A383887 Smallest non-palindromic number that is congruent to its reverse mod n.

10, 13, 10, 15, 16, 13, 18, 19, 10, 1011, 100, 15, 1017, 1027, 16, 1025, 1039, 13, 1048, 1021, 18, 103, 1026, 19, 1026, 1017, 14, 1033, 1013, 1011, 1068, 1049, 100, 1039, 1046, 15, 1000, 1055, 1017, 1041, 1066, 1027, 1048, 105, 16, 1077, 1032, 1025, 1014, 1051, 1039, 1017, 1103, 17, 106, 1065

Offset

1,1

Comments

Comparable to A070837, but such a number provably exists: if n is coprime to 10 then take 10^k where k is the order of 10 mod n; else take a>b>0 such that n divides 10^a - 10^b, then the number 10^(a+b) + 10^b + 1 works.

The n-th term in this sequence is equal to the first nonzero term of A070837 that occurs at an index divisible by n/gcd(n,9).

Links

Erick B. Wong, Table of n, a(n) for n = 1..10000

Example

For n=6, a(6)=13 is congruent to 31 mod 6.
For n=10, note that any number of 3 or fewer digits is necessarily a palindrome if the first digit equals the last, and 1011 is the first 4-digit non-palindrome.

Mathematica

seq[len_] := Module[{s = Table[0, {len}], c = 0, k = 9, d}, While[c < len, k++; krev = IntegerReverse[k]; If[k != krev, d = Select[Divisors[Abs[k - krev]], # <= len &]; Do[If[s[[d[[i]]]] == 0, s[[d[[i]]]] = k; c++], {i, 1, Length[d]}]]]; s]; seq[60] (* Amiram Eldar, May 31 2025 *)

Prog

(Python)
from functools import partial
from itertools import count
def accept(mod, k):
    r = int(str(k)[::-1])
    return r != k and (r-k) % mod == 0
def a(n):
    return next(filter(partial(accept, n), count(10)))
print([a(n) for n in range(1, 57)])
(PARI) a(n) = my(k=1, d=digits(k), rd=Vecrev(d)); while(!((d != rd) && Mod(fromdigits(rd), n) == k), k++; d=digits(k); rd=Vecrev(d)); k; \\ Michel Marcus, May 30 2025

Crossrefs

Cf. A004086, A056965, A070837.

Sequence in context: A332478 A144814 A378488 * A241174 A092887 A131365

Adjacent sequences: A383884 A383885 A383886 * A383888 A383889 A383890

Keyword

nonn,base,easy

Author

Erick B. Wong, May 29 2025, at the suggestion of Lanny Wong

Status

approved