A023109 a(0) = 0. For n > 0, smallest non-palindromic number k such that the smallest palindrome in the Reverse and Add! trajectory of k is reached after exactly n iterations.

0, 10, 19, 59, 69, 166, 79, 188, 193, 1397, 829, 167, 2069, 1797, 849, 177, 1496, 739, 1798, 10777, 6999, 1297, 869, 187, 89, 10797, 10853, 10921, 10971, 13297, 10548, 13293, 17793, 20889, 700269, 106977, 108933, 80359, 13697, 10794, 15891, 1009227, 1007619, 1009246, 1008628, 600259, 131996, 70759, 1007377, 1001699, 600279, 141996, 70269, 10677, 10833, 10911

Offset: 0

Author

David W. Wilson

Keywords

• nonn
• base
• nice

Comments

From Felix Fröhlich, May 28 2022: (Start)

Variant of A015994 not allowing palindromes as starting values.

Smallest non-palindromic k such that A033665(k) = n. (End)

Links

• Chai Wah Wu, Table of n, a(n) for n = 0..100
• Jason Doucette, World Records
• Eric Weisstein's World of Mathematics, 196-Algorithm.
• Index entries for sequences related to Reverse and Add!

Crossrefs

Cf. A015994, A033665, A006960.

Programs

Mathematica

Table[ SelectFirst[Range[0, 20000], (np = #; i = 0;
    While[ ! PalindromeQ[np] && i <= n, np = np + IntegerReverse[np];
     i++]; i == n ) &] , {n, 0, 32}] (* Robert Price, Oct 16 2019 *)

PARI

rev(n)={d=digits(n);p="";for(i=1,#d,p=concat(Str(d[i]),p));return(eval(p))}
nbs(n)=if(n==rev(n),return(0));for(k=1,10^3,i=n+rev(n);if(rev(i)==i,return(k));n=i) \\ A033665
a(n)=for(k=1,10^8,if(nbs(k)==n,return(k)))
n=0;while(n<100,print1(a(n),", ");n++) \\ Derek Orr, Jul 28 2014

PARI

revadd(n) = n+eval(concat(Vecrev(Str(n))))
iterationstosmallestpalindrome(n, bound) = my(x=n, i=0, d); while(1, if(i > bound, return(-1)); x=revadd(x); i++; d=digits(x); if(d==Vecrev(d), return(i)))
a(n) = if(n==0, return(0)); for(k=1, oo, my(d=digits(k)); if(d!=Vecrev(d), if(iterationstosmallestpalindrome(k, n)==n, return(k)))) \\ Felix Fröhlich, May 28 2022

Python

def A023109(n):
    if n > 0:
        k = 0
        while True:
            m = k
            for i in range(n):
                if str(m) == str(m)[::-1]:
                    break
                m += int(str(m)[::-1])
            else:
                if str(m) == str(m)[::-1]:
                    return k
            k += 1
    else:
        return 0
# Chai Wah Wu, Feb 08 2015

Extensions

a(41)-a(55) verified and added by Aldo González Lorenzo, May 15 2011

Name edited by Felix Fröhlich, May 28 2022