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, 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

Offset: 1

Author

Martins Opmanis, Mar 16 2011

Keywords

• nonn
• base

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^((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

Examples

			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.
		

Links

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

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.

Programs

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 *)

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

Extensions

Name corrected by Michel Marcus, Dec 24 2017