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

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

Offset: 1

Author

Chai Wah Wu, Mar 28 2024

Keywords

• nonn
• base

Comments

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.)

Examples

			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.
		

Links

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

Crossrefs

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

Programs

Python

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