A063375 -id:A063375 - cp's OEIS frontend

A366771 Number of divisors of A001045(n) (Jacobsthal numbers).

1, 1, 2, 2, 2, 4, 2, 4, 6, 4, 2, 16, 2, 4, 8, 8, 2, 24, 2, 24, 8, 8, 2, 64, 8, 4, 16, 32, 4, 64, 2, 16, 16, 4, 16, 384, 4, 4, 8, 96, 4, 96, 2, 64, 48, 8, 4, 512, 4, 64, 32, 64, 4, 128, 24, 128, 16, 32, 8, 3072, 2, 4, 48, 64, 32, 256, 4, 64, 16, 256, 4, 6144, 4

Offset: 1

Sean A. Irvine, Oct 21 2023

nonn

			a(9) = 6 because Jacobsthal(9) = 171 has divisors {1, 3, 9, 19, 57, 171}.

Amiram Eldar, Table of n, a(n) for n = 1..1122

Cf. A000005, A001045, A366769, A366770, A366772, A063375.

a(n) = sigma0(Jacobsthal(n)) = A000005(A001045(n)).

A074698 Numbers k that divide the number of divisors of Fibonacci(k).

Original entry on oeis.org

1, 24, 48, 60, 64, 96, 128, 192, 256, 336, 384, 512, 576, 768, 1024, 1536, 1920, 2048, 3072

Offset: 1

Benoit Cloitre, Sep 03 2002

Are all numbers of the form 2^m*24 and 2^m*64, m >= 0, in the sequence?

This sequence is infinite (Luca, 2002). - Amiram Eldar, Jan 12 2022

Florian Luca, Problem H-590, Advanced Problems and Solutions, The Fibonacci Quarterly, Vol. 40, No. 5 (2002), p. 472; Arithmetic Functions of Fibonacci Numbers, Solution to Problem H-590 by J.-Ch. Schlage-Puchta and J. Spilker, ibid., Vol. 41, No. 4 (2002), pp. 382-384.

Cf. A000005, A000045, A063375.

with(combinat): with(numtheory): a:=proc(n) if type(tau(fibonacci(n))/n,integer) then n fi end: seq(a(n),n=1..200); # Emeric Deutsch, Jan 30 2006

With[{nn=200},Select[Thread[{DivisorSigma[0,Fibonacci[Range[nn]]],Range[nn]}],Divisible[#[[1]],#[[2]]]&]][[All,2]] (* The program generates the first 8 terms of the sequence. To generate more, increase the value of nn but the program may take a long time to run. *) (* Harvey P. Dale, Feb 17 2021 *)

isok(n) = ! (numdiv(fibonacci(n)) % n); \\ Michel Marcus, Sep 10 2017

a(9) from Emeric Deutsch, Jan 30 2006

a(10)-a(19) from Charles R Greathouse IV, Nov 07 2016

A080651 Numbers n such that n and Fibonacci(n) have the same number of divisors.

Original entry on oeis.org

1, 3, 5, 6, 7, 8, 10, 11, 13, 14, 17, 22, 23, 26, 29, 34, 43, 47, 83, 94, 131, 137, 359, 431, 433, 449, 509, 569, 571, 2971, 4723, 5387, 9311, 9677, 14431

Offset: 1

Joseph L. Pe, Feb 28 2003

Except for A001605(2) = 4, all terms of A001605 are terms of this sequence. - Chai Wah Wu, Dec 30 2019

Blair Kelly, Fibonacci and Lucas Factorizations.

Cf. A001605, A063375.

Select[Range[2*10^2], DivisorSigma[0, Fibonacci[ # ]] == DivisorSigma[0, # ] &]

isok(n) = numdiv(n) == numdiv(fibonacci(n)); \\ Michel Marcus, Feb 25 2016

a(n) = A001605(n-8) for n >= 21 (conjectured). - Chai Wah Wu, Dec 30 2019

More terms from Ryan Propper, May 31 2006
a(30)-a(34) from Chai Wah Wu, Dec 30 2019
a(35) from Chai Wah Wu, Dec 31 2019

A152774 Number of proper divisors of the Fibonacci number A000045(n).

Original entry on oeis.org

0, 0, 1, 1, 1, 3, 1, 3, 3, 3, 1, 14, 1, 3, 7, 7, 1, 15, 3, 15, 7, 3, 1, 71, 5, 3, 15, 15, 1, 63, 3, 15, 7, 3, 7, 159, 7, 7, 7, 63, 3, 63, 1, 31, 31, 7, 1, 335, 7, 47, 7, 15, 3, 127, 15, 95, 31, 7, 3, 959, 3, 7, 31, 63, 7, 63, 7, 31, 31, 127, 3, 1535, 3, 15, 47, 31, 15, 127, 3

Offset: 1

Omar E. Pol, Jan 17 2009

nonn

Amiram Eldar, Table of n, a(n) for n = 1..1422

Cf. A000005, A032741, A063375, A074283, A152990.

a[n_] := DivisorSigma[0, Fibonacci[n]] - 1; Array[a, 100] (* Amiram Eldar, Apr 07 2024 *)

a(n) = numdiv(fibonacci(n)) - 1; \\ Amiram Eldar, Apr 07 2024

a(n) = A000005(A000045(n))-1 = A032741(A000045(n)) = A063375(n)-1.

A160686 Numbers n such that n/A000005(A000045(n)) is an integer.

Original entry on oeis.org

1, 2, 4, 8, 16, 32, 64

Offset: 1

Ctibor O. Zizka, May 23 2009

The first power of 2 not in this sequence is 2^7 = 128, because 128/A000005(A000045(128)) = 1/2, which is not an integer. - Nathaniel Johnston, May 08 2011

Next term, if it exists, is greater than 3000. Conjecture: the sequence is finite and complete. - Max Alekseyev, May 21 2011

Cf. A000005, A000045, A000079, A160683.

with(combinat):with(numtheory): A160686 := proc(n) option remember: local k: if(n=1)then return 1:fi: for k from procname(n-1)+1 do if(k mod tau(fibonacci(k))=0)then return k:fi: od: end: seq(A160686(n),n=1..7); # Nathaniel Johnston, May 08 2011

{n: A063375(n) | n}. - R. J. Mathar, May 25 2009

Inverted division in the definition - R. J. Mathar, May 25 2009

Erroneous term a(5) = 12 removed by Nathaniel Johnston, May 08 2011

A272122 a(n) is the number of positive divisors of A003266(n).

Original entry on oeis.org

1, 1, 2, 4, 8, 20, 40, 120, 288, 864, 1728, 4800, 9600, 28800, 84480, 304128, 608256, 2322432, 9289728, 40642560, 116121600, 348364800, 696729600, 3185049600, 8918138880, 26754416640, 149824733184, 624269721600, 1248539443200, 6522981580800, 26091926323200, 107629196083200

Offset: 1

Altug Alkan, Apr 28 2016

Amiram Eldar, Table of n, a(n) for n = 1..794

Cf. A000005, A000045, A003266, A027423, A063375, A069744, A260622.

a[n_] := DivisorSigma[0, Fibonorial[n]]; Array[a, 32] (* Amiram Eldar, Aug 09 2022 *)

a(n) = numdiv(prod(k=1, n, fibonacci(k)));

a(n) = A000005(A003266(n)).

a(n+1) = 2*a(n) when n is in A069744.

A272377 Number of positive divisors of A000032(n).

Original entry on oeis.org

2, 1, 2, 3, 2, 2, 6, 2, 2, 6, 4, 2, 8, 2, 4, 12, 2, 2, 16, 2, 4, 12, 8, 4, 8, 8, 4, 12, 6, 4, 24, 2, 4, 12, 8, 16, 16, 2, 4, 24, 8, 2, 48, 4, 16, 96, 8, 2, 16, 4, 16, 24, 8, 2, 40, 24, 4, 24, 8, 8, 64, 2, 4, 96, 4, 32, 48, 4, 4, 48, 16, 2, 16, 4, 8, 192, 4, 16, 24, 2, 4, 48, 16, 4, 96, 16, 4, 48, 8, 4, 128

Offset: 0

Altug Alkan, Apr 28 2016

			a(3) = 3 because A000032(3) = 4 and 4 is divisible by 1, 2 and 4.

Amiram Eldar, Table of n, a(n) for n = 0..1000 (using Blair Kelly's data)
Blair Kelly, Fibonacci and Lucas Factorizations

Cf. A000005, A000032, A063375.

DivisorSigma[0, LucasL /@ Range[0, 90]] (* Michael De Vlieger, Apr 28 2016 *)

a(n) = numdiv(fibonacci(n-1)+fibonacci(n+1));

a(n) = A000005(A000032(n)).

A335001 Integers m such that d(F(m)) = d(L(m)) where d is the number of divisors function, F(n) and L(n) are respectively the n-th Fibonacci and n-th Lucas numbers.

Original entry on oeis.org

1, 4, 5, 7, 10, 11, 13, 14, 17, 18, 26, 46, 47, 58, 73, 77, 85, 89, 103, 107, 121, 139, 167, 179, 181, 187, 205, 221, 233, 241, 247, 253, 257, 262, 269, 273, 281, 293, 295, 317, 329, 335, 337, 341, 371, 377, 397, 407, 409, 427, 442, 454, 466, 491, 506, 563, 611

Offset: 1

Michel Marcus, May 19 2020

nonn

Numbers m such that A063375(m) = A272377(m).

It appears that this is a subsequence of A335002, so that terms also satisfy omega(F(m)) = omega(L(m))

Amiram Eldar, Table of n, a(n) for n = 1..87
Prapanpong Pongsriiam, Fibonacci and Lucas Numbers which have Exactly Three Prime Factors and Some Unique Properties of F18 and L18, Fibonacci Quart. 57 (2019), no. 5, 130-144.

Cf. A000005, A000032, A000045, A063375, A272377, A335002.

lucas(n) = fibonacci(n+1)+fibonacci(n-1);
isok(m) = numdiv(fibonacci(m))==numdiv(lucas(m));

A074697 Fibonacci(k) has more than k divisors (k such that A000005(A000045(k)) > k).

Original entry on oeis.org

12, 24, 30, 36, 40, 42, 48, 54, 56, 60, 70, 72, 78, 80, 81, 84, 88, 90, 96, 100, 102, 104, 105, 108, 110, 112, 114, 120, 126, 128, 130, 132, 135, 138, 140, 144, 147, 150, 154, 156, 160, 162, 165, 168, 170, 171, 174, 176, 180, 182, 184, 186, 189, 190, 192, 196

Offset: 1

Benoit Cloitre, Sep 03 2002

nonn

Amiram Eldar, Table of n, a(n) for n = 1..463

Cf. A000005, A000045, A063375.

Select[Range[200],DivisorSigma[0,Fibonacci[#]]>#&] (* Harvey P. Dale, Aug 17 2019 *)

from sympy import divisor_count,fibonacci
def ok(n): return divisor_count(fibonacci(n)) > n
print([k for k in range(197) if ok(k)]) # Michael S. Branicky, Dec 24 2021

A074700 a(n) = tau(F(2^n)) where tau(x) is the number of divisors of x (A000005(x)) and F(k) the k-th Fibonacci number (A000045(k)).

Original entry on oeis.org

1, 2, 4, 8, 16, 64, 256, 1024, 8192, 131072, 1048576

Offset: 1

Benoit Cloitre, Sep 03 2002

Is there any pattern in this sequence? It seems also that tau(F(m^n)) is a power of 2 for any m, any n > 0.
F(2^n) = L(2)L(4)L(8)...L(2^(n-1)) where L(x) is the Lucas numbers. This greatly reduces the difficulty of factoring these numbers. To find a(9) one needs the factorization of F(512); this was done long ago: F(2^9) = 3 * 7 * 47 * 127 * 1087 * 2207 * 4481 * 34303 * 119809 * 73327699969 * 186812208641 * 4698167634523379875583 * 125960894984050328038716298487435392001; hence a(9) = 2^13 = 8192. Since L(512), L(1024) are completed factored the next few terms are also known. L(2048) has 1 known factor and a C411, thus the next term is at least 2^23. - Sean A. Irvine, Jun 02 2005
If no member of A037917 is a power of 2, then a(n) is a power of 2 for all n. - Charles R Greathouse IV, Apr 09 2012

Blair Kelly, Lucas number factorizations
Index to divisibility sequences

Cf. A000005, A000045, A063375, A074715.

[NumberOfDivisors(Fibonacci(2^n)): n in [1..11]]; // Vincenzo Librandi, Jun 18 2017

Table[DivisorSigma[0, Fibonacci[2^n]], {n, 11}] (* Vincenzo Librandi, Jun 18 2017 *)

a(n)=numdiv(fibonacci(2^n)) \\ Charles R Greathouse IV, Apr 09 2012

More terms from Sean A. Irvine, Jun 02 2005

cp's OEIS Frontend

A366771 Number of divisors of A001045(n) (Jacobsthal numbers).

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A074698 Numbers k that divide the number of divisors of Fibonacci(k).

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A080651 Numbers n such that n and Fibonacci(n) have the same number of divisors.

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A152774 Number of proper divisors of the Fibonacci number A000045(n).

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A160686 Numbers n such that n/A000005(A000045(n)) is an integer.

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A272122 a(n) is the number of positive divisors of A003266(n).

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A272377 Number of positive divisors of A000032(n).

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A335001 Integers m such that d(F(m)) = d(L(m)) where d is the number of divisors function, F(n) and L(n) are respectively the n-th Fibonacci and n-th Lucas numbers.

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