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
Keywords
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
Programs
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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
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Mathematica
t[n_] := FactorInteger[n][[All, 2]]; Select[Range[400], Union[t[#]] == Sort[t[#]] &] (* Clark Kimberling, Mar 12 2015 *)
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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
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PARI
isA130091(n) = issquarefree(factorback(apply(e->prime(e), (factor(n)[, 2])))); \\ Antti Karttunen, Apr 03 2022
Comments