A063375 -id:A063375 - cp's OEIS frontend

A114306 Numbers k such that Fibonacci(k) has more divisors than k does.

9, 12, 15, 16, 18, 19, 20, 21, 24, 25, 27, 28, 30, 31, 32, 33, 35, 36, 37, 38, 39, 40, 41, 42, 44, 45, 46, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 84, 85, 86, 87, 88, 89, 90, 91

Offset: 1

Shyam Sunder Gupta, Feb 05 2006

nonn

			15 is in the sequence because 610 (= Fibonacci(15)) has more divisors than 15.

Amiram Eldar, Table of n, a(n) for n = 1..1377 (terms 1..500 from Jinyuan Wang)

Cf. A000005, A000045, A063375.

with(combinat): with(numtheory): b:=proc(n) if tau(fibonacci(n))>tau(n) then n else fi end: seq(b(n),n=1..100); # Emeric Deutsch, May 13 2006

Select[Range[1, 100], DivisorSigma[0, Fibonacci[#]] > DivisorSigma[0, #] &] (* Vaclav Kotesovec, Feb 13 2019 *)

isok(n) = numdiv(fibonacci(n)) > numdiv(n); \\ Michel Marcus, Feb 13 2019

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}.

A160680 a(n) = sigma_0(F(n-1)) + sigma_0(F(n-2)) where F=A000045 and sigma_0=A000005.

Original entry on oeis.org

2, 3, 4, 4, 6, 6, 6, 8, 8, 6, 17, 17, 6, 12, 16, 10, 18, 20, 20, 24, 12, 6, 74, 78, 10, 20, 32, 18, 66, 68, 20, 24, 12, 12, 168, 168, 16, 16, 72, 68, 68, 66, 34, 64, 40, 10, 338, 344, 56, 56, 24, 20, 132, 144, 112, 128, 40, 12, 964, 964, 12, 40, 96, 72, 72, 72, 40, 64, 160

Offset: 3

Ctibor O. Zizka, May 23 2009

nonn

Amiram Eldar, Table of n, a(n) for n = 3..1409

Cf. A000005, A000045, A063375.

A063375 := proc(n) A000005(A000045(n)) ; end: A000045 := proc(n) combinat[fibonacci](n) ; end: A000005 := proc(n) numtheory[tau](n) ; end: for n from 3 to 120 do printf("%d,",A063375(n-1)+A063375(n-2)) ; od: # R. J. Mathar, May 25 2009

Total[DivisorSigma[0,#]]&/@Partition[Fibonacci[Range[70]],2,1] (* Harvey P. Dale, Mar 07 2012 *)

a(n) = A063375(n-1) + A063375(n-2). - R. J. Mathar, May 25 2009, Jun 01 2009

More terms from R. J. Mathar, May 25 2009

Offset changed to 3 by R. J. Mathar, Jun 19 2021

A160685 Numbers k >= 1 such that A000005(A000045(k))/A000005(k) is an integer.

Original entry on oeis.org

1, 3, 5, 6, 7, 8, 10, 11, 13, 14, 15, 17, 19, 21, 22, 23, 24, 25, 26, 27, 29, 30, 31, 33, 34, 35, 37, 38, 39, 40, 41, 42, 43, 46, 47, 50, 51, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 65, 66, 67, 69, 70, 71, 72, 73, 74, 75, 77, 78, 79, 82, 83, 84, 85, 86, 87, 88, 89, 91, 93, 94

Offset: 1

Ctibor O. Zizka, May 23 2009

A080651 is a subsequence. - R. J. Mathar, May 25 2009

Robert Israel, Table of n, a(n) for n = 1..1073

Cf. A000045, A000005.

A000045 := proc(n) combinat[fibonacci](n) ; end: A000005 := proc(n) numtheory[tau](n) ; end: A063375 := proc(n) A000005(A000045(n)) ; end: for n from 1 to 130 do if A063375(n) mod A000005(n) = 0 then printf("%d,",n) ; fi; od: # R. J. Mathar, May 25 2009

okQ[n_] := With[{tau = DivisorSigma[0, #]&},
  IntegerQ[tau[Fibonacci[n]]/tau[n]]];
Select[Range[100], okQ] (* Jean-François Alcover, May 17 2023 *)

{n: A000005(n) | A063375(n)}. - R. J. Mathar, May 25 2009

2, 12 removed and sequence extended by R. J. Mathar, May 25 2009

A174280 Smallest k such that tau(Fibonacci(k)) = tau(Fibonacci(n+k)).

Original entry on oeis.org

1, 3, 4, 3, 9, 5, 4, 3, 4, 3, 8, 5, 4, 3, 19, 6, 9, 5, 4, 3, 10, 7, 8, 5, 4, 3, 14, 6, 33, 13, 10, 9, 8, 13, 6, 7, 18, 5, 4, 3, 21, 5, 4, 3, 8, 16, 6, 31, 10, 9, 8, 9, 6, 19, 6, 18, 14, 27, 14, 19, 10, 9, 8, 9, 6, 16, 6, 26, 10, 9, 8, 11, 6, 42, 14, 7, 20, 5, 4, 3

Offset: 1

Michel Lagneau, Mar 15 2010

nonn

tau(n) is the number of divisors of n (A000005).

			a(2) = 3 because tau(Fibonacci(3)) = tau(2) = 2, tau(Fibonacci(3+2)) = tau(5) = 2.

Chai Wah Wu, Table of n, a(n) for n = 1..1284
Eric Weisstein's World of Mathematics, Divisor Function.

Cf. A063375.

with(numtheory) ;
with(combinat) ;
A174280 := proc(n)
        for k from 1 do
                if tau(fibonacci(k)) = tau(fibonacci(n+k)) then
                        return k;
                end if;
        end do:
end proc:
seq(A174280(n),n=1..80) ; # R. J. Mathar, Jul 06 2012

Table[k = 1; While[DivisorSigma[0, Fibonacci[k]] != DivisorSigma[0, Fibonacci[k + n]], k++]; k, {n, 100}] (* T. D. Noe, Mar 18 2013 *)

A174331 n such that tau(Fibonacci(n)) is a perfect square.

Original entry on oeis.org

1, 2, 6, 8, 9, 10, 14, 18, 19, 20, 22, 26, 27, 28, 30, 31, 32, 34, 40, 41, 42, 52, 53, 55, 59, 61, 64, 66, 71, 73, 74, 77, 79, 85, 87, 89, 92, 93, 94, 95, 97, 99, 101, 107, 109, 113, 115, 116, 117, 120, 121, 123, 125, 127, 128, 129, 130, 133, 135, 138, 143, 146, 147, 149

Offset: 1

Michel Lagneau, Mar 15 2010

nonn

tau = A000005 is the number of divisors of n.

			40 is in the sequence because tau(Fibonacci(40)) = tau(102334155) = 64 is square.

Amiram Eldar, Table of n, a(n) for n = 1..423

Cf. A000005, A000045, A000290, A036436, A063375.

Select[Range[150], IntegerQ[Sqrt[DivisorSigma[0,Fibonacci[#]]]] &]

is(n) = issquare(numdiv(fibonacci(n))) \\ Felix Fröhlich, Sep 08 2019

{n: A063375(n) in A000290}. - R. J. Mathar, Jul 09 2012

A193733 Start of n consecutive indices k such that Fibonacci(k) contains distinct number of divisors.

Original entry on oeis.org

1, 2, 10, 12, 20, 20, 54, 96, 132, 171, 222, 458, 520, 731, 1083

Offset: 1

Michel Lagneau, Aug 08 2011

a(16) > 1443, if it exists. - Chai Wah Wu, Jan 19 2020

			The 4th number of this sequence, 12, means that:
Fibonacci(12) = 144 = 2 ^ 4 * 3 ^ 2,
Fibonacci(13) = 233 (prime number),
Fibonacci(14) = 377 = 13 * 29,
Fibonacci(15) = 610 = 2 * 5 * 61,
and these Fibonacci numbers have distinct number of divisors: 15, 2, 4 and 8, respectively.

Cf. A000045, A063375.

with(combinat, fibonacci):with(numtheory): for n from 1 to 10 do: i:=0:for k from 1 to 1500 while(i=0) do: lst:={}:for p from 0 to n-1 do :x:= fibonacci(k+p):y:=divisors(x):n1:=nops(y):lst:= lst union {n1}:od:if nops(lst)=n then  printf(`%d, `,k): i:=1:else fi:od:od:

a(12)-a(15) from Amiram Eldar, Oct 18 2019

A278637 Numbers k such that Fibonacci(k) is either prime or semiprime.

Original entry on oeis.org

3, 4, 5, 7, 8, 9, 10, 11, 13, 14, 17, 19, 22, 23, 26, 29, 31, 34, 41, 43, 47, 53, 59, 61, 71, 73, 79, 83, 89, 94, 101, 107, 109, 113, 121, 127, 131, 137, 151, 167, 173, 191, 193, 199, 227, 251, 271, 277, 293, 331, 353, 359, 397, 401, 431, 433, 449, 467, 509, 569, 571, 587, 599, 601, 613, 631, 653, 743, 991, 1091, 1223, 1373, 1487

Offset: 1

Bobby Jacobs, Jan 04 2017

nonn

Numbers k such that 2 <= A063375(k) <= 4.
All numbers in the first 4 rows of A279021 are in this sequence (3, 5, 11, 17, 353, 431, 509, and 587).
Are all numbers of A279021 in this sequence?

Cf. A000045, A001605, A063375, A072381, A136341, A279021.

A001605 U A072381.

a(31)-a(64) from Charles R Greathouse IV, Jan 04 2017

a(65)-a(73) from Max Alekseyev, Feb 26 2023

A074699 a(n) = tau(Fibonacci(242^n))/(242^n) where tau(x) is the number of divisors of x (A000005(x)).

Original entry on oeis.org

3, 7, 32, 144, 5120, 180224, 3145728, 3489660928

Offset: 0

Benoit Cloitre, Sep 03 2002

Are terms always integers?

Cf. A063375, A074698.

[NumberOfDivisors(Fibonacci(24*2^n))/(24*2^n): n in [0..5]]; // Vincenzo Librandi, Sep 11 2017

with(numtheory): with(combinat): a:=n->tau(fibonacci(24*2^n))/(24*2^n): seq(a(n),n=0..4); # Emeric Deutsch, Jan 30 2006

Table[DivisorSigma[0, Fibonacci[24 2^n]] / (24 2^n), {n, 0, 5}] (* Vincenzo Librandi, Sep 11 2017 *)

a(n) = numdiv(fibonacci(24*2^n))/(24*2^n); \\ Michel Marcus, Sep 10 2017

a(5) from Eric Rowland, Jun 18 2017

a(6)-a(7) from Amiram Eldar, Sep 03 2019 (using FactorDB)

A174279 Smallest k such that tau(Fibonacci(k)) = 2^n.

Original entry on oeis.org

1, 3, 6, 15, 18, 44, 30, 54, 128, 80, 138, 90, 162, 198, 308, 294, 210, 460, 288, 270, 378, 510, 680, 594, 920, 570, 690, 1280, 1190, 630, 1040, 1386, 810

Offset: 0

Michel Lagneau, Mar 15 2010

Smallest k such that A000005(A000045(k)) = 2^n.

The multiplicative property of the tau-function implies that the Fibonacci(k) has a prime factor representation p_1^e_1*p_2^e_2*... where (e_1+1)*(e_2+1)*... is a power of 2, that is, the exponents are in {1,3,7,15,...}. This adds for example the squarefree Fibonacci numbers with indices from A037918 to the list of candidates. - R. J. Mathar, Oct 11 2011

			a(0) =  1 because tau(Fibonacci(1))  = tau(1)   = 2^0 = 1.
a(1) =  3 because tau(Fibonacci(3))  = tau(2)   = 2^1 = 2.
a(2) =  6 because tau(Fibonacci(6))  = tau(8)   = 2^2 = 4.
a(3) = 15 because tau(Fibonacci(15)) = tau(610) = 2^3 = 8.

Majorie Bicknell and Verner E Hoggatt, Fibonacci's Problem Book, Fibonacci Association, San Jose, Calif., 1974.

Cf. A085077, A063375.

with(numtheory):for p from 1 to 100 do:indic:=0:u0:=0:u1:=1:for n from 2 to 1000 while(indic=0)do:s:=u0+u1:u0:=u1:u1:=s:if tau(s)= 2^p and indic=0 then print(p): print(n): indic:=1:else fi:od:od:

a(27)-a(32) from Amiram Eldar, Oct 14 2019

cp's OEIS Frontend

A114306 Numbers k such that Fibonacci(k) has more divisors than k does.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A160680 a(n) = sigma_0(F(n-1)) + sigma_0(F(n-2)) where F=A000045 and sigma_0=A000005.

Views

Author

Keywords

Links

Crossrefs

Programs

Maple

Mathematica

Formula

Extensions

A160685 Numbers k >= 1 such that A000005(A000045(k))/A000005(k) is an integer.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

Formula

Extensions

A174280 Smallest k such that tau(Fibonacci(k)) = tau(Fibonacci(n+k)).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

A174331 n such that tau(Fibonacci(n)) is a perfect square.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A193733 Start of n consecutive indices k such that Fibonacci(k) contains distinct number of divisors.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Maple

Extensions

A074699 a(n) = tau(Fibonacci(242^n))/(242^n) where tau(x) is the number of divisors of x (A000005(x)).