[seqfan] Re: Prime partial sum of digits

Bob Selcoe rselcoe at entouchonline.net
Mon Nov 9 20:48:39 CET 2015


Lars, Eric & Chris,

Chris - you repeated 46.  I computed Lars' sequence by hand and  I think 
it's correct:

2, 1, 4, 6, 40, 20, 46, 8, 42, 400, 60, 62, 64, 26, 406, 80, 48, 4000.

Again, the sequence terminates here (per my previous post, partial sum at 
a(18)= 4000 is 113 and next prime 127 > 113+9).

Cheers,
Bob

--------------------------------------------------
From: "Chris Thompson" <cet1 at cam.ac.uk>
Sent: Monday, November 09, 2015 11:39 AM
To: "Sequence Fanatics Discussion list" <seqfan at list.seqfan.eu>
Subject: [seqfan] Re: Prime partial sum of digits

> On Nov 9 2015, Eric Angelini wrote:
>
>>Hello Bob,
>>
>>> S = 2,1,4,6,40,20,46,8,4,24,60,...
>>> P = 2 3 7 1 11 11 22 3 4 44 55
>>>           3 77 99 39 7 1 37 33
>>
>>... a(n) is the nth term of S [I hesitated naming it S(n)]
>>... P(4) is 13 (cumulative sum 2+1+4+6 = 13 --> to be read vertically 
>>under "6")
>>
>>> Why does S(5) = 40 rather than 13?
>>
>>... With S(5) = 13 we would have a non-prime cumulative sum for P(5) as 
>>the "1" of "13" summated to P(4) would give the composite 14.
>>Best (and sorry to be unclear),
>
> But why are you allowing a(9) to equal a(3)=4? If I understand your
> rules correctly, shouldn't you have
>
> S = 2,1,4,6,40,20,46,8,42,46,60,26,48,400,600, ???
> P = 2 3 7 1 11 11 22 3 44 45 55 66 77 888 888
>           3 77 99 39 7 13 73 99 17 19 333 999
>
> after which the sequence stops, as you can't get from 89 to 97 via
> a single digit?
>
> -- 
> Chris Thompson
> Email: cet1 at cam.ac.uk
 




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