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 1-2 of 2 results.

A001292 Concatenations of cyclic permutations of initial positive integers.

Original entry on oeis.org

1, 12, 21, 123, 231, 312, 1234, 2341, 3412, 4123, 12345, 23451, 34512, 45123, 51234, 123456, 234561, 345612, 456123, 561234, 612345, 1234567, 2345671, 3456712, 4567123, 5671234, 6712345, 7123456
Offset: 1

Views

Author

R. Muller

Keywords

Comments

Entries are sorted numerically, so after a(45) = 912345678 we have a(46) = 10123456789 instead of a(46) = 12345678910. - Giovanni Resta, Mar 21 2017
From Marco Ripà, Apr 21 2022: (Start)
In 1996, Kenichiro Kashihara conjectured that there is no prime power of an integer (A093771) belonging to this sequence (disregarding the trivial case 1); a direct search from 12 to a(100128) has confirmed the conjecture up to 10^1035. There are no perfect powers among terms t which are permutations of 123_...(m - 1)_m for m == {2, 3, 5, 6} (mod 9). This is since 10 == 1 (mod 9) and also (1 + 0) == 1 (mod 9), so digit position has no effect. Hence, t == A134804(m) (mod 9). Now, if m is such that A134804(m) = {3, 6}, there is a lone factor of 3, which is not a perfect power (indeed).
Therefore, any perfect power in this sequence is necessarily congruent modulo 9 to 0 or 1.
(End)

Crossrefs

Programs

  • Mathematica
    Sort@ Flatten@ Table[ FromDigits[ Join @@ IntegerDigits /@ RotateLeft[Range[n], i - 1]], {n, 11}, {i, n}] (* Giovanni Resta, Mar 21 2017 *)
  • Python
    from itertools import count, islice
    def A001292gen():
        s = []
        for i in count(1):
            s.append(str(i))
            yield from sorted(int("".join(s[j:]+s[:j])) for j in range(i))
    print(list(islice(A001292gen(), 46))) # Michael S. Branicky, Jul 01 2022

A353025 Terms of A352991 which are perfect powers.

Original entry on oeis.org

1, 13527684, 34857216, 65318724, 73256481, 81432576, 139854276, 152843769, 157326849, 215384976, 245893761, 254817369, 326597184, 361874529, 375468129, 382945761, 385297641, 412739856, 523814769, 529874361, 537219684, 549386721, 587432169, 589324176, 597362481, 615387249, 627953481, 653927184
Offset: 1

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Author

Marco Ripà, Apr 17 2022

Keywords

Comments

It appears that all terms are terms of A062503.
We note that a(n)=A352329(n) up to a(36)=A352329(36)=923187456, while the mentioned match does not hold starting from a(37)=14102987536 (since A352329(37)=1234608769).
There are no perfect powers among terms t which are permutations of 123_...(m - 1)_m for m == {2, 3, 5, 6} (mod 9). This is since 10 == 1 (mod 9) and also (1 + 0) == 1 (mod 9), so digit position has no effect. Hence, t == A134804(m) (mod 9). Now, if m is such that A134804(m) = {3, 6}, there is a lone factor of 3, which is not a perfect power (indeed).
Therefore, all terms are necessarily congruent modulo 9 to 0 or 1 (see Marco Ripà link).
All terms up to 10^34 are squares (in particular, there are 67 squares with no more than 17 digits). - Aldo Roberto Pessolano, May 12 2022

Examples

			75910168324 is a term since 75910168324 = 275518^2.
		

Crossrefs

Programs

  • Mathematica
    z = 1; Do[r = Range[k];
    n = ToExpression[StringJoin[ToString[#] & /@ r]];
    If[And[Mod[n, 9] != 3, Mod[n, 9] != 6], d = DigitCount[n];
      s = IntegerPart[Sqrt[10^(IntegerLength[n] - 1)]];
      f = IntegerPart[Sqrt[10^(IntegerLength[n])]];
      Do[y = x^2;
       If[DigitCount[y] == d, c = True;
        Do[If[Not[StringContainsQ[ToString[y], ToString[i]]],
          c = False], {i, 10, k}]; If[c, Print[z, " ", y]; z++]], {x, s,
        f}]], {k, 1, 10}] (* Aldo Roberto Pessolano, May 12 2022 *)

Formula

Digit sum of a(n) is always congruent to 0 or 1 modulo 9.
a(n) = m^2, where the integer m := m(n) is not a perfect power itself (conjectured).
Showing 1-2 of 2 results.