This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
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).
niv[n_]:=Module[{k=1},While[!Divisible[k*n,Total[IntegerDigits[ k*n]]], k++]; k]; Array[niv,100] (* Harvey P. Dale, Jul 23 2016 *)
digitsum(n) = {local(s=0); while(n, s+=n%10; n\=10); s}
{for(n=1, 100, k=1; while((p=k*n)%digitsum(p)>0, k++); print1(k, ","))} /* Klaus Brockhaus, Sep 19 2008 */
from itertools import count 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
Example: 336 is in the sequence because the sum of digits of 336 when expressed in bases 1 through 10 is: 336, 3, 4, 3, 8, 6, 12, 7, 8, 12; and 336 is divisible by all these numbers. In this particular example 336 keeps this property in bases 11, 12 and 13, but not 14.
Select[Range[10^4], Catch[Do[If[Mod[#, Total@IntegerDigits[#, b]] > 0, Throw@ False], {b, 2, 10}]; True] &] (* Giovanni Resta, May 29 2013 *)
t = Table[b = 2; While[s = Total[IntegerDigits[n, b]]; s < n && Mod[n, s] == 0, b++]; If[s == n, b = 0]; b, {n, 2000}]; Flatten[Position[t, ?(# == 0 || # > 10 &)]] (* _T. D. Noe, May 30 2013 *)
a[n_] := Module[{m = n, k = 1}, While[Divisible[m, DigitSum[m, 2]], m += 2*n; k += 2]; k]; Array[a, 100]
a(n) = {my(m = n, k = 1); while(!(m % hammingweight(m)), m += 2*n; k += 2); k;}
from itertools import count def a(n): return next(k for k in count(1) if (m:=k*n)%m.bit_count() != 0) print([a(n) for n in range(1, 86)]) # Michael S. Branicky, Jun 30 2025
The sum of digits of 24 in bases 1 through 14 are: 24, 2, 4, 3, 8, 4, 6, 3, 8, 6, 4, 2, 12, 11. 24 is divisible by all these numbers except the last one; therefore a(24) = 14.
Table[b = 2; While[s = Total[IntegerDigits[n, b]]; s < n && Mod[n, s] == 0, b++]; If[s == n, b = 0]; b, {n, 100}] (* T. D. Noe, May 30 2013 *)
a(n) = {for (b=2, n-1, if (frac(n/sumdigits(n,b)), return(b));); 0;} \\ Michel Marcus, Oct 23 2018
For n = 3, we have:
.
k 3*k Niven?
- --- ------
1 3 Yes
2 6 Yes
3 9 Yes
4 12 Yes
5 15 No
so a(3) = 15.
a[n_]:=Module[{k=1}, While[Divisible[k*n, Total[IntegerDigits[k*n]]], k++]; k*n]; Array[a, 61]
a(n, base=10) = forstep (m=n, oo, n, if (m%sumdigits(m, base), return (m)))
a(5) = 13 since 13, 14, 15, 16 and 17 are all non-Niven numbers and this is the first occurrence of exactly 5 non-Niven numbers.
digitsum(n) = {local(s=0); while(n, s+=n%10; n\=10); s}
{m=120000; z=42; w=vector(z); n=1; while(n<=m, while(n%digitsum(n)==0, n++); a=n; c=0; while(n%digitsum(n)>0, n++; c++); if(c<=z&&w[c]==0, w[c]=a)); j=1; while(j<=z&&w[j]>0, print1(w[j], ","); j++)} /* Klaus Brockhaus, Sep 24 2008 */
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