A333666 Smallest k > 0 with gcd(k, rev(k)) = n, where rev(k) is digit reversal of k and with sum of digits of k = n, or 0 if no such k exists.

1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 209, 48, 4000009, 21182, 5055, 21184, 13328, 288, 12844, 0, 1596, 2398, 13892, 2976, 52675, 45890, 2889, 61768, 178292, 0, 177475, 29984, 42999, 279718, 529865, 29988, 1009009009009, 485678, 1951599, 0, 694499, 655998, 1677688, 658988

Offset: 1

Author

Ruediger Jehn and Kester Habermann, Sep 03 2020

Keywords

• nonn
• base

Comments

This differs from A069554 in that, additionally, the sum of the digits of a(n) must be equal to n. This is not required in A069554.

As gcd(k, rev(k)) = n, n | k and n | rev(k). - David A. Corneth, Sep 03 2020

Since the sum of the digits of k is n and n | k, all the terms that are not 0 are Niven numbers (A005349). - Amiram Eldar, Sep 03 2020

The first digit of any number of this sequence is less than or equal to the last digit of this number (provided that the last digit is nonzero), because if k satisfies all requirements, also rev(k) does. This means that numbers starting with a "9" are quite rare. So far we have found only 9. But numbers ending with a "1" seem to be even less frequent. Amongst the first 303 terms of this sequence there is none except the trivial solution a(1) = 1. The second term of this sequence ending with a "1", if it exists, is still to be found. - Ruediger Jehn, Sep 20 2020 [Corrected by Pontus von Brömssen, Oct 07 2021]

Examples

			a(11) = 209. The sum of the digits is 11 and gcd(209,902) = 11.
a(12) = 48. The sum of the digits is 12 and gcd(48,84) = 12.
		

Links

• Ruediger Jehn, Table of n, a(n) for n = 1..303
• Ruediger Jehn, Proofs for difficult terms
• Rüdiger Jehn, Porous Numbers, arXiv:2104.02482 [math.GM], 2021.

Crossrefs

Cf. A004086, A005349, A007953, A069554.

Programs

Mathematica

m = 36; s = Table[0, {m}]; c = 0; n = 1; While[c < m - Quotient[m, 10], g = GCD[n, FromDigits @ Reverse @ (d = IntegerDigits[n])]; If[g <= m && g == Plus @@ d && s[[g]] == 0, c++; s[[g]] = n]; n++]; s (* Amiram Eldar, Sep 03 2020 *)

PARI

a(n) = {if ((n % 10) == 0, return(0)); my(k=n); while (! ((sumdigits(k)==n) && (gcd(k, fromdigits(Vecrev(digits(k)))) == n)), k+=n); k;} \\ Michel Marcus, Sep 03 2020

Python

for n in range(11, 20):
    for k in range(n, 1000000000, n):
       s = str(k)
       revk = "" # digit reversal of k
       sum = 0
       for i in range(len(s)):
          revk = revk + s[len(s) - i - 1]
          sum = sum + int(s[i])
       g = gcd(k,int(revk))
       if g == n and sum == n:
          print(n, k, revk, g)
          break

Formula

a(10*n) = 0 since all multiples of 10 have a 0 at the end, but their reverse numbers have no 0 at the end and therefore 10*n cannot be their gcd.