1,3

n is in the sequence iff the palindromic number 1(n).3.1(n) is prime (dot between numbers means concatenation). If n is a positive term of the sequence then n is not of the forms 3m, 6m+4, 12m+10, 28m+5, 28m+8, etc. (the proof is easy). 11 divides each palindromic number of the form 1(n).2.1(n) so there is no prime of this form.

C. Caldwell and H. Dubner, "Journal of Recreational Mathematics", Volume 28, No. 1, 1996-97, pp. 1-9.

Makoto Kamada, Factorizations of 11...11311...11

19 is in the sequence because the palindromic number (10^(2*19+1)+18*10^19-1)/9 = 1(19).3.1(19) = 111111111111111111131111111111111111111 is prime.

Do[If[PrimeQ[(10^(2n + 1) + 18*10^n - 1)/9], Print[n]], {n, 2500}]

Cf. A004023, A077775-A077798, A107123-A107127, A107648, A107649, A114633-A114647.

Sequence in context: A056005 A034572 A041393 this_sequence A055875 A089659 A101253

Adjacent sequences: A107120 A107121 A107122 this_sequence A107124 A107125 A107126

nonn

Farideh Firoozbakht (mymontain(AT)yahoo.com), May 19 2005