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A130091 Numbers having in their canonical prime factorization mutually distinct exponents. 24
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

COMMENTS

Complement of A130092; A006939 and A000961 are subsequences;

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

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

LINKS

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

MathOverflow, Consecutive numbers with mutually distinct exponents in their canonical prime factorization

Eric Weisstein's World of Mathematics, Prime Factorization

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

Sequence in context: A246716 A212165 A319161 * A119848 A265640 A268375

Adjacent sequences:  A130088 A130089 A130090 * A130092 A130093 A130094

KEYWORD

nonn

AUTHOR

Reinhard Zumkeller, May 06 2007

STATUS

approved

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Last modified November 16 21:37 EST 2018. Contains 317275 sequences. (Running on oeis4.)