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%I #8 Oct 03 2015 22:58:56
%S 2,3,5,9,11,14,17,27,49,51,53,63,67,69,74,81,83,92,97,98,106,111,113,
%T 135,169,177,343,356,359,366,367,3125,3127,3131,3133,3249,3251,3254,
%U 3261,3272,3299,3302,3307,3308,3316,3317,3319,3321,3481,3501,3503,3508
%N a(1)=2. For n >= 2, a(n) = the smallest integer > a(n-1) that has the same multiset of prime-factorization exponents as n has.
%C Starting the sequence at a(1)=1 instead leads to a(n) = n for every positive integer n.
%e 12 = 2^2 * 3^1. So the multiset of exponents in the prime factorization of 12 is {1,2}. For a(12), we want the smallest integer > a(11)=53 of the form p^1 * q^2, where p and q are distinct primes. Checking: 54 = 2^1 *3^3, so 54 fails. 55 = 5^1*11^1. 56 = 2^3*7^1. 57 = 3^1*19^1. 58 = 2^1*29^1. 59=59^1. 60 = 2^2*3^1*5^1. 61 = 61^1. 62 = 2^1 *31^1. So 54 through 62 all fail. But 63 = 3^2 * 7^1, which has the same multiset of prime exponents, {1,2}, as 12 has. Therefore a(12) = 63.
%p pmset := proc(n) local e,a ; a := [] ; for e in ifactors(n)[2] do a := [op(a),e[2]] ; od: sort(a) ; end: A137509 := proc(n) option remember ; local nset,a ; if n = 1 then RETURN(2) ; fi ; nset := pmset(n) ; for a from A137509(n-1)+1 do if pmset(a) = nset then RETURN(a) ; fi ; od: end: seq(A137509(n),n=1..120) ; # _R. J. Mathar_, May 23 2008
%Y Cf. A081761.
%K nonn
%O 1,1
%A _Leroy Quet_, Apr 23 2008
%E More terms from _R. J. Mathar_, May 23 2008
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