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A063647 Number of ways to write 1/n as a difference of exactly 2 unit fractions. 29
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 (list; graph; refs; listen; history; text; internal format)
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/(c-b) and 1/n = 1/(2n-b) + 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

Number of elements in the set {(x,y): x|n, y|n, 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) = omega(n) <= 2, then a(n) = A062799(n); if A001221(n) > 2, then a(n) > A062799(n). - Matthew Vandermast, Aug 25 2004

Number of r X s integer-sided rectangles such that r + s = 4n, r < s and (s - r) | (s * r). - Wesley Ivan Hurt, Apr 24 2020

Also number of integer-sided right triangles with 2n as a leg. Equivalent to the even indices of A046079. - Nathaniel C Beckman, May 14 2020; Jun 26 2020

REFERENCES

Amarnath Murthy, Decomposition of the divisors of a natural number into pairwise coprime sets, Smarandache Notions Journal, vol. 12, No. 1-2-3, Spring 2001. pp. 303-306.

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 443-444.

Umberto Cerruti, Percorsi tra i numeri (in Italian), pages 3-4.

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*n-1} (1 - ceiling(i*(4*n-i)/(4*n-2*i)) + floor(i*(4*n-i)/(4*n-2*i))). - 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 twenty-nine terms identical to those of A062799.

Cf. A063717, A063718, A048691.

Cf. A046079.

Sequence in context: A353359 A113901 A062799 * A353379 A263653 A330328

Adjacent sequences:  A063644 A063645 A063646 * A063648 A063649 A063650

KEYWORD

nonn,easy,nice

AUTHOR

Henry Bottomley, Jul 23 2001

STATUS

approved

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Last modified September 30 21:21 EDT 2022. Contains 357106 sequences. (Running on oeis4.)