A241173 Start with n; add to it any of its digits; repeat; a(n) = minimal number of steps needed to reach a palindrome.

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 3, 2, 3, 2, 1, 4, 3, 2, 1, 1, 0, 3, 3, 2, 3, 3, 2, 4, 1, 3, 3, 0, 2, 2, 2, 1, 3, 2, 1, 2, 1, 3, 0, 2, 2, 3, 4, 2, 1, 4, 3, 2, 2, 0, 4, 3, 1, 3, 1, 4, 3, 1, 2, 2, 0, 3, 4, 3, 1, 3, 2, 2, 4, 2, 3, 0, 3, 1, 1, 3, 2, 3, 1, 2, 2, 3, 0, 5, 1, 2, 1, 4, 5, 2, 3, 4, 5, 0

Offset: 0

Author

N. J. A. Sloane, Apr 23 2014

Keywords

• nonn
• base

Comments

a(n) = 0 iff n is already a palindrome (A002113).

Is it a theorem that a(n) always exists?

a(n) always exists. Proof: A palindrome can be reached by simply adding the initial digit until a palindrome with the same number of digits as the initial number is reached: If no palindrome is reached by then, this will yield a number with initial digit '1'. Thereafter, this procedure will yield the next larger palindrome - either not larger than 19...91 or, after 19...9 + 1 = 20...0, at 20...02. - M. F. Hasler, Apr 26 2014

Examples

			Examples for a(10) through a(23):
a(10) = 1 via 10 -> 11
a(11) = 0 via 11
a(12) = 3 via 12 -> 13 -> 16 -> 22
a(13) = 2 via 13 -> 16 -> 22
a(14) = 3 via 14 -> 15 -> 16 -> 22
a(15) = 2 via 15 -> 16 -> 22
a(16) = 1 via 16 -> 22
a(17) = 4 via 17 -> 18 -> 19 -> 20 -> 22
a(18) = 3 via 18 -> 19 -> 20 -> 22
a(19) = 2 via 19 -> 20 -> 22
a(20) = 1 via 20 -> 22
a(21) = 1 via 21 -> 22
a(22) = 0 via 22
a(23) = 3 via 23 -> 25 -> 30 -> 33
		

References

• Eric Angelini, Posting to Sequence Fans Mailing List, Apr 20 2014

Links

• David A. Corneth, Table of n, a(n) for n = 0..9999
• David A. Corneth, PARI program
• David A. Corneth, Steps taken from n as described in name to reach a palindrome

Crossrefs

Cf. A002113.

A241174 gives the smallest number that takes n steps to reach a palindrome.

Related sequences: A241173, A241174, A241175, A241176, A241177, A241178, A241179, A241180, A241181, A241182, A241183.

Programs

Mathematica

A241173[n_] := Module[{c, nx},
   If[n == IntegerReverse[n], Return[0]];
   c = 1; nx = n;
   While[ ! AnyTrue[nx = Flatten[nx + IntegerDigits[nx]], # == IntegerReverse[#] &], c++];
   Return[c]];
Table[A241173[i], {i, 0, 100}] (* Robert Price, Mar 17 2019 *)

PARI

a(n,m=0)={ if( m, my(d); for(i=1,#d=vecsort(digits(n),,12), d[i] && if( m>1, a(n+d[i],m-1) /*&& !print1("/*",[n,d[i],m],"* /")*/, is_A002113(n+d[i])) && return(m)), is_A002113(n) || until(a(n,m++),); m)} \\ Memoization should be implemented to improve performance which remains poor esp. for terms just above 10^k+1. - M. F. Hasler, Apr 26 2014

PARI

\\ See Corneth link; faster than above. David A. Corneth, Mar 21 2019

Extensions

More terms from M. F. Hasler, Apr 26 2014