[seqfan] Re: A159559

Benoît Jubin benoit.jubin at gmail.com
Thu Jul 30 16:23:34 CEST 2009

There is something interesting: for a positive integer i, let b_i be
the sequence defined by:
b_i(1)=i and for n>1, if n is prime (resp. composite) then b_i(n) is
the smallest prime (resp. composite) greater than b_i(n-1).
b_1 is the sequence of positive integers
b_2 is A159559 (the sequence in the object of this thread)
b_4 is A159698 (thanks Ray for pointing this out to me)

Now, if i is composite, b_i agrees with b_{i-1} except for the first term. Also,
b_3 agrees with b_2 from rank 11
b_5 agrees with b_3 from rank 47
b_7 agrees with b_5 from rank 11
b_11 agrees with b_7 from rank 17
b_13 agrees with b_11 from rank 683
b_17 agrees with b_13 from rank 11
Does this inspire anyone?


2009/7/28 Benoît Jubin <benoit.jubin at gmail.com>:
> On Tue, Jul 28, 2009 at 9:10 PM, Ray Chandler<rayjchandler at sbcglobal.net> wrote:
>> You don't need a special case for a(1)=2 if you define as:
>> a(n) is non-composite iff n is non-composite.
>> Or replace "prime" with "non-composite" in your definition.
> I'm not sure I understand: your definition would give the sequence
> a(n)=n.  In a way or another, we need an initialization.
> But this makes me think that the current comment that n is prime iff
> a(n) is should be changed to n is composite iff a(n) is (so that it
> still holds for a(1)=2).
> Benoit
>> Ray
>>> I agree that setting a(1)=2 is natural (taking a(1)=1 gives
>>> the positive integers, of course).
>>> What do you think of this definition:
>>> a(1)=2 and for n>1, if n is prime (resp. composite) then a(n)
>>> is the smallest prime (resp. composite) greater than a(n-1).
>>> Benoit
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