A229545 Numbers n such that n + (sum of digits of n) is a palindrome.

0, 1, 2, 3, 4, 10, 20, 30, 40, 50, 60, 70, 80, 90, 91, 96, 100, 105, 124, 129, 143, 148, 162, 167, 181, 191, 196, 200, 205, 224, 229, 243, 248, 262, 267, 281, 291, 296, 300, 305, 324, 329, 343, 348, 362, 367, 381, 391, 396, 400, 405, 424, 429, 443, 448, 462

Offset: 1

Author

Derek Orr, Sep 25 2013

Keywords

• nonn
• base
• easy

Comments

It appears the ones and tens digits in the 3-digit numbers have a pattern to them (00-->05-->24-->29-->43-->48-->62-->67-->81-->91-->96-->00).

Analyzing a(n) mod 10^e, n<100000, for e=2: starting at n=15 there are 9 cycles of length 11 [91,96,0,5,24,29,43,48,62,67,81], followed by 9 cycles of length 10 [82,0,9,18,27,36,45,54,63,72], then 9 of length 101, 9 of 102, 9 of 1011, 9 of 1012, and at least 7 of length 10103. For e=1 the cycles have the same position and length, for e>2 the shorter cycles successively disappear. [Lars Blomberg, Jan 05 2013]

Examples

			196 + (1+9+6) = 212 (a palindrome). So, 196 is in this sequence.
		

Links

• Lars Blomberg, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A062028.

Programs

Mathematica

palQ[n_] := Block[{d = IntegerDigits@ n}, d == Reverse@ d]; Select[Range[0, 462], palQ[# + Plus @@ IntegerDigits@ #] &] (* Michael De Vlieger, Apr 12 2015 *)

PARI

for(n=0,10^3,D=digits(n+sumdigits(n));if(Vecrev(D)==D,print1(n,", "))) \\ Derek Orr, Mar 22 2015

Python

def ispal(n):
  r = ''
  for i in str(n):
    r = i + r
  return n == int(r)
def DS(n):
  s = 0
  for i in str(n):
    s += int(i)
  return s
{print(n,end=', ') for n in range(10**3) if ispal(n+DS(n))}
# Simplified by Derek Orr, Mar 22 2015