Paolo Lava paoloplava at gmail.com
Mon Nov 19 14:40:52 CET 2012

```There was an error(a(38)=35 was not correct). The first terms should be:

1,2,3,4,5,8,6,10,7,12,9,11,18,16,13,22,15,14,
17,23,19,20,34,21,24,25,26,33,27,40,32,42,29,

28,30,46,31,*37*,35,39,36,38,41,42,52,44,45,47,62,48,53,....

Paolo

2012/11/19 Paolo Lava <paoloplava at gmail.com>

> Dear Seqfans,
>
>
>
> I am thinking to a sort of headache sequence (permutation of natural
> numbers) like the following:
>
>
>
> “Starting from a(1)=1, for any n the sum of the next a(n) numbers is a
> prime. No repetition of numbers is allowed. At any step the minimum integer
> not yet used  and not leading to a contradiction should be choosen.”
>
>
>
> The squence starts with 1, 2, 3, 4, 5, 8, 6, 10, 7, 12…
>
>
>
> a(1)=1 -> Next term is 2, prime.
>
> a(2)=2 -> the sum of the next two terms 3+4=7, prime.
>
> a(3)=3 -> 4+5+8=17, prime.
>
> a(4)=4 -> 5+8+6+10=29, prime.
>
> a(5)=5 -> 8+6+10+7+12=43, prime.
>
> a(6)=8 but I have also a(7)=6 that must be covered before a(6). Therefore
> the sequence became:
>
>
>
> 1, 2, 3, 4, 5, 8, 6, 10, 7, 12, 9, 11, 18,… with 10+7+12+9+11+18=67, prime.
>
> Then coming back to a(6)=8 :  1, 2, 3, 4, 5, 8, 6, 10, 7, 12, 9, 11, 18,
> 16… with 6+10+7+12+9+11+18+16=89.
>
>
>
> I calculated:
>
> 1,2,3,4,5,8,6,10,7,12,9,11,18,16,13,22,14,15,17,23,19,20,34,21,24,25,26,33,27,40,
> x…
>
>
>
> The problem is that the next term “x”, that is a(31), must satisfy at the
> same time a(13)=18 and a(17)=14 but
>
>
>
> 16+13+22+14*+*15+17+23+19+20+34+21+23+24+25+26+27+30=369, that is odd,
> and 15+17+23+19+20+34+21+23+24+25+26+27+30=304, that is even and therefore
> no integer “x” can satisfy the system 369+x=p1 and 304+x=p2, with p1 and p2
> both prime.
>
>
>
> To avoid the incongruence I changed a(17)=14 (because it came after the
> choice of a(13)=18) with the minimum available integer greater than 14,
> that is 15.
>
> Of course this problem will arise again in the contruction process (two or
> more terms to be satisfied at the same point).
>
>
>
> I calculated 42 terms that should be OK:
>
>
> 1,2,3,4,5,8,6,10,7,12,9,11,18,16,13,22,15,14,17,23,19,20,34,21,24,25,26,33,27,40,32,42,29,28,30,46,31,35,36,48,37,49
>
>
>
> The sequence I found was:
>
> 1,2,3,4,5,8,6,10,7,12,9,11,18,16,13,22,15,14,17,23,19,20,34,21,24,25,26,33,27,40,32,42,29,28,30,46,31,35,36,48,37,49,39,38,44,41,43,45,63,47,54,50,56,51,52,57,62,53,55,58,65,96,69,59,
> 64, 60, 61, 70, 65, 67, 66, 68, 85, 71, 73, 72, 74, 79, 75, 76, 77, x…
>
>
>
> Again, the choice of x depends on a(36)=46 and a(43)=39 but
>
> 31+35+36+48+37+49+39+38+44+41+43+45+63+47+54+50+56+51+52+57+62+53+55+58+65+96+69+59+64+60+61+70+65+67+66+68+85+71+73+72+74+79+75+76+77=2636
> (even) and
> 38+44+41+43+45+63+47+54+50+56+51+52+57+62+53+55+58+65+96+69+59+64+60+61+70+65+67+66+68+85+71+73+72+74+79+75+76+77=2361
> (odd).
>
>
>
> Therefore a(43)=39 must be changed into a(43)=45. And so on.
>
>
>
> Is anybody willing to check/extend the numbers in this sequence? Thank in
>
>
>
> Paolo
>
>
> P.S. Variant for masochists: “Starting from a(1)=1, for any n the sum of
> the next a(n) digits is a prime. No repetition of numbers is allowed. At
> any step the minimum integer not yet used  and not leading to a
> contradiction should be choosen.”. 1,2,3,4,5,8,6,40,10,7,14,...
>
>
>

```