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A067742
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Number of middle divisors of n, i.e., divisors in the half-open interval [sqrt(n/2), sqrt(n*2)).
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69
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1, 1, 0, 1, 0, 2, 0, 1, 1, 0, 0, 2, 0, 0, 2, 1, 0, 1, 0, 2, 0, 0, 0, 2, 1, 0, 0, 2, 0, 2, 0, 1, 0, 0, 2, 1, 0, 0, 0, 2, 0, 2, 0, 0, 2, 0, 0, 2, 1, 1, 0, 0, 0, 2, 0, 2, 0, 0, 0, 2, 0, 0, 2, 1, 0, 2, 0, 0, 0, 2, 0, 3, 0, 0, 0, 0, 2, 0, 0, 2, 1, 0, 0, 2, 0, 0, 0, 2, 0, 2, 2, 0, 0, 0, 0, 2, 0, 1, 2, 1, 0, 0, 0, 2, 0
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OFFSET
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1,6
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COMMENTS
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Theorem 1: (i) a(n) = (number of odd divisors of n <= sqrt(2*n)) - (number of odd divisors of n > sqrt(2*n)).
(ii) Let r(n) = A003056(n). Then a(n) = (number of odd divisors of n <= r(n)) - (number of odd divisors of n > r(n)).
(iii) a(n) = Sum_{k=1..r(n)} (-1)^(k+1)*A237048(n,k).
(iv) a(n) is odd iff n is a square or twice a square (cf. A053866). Indices of odd terms give A028982. Indices of even terms give A028983.
The proofs are straightforward. These results were conjectured by Omar E. Pol in 2017. (End)
Theorem 2: a(n) is equal to the difference between the number of partitions of n into an odd number of consecutive parts and the number of partitions of n into an even number of consecutive parts. [Chapman et al., 2001; Hirschhorn and Hirschhorn, 2005]. - Omar E. Pol, Feb 06 2017
Conjecture 1: This is the central column of the isosceles triangle of A249351.
Conjecture 2: a(n) is also the width of the terrace at the n-th level in the main diagonal of the pyramid described in A245092.
Conjecture 3: a(n) is also the number of central subparts of the symmetric representation of sigma(n). For more information see A279387.
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REFERENCES
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Robin Chapman, Kimmo Eriksson and Richard Stanley, On the Number of Divisors of n in a Special Interval: Problem 10847, Amer. Math. Monthly 108:1 (Jan 2001), p. 77 (Proposal); 109:1 (Jan 2002), p. 80 (Solution). [Please do not delete this reference. - N. J. A. Sloane, Dec 16 2020]
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LINKS
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FORMULA
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G.f.: Sum_{k>=1} (-1)^(k-1)*x^(k*(k+1)/2)/(1-x^k). (This g.f. corresponds to the assertion in Theorem 2.)
Another g.f., corresponding to the definition: Sum_{k>=1} x^(2*k*(k+1))*(1-x^(6*k^2))/(1-x^(2*k)) + Sum_{k>=0} x^((k+1)*(2*k+1))*(1-x^((2*k+1)*(3*k+2))/(1-x^(2*k+1). - N. J. A. Sloane, Jan 04 2021
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EXAMPLE
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a(6)=2 because the 2 middle divisors of 6 (2 and 3) are between sqrt(3) and sqrt(12).
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MATHEMATICA
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(* number of middle divisors *)
a067742[n_] := Select[Divisors[n], Sqrt[n/2] <= # < Sqrt[2n] &]
a067742[115] (* data *)
a[ n_] := If[ n < 1, 0, DivisorSum[ n, 1 &, n/2 <= #^2 < 2 n &]]; (* Michael Somos, Jun 04 2015 *)
(* support function a240542[] is defined in A240542 *)
b[n_] := a240542[n] - a240542[n-1]
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PROG
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(PARI) A067742(n) = {sumdiv(n, d, d2 = d^2; n / 2 < d2 && d2 <= n << 1)} \\ M. F. Hasler, May 12 2008
(PARI) a(n) = A067742(n) = {my(d = divisors(n), iu = il = #d \ 2); if(#d <= 2, return(n < 3)); while(d[il]^2 > n>>1, il--); while(d[iu]^2 < (n<<1), iu++);
(Python)
from sympy import divisors
def A067742(n): return sum(1 for d in divisors(n, generator=True) if n <= 2*d**2 < 4*n) # Chai Wah Wu, Jun 09 2022
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CROSSREFS
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Cf. also A071561 (positions of zeros), A071562 (positions of nonzeros), A299761 (middle divisors of n), A355143 (products of middle divisors).
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KEYWORD
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easy,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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