"More"-Cam

[David W. Wilson]

Richard Guy wrote:
> 
> Should A062061 be
> 
> 1  2  3  5  7  8  9  11  13  14  15  17  19  20  21
>                                         (not     23) ?
> 23  25  26  27  29  31  32  33  35  37  38  39  41
> 
> 43  44  45  47  49  50  51  53  55  56  57  59  61
>    (44 missing ?)
> 62  63  65  67  68  69  71  73  74  ...
> 
>  6k-1  6k+1  6k+2  6k+3 ...   ?
> 

In regard to A062061-3:

For each k >= 1, let sequence Sk be formed by starting with the empty
sequence, then adding successively the next positive integer which is
coprime to the last k elements of the sequence, or to all elements if
fewer than k exist in the sequence.

We then have

    S1 = A000027 = natural numbers
    S2 = A062061
    S3 = A062062
    S4 = A062063

Empirically, I obtained the following:

    S1(n + 1)       = S1(n) + 1       (n >= 1)
    S2(n + 4)       = S2(n) + 6       (n >= 1)
    S3(n + 125)     = S3(n) + 210     (n >= 7)

I was able to prove these, but only by considerations specific to each
sequences.  The S3 identity above answers

%C A062062 Is a(n+1) - a(n) bounded?

in the affirmative.  The maximum term difference is 4 = a(68)-a(67).

I was initially convinced that the above forms would generalize to
all Sk < inf (but obviously not to Sinf = (1, primes)).  However, 

    S4(n + 1111)   = S4(n) + 2310
    S4(n + 14447)  = S4(n) + 30030

both failed to be true, and now I am not so sure.  

Anyway, I suggest the following improvements(?) to the existing sequences:

%N A062061 Each element coprime to preceeding two elements
%F A062061 a(n) = {n | n == 1, 2, 3, 5 (mod 6)}.

%N A062062 Each element coprime to preceeding three elements
%C A062062 a(n+125) = a(n)+210 (n >= 7).

%N A062063 Each element coprime to preceeding four elements