

A080221


n is Harshad (divisible by the sum of its digits) in a(n) bases from 1 to n.


9



1, 2, 2, 4, 2, 6, 2, 7, 5, 7, 2, 11, 2, 5, 8, 11, 2, 13, 2, 13, 10, 5, 2, 19, 7, 6, 10, 14, 2, 18, 2, 16, 9, 6, 11, 23, 2, 5, 8, 23, 2, 20, 2, 11, 19, 5, 2, 30, 7, 16, 9, 14, 2, 21, 10, 21, 9, 5, 2, 34, 2, 5, 19, 23, 13, 23, 2, 12, 9, 22, 2, 39, 2, 5, 20, 13, 13, 21, 2, 34, 18, 7, 2, 37, 12, 5
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OFFSET

1,2


COMMENTS

For noncomposite integers, a(n)=d(n) (cf. A000005); for composite integers, a(n)> d(n). a(n) < n for all n > 6.
It appears that a(n) never takes on the value 3. Is there a proof of this? See A100263 for the sequence of values of n for which a(n)=5. It appears that, except for n=9, all values of n such that a(n) is 5 or 6 are twice a prime.  John W. Layman, Nov 10 2004
a(n) is never 3. As noted, 1 or any prime has a(n) = d(n) < 3. The only composites with d(n) <= 3 are squares of primes, for which d(n) = 3. But p^2 has the representation (p1)(1) in base (p+1), so a(p^2) >= 4. Any product of two distinct odd numbers n = ab with 1<a<b can be written as a,0 in base b; 1,(a1) in base aba+1; 1,(b1) in base abb+1; 2,a2 in base a(b1)/2+1; and 2,b2 in base (a1)b/2+1; plus 1 and n work for any n, so a(n)>6. If n = a^2, with a>3, we have 1,0 in base a; (a1)1 in base a+1; 1,(a1) in base a^2a+1; 2,(a2) in base a(a1)/2+1; and (a1)/2,(a+1)/2 in base 2a+1; together with 1 and n this means a(n)>6 for this form, too. Similar considerations eliminate other forms, leaving only 2p as possible values to have a(n) = 5 or 6.  Franklin T. AdamsWatters, Aug 03 2006
It is easy to prove that only 1, 2, 4 and 6 are allHarshad numbers (numbers that are divisible by the sum of their digits in every base).  Adam Kertesz, Feb 04 2008


REFERENCES

Eric W. Weisstein, CRC Concise Encyclopedia of Mathematics, Second ed., Chapman & Hall/CRC, 2003, p. 1310.


LINKS



EXAMPLE

6 is represented by the numeral 111111 in unary, 110 in binary, 20 in base 3, 12 in base 4, 11 in base 5 and 10 in base 6. The sums of the digits are 6, 2, 2, 3, 2 and 1 respectively, all divisors of 6; therefore a(6)=6.


MATHEMATICA

nivenQ[n_, b_] := Divisible[n, Total @ IntegerDigits[n, b]]; a[n_] := 1 + Sum[Boole @ nivenQ[n, b], {b, 2, n}]; Array[a, 100] (* Amiram Eldar, Jan 01 2020 *)


PROG

(Python)
from sympy.ntheory.factor_ import digits
def A080221(n): return nsum(1 for b in range(2, n) if n%sum(digits(n, b)[1:])) # Chai Wah Wu, Oct 19 2022


CROSSREFS

See A005349 for numbers that are Harshad in base 10.


KEYWORD

nonn,base


AUTHOR



EXTENSIONS



STATUS

approved



