|
%I #30 Feb 18 2024 02:53:49
%S 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,
%T 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,
%U 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
%N n is Harshad (divisible by the sum of its digits) in a(n) bases from 1 to n.
%C For noncomposite integers, a(n)=d(n) (cf. A000005); for composite integers, a(n)> d(n). a(n) < n for all n > 6.
%C 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
%C 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 (p-1)(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,(a-1) in base ab-a+1; 1,(b-1) in base ab-b+1; 2,a-2 in base a(b-1)/2+1; and 2,b-2 in base (a-1)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; (a-1)1 in base a+1; 1,(a-1) in base a^2-a+1; 2,(a-2) in base a(a-1)/2+1; and (a-1)/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. Adams-Watters_, Aug 03 2006
%C It is easy to prove that only 1, 2, 4 and 6 are all-Harshad numbers (numbers that are divisible by the sum of their digits in every base). - _Adam Kertesz_, Feb 04 2008
%D Eric W. Weisstein, CRC Concise Encyclopedia of Mathematics, Second ed., Chapman & Hall/CRC, 2003, p. 1310.
%H Chai Wah Wu, <a href="/A080221/b080221.txt">Table of n, a(n) for n = 1..10000</a>
%e 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.
%t 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 *)
%o (Python)
%o from sympy.ntheory.factor_ import digits
%o def A080221(n): return n-sum(1 for b in range(2,n) if n%sum(digits(n,b)[1:])) # _Chai Wah Wu_, Oct 19 2022
%Y See A005349 for numbers that are Harshad in base 10.
%Y Cf. A100263.
%K nonn,base
%O 1,2
%A _Matthew Vandermast_, Mar 16 2003
%E More terms from _John W. Layman_, Nov 10 2004
|
