A329407 Among the pairwise sums of any five consecutive terms there is exactly one prime sum; lexicographically earliest such sequence of distinct positive numbers.
1, 2, 7, 8, 13, 12, 14, 4, 20, 21, 6, 18, 15, 10, 3, 17, 5, 11, 16, 25, 9, 19, 23, 30, 26, 32, 22, 33, 24, 27, 28, 36, 29, 34, 35, 40, 31, 41, 37, 44, 38, 43, 39, 42, 45, 46, 47, 48, 49, 68, 51, 57, 54, 53, 61, 58, 62, 50, 52, 59, 56, 60, 55, 67, 63, 65, 66, 69, 75, 77, 64, 71, 70, 72, 73, 76, 74, 80
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 already have our prime sum. 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) = 12 and we have the single prime sum we need among the last 5 integers {2,7,8,13,12}, which is 19 = 12 + 7. And so on.
Links
- Jean-Marc Falcoz, Table of n, a(n) for n = 1..10000