A050782 Smallest positive multiplier m such that m*n is palindromic (or zero if no such m exists).

0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 21, 38, 18, 35, 17, 16, 14, 9, 0, 12, 1, 7, 29, 21, 19, 37, 9, 8, 0, 14, 66, 1, 8, 15, 7, 3, 13, 15, 0, 16, 6, 23, 1, 13, 9, 3, 44, 7, 0, 19, 13, 4, 518, 1, 11, 3, 4, 13, 0, 442, 7, 4, 33, 9, 1, 11, 4, 6, 0, 845, 88, 4, 3, 7, 287, 1, 11, 6, 0, 12345679, 8

Offset: 0

Author

Patrick De Geest, Oct 15 1999

Keywords

• nonn
• base
• nice

Comments

Multiples of 81 require the largest multipliers.

From Jon E. Schoenfield, Jan 15 2015: (Start)

In general, a(n) is large when n is a multiple of 81. E.g., for n in [1..10000], of the 9000 terms where a(n)>0, 111 are at indices n that are multiples of 81; of the remaining 8889 terms,

755 are in [1..9],

1760 are in [10..99],

3439 are in [100..999],

2180 are in [1000..9999],

708 are in [10000..99999],

36 are in [100000..999999],

6 are in [1000000..9999999],

2 are in [10000000..99999999],

2 are in [100000000..999999999],

and 1 (the largest) is a(8891) = 8546948927,

but the smallest of the 111 terms whose indices are multiples of 81 is a(2997)=333667. (End)

a(n) = 0 iff 10 | n. a(n) = 1 iff n is a palindrome. If k | a(n) then a(k*n) = a(n)/k. - Robert Israel, Jan 15 2015

Examples

			E.g., a(81) -> 81 * 12345679 = 999999999 and a palindrome.
		

Links

• Giovanni Resta, Table of n, a(n) for n = 0..10000 (first 8181 terms from Chai Wah Wu)
• Patrick De Geest, World!Of Numbers

Crossrefs

Cf. A002113, A020485, A050810.

Programs

Maple

digrev:= proc(n) local L,d,i;
  L:= convert(n,base,10);
  d:= nops(L);
  add(L[i]*10^(d-i),i=1..d);
end proc:
f:= proc(n)
local d,d2,x,t,y;
if n mod 10 = 0 then return 0 fi;
if n < 10 then return 1 fi;
for d from 2 do
  if d::even then
    d2:= d/2;
    for x from 10^(d2-1) to 10^d2-1 do
       t:= x*10^d2 + digrev(x);
       if t mod n = 0 then return(t/n) fi;
    od
  else
    d2:= (d-1)/2;
    for x from 10^(d2-1) to 10^d2-1 do
      for y from 0 to 9 do
        t:= x*10^(d2+1)+y*10^d2+digrev(x);
        if t mod n = 0 then return(t/n) fi;
      od
    od
  fi
od;
end proc:
seq(f(n),n=0 .. 100); # Robert Israel, Jan 15 2015

Mathematica

t={0}; Do[i=1; If[IntegerQ[n/10],y=0,While[Reverse[x=IntegerDigits[i*n]]!=x,i++]; y=i]; AppendTo[t,y],{n,80}]; t (* Jayanta Basu, Jun 01 2013 *)

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 # Chai Wah Wu, Dec 30 2014