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Search: a236295 -id:a236295
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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
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

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