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%I A329413 #31 Apr 11 2022 22:21:26
%S A329413 1,2,3,7,13,5,8,9,17,16,4,6,11,12,10,14,15,21,18,19,20,30,22,24,29,26,
%T A329413 23,25,36,32,33,27,28,37,31,39,35,34,38,42,44,41,40,43,45,46,47,50,52,
%U A329413 49,65,53,51,54,55,57,48,60,56,59,61,71,70,67,58,64,62,63,68,66,73,72,69,76,75,74,78,80
%N A329413 Lexicographically earliest sequence of distinct positive numbers such that among the pairwise sums of any five consecutive terms there are exactly two prime sums.
%C A329413 Conjectured to be a permutation of the positive integers: a(10^6) = 10^6 + 9 and all numbers below 10^6 - 7 are used at that point. - _M. F. Hasler_, Nov 15 2019
%H A329413 Jean-Marc Falcoz, <a href="/A329413/b329413.txt">Table of n, a(n) for n = 1..10000</a>
%e A329413 a(1) = 1 is the smallest possible choice for the first term.
%e A329413 a(2) = 2 as 2 is the smallest available integer not leading to a contradiction. Note that as 1 + 2 = 3 we already have one prime sum (on the required two) with the 5-tuple {1,2,a(3),a(4),a(5)}.
%e A329413 a(3) = 3 as 3 is the smallest available integer not leading to a contradiction. Note that as 2 + 3 = 5 we now have the two prime sums required with the 5-tuple {1,2,3,a(4),a(5)}.
%e A329413 a(4) = 7 as a(4) = 4, 5 or 6 would lead to a contradiction: indeed, the quintuplets {1,2,3,4,a(5)}, {1,2,3,5,a(5)} and {1,2,3,6,a(5)} will produce more than the two required prime sums. With a(4) = 7 we have no contradiction as the 5-tuple {1,2,3,7,a(5)} has now exactly two prime sums: 1 + 2 = 3 and 2 + 3 = 5.
%e A329413 a(5) = 13 as a(5) = 4, 5, 6, 8, 9, 10, 11 or 12 would again lead to a contradiction (more than 2 prime sums with the 5-tuple); in combination with any other term before it, a(5) = 13 will produce only composite sums.
%e A329413 a(6) = 5 as 5 is the smallest available integer not leading to a contradiction: indeed, the 5-tuple {2,3,7,13,5} shows exactly the two prime sums we are looking for: 2 + 3 = 5 and 2 + 5 = 7.
%e A329413 And so on.
%o A329413 (PARI) A329413(n,show=0,o=1,N=2,M=4,p=[],U,u=o)={for(n=o,n-1, show&&print1(o","); U+=1<<(o-u); U>>=-u+u+=valuation(U+1,2); p=concat(if(#p>=M,p[^1],p),o); my(c=N-sum(i=2,#p,sum(j=1,i-1,isprime(p[i]+p[j])))); if(#p<M && sum(i=1,#p,isprime(p[i]+u))<=c, o=u)|| for(k=u,oo, bittest(U,k-u) || sum(i=1,#p,isprime(p[i]+k))!=c || [o=k, break]));o} \\ Optional args: show=1: print a(o..n-1); o=0: start with a(0) = 0 (A329453), N, M: produce N primes using M+1 consecutive terms. - _M. F. Hasler_, Nov 15 2019
%Y A329413 Cf. A329333 (3 consecutive terms, exactly 1 prime sum). See also A329450, A329452 onwards.
%K A329413 nonn
%O A329413 1,2
%A A329413 _Eric Angelini_ and _Jean-Marc Falcoz_, Nov 14 2019

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