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 1-3 of 3 results.

A180935 A180934(n)^a(n) has A180934(n) divisors.

Original entry on oeis.org

1, 1, 2, 4, 6, 4, 10, 12, 16, 18, 22, 12, 3, 28, 30, 36, 3, 40, 42, 4, 46, 24, 52, 58, 60, 66, 70, 72, 78, 20, 82, 88, 96, 100, 102, 106, 108, 112, 60, 126, 130, 136, 138, 148, 150, 8, 156, 162, 166, 84, 172, 178, 180, 190, 192, 196, 198, 210, 222, 7, 226, 228, 232, 238
Offset: 1

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Author

David W. Wilson, Sep 26 2010

Keywords

Comments

For n > 1, a(n) gives the unique solution k of d(m^k) = m where d = A000005. For m = 1, any integer k will do, we choose the smallest positive solution a(1) = 1.
For prime p, p-1 is in this sequence.
For odd semiprime s, (s-1)/2 is in this sequence.

Examples

			11^10 has 11 divisors, so a(n) = 10 where A180934(n) = 11.
225^7 has 225 divisors, so a(n) = 7 where A180934(n) = 225.

Links

Crossrefs

Cf. A000005, A006093, A046315, A180934, A180936.

Programs

  • Mathematica
    f[n_] := Module[{e = FactorInteger[n][[;; , 2]], k = 1}, While[n > Times @@ (k*e + 1), k++]; If[n == Times @@ (k*e + 1), k, Nothing]]; f[1] = 1; Array[f, 250] (* Amiram Eldar, Apr 09 2024 *)

A143026 Positive integers k such that the fourth power of the number of positive divisors of k equals k.

Original entry on oeis.org

1, 625, 6561, 4100625
Offset: 1

Views

Author

Emeric Deutsch, Aug 11 2008

Keywords

Comments

625=5^4, 6561=3^8, 4100625=(3^8)(5^4).
There are no more terms in the sequence.

Examples

			625 has 5 divisors (1, 5, 25, 125 and 625) and 5^4 = 625.

References

  • T. Andreescu, D. Andrica and Z. Feng, 104 Number Theory Problems (from the training of the USA IMO team), Birkhäuser, Boston, 2007, Advanced problem # 19, pp. 85, 145, 146.
  • Roozbeh Hazrat, Mathematica: A Problem-Centered Approach, Springer 2010, p. 39.

Crossrefs

Cf. A066693, A180936.

Programs

  • Mathematica
    Select[Range[4200000],DivisorSigma[0,#]^4==#&] (* Harvey P. Dale, Oct 17 2011 *)

Extensions

Second reference added by Harvey P. Dale, Oct 17 2011

A219338 Numbers n for which n = (tau(n) - 1)^k with integer k.

Original entry on oeis.org

4, 16, 27, 3125, 3375, 65536, 823543, 3748096, 52521875, 285311670611, 7625597484987, 302875106592253, 1156831381426176, 66182427701415936, 827240261886336764177, 511324276025564512546607, 1978419655660313589123979, 281633339785852578930098176
Offset: 1

Views

Author

Zdenek Cervenka, Nov 18 2012

Keywords

Comments

tau(n) is the number of positive divisors of n.

Examples

			a(1) = 4 because (tau(4) - 1)^2 = (3 - 1)^2 = 4 and this is the first number satisfying this condition.
a(2) = 16 because (tau(16) - 1)^2 = (5 - 1)^2 = 16 and this is the second number satisfying this condition.
a(3) = 27 because (tau(27) - 1)^3 = (4 - 1)^3 = 27 and this is the third number satisfying this condition.

Crossrefs

Cf. A180936.

Programs

  • Mathematica
    Select[Range[10^4], IntegerQ[Log[DivisorSigma[0, #] - 1, #]] &] (* Alonso del Arte, Nov 18 2012 *)
  • PARI
    v=vector(18); mx=3*10^26; c=0; for(m=2, 3440639, for(k=2, 87, n=m^k; if(n>mx, next(2)); if(m==numdiv(n)-1, c++; v[c]=n))); v=vecsort(v); for(i=1, c, print1(v[i]", ")) /* Donovan Johnson, Nov 19 2012 */

Formula

Numbers n for which n = (tau(n) - 1)^k with integer k.

Extensions

a(10)-a(18) from Donovan Johnson, Nov 19 2012
Showing 1-3 of 3 results.

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