%I #16 Jul 10 2018 14:00:00
%S 10,10,10,10,10,10,10,10,10,100,51,40,31,24,23,22,21,20,19,100,70,52,
%T 43,40,34,32,30,28,27,100,73,62,53,46,43,40,37,35,33,100,81,70,61,54,
%U 50,46,43,40,310,100,83,73,65,60,55,51,49,46,43,100,85,76,70,64,510,56,52,412,49,100,87
%N Result when n is divided by the sum of its digits and the resulting integer is concatenated to the remainder.
%C When the remainder is zero, this 0 is still concatenated to the result (12/3 becomes 40).
%C All terms of A052224 are fixed points; more generally, if n has digital sum 10^k for some k > 0 and the remainder of n when divided by 10^k has k decimal digits, then n is a fixed point (corrected by _Rémy Sigrist_, Jul 10 2018).
%H Jean-Marc Falcoz, <a href="/A316650/b316650.txt">Table of n, a(n) for n = 1..20000</a>
%F a(10^k) = 10^(k+1) for any k >= 0. - _Rémy Sigrist_, Jul 10 2018
%e 1 divided by 1 is 1 with the remainder 0, thus a(1) = 10;
%e 2 divided by 2 is 1 with the remainder 0, thus a(2) = 10;
%e 3 divided by 3 is 1 with the remainder 0, thus a(3) = 10;
%e 4 divided by 4 is 1 with the remainder 0, thus a(4) = 10;
%e ...
%e 10 divided by (1+0) is 10 with the remainder 0, thus a(10) = 100;
%e 11 divided by (1+1) is 5 with the remainder 1, thus a(11) = 51;
%e 12 divided by (1+2) is 4 with the remainder 0, thus a(12) = 40;
%e 13 divided by (1+3) is 3 with the remainder 1, thus a(13) = 31;
%e ...
%e 2018 divided by (2+0+1+8) is 183 with the remainder 5, thus a(2018) = 1835.
%e Etc.
%t Array[FromDigits@ Flatten[IntegerDigits@ # & /@ QuotientRemainder[#, Total[IntegerDigits@ #]]] &, 71] (* _Michael De Vlieger_, Jul 10 2018 *)
%o (PARI) a(n, base=10) = my (ds=sumdigits(n, base), q=n\ds, r=n%ds); q * base^max(1, #digits(r, base)) + r \\ _Rémy Sigrist_, Jul 10 2018
%Y Cf. A007953, A052224.
%K base,nonn,look
%O 1,1
%A _Eric Angelini_ and _Jean-Marc Falcoz_, Jul 09 2018
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