[seqfan] Headache sequence

[Paolo Lava]

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
advance.



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,...