

A329416


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


14



1, 2, 3, 7, 13, 19, 23, 25, 31, 32, 17, 8, 26, 37, 43, 49, 14, 38, 55, 61, 11, 20, 35, 67, 73, 79, 57, 9, 5, 15, 21, 42, 27, 12, 33, 30, 39, 45, 47, 18, 48, 6, 51, 24, 63, 69, 72, 75, 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
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OFFSET

1,2


COMMENTS

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


LINKS

JeanMarc Falcoz, Table of n, a(n) for n = 1..10000


EXAMPLE

a(1) = 1 is the smallest possible choice, there's no restriction on the first term.
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 10set {1,2,a(3),a(4),a(5),a(6),a(7),a(8),a(9),a(10)}.
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 10set {1,2,a(3),a(4),a(5),a(6),a(7),a(8),a(9),a(10)}.
a(4) = 7 as a(4) = 4, 5 or 6 would lead to a contradiction: indeed, the 10sets {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 10set {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.
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 10set); in combination with any other term before it, a(5) = 13 will produce only composite sums.
a(6) = 19 as 19 is the smallest available integer not leading to a contradiction: indeed, the 10set {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.
a(7) = 23 as 23 is the smallest available integer not leading to a contradiction; indeed, the 10set {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.
a(8) = 25 as 25 is the smallest available integer not leading to a contradiction and producing two prime sums so far with the 10set {1,2,3,7,13,19,23,25,a(9),a(10)}; etc.


PROG

(PARI) A329416(n, show=0, o=1, N=2, M=9, p=[], U, u=o)={for(n=o, n1, show&&print1(o", "); U+=1<<(ou); U>>=u+u+=valuation(U+1, 2); p=concat(if(#p>=M, p[^1], p), o); my(c=Nsum(i=2, #p, sum(j=1, i1, 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, ku) sum(i=1, #p, isprime(p[i]+k))!=c[o=k, break])); print([u]); o} \\ Optional args: show=1: print terms a(o..n1); o=0: start with a(0)=0; N, M: produce N primes using M+1 consecutive terms.  M. F. Hasler, Nov 15 2019


CROSSREFS

Cf. A329333 (3 consecutive terms, exactly 1 prime sum).
Cf. A329405 (no prime among the pairwise sums of 3 consecutive terms).
Cf. A329406 .. A329410 (exactly 1 prime sum using 4, ..., 10 consecutive terms).
Cf. A329411 .. A329415 (exactly 2 prime sums using 3, ..., 7 consecutive terms).
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).
Sequence in context: A068948 A329414 A329415 * A155479 A019411 A105792
Adjacent sequences: A329413 A329414 A329415 * A329417 A329418 A329419


KEYWORD

nonn


AUTHOR

Eric Angelini and JeanMarc Falcoz, Nov 14 2019


STATUS

approved



