A225538 Let r(n) denote the reverse of n. For every n, consider the sequence n_1 = n + 1 + r(n+1), and for m >= 2, n_m = n_(m-1) + 1 + r(n_(m-1) + 1). a(n) is the least m for which n_m is a palindrome, or 0 if there is no such m.

1, 1, 1, 1, 2, 2, 2, 2, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 2, 2, 1, 1, 1, 1, 1, 2, 1, 2, 2, 4, 1, 1, 1, 1, 2, 1, 2, 2, 4, 7, 1, 1, 1, 2, 1, 2, 2, 4, 7, 10, 1, 1, 2, 1, 2, 2, 4, 7

Offset: 0

Author

Vladimir Shevelev, May 10 2013

Keywords

• nonn
• base

Comments

Conjecture: the least n's for which a(n) = 0 are 1895, 1985, 2894, 2984, 3893, and 3983. - Peter J. C. Moses, May 10 2013

See analogous numbers in A023108 for which the so-called Lychrel process "Reverse and Add!", apparently, never leads to a palindrome.

Examples

			For n=8, 9 + 9 = 18, 19 + 91 = 110, 111 + 111 = 222 is a palindrome. Thus a(8)=3.
		

Links

• Peter J. C. Moses, Table of n, a(n) for n = 0..5000

Crossrefs

Cf. A023108.