A130091 Numbers having in their canonical prime factorization mutually distinct exponents.

1, 2, 3, 4, 5, 7, 8, 9, 11, 12, 13, 16, 17, 18, 19, 20, 23, 24, 25, 27, 28, 29, 31, 32, 37, 40, 41, 43, 44, 45, 47, 48, 49, 50, 52, 53, 54, 56, 59, 61, 63, 64, 67, 68, 71, 72, 73, 75, 76, 79, 80, 81, 83, 88, 89, 92, 96, 97, 98, 99, 101, 103, 104, 107, 108, 109, 112, 113, 116

Offset: 1

Author

Reinhard Zumkeller, May 06 2007

Keywords

• nonn

Comments

This sequence does not contain any number of the form 36n-6 or 36n+6, as such numbers are divisible by 6 but not by 4 or 9. Consequently, this sequence does not contain 24 consecutive integers. The quest for the greatest number of consecutive integers in this sequence has ties to the ABC conjecture (see the MathOverflow link). - Danny Rorabaugh, Sep 23 2015

The Heinz number of an integer partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k), so these are Heinz numbers of integer partitions with distinct multiplicities. The enumeration of these partitions by sum is given by A098859. - Gus Wiseman, May 04 2019

Aktaş and Ram Murty (2017) called these terms "special numbers" ("for lack of a better word"). They prove that the number of terms below x is ~ c*x/log(x), where c > 1 is a constant. - Amiram Eldar, Feb 25 2021

Sequence A005940(1+A328592(n)), n >= 1, sorted into ascending order. - Antti Karttunen, Apr 03 2022

Examples

			From _Gus Wiseman_, May 04 2019: (Start)
The sequence of terms together with their prime indices begins:
   1: {}
   2: {1}
   3: {2}
   4: {1,1}
   5: {3}
   7: {4}
   8: {1,1,1}
   9: {2,2}
  11: {5}
  12: {1,1,2}
  13: {6}
  16: {1,1,1,1}
  17: {7}
  18: {1,2,2}
  19: {8}
  20: {1,1,3}
  23: {9}
  24: {1,1,1,2}
  25: {3,3}
  27: {2,2,2}
(End)
		

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
• Kevser Aktaş and M. Ram Murty, On the number of special numbers, Proceedings - Mathematical Sciences, Vol. 127, No. 3 (2017), pp. 423-430; alternative link.
• MathOverflow, Consecutive numbers with mutually distinct exponents in their canonical prime factorization
• Carlo Sanna, On the number of distinct exponents in the prime factorization of an integer, arXiv:1902.09224 [math.NT], 2019.
• Eric Weisstein's World of Mathematics, Prime Factorization

Crossrefs

Complement of A130092. A351564 is the characteristic function.

Subsequence of A351294.

Cf. A000961, A006939, A181818, A304686, A319161, A342028, A342029, A342030, A342031, A342032 (subsequences).

Cf. A005940, A048767, A048768, A056239, A098859, A112798, A118914, A181796, A217605, A325326, A325337, A325368, A327498, A327523, A328592, A336423, A336424, A336569, A336570, A336571, A343012, A343013.

Programs

Maple

filter:= proc(t) local f;
f:= map2(op,2,ifactors(t)[2]);
nops(f) = nops(convert(f,set));
end proc:
select(filter, [$1..1000]); # Robert Israel, Mar 30 2015

Mathematica

t[n_] := FactorInteger[n][[All, 2]]; Select[Range[400],  Union[t[#]] == Sort[t[#]] &]  (* Clark Kimberling, Mar 12 2015 *)

PARI

isok(n) = {nbf = omega(n); f = factor(n); for (i = 1, nbf, for (j = i+1, nbf, if (f[i, 2] == f[j, 2], return (0)););); return (1);} \\ Michel Marcus, Aug 18 2013

PARI

isA130091(n) = issquarefree(factorback(apply(e->prime(e), (factor(n)[, 2])))); \\ Antti Karttunen, Apr 03 2022

Formula

a(n) < A130092(n) for n<=150, a(n) > A130092(n) for n>150.