cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-4 of 4 results.

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

Views

Author

Gonzalo Martínez, Feb 08 2024

Keywords

Comments

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

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

Crossrefs

Cf. A000045, A227875, A001597, A118715.

Programs

  • 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
  • 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

A118895 Numbers n with property that both n and its digit reversal are perfect powers (i.e., in A001597).

Original entry on oeis.org

1, 4, 8, 9, 100, 121, 144, 169, 343, 400, 441, 484, 676, 900, 961, 1000, 1089, 1331, 8000, 9801, 10000, 10201, 10404, 10609, 12100, 12321, 12544, 12769, 14400, 14641, 14884, 16900, 40000, 40401, 40804, 44100, 44521, 44944, 48400, 48841, 67600
Offset: 1

Views

Author

Joshua Zucker, May 04 2006

Keywords

Comments

An alternate version of A110811, which doesn't include the terms with trailing zeros.

Examples

			100 is in the sequence because 100 and 001 = 1 are both perfect powers;
44521 and 12544 are both in the sequence because both are perfect powers, 221^2 and 112^2.

Crossrefs

This sequence is a subset of both A115656 and A001597 and contains A110811.
Cf. A118715.

Extensions

Edited by N. J. A. Sloane, Oct 11 2008

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

Views

Author

Chai Wah Wu, Mar 28 2024

Keywords

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

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

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

Views

Author

Chai Wah Wu, Mar 28 2024

Keywords

Comments

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

Examples

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

Crossrefs

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

Programs

  • Python
    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

Extensions

a(15)-a(23) from Bert Dobbelaere, Apr 10 2024
Showing 1-4 of 4 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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