A226169 -id:A226169 - cp's OEIS frontend

A226320 a(n) is the smallest k > 6 such that k is a Niven number at least in all the bases from 1 to n.

7, 8, 8, 8, 8, 12, 12, 24, 24, 24, 24, 24, 24, 432, 720, 720, 720, 720, 720, 840, 840, 840, 3360, 13860, 13860, 13860, 13860, 13860, 40320, 100800, 100800, 2106720, 7698600, 9028800, 9028800, 9028800, 9028800, 9028800, 9028800, 9028800, 9028800, 9028800

Offset: 1

Giovanni Resta, Jun 03 2013

The bound a(i) > 6 is motivated by the fact that 1, 2, 4 and 6 are Niven numbers in every base.

			a(8) = 24 because 24 is the smallest k > 6 which is Niven in bases 1 (trivial), 2,..., 8. For example, 24 = (33)_7 = (44)_5 = (220)_3.

Giovanni Resta, Table of n, a(n) for n = 1..54

Cf. A005349, A226169, A226319, A225427 (Niven in bases 1,...,n but not in base n+1).

a[n_] := Block[{k=7}, n > 1 && While[ Max@ Mod[k, Total /@ IntegerDigits[k, Range[2, n]]] > 0, k++]; k]; Array[a, 20]

A226171 Smallest base in which n is not Niven (or zero if n is Niven in every base).

Original entry on oeis.org

0, 0, 2, 0, 2, 0, 2, 6, 2, 4, 2, 8, 2, 2, 2, 6, 2, 8, 2, 7, 5, 2, 2, 14, 2, 2, 2, 2, 2, 2, 2, 6, 2, 3, 2, 8, 2, 2, 2, 12, 2, 3, 2, 2, 2, 2, 2, 14, 2, 2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 8, 2, 2, 2, 6, 2, 3, 2, 3, 3, 2, 2, 14, 2, 2, 2, 2, 2, 2, 2, 8, 5, 2, 2, 5, 2, 2

Offset: 1

Sergio Pimentel, May 29 2013

Niven numbers (in base b) are divisible by the sum of their digits (in base b).
Questions: are 1, 2, 4 and 6 the only zeros in this sequence? Where are the records or high water marks?
From Bert Dobbelaere, Oct 08 2018: (Start)
1,2,4,6 are the only numbers that are Niven in every base.
Proof: Suppose n is Niven in every base, then consider the base-b representations of n for (n/2) < b <= n. These are all 2-digit numbers with 1 as 1st digit and (n-b) as last digit. Then 1+n-b is a divisor of n for all b, meaning that all numbers between 1 up to n/2 are divisors of n. Clearly there are no such numbers larger than 6.
a(n) < 60 for n < 10^13.
(End)

			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.

T. D. Noe, Table of n, a(n) for n = 1..10000

Cf. A049445, A118363, A005349, A144261, A144262, A226169.

Cf. A225427 (least Niven number for all bases from 1 to n).

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

A226319 a(n) is the smallest odd k > 1 such that k is a Niven number at least in all the bases from 1 to n.

Original entry on oeis.org

3, 21, 21, 21, 675, 4725, 4725, 98175, 140175, 543375, 543375, 23186625, 35026425, 139264125, 139264125, 608679225, 11553990525, 87662479905, 87662479905, 343947649815, 2383529269275, 4005262262925, 4005262262925

Offset: 1

Giovanni Resta, Jun 03 2013

			a(4) = 21 because 21 is the smallest odd k > 1 which is Niven in bases 1 (trivial), 2, 3, and 4, being equal to (10101)_2, (210)_3 and (111)_4.

Cf. A005349, A226169, A226320, A225427 (Niven in bases 1,...,n but not in base n+1).

a[n_] := Block[{k=3}, n > 1 &&  While[Max@ Mod[k, Total /@ IntegerDigits[k, Range[2, n]]] > 0, k += 2]; k]; Array[a,9]

A330812 Least number >= n that is a Niven number in all bases 1 <= b <= n.

Original entry on oeis.org

1, 2, 4, 4, 6, 6, 12, 24, 24, 24, 24, 24, 24, 432, 720, 720, 720, 720, 720, 840, 840, 840, 3360, 13860, 13860, 13860, 13860, 13860, 40320, 100800, 100800, 2106720, 7698600, 9028800, 9028800, 9028800, 9028800, 9028800, 9028800, 9028800, 9028800, 9028800, 9028800

Offset: 1

Amiram Eldar, Jan 01 2020

			a(4) = 4 since the representations of 4 in bases 1 to 4 are 1111, 100, 11, 10, the corresponding sums of digits are 4, 1, 2, and 1, and all are divisors of 4. Thus 4 is a Niven number in bases 1, 2, 3, and 4, and it is the least number with this property.

Cf. A005349, A049445, A080221, A226169.

A[1]:= 1: m:= 1:
for n from 2 while m < 30 do
   kk:= n;
   for k from 2 to n-1 do
     if n mod convert(convert(n,base,k),`+`) <> 0 then kk:= k-1; break fi;
     od;
   if kk > m then
     for k from m+1 to kk do A[k]:= n od;
     m:= kk;
   fi
od:
seq(A[k],k=1..m); # Robert Israel, Jan 01 2020

nivenQ[n_, b_] := Divisible[n, Total @ IntegerDigits[n,b]]; a[n_] := Module[{k = n}, While[!AllTrue[Range[2, n], nivenQ[k, #] &], k++]; k]; Array[a,30]

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A226320 a(n) is the smallest k > 6 such that k is a Niven number at least in all the bases from 1 to n.

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A226171 Smallest base in which n is not Niven (or zero if n is Niven in every base).

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PARI

A226319 a(n) is the smallest odd k > 1 such that k is a Niven number at least in all the bases from 1 to n.

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Mathematica

A330812 Least number >= n that is a Niven number in all bases 1 <= b <= n.

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