correcting typo: 2552, 4892, ????

[Robert G. Wilson v]

Et al,

        A better sequence would be the same Rule ( a[n]=2*a[n-1]+1 ),
but for each k>0, a[n] is the stopping iteration number as determined
when a[n] is prime. Such a sequence would look like for k= 1 & n=1 since
a[1]=3 (a prime), k=2 & n=1 since a[1]=5 (a prime),  k=3 & n=1 since
a[1]=7 (a prime), k=4 & n=2 since a[1]=9 and a[2]=19 (a prime).
Therefore the sequence begins 1, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 3, 4, 1,
1, 2, 2, 1, 2, 1, 1, 4, 1, 3, 2, 1, 3, 4, 1, 1, 2, 2, 1, 2, 1, 1, 2, 3,
1, 2, 1, 7, 24, 1, 3, 4, 2, 1, 2, 1, 1, 2, 1, 1, 2, 1, 3, 12, 2, 3, 4,
2, 1, 4, 1, 5, 2, 1, 1, 2, ... [1,4]. In[]:= q = ""; Do[a = k; a = 2*a +
1; n = 1; While[ !PrimeQ[a], a = 2*a + 1; n++]; q = Append[q, n], {k, 1,
70}]  Notice that if the initial number is prime, we still proceed with
the iterations.


Sequentially yours,

Robert G. 'Bob' Wilson v


David Broadhurst wrote:

> with apologies, i meant:
> > n>0 such that a[n]=2*a[n-1]+1 is probably prime with a[0]=73
> >              2552, 4892, ????
>                 !