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%I #76 Oct 07 2021 13:50:33
%S 1,2,3,4,5,6,7,8,9,0,209,48,4000009,21182,5055,21184,13328,288,12844,
%T 0,1596,2398,13892,2976,52675,45890,2889,61768,178292,0,177475,29984,
%U 42999,279718,529865,29988,1009009009009,485678,1951599,0,694499,655998,1677688,658988
%N 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.
%C 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.
%C As gcd(k, rev(k)) = n, n | k and n | rev(k). - _David A. Corneth_, Sep 03 2020
%C 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
%C 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]
%H Ruediger Jehn, <a href="/A333666/b333666.txt">Table of n, a(n) for n = 1..303</a>
%H Ruediger Jehn, <a href="/A333666/a333666_2.pdf">Proofs for difficult terms</a>
%H Rüdiger Jehn, <a href="https://arxiv.org/abs/2104.02482">Porous Numbers</a>, arXiv:2104.02482 [math.GM], 2021.
%F 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.
%e a(11) = 209. The sum of the digits is 11 and gcd(209,902) = 11.
%e a(12) = 48. The sum of the digits is 12 and gcd(48,84) = 12.
%t 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 *)
%o (Python)
%o for n in range(11, 20):
%o for k in range(n, 1000000000, n):
%o s = str(k)
%o revk = "" # digit reversal of k
%o sum = 0
%o for i in range(len(s)):
%o revk = revk + s[len(s) - i - 1]
%o sum = sum + int(s[i])
%o g = gcd(k,int(revk))
%o if g == n and sum == n:
%o print(n, k, revk, g)
%o break
%o (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
%Y Cf. A004086, A005349, A007953, A069554.
%K nonn,base
%O 1,2
%A _Ruediger Jehn_ and _Kester Habermann_, Sep 03 2020
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