

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.


1



1, 2, 3, 4, 5, 6, 7, 8, 9, 910, 0, 912, 11713, 6314, 915, 3616, 15317, 918, 17119, 9920, 18921, 9922, 82823, 19824, 9925, 46826, 18927, 18928, 78329, 99930, 585931, 388832, 1098933, 198934, 289835, 99936, 99937, 478838, 198939, 1999840
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OFFSET

1,2


COMMENTS

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.
The conjecture is correct. Let m = (n*10^((nS(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


LINKS

Martins Opmanis, Table of n, a(n) for n = 1..582


EXAMPLE

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.


MATHEMATICA

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 *)


PROG

(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


CROSSREFS

A075154 is similar but limited to equivalence of last two digits; therefore at least the first 99 terms are the same in both sequences.
Sequence in context: A112014 A145461 A075154 * A070938 A070594 A121939
Adjacent sequences: A187921 A187922 A187923 * A187925 A187926 A187927


KEYWORD

nonn,base


AUTHOR

Martins Opmanis, Mar 16 2011


EXTENSIONS

Name corrected by Michel Marcus, Dec 24 2017


STATUS

approved



