A229546 Numbers n such that n + product_of_digits(n) is a palindrome.

0, 1, 2, 3, 4, 16, 28, 39, 43, 64, 89, 101, 127, 163, 166, 174, 179, 188, 202, 214, 236, 247, 252, 296, 303, 329, 341, 348, 354, 359, 366, 372, 385, 387, 393, 404, 426, 442, 445, 455, 463, 465, 489, 505, 525, 536, 546, 567, 568, 571, 578, 589, 591, 606, 618, 622, 629, 658, 659, 664, 667, 707, 734, 749, 753, 808, 812

Offset: 1

Author

Derek Orr, Sep 26 2013

Keywords

• nonn
• base
• easy

Comments

From Derek Orr, Mar 22 2015 (Start):

The density of these numbers is roughly steady for 10^(2*k-1) < a(n) < 10^(2*k+1) for k = 1, 2, 3, ...

Examples:

k = 1: For 10 < a(n) < 1000, n/a(n) ~ 0.08127...

k = 2: For 1000 < a(n) < 10^5, n/a(n) ~ 0.008192...

k = 3: For 10^5 < a(n) < 10^7, n/a(n) ~ 0.0007753...

(End)

Examples

			329 + (3*2*9) = 383 (a palindrome). So, 329 is in this sequence.
		

Links

• Harvey P. Dale, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A007954.

Programs

Mathematica

Select[Range[0,1000],PalindromeQ[#+Times@@IntegerDigits[#]]&] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Jan 08 2019 *)

PARI

for(n=0, 10^3,d=digits(n);D=digits(n+prod(i=1,#d,d[i]));if(Vecrev(D)==D,print1(n,", "))) \\ Derek Orr, Mar 22 2015

Python

def rev(n):
    return int(''.join(reversed(str(n))))
def DP(n):
    p = 1
    for i in str(n):
        p *= int(i)
    return p
{print(n,end=', ') for n in range(10**3) if rev(n+DP(n))==n+DP(n)}
# Simplified by Derek Orr, Mar 22 2015