OFFSET

3,2

COMMENTS

A positive integer that is a multiple of 3 ends with 0 in base 3, so it cannot be a palindrome in base 3.

A positive integer that is a multiple of 9 ends with 0 in base 9, so it cannot be a palindrome in base 9.

From Michael S. Branicky, Aug 15 2024: (Start)

Regarding a(2): To be a palindrome in base 2, it must end with 1, hence odd. To be odd and have digit sum 2 in base 10, it must be of the form t_d = 10^(d-1) + 1, d > 1 (a d-digit base-10 number). t_d is not divisible by 3, and base-2 palindromes with even length (i.e., number of binary digits) are divisible by 3, so, if a(2) exists, it must be a base-2 palindrome with odd length.

Computer search shows no such terms with d <= 10^6, so a(2), if it exists, has > 10^6 decimal digits. (End)

LINKS

Michael S. Branicky, Table of n, a(n) for n = 3..149

Michael S. Branicky, Python programs for OEIS A375387

Jean-Marc Rebert, a375387_1e10

EXAMPLE

a(5) = 41, because 4 + 1 = 5 and 41 = 131_5, and no lesser number has this property.

First terms are:

130 = 2002_4

41 = 131_5

123 = 3323_6

16 = 22_7

170 = 252_8

PROG

(PARI) isok(k, n) = if (sumdigits(k)==n, my(d=digits(k, n)); d==Vecrev(d));

a(n) = if ((n==3) || (n==9), return((-1))); my(k=1); while (!isok(k, n), k++); k; \\ Michel Marcus, Aug 13 2024

(Python) # see Links for faster variants

from itertools import count

from sympy.ntheory import is_palindromic

def a(n):

if n in {3, 9}: return -1

return next(k for k in count(10**(n//9)-1) if sum(map(int, str(k)))==n and is_palindromic(k, n))

print([a(n) for n in range(3, 47)]) # Michael S. Branicky, Aug 13 2024

CROSSREFS

KEYWORD

sign,base

AUTHOR

Jean-Marc Rebert, Aug 13 2024

STATUS

approved