A370071 Fibonacci numbers such that some permutation of their digits is a perfect power.
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
Examples
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).
Links
- Michael S. Branicky, Table of n, a(n) for n = 1..45
Programs
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PARI
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
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Python
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
Extensions
a(19) inserted and a(21)-a(23) by Michael S. Branicky, Feb 17 2024
More terms from David A. Corneth, Feb 17 2024
Comments