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 A067743 Number of divisors of n not in the half-open interval [sqrt(n/2), sqrt(n*2)). 2
 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; refs; listen; history; text; internal format)
 OFFSET 1,3 REFERENCES Problem 10847, Amer. Math. Monthly 109, (2002), p. 80. LINKS FORMULA 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=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) EXAMPLE a(6)=2 because 2 divisors of 6 (i.e. 1 and 6) fall outside sqrt(3) to sqrt(12). PROG (PARI from M. F. Hasler, May 12 2008) A067743(n)=sumdiv( n, d, d*d= 2*n ) CROSSREFS Cf. A067742, A000005. Sequence in context: A001031 A035250 A165054 * A029230 A280945 A196067 Adjacent sequences:  A067740 A067741 A067742 * A067744 A067745 A067746 KEYWORD easy,nonn AUTHOR Marc LeBrun, Jan 29 2002 STATUS approved

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