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A329416 Among the pairwise sums of any ten consecutive terms there are exactly two prime sums: lexicographically earliest such sequence of distinct positive numbers.

%I A329416 #22 Feb 09 2020 16:10:38
%S A329416 1,2,3,7,13,19,23,25,31,32,17,8,26,37,43,49,14,38,55,61,11,20,35,67,
%T A329416 73,79,57,9,5,15,21,42,27,12,33,30,39,45,47,18,48,6,51,24,63,69,72,75,
%U A329416 16,36,54,60,22,66,10,4,40,29,28,34,44,41,46,50,52,58,64,53,70,71,59,62,76,56,82,88,94,65,100
%N A329416 Among the pairwise sums of any ten consecutive terms there are exactly two prime sums: lexicographically earliest such sequence of distinct positive numbers.
%C A329416 Conjectured to be a permutation of the positive integers: a(10^6) = 10^6 + 2 and all numbers up to 10^6 - 7 have appeared at that point. - _M. F. Hasler_, Nov 15 2019
%H A329416 Jean-Marc Falcoz, <a href="/A329416/b329416.txt">Table of n, a(n) for n = 1..10000</a>
%e A329416 a(1) = 1 is the smallest possible choice, there's no restriction on the first term.
%e A329416 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 10-set {1,2,a(3),a(4),a(5),a(6),a(7),a(8),a(9),a(10)}.
%e A329416 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 10-set {1,2,a(3),a(4),a(5),a(6),a(7),a(8),a(9),a(10)}.
%e A329416 a(4) = 7 as a(4) = 4, 5 or 6 would lead to a contradiction: indeed, the 10-sets {1,2,3,4,a(5),a(6),a(7),a(8),a(9),a(10)}, {1,2,3,5,a(5),a(6),a(7),a(8),a(9),a(10)} and {1,2,3,6,a(5),a(6),a(7),a(8),a(9),a(10)} will produce more than the two required prime sums. With a(4) = 7 we have no contradiction as the 10-set {1,2,3,7,a(5),a(6),a(7),a(8),a(9),a(10)} has now two prime sums so far: 1 + 2 = 3 and 2 + 3 = 5.
%e A329416 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 10-set); in combination with any other term before it, a(5) = 13 will produce only composite sums.
%e A329416 a(6) = 19 as 19 is the smallest available integer not leading to a contradiction: indeed, the 10-set {1,2,3,7,13,19,a(7),a(8),a(9),a(10)} shows two prime sums so far: 1 + 2 = 3 and 2 + 3 = 5.
%e A329416 a(7) = 23 as 23 is the smallest available integer not leading to a contradiction; indeed, the 10-set {1,2,3,7,13,19,23,a(8),a(9),a(10)} shows only two prime sums so far, which are 1 + 2 = 3 and 2 + 3 = 5.
%e A329416 a(8) = 25 as 25 is the smallest available integer not leading to a contradiction and producing two prime sums so far with the 10-set {1,2,3,7,13,19,23,25,a(9),a(10)}; etc.
%o A329416 (PARI) A329416(n, show=0, o=1, N=2, M=9, 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])); print([u]); o} \\ Optional args: show=1: print terms a(o..n-1); o=0: start with a(0)=0; N, M: produce N primes using M+1 consecutive terms. - _M. F. Hasler_, Nov 15 2019
%Y A329416 Cf. A329333 (3 consecutive terms, exactly 1 prime sum).
%Y A329416 Cf. A329405 (no prime among the pairwise sums of 3 consecutive terms).
%Y A329416 Cf. A329406 .. A329410 (exactly 1 prime sum using 4, ..., 10 consecutive terms).
%Y A329416 Cf. A329411 .. A329415 (exactly 2 prime sums using 3, ..., 7 consecutive terms).
%Y A329416 See also "nonnegative" variants: A329450 (0 primes using 3 terms), A329452 (2 primes using 4 terms), A329453 (2 primes using 5 terms), A329454 (3 primes using 4 terms), A329449 (4 primes using 4 terms), A329455 (3 primes using 5 terms), A329456 (4 primes using 5 terms).
%K A329416 nonn
%O A329416 1,2
%A A329416 _Eric Angelini_ and _Jean-Marc Falcoz_, Nov 14 2019