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Search: id:A080221
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| A080221 |
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n is Harshad (divisible by the sum of its digits) in a(n) bases from 1 to n. |
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+0
2
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| 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
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1,2
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COMMENT
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For non-composite 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 (layman(AT)math.vt.edu), 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 (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. - Frank Adams-Watters (FrankTAW(AT)Netscape.net), Aug 03 2006
It is easy to prove that only 1, 2, 4 and 6 are all-Harshad numbers (numbers which are divisible by the sum of their digits in every base). - Adam Kertesz (Kertesz.Adam(AT)gmail.com), Feb 04 2008
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REFERENCES
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Eric W. Weisstein: CRC Concise Encyclopedia of Mathematics, Second ed.,Chapman & Hall/CRC, 2003, p. 1310
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EXAMPLE
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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.
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CROSSREFS
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See A005349 for numbers that are Harshad in base 10.
Cf. A100263.
Sequence in context: A128982 A096216 A121599 this_sequence A137849 A118982 A129457
Adjacent sequences: A080218 A080219 A080220 this_sequence A080222 A080223 A080224
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KEYWORD
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nonn
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AUTHOR
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Matthew Vandermast (ghodges14(AT)comcast.net), Mar 16 2003
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EXTENSIONS
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More terms from John W. Layman (layman(AT)math.vt.edu), Nov 10 2004
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