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Search: id:A067743
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| A067743 |
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Number of divisors of n not in the half-open interval [sqrt(n/2), sqrt(n*2)). |
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+0
2
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| 0, 1, 2, 2, 2, 2, 2, 3, 2, 4, 2, 4, 2, 4, 2, 4, 2, 5, 2, 4, 4, 4, 2, 6, 2, 4, 4, 4, 2, 6, 2, 5, 4, 4, 2, 8, 2, 4, 4, 6, 2, 6, 2, 6, 4, 4, 2, 8, 2, 5, 4, 6, 2, 6, 4, 6, 4, 4, 2, 10, 2, 4, 4, 6, 4, 6, 2, 6, 4, 6, 2, 9, 2, 4, 6, 6, 2, 8, 2, 8, 4, 4, 2, 10, 4, 4, 4, 6, 2, 10, 2, 6, 4, 4, 4, 10, 2, 5, 4, 8, 2, 8
(list; graph; listen)
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OFFSET
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1,3
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REFERENCES
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Problem 10847, Amer. Math. Monthly 109, (2002), p. 80.
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FORMULA
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a(n) = A000005(n) - A067742(n).
G.f.: sum(k=1,N, z^(2*k^2)*(1+z^k)/(1-z^k) ). - Joerg Arndt, May 12 2008
Direct proof of Joerg Arndt's g.f., from Max Alekseyev, May 13 2008 (Start):
We need to count divisors d|n such that d^2<=n/2 or d^2>2n. In the latter case, let's switch to co-divisor, replacing d by n/d.
Then we need to find the total count of: 1) divisors d|n such that 2d^2<=n; 2) divisors d|n such that 2d^2<n.
Let d|n and 2d^2<=n. Then n-2d^2 must be a multiple of d, i.e. n-2d^2=td for some integer t>=0.
Moreover it is easy to see that 1) is equivalent to n = 2d^2 + td for some integer t>=0. Therefore the answer for 1) is the coefficient of z^n in SUM[d=1..oo] SUM[t=0..oo] x^(2d^2 + td) = SUM[d=1..oo] x^(2d^2)/(1 - x^d).
Similarly, the answer for 2) is SUM[d=1..oo] x^(2d^2)/(1 - x^d) * x^d.
Therefore the g.f. for A067743 is SUM[d=1..oo] x^(2d^2)/(1 - x^d) + SUM[d=1..oo] x^(2d^2)/(1 - x^d) * x^d = SUM[d=1..oo] x^(2d^2)/(1 - x^d) * (1 + x^d), as proposed. (End)
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EXAMPLE
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a(6)=2 because 2 divisors of 6 (i.e. 1 and 6) fall outside sqrt(3) to sqrt(12).
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PROGRAM
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(PARI from M. F. Hasler, May 12 2008) A067743(n)=sumdiv( n, d, d*d<n/2 || d*d >= 2*n )
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CROSSREFS
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Cf. A067742, A000005.
Sequence in context: A001031 A035250 A165054 this_sequence A029230 A084294 A067752
Adjacent sequences: A067740 A067741 A067742 this_sequence A067744 A067745 A067746
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KEYWORD
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easy,nonn
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AUTHOR
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Marc LeBrun (mlb(AT)well.com), Jan 29 2002
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