1,1

3*a(n) are the indices of the middle terms in the triplets of consecutive zeros in A125162(n) = {1,1,1,1,3,1,4,0,1,1,5,1,3,0,1,1,6,1,7,0,1,1,6,0,1,0,1,1,6,1,9,0,0,0,3,1,11,0,1,1,9,1,5,0,1,1,10,0,2,0,1,1,9,0,2,0,1,1,10,1,9,0,0,0,3,1,9,0,1,1,8,1,9,0,0,0,5,1,9,0,1,1,11,0,1,0,1,1,8,0,3,0,0,0,2,1,10,0,1,1,...} Number of primes of the form k! + n. Indices of zeros in A125162(n) are listed in A125163(n) = {8,14,20,24,26,32,33,34,38,44,48,50,54,56,62,63,64,68,74,75,76,80,84,86,90,92,93,94,98,...} Numbers n such that no prime exists of the form k! + n. Note the triplets of consecutive terms in A125163(n): {32,33,34}, {62,63,64}, {74,75,76}, {92,93,94), {116,117,118}, {122,123,124}, {140,141,142}, {152,153,154}, {158,159,160}, {182,183,184}, {200,201,202}, {206,207,208}, {212,213,214}, {218,219,220}, {242,243,244}, {272,273,274}, {284,285,286}. It appears that all listed primes in a(n) belong to A115058(n) = {2, 11, 31, 41, 47, 53, 61, 67, 71, 73, 101, ...} Primes p that are also the largest prime factor of p(p^2-1)(3p+2)/24.

Cf. A125162 = number of primes of the form k! + n. Cf. A125163 = numbers n such that no prime exists of the form k! + n. Cf. A115058.

Sequence in context: A117841 A105956 A050718 this_sequence A101223 A109686 A077522

Adjacent sequences: A125161 A125162 A125163 this_sequence A125165 A125166 A125167

hard,more,nonn

Alexander Adamchuk (alex(AT)kolmogorov.com), Nov 22 2006