[seqfan] Re: Prime partial sum of digits

Lars Blomberg larsl.blomberg at comhem.se
Mon Nov 9 18:48:13 CET 2015


Hello, 

I get 18 terms =  2, 1, 4, 6, 40, 20, 46, 8, 42, 400, 60, 62, 64, 26, 406, 80, 48, 4000 and no more terms < 10^12.
/Lars B

-----Ursprungligt meddelande-----
Från: SeqFan [mailto:seqfan-bounces at list.seqfan.eu] För Eric Angelini
Skickat: den 9 november 2015 18:00
Till: Bob Selcoe <rselcoe at entouchonline.net>; Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
Ämne: [seqfan] Re: Prime partial sum of digits

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),
É.

> Hi Eric,
>
>What is a(n)? Why does P(4) = 1?  Why does S(5) = 40 rather than 13?
>
>Cheers,
>Bob Selcoe


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