A089233 Number of coprime pairs of divisors > 1 of n.

0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 2, 0, 1, 1, 0, 0, 2, 0, 2, 1, 1, 0, 3, 0, 1, 0, 2, 0, 6, 0, 0, 1, 1, 1, 4, 0, 1, 1, 3, 0, 6, 0, 2, 2, 1, 0, 4, 0, 2, 1, 2, 0, 3, 1, 3, 1, 1, 0, 11, 0, 1, 2, 0, 1, 6, 0, 2, 1, 6, 0, 6, 0, 1, 2, 2, 1, 6, 0, 4, 0, 1, 0, 11, 1, 1, 1, 3, 0, 11, 1, 2, 1, 1, 1, 5, 0, 2, 2, 4, 0, 6, 0, 3, 6

Offset: 1

Author

Reinhard Zumkeller, Dec 11 2003

Keywords

• nonn

Comments

Also the number of divisors of n^2 which do not divide n and which are less than n. See link for proof. - Andrew Weimholt, Dec 06 2009

a(A000961(n)) = 0; a(A006881(n)) = 1; a(A054753(n)) = 2; a(A065036(n)) = 3. - Robert G. Wilson v, Dec 16 2009

First occurrence of k beginning with 0: 1, 6, 12, 24, 36, 96, 30, 384, 144, 216, 288, 60, 432, 24576, 1152, 864, 120, 393216, 1728, 1572864, 180, 240, 18432, 25165824, 5184, 210, 480, 13824, 10368, 360, 15552, 960, 20736, 55296, 1179648, 31104, 900, ..., . Except for 1, each is divisible by 6. Also the first occurrence of k must occur at or before 6*2^(n-1). - Robert G. Wilson v, Dec 16 2009

a(3*2^n) = n; if x = 2^n, then a(x) = a(2*x); and if x is not a power of two, then a(x) = y then a(2*x) > y. - Robert G. Wilson v, Dec 16 2009

a(n) = 0 iff n is a prime power. - Franklin T. Adams-Watters, Aug 20 2013

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
• Andrew Weimholt, Proof of an alternative characterization

Programs

Haskell

a089233 n = sum $ [a063524 $ gcd u v | let ds = tail $ a027750_row n,
                                       u <- ds, v <- dropWhile (<= u) ds]
-- Reinhard Zumkeller, Sep 04 2013

Magma

[(NumberOfDivisors(n^2)-1)/2 - NumberOfDivisors(n)+1: n in [1..100]]; // Vincenzo Librandi, Dec 23 2018

Mathematica

a[n_] := (DivisorSigma[0, n^2] - 1)/2 - DivisorSigma[0, n] + 1; Array[a, 104] (* Robert G. Wilson v, Dec 16 2009 *)

PARI

a(n) = (numdiv(n^2)-1)/2 - numdiv(n) + 1; \\ Michel Marcus, Feb 17 2016

Formula

a(n) = #{(x,y): 1 < x < y, x|n, y|n and gcd(x, y) = 1}.

a(n) = A063647(n) - A000005(n) + 1.

a(n) = A018892(n) - A000005(n). - Franklin T. Adams-Watters, Aug 20 2013