A329410 Among the pairwise sums of any ten consecutive terms there is exactly one prime sum: lexicographically earliest such sequence of distinct positive numbers.
1, 2, 7, 8, 13, 14, 19, 20, 25, 26, 108, 32, 37, 38, 44, 50, 10, 40, 12, 18, 28, 48, 105, 6, 4, 16, 24, 30, 36, 42, 54, 56, 9, 46, 22, 60, 66, 68, 72, 76, 78, 82, 93, 34, 52, 62, 43, 83, 92, 102, 23, 53, 3, 29, 31, 27, 33, 41, 15, 88, 5, 11, 17, 35, 45, 47, 55, 57, 21, 64, 51, 59, 61, 65, 39, 69, 71, 77, 79, 136, 49, 67, 63
Offset: 1
Keywords
Examples
a(1) = 1 by minimality.
a(2) = 2 as 2 is the smallest available integer not leading to a contradiction. Note that as 1 + 2 = 3 we have already the prime sum we need.
a(3) = 7 as a(3) = 3, 4, 5 or 6 would produce at least one prime sum too many.
a(4) = 8 as a(4) = 3, 4, 5 or 6 would again produce at least one prime sum too many.
a(5) = 13 as a(5) = 3, 4, 5, 6, 9, 10, 11 or 12 would also produce at least one prime sum too many.
a(6) = 14 as a(6) = 14 doesn't produce an extra prime sum - only composite sums.
a(7) = 19 as a(7) = 3, 4, 5, 6, 9, 10, 11, 12, 15, 16, 17 or 18 would produce at least a prime sum too many.
a(8) = 20 as a(8) = 20 doesn't produce an extra prime sum - only composite sums.
a(9) = 25 as a(9) = 3, 4, 5, 6, 9, 10, 11, 12, 15, 16, 17, 18, 21, 22, 23 or 24 would produce at least a prime sum too many.
a(10) = 26 as(10) = 26 doesn't produce an extra prime sum - only composite sums.
a(11) = 108 is the smallest available integer that produces the single prime sum we need among the last 10 integers {2,7,8,13,14,19,20,25,26,108}, which is 127 = 108 + 19.
And so on.
Links
- Jean-Marc Falcoz, Table of n, a(n) for n = 1..10000