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a(n) = smallest k such that k*n is a Niven (or Harshad) number.
11

%I #22 Nov 04 2022 10:48:11

%S 1,1,1,1,1,1,1,1,1,1,10,1,9,3,2,3,6,1,6,1,1,5,9,1,2,6,1,3,9,1,12,6,4,

%T 3,2,1,3,3,3,1,10,1,12,3,1,5,9,1,8,1,2,3,18,1,2,2,2,9,9,1,12,6,1,3,3,

%U 2,3,3,3,1,18,1,7,3,2,2,4,2,9,1,1,5,18,1,6,6,3,3,9,1,4,5,4,9,2,2,12,4,2,1

%N a(n) = smallest k such that k*n is a Niven (or Harshad) number.

%C Niven (or Harshad) numbers are numbers that are divisible by the sum of their digits.

%C Does a(n) exist for all n? - _Klaus Brockhaus_, Sep 19 2008

%H David Radcliffe, <a href="/A144261/b144261.txt">Table of n, a(n) for n = 1..10000</a>

%H David Radcliffe, <a href="/A144261/a144261.pdf">Every positive integer divides a Harshad number</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/HarshadNumber.html">Harshad Number</a>

%e a(14) = 3 since neither 1*14 or 2*14 are Niven numbers, but 3*14 = 42 is a Niven number: 42 = 7*(4+2).

%t niv[n_]:=Module[{k=1},While[!Divisible[k*n,Total[IntegerDigits[ k*n]]], k++]; k]; Array[niv,100] (* _Harvey P. Dale_, Jul 23 2016 *)

%o (PARI) digitsum(n) = {local(s=0); while(n, s+=n%10; n\=10); s}

%o {for(n=1, 100, k=1; while((p=k*n)%digitsum(p)>0, k++); print1(k, ","))} /* _Klaus Brockhaus_, Sep 19 2008 */

%o (Python)

%o from itertools import count

%o def A144261(n): return next(filter(lambda k:not (m:=k*n) % sum(int(d) for d in str(m)), count(1))) # _Chai Wah Wu_, Nov 04 2022

%Y Cf. A005349 (Niven numbers), A144262 (smallest k such that k*n is not a Niven number), A144363 (records in A144261), A144364 (where records occur in A144261).

%K base,nonn

%O 1,11

%A _Sergio Pimentel_, Sep 16 2008

%E Edited and extended by _Klaus Brockhaus_, Sep 19 2008