

A060382


In base n, a(n) is the smallest number m that leads to a palindromefree sequence, using the following process: start with m; reverse the digits and add it to m, repeat. Stop if you reach a palindrome.


2



22, 103, 290, 708, 1079, 2656, 1021, 593, 196, 1011, 237, 2701, 361, 447, 413, 3297, 519, 341, 379, 711, 461, 505, 551, 1022, 649, 701, 755, 811, 869, 929, 991, 1055, 1799, 1922, 1259, 1331, 1405, 1481, 1559, 1639, 1595, 1762, 1891, 1934, 2069, 2161
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OFFSET

2,1


COMMENTS

Only a(2) is proved, all the others are conjectured.  Eric Chen, Apr 20 2015 [corrected by A.H.M. Smeets, May 27 2019]
Brown's link reports a(3) as 103 instead of 100. What is the correct value? Dmitry Kamenetsky, Mar 06 2017 [a(3) = 103 is correct as from A077404, A.H.M. Smeets, May 27 2019]
From A.H.M. Smeets, May 27 2019: (Start)
It seems that a(n) < n^2 (i.e., a(n) in base n has two digits) and the least significant digit of a(n) in base n equals n1, for n > 73.
For n <= 73 and the least significant digit of a(n) in base n is unequal to n1, then the most significant digit of a(n) in base n equals 1.
From this it seems that, the least significant digit of a(n) in base n equals n1 or the most significant digit of a(n) in base n equals 1, holds for all n > 1.
For n > 305 it seems that a(n) < n^2  n  1.
It seems that a(n) >= n*floor(3*n/4)1; i.e. for any a(n) which is represented by a twodigit number in base n, the most significant digit is at least floor(3*n/4)1. (End)
From A.H.M. Smeets, May 30 2019: (Start)
a(n) is a 5 digit number in base n representation for n in {2,3,4,5,7}.
a(n) is a 4 digit number in base n representation for n in {6,8,13}.
a(n) is a 3 digit number in base n representation for n in {9,10,11,12,14,15,16,17,18,21,25,34,35,52,71,72,73}.
For all other bases n, a(n) is a 2 digit number in base n representation.
If a(n) = n*floor(3*n/4)1, then n == 0 (mod 4) or n == 3 (mod 4). (End)


LINKS

A.H.M. Smeets, Table of n, a(n) for n = 2..20000 (terms 2..3428 from Karl Hovekamp, with some corrections)
K. S. Brown, Digit Reversal Sums Leading to Palindromes
A.H.M. Smeets, Lists of first 20 Lychrel numbers for bases n <= 1000
A.H.M. Smeets, Scatterplot of log_10(a(n)/n^2) versus n for 73 < n <= 20000


EXAMPLE

a(2) = 22 since A062129(k) > 1 (equivalently, A062131(k) > 1) for k < 22.


PROG

(Python)
def rev(n, base):
....m = 0
....while n > 0:
........n, m = n//base, m*base+n%base
....return m
n, a, steps = 2, 3, 0
while n <= 20000:
....aa = a
....ra = rev(a, n)
....while aa != ra and steps < 1000:
........aa = aa+ra
........ra, steps = rev(aa, n), steps+1
....if aa == ra:
........a, aa, steps = a+1, a+1, 0
....if steps == 1000:
........print(n, a)
........n, a, steps = n+1, n+2, 0 # A.H.M. Smeets, May 27 2019


CROSSREFS

For the first palindrome in nonpalindromefree sequences, cf. A062129/A062131 (base 2), A033865 (base 10), A253241 (base 12).
Sequence in context: A044654 A156795 A095265 * A066450 A231225 A124950
Adjacent sequences: A060379 A060380 A060381 * A060383 A060384 A060385


KEYWORD

nonn,base


AUTHOR

Michel ten Voorde, Apr 03 2001


EXTENSIONS

More terms from Karl Hovekamp, Jan 03 2007


STATUS

approved



