1,1

The sequence is not monotonic and it seems that p=5 may arise at any position > 2. a[2]=0 means that 5 is never the 2nd term in an EM sequence of A000945-type because a[2]=2 or 3.

a(25) > 8000. a(26) = 383. a(27) = 6971. - David Wasserman (dwasserm(AT)earthlink.net), Apr 20 2007

The sequence for 17 is 17, 2, 5, ... where the 5 is at the third place, therefore a(3)=17.

For n=15 we have the sequence 181, 2, 3, 1087, 73, 7, 29, 151, 61, 98689, 11, 10929259909, 678859, 97, 5, ...

a(16) = 449 uses the sequence 449, 2, 29, 3, 7, 349, 190861819, 166273, 16091, 11, 3807491, 53, 17, 313, 23, 5, ...

The sequence for 11 is 11, 2, 23, 3, 7, 13, 10805892983887, 73, 6397, 19, 489407, 2753, 87491, 18618443, 5, ... with the 5 at the 18th place, so a(18)=11.

Cf. A000945, A051308-A051334, A056756, A093777-A093781.

Sequence in context: A167297 A027635 A022920 this_sequence A085105 A094031 A056461

Adjacent sequences: A093779 A093780 A093781 this_sequence A093783 A093784 A093785

more,nonn

Labos E. (labos(AT)ana.sote.hu), May 04 2004

Corrected by R. J. Mathar, Oct 06 2006

a(16) = 449 was conjectured by Mathar and confirmed by Don Reble, Oct 07 2006

a(19)-a(24) from David Wasserman (dwasserm(AT)earthlink.net), Apr 20 2007