a(n) >= a(n-1) and sum of first n terms is prime

[Neil Fernandez]

In message <3E01FF45.8010501 at kspaint.com>, Robert G. Wilson v
<rgwv at kspaint.com> writes

>Neil,
>
>       Would it not be better to ask for the first number to appear exactly n 
>time? 
>i.e.; 2, 294, 12, 6, ...

That would give more info about the sequence, although the first number
to appear exactly twice is 90.

As defined above, the derived sequence may have some holes, and/or stop;
as defined below, it won't have any holes, but may stop (which I don't
think would be disproved easily even if it were proved that there were
arbitrarily long sequences of consecutive primes in arithmetic
progression).

Neil

>Neil Fernandez wrote:
>> 2, 3, 6, 6, 6, 6, 8, 10, 12, 12, 12, 14, 16, 18, 18, 18, 24, 32, 34, 36,
>> 38, 42, 46, 48, 54, 56, 64, 68, 78, 90, 90, 94, 102, 114, 122, 124, 134,
>> 144, 148, 150, 152, 160, 168, 170, 178, 182, 190, 192, 200, 216, 220,
>> 222, 234, 234, 234, 246, 260, 264, 268, 270, 278, 280, 290, 292, 294,
>> 294, 300, 302, 310, 312, 312, 318, 318, 330, 348, 350, 370, 380, 384,
>> 388, 408, 410, 424, 428, 444, 450, 450
>> 
>> a(n) = smallest integer greater or equal to a(n-1) such that the sum of
>> the first n terms is prime
>> 
>> First numbers to appear once, twice, three times, etc. are 2, 6, 6, 6, ?

-- 
Neil Fernandez