|
|
A137509
|
|
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.
|
|
1
|
|
|
2, 3, 5, 9, 11, 14, 17, 27, 49, 51, 53, 63, 67, 69, 74, 81, 83, 92, 97, 98, 106, 111, 113, 135, 169, 177, 343, 356, 359, 366, 367, 3125, 3127, 3131, 3133, 3249, 3251, 3254, 3261, 3272, 3299, 3302, 3307, 3308, 3316, 3317, 3319, 3321, 3481, 3501, 3503, 3508
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,1
|
|
COMMENTS
|
Starting the sequence at a(1)=1 instead leads to a(n) = n for every positive integer n.
|
|
LINKS
|
|
|
EXAMPLE
|
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.
|
|
MAPLE
|
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
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|