

A063647


Number of ways to write 1/n as a difference of exactly 2 unit fractions.


23



0, 1, 1, 2, 1, 4, 1, 3, 2, 4, 1, 7, 1, 4, 4, 4, 1, 7, 1, 7, 4, 4, 1, 10, 2, 4, 3, 7, 1, 13, 1, 5, 4, 4, 4, 12, 1, 4, 4, 10, 1, 13, 1, 7, 7, 4, 1, 13, 2, 7, 4, 7, 1, 10, 4, 10, 4, 4, 1, 22, 1, 4, 7, 6, 4, 13, 1, 7, 4, 13, 1, 17, 1, 4, 7, 7, 4, 13, 1, 13, 4, 4, 1, 22, 4, 4, 4, 10, 1, 22, 4, 7, 4, 4, 4
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OFFSET

1,4


COMMENTS

Also number of ways to write 1/n as sum of exactly two distinct unit fractions.  Thomas L. York, Jan 11 2014
Also number of positive integers m such that 1/n + 1/m is a unit fraction.  Jon E. Schoenfield, Apr 17 2018
If 1/n = 1/b  1/c then n = bc/(cb) and 1/n = 1/(2nb) + 1/(c+2n) (though it is also the case that 1/n = 1/(2n) + 1/(2n) equivalent to b = c = 0).
Also number of divisors of n^2 less than n.  Vladeta Jovovic, Aug 13 2001
Also number of decompositions of divisors of n into coprime pairs.  K. B. Subramaniam (kb_subramaniambalu(AT)yahoo.com) and Amarnath Murthy, Dec 24 2001
Number of elements in the set {(x,y): xn, yn, x<y, gcd(x,y)=1}.  Vladeta Jovovic, May 03 2002
Also number of positive integers of the form k*n/(k+n).  Benoit Cloitre, Jan 04 2002
This is similar to A062799, having the same first 29 terms. But they are different sequences.
If A001221(n) <= 2, then a(n)=A062799(n); if A001221(n) > 2, then a(n) > A062799(n). A001221(n), or omega(n), is the number of distinct primes dividing n.  Matthew Vandermast, Aug 25 2004
Number of r X s integersided rectangles such that r + s = 4n, r < s and (s  r)  (s * r).  Wesley Ivan Hurt, Apr 24 2020
Also number of integersided right triangles with 2n as a side length. Equivalent to the even indices of A046081.  Nathaniel C Beckman, May 14 2020


REFERENCES

Amarnath Murthy, Decomposition of the divisors of a natural number into pairwise coprime sets, Smarandache Notions Journal, vol. 12, No. 123, Spring 2001. pp. 303306.


LINKS

T. D. Noe, Table of n, a(n) for n = 1..10000
Christopher J. Bradley, Solution to Problem 2175, Crux Mathematicorum, Vol. 23, No. 7, (Nov 1997), pp 4434
Umberto Cerruti, Percorsi tra i numeri (in Italian), pages 34.
Roger B. Eggleton, Unitary Fractions: 10501, The American Mathematical Monthly, Vol. 105, No. 4 (Apr., 1998), p. 372.


FORMULA

a(n) = (tau(n^2)1)/2.
a(n) = A018892(n)1. If n = (p1^a1)(p2^a2)...(pt^at), a(n) = ((2*a1+1)(2*a2+1)...(2*at+1)1)/2.
If n is prime a(n)=1. Conjecture: (1/n)*Sum_{i=1..n} a(i) = C*log(n)*log(log(n)) + o(log(n)) with C=0.7...
Bisection of A046079.  Lekraj Beedassy, Jul 09 2004
a(n) = Sum_{i=1..2*n1} chi(i*(4*ni)/(4*n2*i)), where chi is the integer characteristic.  Wesley Ivan Hurt, Apr 24 2020


EXAMPLE

a(10) = 4 since 1/10 = 1/5  1/10 = 1/6  1/15 = 1/8  1/40 = 1/9  1/90.
a(12) = 7: the divisors of 12 are 1, 2, 3, 4, 6 and 12 and the decompositions are (1, 2), (1, 3), (1, 4), (1, 6), (1, 12), (2, 3), (3, 4).


MATHEMATICA

Table[(Length[Divisors[n^2]]  1)/2, {n, 1, 100}]
(DivisorSigma[0, Range[100]^2]1)/2 (* Harvey P. Dale, Apr 15 2013 *)


PROG

(PARI) for(n=1, 100, print1(sum(i=1, n^2, if((n*i)%(i+n), 0, 1)), ", "))
(PARI) a(n)=numdiv(n^2)\2 \\ Charles R Greathouse IV, Oct 03 2016
(MAGMA) [(NumberOfDivisors(n^2)1)/2 : n in [1..100]]; // Vincenzo Librandi, Apr 18 2018


CROSSREFS

Cf. A018892, A063427, A063428. First twentynine terms identical to those of A062799.
Cf. A063717, A063718, A048691.
Sequence in context: A067614 A113901 A062799 * A263653 A330328 A269427
Adjacent sequences: A063644 A063645 A063646 * A063648 A063649 A063650


KEYWORD

nonn,easy,nice,changed


AUTHOR

Henry Bottomley, Jul 23 2001


STATUS

approved



