[seqfan] Re: semiprime (A001358) analogue of A181503 Slowest-growing sequence of primes where 1/(p+1) sums to 1...

Maximilian Hasler maximilian.hasler at gmail.com
Fri Oct 29 15:43:42 CEST 2010

```Random comment (don't take it too serious) :
I think that if I was to ask s/o in the street which is the "smaller
growing" among the following two  sequences of semiprimes,

4,6,9,10,14,15,21,22,25,26,33,34,355,16627,76723511,17218740226618333,
374886275842473712491638217368219

or

4,4,9,21,33,65,129,259,514,1027,2049,4097,8193,16387,32773,65542,
131073,262149,524289,1048577,2097157,4194311,8388609,16777219,
33554435,67108867,134217731,268435457,536870918,1073741846,

then she would elect the second one...

Maximilian

%I A000001
%S A000001 4, 4, 4, 9, 21, 33, 65, 129, 259, 514, 1027, 2049, 4097,
8193, 16387, 32773, 65542, 131073, 262149, 524289, 1048577, 2097157,
4194311, 8388609, 16777219, 33554435, 67108867, 134217731, 268435457,
536870918, 1073741846
%O A000001 0,1
%N A000001 Least semiprime >= 2^n
%C A000001 Might also be considered as the least growing sequence of
semiprimes such that sum( 1/a(k), k=0 ... oo ) < 1
%F A000001 a(k) = min{ x >= 2^k | A001222(x)=2 }
%o A000001 (PARI) a(n)={ n=2^n; while( bigomega(n) != 2, n++); n }
%K A000001 nonn,new
%A A000001 M. F. Hasler (www.univ-ag.fr/~mhasler), Oct 29 2010

On Fri, Oct 29, 2010 at 7:31 AM, Richard Mathar
<mathar at strw.leidenuniv.nl> wrote:
>
> More terms on behalf of http://list.seqfan.eu/pipermail/seqfan/2010-October/006328.html
>
> %I A000001
> %S A000001 4,6,9,10,14,15,21,22,25,26,33,34,355,16627,76723511,17218740226618333,
> %T A000001 374886275842473712491638217368219
> %N A000001 Smallest growing sequence of semiprimes A001358 such that sum_{i=1..n} 1/a(i) < 1 for all n.
> %C A000001 Semiprime variant of A075442.
> %C A000001 The first semiprime that is not in the sequence is 35, because 1/4+1/6+1/9+..+1/34+1/35 > 1.
> %p A000001 A := proc(n) option remember; local a,psum; if n = 1 then A001358(1); else psum := add(1/procname(i),i=1..n-1) ;
>                for a from max(procname(n-1)+1,ceil(1/(1-psum)) ) do if isA001358(a) then if psum+1/a < 1 then
>                return a; end if; end if; end do: end if; end proc: # R. J. Mathar, Oct 29 2010
> %K A000001 nonn,new
> %O A000001 1,1
> %A A000001 Jonathan Vos Post (jvospost3(AT)gmail.com), Oct 29 2010
>

```