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A130091
Numbers having in their canonical prime factorization mutually distinct exponents.
165
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
1,2
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
LINKS
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.
Eric Weisstein's World of Mathematics, Prime Factorization
FORMULA
a(n) < A130092(n) for n<=150, a(n) > A130092(n) for n>150.
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
(PARI) isA130091(n) = issquarefree(factorback(apply(e->prime(e), (factor(n)[, 2])))); \\ Antti Karttunen, Apr 03 2022
CROSSREFS
KEYWORD
nonn
AUTHOR
Reinhard Zumkeller, May 06 2007
STATUS
approved