_Amarnath Murthy (amarnath_murthy(AT)yahoo.com), _, Sep 13 2003

For sufficiently large n, a(n) = 7^floor(log(n)/log(3)) because log(prime(2m))/log(prime(m)) is largest for m = 2. - _David Wasserman (wasserma(AT)spawar.navy.mil), _, Jun 03 2005

More terms from _David Wasserman (wasserma(AT)spawar.navy.mil), _, Jun 03 2005

For sufficiently large n, a(n) = 7^floor(log(n)/log(3)), because log(prime(2m))/log(prime(m)) is largest for m = 2. - David Wasserman (wasserma(AT)spawar.navy.mil), Jun 03 2005

nonn,new

nonn

a(n) = smallest k such that for every each r, 2= <= r= <= n , there exists an a distinct s, n < s <= k , with the same prime signature of as r.

3, 5, 7, 9, 11, 13, 13, 19, 27, 49, 49, 49, 49, 49, 49, 49, 81, 81, 81, 81, 81, 81, 81, 81, 81, 169, 169, 343, 343, 343, 343, 343, 343, 343, 343, 343, 343, 343, 343, 343, 343, 343, 343, 343, 343, 343, 343, 343, 343, 361, 361, 361, 361, 361, 361, 361, 361, 361, 361, 361

Question: Find an ( the best possible) estimate of a(n) in terms of n. Conjecture: a(prime(r)^k) = prime(2r)^k.

For sufficiently large n, a(n) = 7^floor(log(n)/log(3)), because log(prime(2m))/log(prime(m)) is largest for m = 2. - David Wasserman (wasserma(AT)spawar.navy.mil), Jun 03 2005

a(7) = 19 and For numbers ( 2,3,4,5,6,7) we have the set of numbers ( 11,13,9,17,10,19) with matching prime signatures.

( 11,13,9,17,10,19) with matching prime signatures.

more,nonn,new

nonn

More terms from David Wasserman (wasserma(AT)spawar.navy.mil), Jun 03 2005

a(n) = smallest k such that for every r, 2=<r=<n there exists an s, n<s<=k with the prime signature of r.

3, 5, 9, 11, 13, 19, 27, 49, 49, 49, 49, 49, 49, 49, 81, 81, 81, 81, 81, 81, 81, 81

2,1

Question: Find an ( the best possible) estimate of a(n) in terms of n. Conjecture: a(prime(r)^k) = prime(2r)^k.

a(7) = 19 and For numbers ( 2,3,4,5,6,7) we have the set of numbers

( 11,13,9,17,10,19) with matching prime signatures.

more,nonn

Amarnath Murthy (amarnath_murthy(AT)yahoo.com), Sep 13 2003

approved