|
%I #48 Apr 03 2022 09:56:11
%S 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,
%T 40,41,43,44,45,47,48,49,50,52,53,54,56,59,61,63,64,67,68,71,72,73,75,
%U 76,79,80,81,83,88,89,92,96,97,98,99,101,103,104,107,108,109,112,113,116
%N Numbers having in their canonical prime factorization mutually distinct exponents.
%C 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
%C 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
%C 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
%C Sequence A005940(1+A328592(n)), n >= 1, sorted into ascending order. - _Antti Karttunen_, Apr 03 2022
%H Reinhard Zumkeller, <a href="/A130091/b130091.txt">Table of n, a(n) for n = 1..10000</a>
%H Kevser Aktaş and M. Ram Murty, <a href="https://doi.org/10.1007/s12044-016-0326-z">On the number of special numbers</a>, Proceedings - Mathematical Sciences, Vol. 127, No. 3 (2017), pp. 423-430; <a href="https://www.ias.ac.in/article/fulltext/pmsc/127/03/0423-0430">alternative link</a>.
%H MathOverflow, <a href="http://mathoverflow.net/questions/201489">Consecutive numbers with mutually distinct exponents in their canonical prime factorization</a>
%H Carlo Sanna, <a href="https://arxiv.org/abs/1902.09224">On the number of distinct exponents in the prime factorization of an integer</a>, arXiv:1902.09224 [math.NT], 2019.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/PrimeFactorization.html">Prime Factorization</a>
%F a(n) < A130092(n) for n<=150, a(n) > A130092(n) for n>150.
%e From _Gus Wiseman_, May 04 2019: (Start)
%e The sequence of terms together with their prime indices begins:
%e 1: {}
%e 2: {1}
%e 3: {2}
%e 4: {1,1}
%e 5: {3}
%e 7: {4}
%e 8: {1,1,1}
%e 9: {2,2}
%e 11: {5}
%e 12: {1,1,2}
%e 13: {6}
%e 16: {1,1,1,1}
%e 17: {7}
%e 18: {1,2,2}
%e 19: {8}
%e 20: {1,1,3}
%e 23: {9}
%e 24: {1,1,1,2}
%e 25: {3,3}
%e 27: {2,2,2}
%e (End)
%p filter:= proc(t) local f;
%p f:= map2(op,2,ifactors(t)[2]);
%p nops(f) = nops(convert(f,set));
%p end proc:
%p select(filter, [$1..1000]); # _Robert Israel_, Mar 30 2015
%t t[n_] := FactorInteger[n][[All, 2]]; Select[Range[400], Union[t[#]] == Sort[t[#]] &] (* _Clark Kimberling_, Mar 12 2015 *)
%o (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
%o (PARI) isA130091(n) = issquarefree(factorback(apply(e->prime(e), (factor(n)[, 2])))); \\ _Antti Karttunen_, Apr 03 2022
%Y Complement of A130092. A351564 is the characteristic function.
%Y Subsequence of A351294.
%Y Cf. A000961, A006939, A181818, A304686, A319161, A342028, A342029, A342030, A342031, A342032 (subsequences).
%Y Cf. A005940, A048767, A048768, A056239, A098859, A112798, A118914, A181796, A217605, A325326, A325337, A325368, A327498, A327523, A328592, A336423, A336424, A336569, A336570, A336571, A343012, A343013.
%K nonn
%O 1,2
%A _Reinhard Zumkeller_, May 06 2007
|
