cp's OEIS Frontend

A187924 a(n) is the smallest multiple of n such that a(n) ends with n and S(a(n))=n where S(m) is the sum of the base ten digits of m, or 0 if no such a(n) exists.

%I A187924 #25 Jan 19 2018 03:32:07
%S A187924 1,2,3,4,5,6,7,8,9,910,0,912,11713,6314,915,3616,15317,918,17119,9920,
%T A187924 18921,9922,82823,19824,9925,46826,18927,18928,78329,99930,585931,
%U A187924 388832,1098933,198934,289835,99936,99937,478838,198939,1999840
%N A187924 a(n) is the smallest multiple of n such that a(n) ends with n and S(a(n))=n where S(m) is the sum of the base ten digits of m, or 0 if no such a(n) exists.
%C A187924 It can be proved that a(11)=0 and, for infinitely many n, a(n) is the least integer with S(n)=n. Conjecture: 11 is the only n for which a(n)=0.
%C A187924 The conjecture is correct. Let m = (n*10^((n-S(n))/9) - n) * 10^floor(1+log_10(n)) + n. If n != 11, then it can be proved that m has all the required properties of a(n) except that it may not be the smallest candidate. If n=11, then S(m)=20 instead of the required 11. - _Ørjan Johansen_, Dec 08 2017
%H A187924 Martins Opmanis, <a href="/A187924/b187924.txt">Table of n, a(n) for n = 1..582</a>
%e A187924 For n=13 11713 is the least integer which is multiple of 13, ends with "13" and sum of digits in decimal notation also is 13.
%t A187924 Table[If[n == 11, 0, Block[{k = 1}, While[Nand[FromDigits@ Take[#, -IntegerLength@ n] == n, Total@ # == n] &@ IntegerDigits[k n], k++]; k n]], {n, 40}] (* _Michael De Vlieger_, Dec 09 2017 *)
%o A187924 (PARI) a(n) = {if (n == 11, return (0)); my(k = 1); while (!((sumdigits(k*n) == n) && (nd = #digits(n)) && !((k*n - n) % 10^nd)), k++); k*n;} \\ _Michel Marcus_, Dec 23 2017
%Y A187924 A075154 is similar but limited to equivalence of last two digits; therefore at least the first 99 terms are the same in both sequences.
%K A187924 nonn,base
%O A187924 1,2
%A A187924 _Martins Opmanis_, Mar 16 2011
%E A187924 Name corrected by _Michel Marcus_, Dec 24 2017