A259316 Numbers n such that the result of n multiplied by the reversal of n can be split into two numbers a and b of equal length (if the length is odd a leading zero is allowed), where a + b equals n (b can also have a leading zero).
1, 9, 54, 55, 99, 999, 2727, 3222, 7777, 8272, 9999, 12466, 22222, 25912, 39114, 75880, 87777, 87804, 93357, 99999, 124660, 142857, 181818, 185185, 189189, 230769, 231868, 324675, 390313, 412587, 428274, 443926, 503866, 513513, 533169, 568468
Offset: 1
Examples
124660 is a term. Indeed 124660*66421 = 8280041860 and 82800 + 41860 = 124660.
Links
- Pieter Post and Giovanni Resta, Table of n, a(n) for n = 1..187 (terms < 10^12, first 84 terms from Pieter Post)
Programs
-
Mathematica
fQ[n_] := Block[{c, d, len}, c = n FromDigits@ Reverse@ IntegerDigits@ n; d = IntegerDigits@ c; len = Length@ d; If[OddQ@ len, d = PadLeft[d, len + 1]; len++]; n == FromDigits@ Take[d, len/2] + FromDigits@ Take[d, -len/2]]; Select[Range@ 1000000, fQ] (* Michael De Vlieger, Jul 20 2015 *) -
Python
def sod(n,m): kk = 0 while n > 0: kk= kk+(n%m) n =int(n//m) return kk uu=1 for a in range (1,9): for n in range (10**(a-1)+1, 10**a): y=int(str(n)[::-1]) ll=int(len(str(n*y))/2+0.5) u=sod(n*y,10**ll) if n==u: print (n) -
Python
# for rep2-digit for f in range (12,98): aa=1 for i in range(1,200): aa=10**(2*i)+aa c=f*aa cc=str(c*int(str(c)[::-1])) l=int(len(cc)/2) cc1,cc2=int(cc[0:l]),int(cc[l:2*l+1]) if c==cc1+cc2: print (c)
Extensions
Missing a(21) from Giovanni Resta, Jul 19 2015
Comments