequence).
Numbers of the form 2^k, with k>=3, appear to be part of the sequence.
The file "List of indexes and steps (k, x, y)" (see Links) for k = 1, 2, 3, 4, ... consecutive Fibonacci numbers gives the minimum index to start to calculate the average ( x ) and the step to add to get all the other averages ( y ).
E.g.: for k = 7 we have 7, 6, 8. This means that we must start from the 6th Fibonacci number to add 7 consecutive Fibonacci numbers and get an average that is an integer. Fibonacci(6) + Fibonacci(7) + ... + Fibonacci(12) = 8 + 13 + 21 + 34 + 55 + 89 + 144 = 364 and 364 / 7 = 52.
Then 6 + 1*8 = 14, 6 + 2*8 = 22, 6 + 3*8 = 30, etc. are the other indexes:
Fibonacci(14) + Fibonacci (15) + ... + Fibonacci(20) = 377 + 610 + 987 + 1597 + 2584 + 4181 + 6765 = 17101 and 17101 / 7 = 2443;
Fibonacci(22) + Fibonacci(23) + ... + Fibonacci(28) = 17711 + 28657 + 46368 + 75025 + 121393 + 196418 + 317811 = 803383 and 803383 / 7 = 114769;
Fibonacci(30) + Fibonacci(31) + ... + Fibonacci(36) = 832040 + 1346269 + 2178309 + 3524578 + 5702887 + 9227465 + 14930352 = 37741900 and 37741900 / 7 = 5391700; etc.
In particular we note that:
x = 0 is A219612; x = 1 is A124456; x = 0 and y = k - 1 is A106535;
y = 1 is A141767; x = k - 1 and y = k + 1 is A000057;
x = y - 1 or y|k is A023172; y = k is A000351;
x = y - k + 1 appears to give only prime numbers: 3,11,19,31,59,71,79,131,179,191,239,251,271,311,359,379,419,431,439,479,491,499,571,599,631,659,719,739,751,839,971, etc.

Paolo P. Lava, Table of n, a(n) for n = 1..200
Paolo P. Lava, List of indexes and steps (k, x, y)

Cf. A000045, A000057, A000071, A023172, A033949, A106535, A124456, A141767, A175594, A219612.

with(numtheory); with(combinat):P:=proc(q) local a,b,k,j,n,ok;
for j from 1 to q do b:=0; ok:=1;
for n from 0 to q do a:=add(fibonacci(n+k),k=0..j-1)/j;
if type(a,integer) then ok:=0; break; fi; od;
if ok=1 then print(j); fi; od; end: P(20000);

A299155 Numbers i such that Fibonacci(i) is divisible by i+k for k=0..6.

Original entry on oeis.org

329400, 175472640, 808214400, 1367566200, 1928871000, 3433706640, 16109533440, 16916445000, 17847413640, 19507068000, 23163310800, 24961167000, 32645743200, 37263088800, 39803400000, 44915270400, 50238568800, 66072283200, 81924990840, 85368457440

Offset: 1

Chai Wah Wu, Feb 03 2018

nonn

A subsequence of A298687.

Cf. A000045, A023172, A217738, A221018, A225219, A298684, A298685, A298686, A298687.

fibomod(n,m) = lift(Mod([0,1;1,1],m)^(n+1))[1,1];
isok(n) = prod(k=0,6,fibomod(n,n+k)==0); \\ Silvester Resnik, May 18 2018

a(7)-a(20) from Silvester Resnik, May 18 2018

A325630 Numbers k such that A000110(k) is divisible by k.

Original entry on oeis.org

1, 2, 35, 16833, 16989, 23684

Offset: 1

Vaclav Kotesovec, Sep 07 2019

No other terms below 50000.
From Amiram Eldar, Jun 20 2024: (Start)
Numbers k such that A166226(k) = 0.
All the terms above 2 are composites since A166226(p) == 2 (mod p) for prime p. (End)
No other terms below 90000. - Michael S. Branicky, Jan 09 2025

			35 is in the sequence because A000110(35) = 35 * 8045720086273150473238297902.

Cf. A000110, A051130, A066562, A073877, A113883, A166226, A280727, A325935.

Cf. A014847, A016089, A023172, A051177, A056848.

Select[Range[1000], Divisible[BellB[#], #] &]

A078345 Numbers k such that F(k) mod k divides F(F(k) mod k) where F(k) denotes the k-th Fibonacci number.

Original entry on oeis.org

1, 2, 5, 8, 10, 11, 12, 13, 19, 20, 21, 22, 24, 25, 26, 29, 31, 32, 36, 37, 38, 41, 44, 48, 49, 50, 55, 58, 59, 60, 61, 62, 65, 71, 72, 73, 79, 80, 82, 84, 89, 95, 96, 97, 101, 104, 108, 109, 118, 120, 122, 125, 131, 132, 139, 140, 142, 144, 145, 149, 151, 155, 156

Offset: 1

Benoit Cloitre, Nov 22 2002

nonn

			F(44) = 701408733; 701408733 mod 44 = 25, F(25)=75025 and 25 divides 75025, hence 44 is in the sequence.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A000045, A002708, A023172.

fmod:= proc(n,m) local M,t; uses LinearAlgebra:-Modular;
    if m <= 1 then return 0 fi;
    if m < 2^25 then t:= float[8] else t:= integer fi;
    M:= Mod(m,<<1,1>|<1,0>>,t);
    round(MatrixPower(m,M,n)[1,2])
end proc:
filter:= proc(n) local s;
  s:= fmod(n,n);
  fmod(s,s) = 0
end proc:
select(filter, [$1..200]); # Robert Israel, May 10 2016

Unprotect[Divisible];
Divisible[0, 0] = True;
okQ[n_] := Module[{F = Fibonacci, m}, m = Mod[F[n], n];  Divisible[F[m], m]];
Select[Range[75000], okQ] (* Jean-François Alcover, Jul 09 2024 *)

Conjecture: a(n) is asymptotic to c*n*log(n) with c>0.7

A119588 Numbers k such that the number of divisors of Fibonacci(k), tau(Fibonacci(k)), is not a perfect power of 2.

Original entry on oeis.org

12, 24, 25, 36, 48, 50, 56, 60, 72, 75, 84, 91, 96, 100, 108, 110, 112, 120, 132, 144, 150, 153, 156, 168, 175, 180, 182, 192, 200, 204, 216, 220, 224, 225, 228, 240, 252, 264, 273, 275, 276, 280, 300, 306, 312, 324, 325, 330, 336, 342, 348, 350, 360, 364, 372

Offset: 1

Ryan Propper, Jun 01 2006

nonn

Has many terms in common with A023172 (41 below 1000), but neither is a subsequence of the other since 125 is not in this sequence.

			F(12) = 144 has 15 divisors: {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144}. Since 15 is not a power of 2, 12 is in the sequence.
F(24) = 46368 has 72 divisors. Since 72 is not a power of 2, 24 is in the sequence.

Amiram Eldar, Table of n, a(n) for n = 1..212
Blair Kelly, Fibonacci and Lucas Factorizations.

Cf. A000005, A000045, A063375, A162643.

Do[If[ !IntegerQ[Log[2, DivisorSigma[0, Fibonacci[n]]]], Print[n]], {n, 10^3}]

is(k) = {my(d = numdiv(fibonacci(k))); d >> valuation(d, 2) > 1;} \\ Amiram Eldar, Aug 12 2024

a(n) = {k: tau(Fibonacci(k)) != 2^i for all i}.

A132365 Least number k such that the Lucas number A000032(k) contains n.

Original entry on oeis.org

1, 0, 2, 3, 13, 9, 4, 6, 7, 24, 5, 10, 15, 26, 20, 25, 49, 6, 11, 16, 13, 12, 10, 21, 45, 40, 20, 36, 7, 31, 50, 12, 35, 19, 17, 15, 41, 36, 22, 23, 39, 39, 14, 21, 41, 60, 8, 32, 19, 56, 20, 13, 45, 37, 51, 44, 17, 56, 42, 22, 25, 62, 35, 15, 71, 47, 25, 24, 43, 32, 17, 45, 49, 38

Offset: 1

Jonathan Vos Post, Nov 08 2007

Values such that a(n)=n (fixed points) are 1, 62. I don't know if there are any other fixed points. The first time a(n)=a(n+1) occurs because L(39)=141422324 which includes both 41 and 42 (and later on in the sequence, because it contains 141 and 142). [Sean A. Irvine, Nov 30 2009]

Chai Wah Wu, Table of n, a(n) for n = 1..10000

Cf. A000032, A000045, A000350, A050816, A038546, A052000, A023172.

def A132365(n):
    a, b, m, s = 2, 1, 0, str(n)
    while True:
        if s in str(a):
            return m
        m += 1
        a, b = b, a+b # Chai Wah Wu, Jun 06 2017

a(n) = Min{k such that A000032(k) contains the decimal digit substring which represents the integer n}.

Incorrect comment removed by Sean A. Irvine, Nov 30 2009

Corrected and extended by Sean A. Irvine, Nov 30 2009

A270312 Numerator of Fibonacci(n)/n.

Original entry on oeis.org

1, 1, 2, 3, 1, 4, 13, 21, 34, 11, 89, 12, 233, 377, 122, 987, 1597, 1292, 4181, 1353, 10946, 17711, 28657, 1932, 3001, 121393, 196418, 317811, 514229, 83204, 1346269, 2178309, 3524578, 5702887, 1845493, 414732, 24157817, 39088169, 63245986, 20466831, 165580141

Offset: 1

Jean-François Alcover and Paul Curtz, Mar 15 2016

The fractions are an autosequence of the second kind. See the link.
Array of fractions and successive differences:
   1,   1/2,   2/3,   3/4,      1, ...
-1/2,   1/6,  1/12,   1/4,    1/3, ...
2 /3, -1/12,   1/6,  1/12,   4/21, ...
-3/4,   1/4, -1/12,  3/28,   3/56, ...
   1,  -1/3,  4/21, -3/56, 11/126, ...
...
The sequence of fractions being an autosequence, it can be noticed that first column, which is the inverse binomial transform of first row, is identical to the sequence, up to alternating signs.
In addition, main diagonal is twice the first upper diagonal (autosequence of the second kind).

			Fractions begin:
1, 1/2, 2/3, 3/4, 1, 4/3, 13/7, 21/8, 34/9, 11/2, 89/11, 12, ...

OEIS Wiki, Autosequence

Cf. A000045, A023172, A127787, A270313 (denominators).

Table[Fibonacci[n]/n, {n, 1, 50}] // Numerator

a(n) = numerator(fibonacci(n)/n); \\ Michel Marcus, Mar 15 2016

A289586 Numbers k whose smallest multiple that is a Fibonacci number is Fibonacci(k).

Original entry on oeis.org

1, 5, 12, 25, 60, 125, 300, 625, 1500, 3125, 7500, 15625, 37500, 78125, 187500, 390625, 937500, 1953125, 4687500, 9765625, 23437500, 48828125, 117187500, 244140625, 585937500, 1220703125, 2929687500, 6103515625, 14648437500, 30517578125, 73242187500, 152587890625, 366210937500

Offset: 1

Jon E. Schoenfield, Aug 06 2017

Alternative names:
Numbers k such that Fibonacci(k) is the smallest positive Fibonacci number that is divisible by k.
Numbers that are their own Fibonacci entry points.
Numbers k such that k = A001177(k).
Numbers that are either a power of 5 or 12 times a power of 5. - Robert Israel, Aug 07 2017

			Fibonacci(25) = 75025 = 25*3001 is the smallest Fibonacci number that is divisible by 25, so 25 is in the sequence.
Although Fibonacci(24) = 46368 = 24*1932 is divisible by 24, it is not the smallest Fibonacci number that is divisible by 24, so 24 is not in the sequence.

Robert Israel, Table of n, a(n) for n = 1..2858
Index entries for linear recurrences with constant coefficients, signature (0,5).

Subsequence of A023172 ("Self-Fibonacci numbers").
Cf. A000045, A001177, A000351 (bisection), A216491 (bisection)
(Cf. A001602 for a different definition of "Fibonacci entry point".)

1,seq(op([5^k,12*5^(k-1)]), k=1..100); # Robert Israel, Aug 07 2017

From Robert Israel, Aug 07 2017: (Start)
a(2*k) = 5^k for k >= 1.
a(2*k-1) = 12*5^(k-2) for k >= 2.
G.f.: (1+5*x+7*x^2)/(1-5*x^2). (End)

More terms from Robert Israel, Aug 07 2017

cp's OEIS Frontend

A168041 Fibonacci numbers that are divisible by their index.

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A220168 Numbers k that divide Fibonacci(k+2).

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A254141 The average of a(n) consecutive Fibonacci numbers is never an integer.

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A299155 Numbers i such that Fibonacci(i) is divisible by i+k for k=0..6.

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A325630 Numbers k such that A000110(k) is divisible by k.

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A078345 Numbers k such that F(k) mod k divides F(F(k) mod k) where F(k) denotes the k-th Fibonacci number.

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A119588 Numbers k such that the number of divisors of Fibonacci(k), tau(Fibonacci(k)), is not a perfect power of 2.

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A132365 Least number k such that the Lucas number A000032(k) contains n.

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