A061906 Obtain m by omitting trailing zeros from n; a(n) = smallest k such that k*m is a palindrome.

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 21, 38, 18, 35, 17, 16, 14, 9, 1, 12, 1, 7, 29, 21, 19, 37, 9, 8, 1, 14, 66, 1, 8, 15, 7, 3, 13, 15, 1, 16, 6, 23, 1, 13, 9, 3, 44, 7, 1, 19, 13, 4, 518, 1, 11, 3, 4, 13, 1, 442, 7, 4, 33, 9, 1, 11, 4, 6, 1, 845, 88, 4, 3, 7, 287, 1, 11, 6, 1, 12345679, 8

Offset: 0

Author

Klaus Brockhaus, Jun 25 2001

Keywords

• base
• easy
• nonn

Comments

Every positive integer is a factor of a palindrome, unless it is a multiple of 10 (D. G. Radcliffe, see Links).

Every integer n has a multiple of the form 99...9900...00. To see that n has a multiple that's a palindrome (allowing 0's on the left) with even digits, let 9n divide 99...9900...00; then n divides 22...2200...00. - Dean Hickerson, Jun 29 2001

Examples

			For n = 30 we have m = 3, 1*m = 3 is a palindrome, so a(30) = 1. For n = m = 12 the smallest palindromic multiple is 21*m = 252, so a(12) = 21.
		

Links

• Chai Wah Wu, Table of n, a(n) for n = 0..8180
• P. De Geest, Smallest multipliers to make a number palindromic.

Crossrefs

Cf. A050782, A062293, A061915, A061916, A061816. Values of k*m are given in A061906.

Programs

ARIBAS

stop := 20000000; for n := 0 to maxarg do k := 1; test := true; while test and k < stop do mp := omit_trailzeros(n)*k; if test := mp <> int_reverse(mp) then inc(k); end; end; if k < stop then write(k," "); else write(-1," "); end; end;

Mathematica

skp[n_]:=Module[{m=n/10^IntegerExponent[n,10],k=1},While[!PalindromeQ[k*m],k++];k]; Array[ skp,90,0] (* Harvey P. Dale, Jul 04 2024 *)

Python

from _future_ import division
def palgen(l, b=10): # generator of palindromes in base b of length <= 2*l
    if l > 0:
        yield 0
        for x in range(1, l+1):
            n = b**(x-1)
            n2 = n*b
            for y in range(n, n2):
                k, m = y//b, 0
                while k >= b:
                    k, r = divmod(k, b)
                    m = b*m + r
                yield y*n + b*m + k
            for y in range(n, n2):
                k, m = y, 0
                while k >= b:
                    k, r = divmod(k, b)
                    m = b*m + r
                yield y*n2 + b*m + k
def A050782(n, l=10):
    if n % 10:
        x = palgen(l)
        next(x)  # replace with x.next() in Python 2.x
        for i in x:
            q, r = divmod(i, n)
            if not r:
                return q
        else:
            return 'search limit reached.'
    else:
        return 0
def A061906(n, l=10):
    return A050782(int(str(n).rstrip('0')),l) if n > 0 else 1
# Chai Wah Wu, Dec 30 2014