A098834 Palindromic Smith numbers.

4, 22, 121, 202, 454, 535, 636, 666, 1111, 1881, 3663, 7227, 7447, 9229, 10201, 17271, 22522, 24142, 28182, 33633, 38283, 45054, 45454, 46664, 47074, 50305, 51115, 51315, 54645, 55055, 55955, 72627, 81418, 82628, 83038, 83938, 90409, 95359, 96169, 164461

Offset: 1

Author

Shyam Sunder Gupta, Oct 10 2004

Keywords

• base
• nonn

Examples

			a(3) = 121 because 121 is a Smith number as well as a palindromic number.
		

Links

• Donovan Johnson, Table of n, a(n) for n = 1..10000
• Shyam Sunder Gupta, Smith Numbers, Exploring the Beauty of Fascinating Numbers, Springer (2025) Ch. 4, 127-157.

Crossrefs

Cf. A006753.

Subsequence of A104171. Supersequence of A104166.

Programs

Mathematica

d[n_] := IntegerDigits[n]; tr[n_] := Transpose[FactorInteger[n]]; Select[Range[2, 1.7*10^5], !PrimeQ[#] && Reverse[x=d[#]] == x && Total[x] == Total[d@tr[#][[1]]*tr[#][[2]], 2]&] (* Jayanta Basu, Jun 04 2013 *)

Python

from sympy import factorint
from itertools import product
def sd(n): return sum(map(int, str(n)))
def smith(n):
  f = factorint(n)
  return sum(f[p] for p in f) > 1 and sd(n) == sum(sd(p)*f[p] for p in f)
def palsto(limit):
  yield from range(min(limit, 9)+1)
  midrange = [[""], [str(i) for i in range(10)]]
  for digs in range(2, 10**len(str(limit))):
    for p in product("0123456789", repeat=digs//2):
      left = "".join(p)
      if left[0] == '0': continue
      for middle in midrange[digs%2]:
        out = int(left + middle + left[::-1])
        if out > limit: return
        yield out
print(list(filter(smith, palsto(164461)))) # Michael S. Branicky, Apr 22 2021