A227875 -id:A227875 - cp's OEIS frontend

A235383 Fibonacci numbers that are the product of other Fibonacci numbers.

8, 144

Offset: 1

Robert C. Lyons, Jan 08 2014

This sequence and A229037 and A235265 are winners in the contest held at the 2014 AMS/MAA Joint Mathematics Meetings. - T. D. Noe, Jan 20 2014
Carmichael's theorem implies that 8 and 144 are the only terms of this sequence.
First two terms of A061899, A111687, A172150, A212703, and A231851. - Omar E. Pol, Jan 21 2014
Saha and Karthik conjectured (without reference to Carmichael's theorem) that the only positive integers k for which A001175(k^2) = A001175(k) are 6 and 12. (A000045(6) = 8 and A000045(12) = 144.) - L. Edson Jeffery, Feb 13 2014
Y. Bugeaud, M. Mignotte, and S. Siksek proved that 8 and 144 are the only nontrivial perfect power Fibonacci numbers. - Robert C. Lyons, Dec 23 2015

			The Fibonacci number 8 is in the sequence because 8=2*2*2, and 2 is a Fibonacci number that is not equal to 8. The Fibonacci number 144 is in the sequence because 144=3*3*2*2*2*2, and both 2 and 3 are Fibonacci numbers that are not equal to 144.

Yann Bugeaud, Maurice Mignotte, and Samir Siksek, Classical and modular approaches to exponential Diophantine equations I. Fibonacci and Lucas perfect powers, Annals of Mathematics, 163 (2006), pp. 969-1018.
Arpan Saha and Karthik C S, A few equivalences of Wall-Sun-Sun prime conjecture, arXiv:1102.1636 [math.NT], 2011.
Wikipedia, Carmichael's theorem.

Cf. A000045, A061899, A065108, A227875.

A370071 Fibonacci numbers such that some permutation of their digits is a perfect power.

Original entry on oeis.org

0, 1, 8, 144, 610, 46368, 1346269, 14930352, 63245986, 165580141, 267914296, 2971215073, 4807526976, 7778742049, 12586269025, 139583862445, 365435296162, 591286729879, 956722026041, 1548008755920, 2504730781961, 4052739537881, 6557470319842, 17167680177565, 27777890035288, 190392490709135

Offset: 1

Gonzalo Martínez, Feb 08 2024

0, 1, 8 and 144 are the only Fibonacci numbers that are themselves perfect powers (see A227875).
A term might have multiple permutations which are perfect powers.
Leading 0 digits are allowed in the perfect power, so that 610 is a term since 016 = 2^4.
From David A. Corneth, Feb 17 2024: (Start)
Four ideas to find terms:
1. For each Fibonacci number, check each anagram to determine whether it is a perfect power. A Fibonacci number is a term if and only if such an anagram exists.
2. Let q be the number of digits of some Fibonacci number F. Then check all perfect powers m < 10^q if the frequency of positive digits matches the corresponding positive digits of F and F has at least as many 0's as m. E.g., 610 is a term as the perfect power 16 has the same number of 1's, and the same number of 6's, and 610 has at least as many 0's as 16.
3. Match the last digits of perfect powers and whittle down the number of candidates. So for 46368, a perfect square must end in 4 or 6 for the last digit, 36, 64 or 84 for the last two digits and so on.
4. If some Fibonacci number F has q digits then we could list the possible strings the first floor((q + 1)/2) such numbers could form and see what squares start with those digits. (End)
a(46) = 1500520536206896083277. Some other terms: 3928413764606871165730, 6356306993006846248183, 10284720757613717413913, 16641027750620563662096, 26925748508234281076009, 43566776258854844738105, 3311648143516982017180081, 5358359254990966640871840, 8670007398507948658051921, 14028366653498915298923761, 155576970220531065681649693, 659034621587630041982498215, 30960598847965113057878492344. - Chai Wah Wu, Mar 27 2024

			46368 is a term because it is a Fibonacci number whose digits can be permuted to 36864 = 192^2 (and also to 86436 = 294^2).

Michael S. Branicky, Table of n, a(n) for n = 1..45

Cf. A000045, A227875, A001597, A118715.

isok(f) = my(d=digits(f), n=#d); for (i=1, n!, my(p=numtoperm(n, i), dd=vector(#d, i, d[p[i]])); if (ispower(fromdigits(dd)), return(1)););
lista(nn) = my(list = List([0, 1])); for (n=3, nn, my(f=fibonacci(n)); if (isok(f), listput(list, f));); Vec(list); \\ Michel Marcus, Feb 17 2024

from itertools import count, islice
from sympy import integer_log
def A370071_gen(): # generator of terms
    a, b = 1, 2
    yield from (0,1)
    while True:
        s = sorted(str(b))
        l = len(s)
        m = int(''.join(s[::-1]))
        u = int(''.join(s))
        for i in count(2):
            if i**2 > m:
                break
            for j in count(max(2,integer_log(u,i)[0])):
                if (k:=i**j) > m:
                    break
                t = sorted(str(k))
                if ['0']*(l-len(t))+t == s:
                    yield b
                    break
            else:
                continue
            if k<=m:
                break
        a, b = b, a+b
A370071_list = list(islice(A370071_gen(),20)) # Chai Wah Wu, Mar 27 2024

a(19) inserted and a(21)-a(23) by Michael S. Branicky, Feb 17 2024

More terms from David A. Corneth, Feb 17 2024

A081979 Smallest Fibonacci number with 2n divisors, or 0 if no such number exists.

Original entry on oeis.org

2, 8, 75025, 610

Offset: 1

Amarnath Murthy, Apr 03 2003

Further known terms are a(12)=A000045(91); a(16)=A000045(44); a(24)=A000045(50); a(32)=A000045(30); a(36)=A000045(24); a(48)=A000045(56); a(64)=A000045(54); a(80)=A000045(36); a(96)=A000045(182); a(128)=A000045(128); a(168)=A000045(48); a(192)=A000045(110); a(256)=A000045(80), ..., a(688128)=A000045(240) from the Kelly factorizations. - R. J. Mathar, Apr 05 2007
For n prime, a(n) = q*p^(n-1) or p^(2n-1) for some primes p and q since those are the only numbers with 2*n divisors. a(8) = 2584. - Chai Wah Wu, Dec 08 2014
The sequence is restricted to even numbers of divisors since 1 and 144 are the only Fibonacci numbers with an odd number of divisors (because they are the only positive Fibonacci numbers that are squares, see A227875). - Amiram Eldar, Jul 02 2023

			a(2) = 8 because 8 is the smallest Fibonacci number with 4 divisors (1,2,4,8).

Blair Kelly, Fibonacci Factorizations.

Cf. A000045, A081978, A227875.

Corrected by Emeric Deutsch, Apr 19 2005

A371588 Smallest Fibonacci number > 1 such that some permutation of its digits is a perfect n-th power.

Original entry on oeis.org

2, 144, 8, 610, 5358359254990966640871840, 68330027629092351019822533679447, 15156039800290547036315704478931467953361427680642, 23770696554372451866815101694984845480039225387896643963981, 119447720249892581203851665820676436622934188700177088360

Offset: 1

Chai Wah Wu, Mar 28 2024

Subsequence of A370071 after reordering (as the sequence is not monotonic; e.g., a(2) > a(3) and a(8) > a(9)). Leading 0 digits are allowed in the perfect power. For example, a(4) = 610 since 016 = 2^4. (If leading 0 digits were not allowed, a(4) would be 160500643816367088.)

			a(1) = 2 since 2 = 2^1.
a(2) = 144 since 144 = 12^2.
a(3) = 8 since 8 = 2^3.
a(4) = 610 since 016 = 2^4.
a(5) = 5358359254990966640871840 since 0735948608251696955804943 = 59343^5
a(6) = 68330027629092351019822533679447 since 00059398947526192142327360782336 = 62464^6.

Chai Wah Wu, Table of n, a(n) for n = 1..16

Cf. A000045, A227875, A001597, A118715, A370071.

from itertools import count
from sympy import integer_nthroot
def A371588(n):
    a, b = 1, 2
    while True:
        s = sorted(str(b))
        l = len(s)
        m = int(''.join(s[::-1]))
        u = int(''.join(s))
        for i in count(max(2,integer_nthroot(u,n)[0])):
            if (k:=i**n) > m:
                break
            t = sorted(str(k))
            if ['0']*(l-len(t))+t == s:
                return b
                break
        a, b = b, a+b

A371589 Smallest number m > 1 such that some permutation of the digits of m^n is a Fibonacci number.

Original entry on oeis.org

2, 12, 2, 19002, 433153, 472133, 10064513, 61054259, 67878079, 8152101, 46077414, 11395185, 28556455, 11730986, 179311318, 1542839498, 443163383, 2426412518, 433059953, 443302473, 2654438078, 2764480203, 5945916934

Offset: 1

Chai Wah Wu, Mar 28 2024

Unlike in A370071 or A371588, no leading 0's are allowed in m^n or the Fibonacci number.

			a(4) = 19002 since 19002^4 = 130375880664608016 and a permutation of its digits results in 160500643816367088, a Fibonacci number.

Cf. A000045, A227875, A001597, A118715, A370071, A371588.

from itertools import count
from sympy import integer_nthroot
def A371589(n):
    a, b = 1, 2
    while True:
        s = sorted(str(b))
        m = int(''.join(s[::-1]))
        u = int(''.join(s))
        for i in count(max(2,integer_nthroot(u,n)[0])):
            if (k:=i**n) > m:
                break
            t = sorted(str(k))
            if t == s:
                return i
                break
        a, b = b, a+b

a(15)-a(23) from Bert Dobbelaere, Apr 10 2024

A307991 Fibonacci numbers of the form k^2 - k - 1 with integer k.

Original entry on oeis.org

1, 5, 55, 89

Offset: 1

Amiram Eldar, May 09 2019

The corresponding values of k are 2, 3, 8, 10.
Intersection of A000045 and A028387.
Also Fibonacci numbers whose reciprocals equal to Sum_{i>=1} F(i)/k^(i+1), where F(i) is the i-th Fibonacci number.
de Weger proved that there are no other terms.

			89 is in the sequence since 89 = 10^2 - 10 - 1 or equivalently 1/89 = 1/10^2 + 1/10^3 + 2/10^4 + 3/10^5 + 5/10^6 + ... This is why the first digits of the decimal expansion of 1/89 = 0.011235... are the first terms of the Fibonacci sequence.

Fenton Stancliff, A curious property of a_11, Scripta Math., Vol. 19 (1953), p. 126.

B. M. M. de Weger, A Curious Property of the Eleventh Fibonacci Number, The Rocky Mountain Journal of Mathematics, Vol. 25, No. 3 (1995), pp. 977-994.

Cf. A000045, A021093, A028387, A039595, A165900, A227875.

Select[Fibonacci[Range[2, 20]], IntegerQ[Sqrt[4# + 5]] &]

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A235383 Fibonacci numbers that are the product of other Fibonacci numbers.

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A370071 Fibonacci numbers such that some permutation of their digits is a perfect power.

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A081979 Smallest Fibonacci number with 2n divisors, or 0 if no such number exists.

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A371588 Smallest Fibonacci number > 1 such that some permutation of its digits is a perfect n-th power.

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A371589 Smallest number m > 1 such that some permutation of the digits of m^n is a Fibonacci number.

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A307991 Fibonacci numbers of the form k^2 - k - 1 with integer k.

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