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A063647
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Number of ways to write 1/n as a difference of exactly 2 unit fractions.
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23
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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
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1,4
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COMMENTS
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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
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): 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) <= 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 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 side length. Equivalent to the even indices of A046081. - Nathaniel C Beckman, May 14 2020
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REFERENCES
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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.
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LINKS
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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 443-4
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.
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FORMULA
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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} chi(i*(4*n-i)/(4*n-2*i)), where chi is the integer characteristic. - Wesley Ivan Hurt, Apr 24 2020
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EXAMPLE
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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).
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MATHEMATICA
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Table[(Length[Divisors[n^2]] - 1)/2, {n, 1, 100}]
(DivisorSigma[0, Range[100]^2]-1)/2 (* Harvey P. Dale, Apr 15 2013 *)
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PROG
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(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
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CROSSREFS
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Cf. A018892, A063427, A063428. First twenty-nine 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
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KEYWORD
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nonn,easy,nice
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AUTHOR
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Henry Bottomley, Jul 23 2001
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STATUS
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approved
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