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A114440
Numbers which divided by the sum of their digits (Harshad or Niven numbers) give integers which are also divisible by the sum of their digits (until a single-digit Harshad remains).
10
1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 18, 21, 24, 27, 36, 42, 45, 48, 54, 63, 72, 81, 84, 108, 162, 216, 243, 324, 378, 405, 432, 486, 648, 756, 864, 972, 1296, 1458, 1944, 2916, 3402, 4374, 5832, 6804, 7290, 8748, 11664, 13122, 13608, 15552, 17496, 23328, 26244
OFFSET
1,2
COMMENTS
The sequence is finite with a(15095), a 1434-digit number, being the final term. - Hans Havermann and Ray Chandler, Jan 21 2014
LINKS
Donovan Johnson, Table of n, a(n) for n = 1..235 (terms < 10^17)
Hans Havermann and Ray Chandler, Table of n, a(n) for n = 1..15095 (9.3 MB file)
Kornel, Ojciec i Syn.
EXAMPLE
The number 216 is a term of the sequence because it is divisible by the sum of its digits: 2+1+6=9; 216/9=24. Also, the successive quotients are divisible by the sum of their digits, until a single-digit Harshad remains: 24: 2+4=6; 24/6=4 and 4: 4/4=1.
MATHEMATICA
s=w={1}; Do[t={}; Do[v=s[[k]]; u={}; Do[If[Total[IntegerDigits[c*v]]==c, AppendTo[u, c*v]], {c, 2, 7000}]; t=Join[t, u], {k, Length[s]}]; s=Sort[t]; w=Join[w, s], {440}]; Union[w] (* Hans Havermann, Jan 21 2014 *)
PROG
(PARI) v=vector(118); for(n=1, 9, v[n]=n; print1(n ", ")); c=9; for(n=10, 10^9, d=length(Str(n)); m=n; s=0; for(j=1, d, s=s+m%10; m=m\10); if(s==1, next); if(n%s==0, m=n/s, next); forstep(j=c, 1, -1, if(v[j]<=m, if(v[j]==m, c++; v[c]=n; print1(n ", ")); next(2)))) /* Donovan Johnson, Apr 09 2013 */
KEYWORD
nonn,base,fini,full
AUTHOR
Piotr K. Olszewski (piotrkornelolszewski(AT)poczta.onet.pl), Feb 14 2006
EXTENSIONS
Offset corrected by Donovan Johnson, Apr 09 2013
a(54)-a(235) from Donovan Johnson, Apr 09 2013
a(236)-a(15095) from Hans Havermann and Ray Chandler, Jan 21 2014
STATUS
approved