1,2

a(n+1) = (n+1)(10^a(n)-1)/9.

a(1) says: there will be one 2 in a(2).

a(2)=2 because a(1) said so; and a(2)=2 says: there will be two 3's in a(3).

a(3)=33 because a(2) said so; and also a(3) says: there will be thirty three 4's in a(4).

Therefore a(4)= 444444444444444444444444444444444 (33 times the digit 4).

And a(5)= 555555555555555...555 (with 444444444444444444444444444444444 '5s).

Cf. A001462, A001463, A103320, A102357, A076782.

Adjacent sequences: A109333 A109334 A109335 this_sequence A109337 A109338 A109339

Sequence in context: A117969 A003820 A112980 this_sequence A098869 A131544 A123284

base,nonn

Alexandre Wajnberg (alexandre.wajnberg(AT)ulb.ac.be), Aug 23 2005