

A130091


Numbers having in their canonical prime factorization mutually distinct exponents.


50



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
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OFFSET

1,2


COMMENTS

a(n) < A130092(n) for n<=150, a(n) > A130092(n) for n>150.
This sequence does not contain any number of the form 36n6 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


LINKS

R. Zumkeller, Table of n, a(n) for n = 1..10000
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


EXAMPLE

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)


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 *)


PROG

(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


CROSSREFS

Complement of A130092; A006939 and A000961 are subsequences.
Cf. A048767, A048768, A056239, A098859, A112798, A118914, A217605, A325326, A325337, A325368.
Sequence in context: A212165 A319161 A325370 * A119848 A265640 A268375
Adjacent sequences: A130088 A130089 A130090 * A130092 A130093 A130094


KEYWORD

nonn


AUTHOR

Reinhard Zumkeller, May 06 2007


STATUS

approved



