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%I #4 Dec 05 2013 19:56:53
%S 1,3,7,9,21,13
%N a(1) = 1, a(2) = 3; for n > 2: a(n) = smallest (odd) number not occurring earlier such that the sum of each section of odd length >=3 is prime.
%C If 1, 3, 7, 13 are taken (rather arbitrarily) as starting terms, then the continuation is 17, 31, 11, 25, 5, 37, 341, 163, 647, 571, 989, 3451, 17669, 206413, 6767, 252289, but no number < 10000000 is suited to continue this sequence further.
%C There are no further terms. For k to qualify as next term the sums 21+13+k, 7+9+21+13+k and 1+3+7+9+21+13+k have to be prime. One of these sums however is divisible by 3, since 34+k = k+1 (mod 3), 50+k = k+2 (mod 3) and 54+k = k (mod 3). - Klaus Brockhaus, Jul 02 2004
%e 1+3+7 = 11, 3+7+9 = 19, 7+9+21 = 37, 9+21+13 = 43, 1+3+7+9+21 = 41, 3+7+9+21+13 = 53 are all prime.
%Y Cf. A096100, A096101.
%K nonn
%O 1,2
%A _Amarnath Murthy_, Jun 24 2004
%E Edited and corrected by _Klaus Brockhaus_, Jun 29 2004
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