Abundant numbers

[hv at crypt.org]

Earlier I wrote:
:franktaw at netscape.net wrote:
::Based on the recent discussion of "First even term", I have been inspired
::to submit the following sequence:
:: 
::%S A000001 12,945,20,945,12,5391411025,12,945,20,81081,12,5391411025,12,6435,56,945,12,
::5391411025,12,81081,20,945,12,5391411025,12,945,20,6435,12,
::20169691981106018776756331,12,945,20,945,12,5391411025,12,945,20,81081,12,
::366005822969340125,12,945,56,945,12,5391411025,12,81081
::%N A000001 Smallest abundant number relatively prime to n.
:[...]
::I would appreciate it if someone would check these before I actually submit
::the sequence.
:
:I can confirm the figures; the code below takes me 22s to calculate
:up to a(210) = 49061132957714428902152118459264865645885092682687973.

It is also interesting to consider "a(n) relatively prime to n" as a
special case of "a(n) fixes certain prime powers".

a(n) is uninteresting for squareful n: a(xy^2) = a(xy) for all x, y.
If b(n), n = \prod{ p_i^x_i } is defined as the least abundant
k = \prod{ p_i^y_i } such that y_i = x_i - 1 whenever x_i > 0,
we get a new sequence identical to a(n) above for squarefree n,
but with the additional property that every abundant k appears
in the sequence at least at a(k * \prod{ p | k }) = k.

I'll submit this once Franklin's sequence is in, so I can correct the
crossrefs (or Franklin can submit the two together if that is convenient).

It may also be interesting to consider the sequence of abundant numbers
that appear earlier than their guaranteed spot in b(n).

%I A000002
%S A000002 12,945,20,18,12,5391411025,12,12,12,81081,12,70,12,6435,56,24,12
%T A000002 5775,12,18,20,945,12,20,20,945,18,18,12,20169691981106018776756331
%U A000002 12,48,20,945,12,30,12,945,20,12,12,366005822969340125,12,18,12,945
%N A000002 Smallest abundant number with some prime powers fixed by n.
%C A000002 If n = \prod{ p_i^x_i }, a(n) = \prod{ p_i^y_i } then we require y_i = x_i - 1 whenever x_i > 0.
%C A000002 Same as A000001 for squarefree n.
%C A000002 If k is abundant then k = a(k * \prod{ p | k })
%e A000002 12 = 2^2.3^1, so a(12)=70 is the least abundant 2^1.3^0.k with (k,2.3)=1
%Y A000002 Cf A005101, A005231, A047802, A000001.
%O A000002 1,1
%K A000002 nonn
%A A000002 Hugo van der Sanden (hv at crypt.org), Feb 9 2006

Hugo