A063647 Number of ways to write 1/n as a difference of exactly 2 unit fractions.
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
Offset: 1
Examples
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).
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.
- Amiram Eldar, Plot of (Sum_{i=1..n} a(i))/f(n) - 1, where f(n) is an asymptotic function, for n = 2^(1..28).
- Amiram Eldar, Plot of Sum_{i=1..n} a(i)/(n*log(n)*log(log(n))) for n = 2^(1..28).
- 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.
Crossrefs
Programs
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Magma
[(NumberOfDivisors(n^2)-1)/2 : n in [1..100]]; // Vincenzo Librandi, Apr 18 2018
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Mathematica
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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PARI
for(n=1,100,print1(sum(i=1,n^2,if((n*i)%(i+n),0,1)),","))
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PARI
a(n)=numdiv(n^2)\2 \\ Charles R Greathouse IV, Oct 03 2016
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... [The conjecture is false. See the plot and the asymptotic formula below. - Amiram Eldar, Oct 03 2024]
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
Sum_{k=1..n} a(k) ~ (n/(2*zeta(2)))*(log(n)^2/2 + log(n)*(3*gamma - 1) + 1 - 3*gamma + 3*gamma^2 - 3*gamma_1 - zeta(2) + (2 - 6*gamma - 2*log(n))*zeta'(2)/zeta(2) + (2*zeta'(2)/zeta(2))^2 - 2*zeta''(2)/zeta(2)), where gamma is Euler's constant (A001620) and gamma_1 is the first Stieltjes constant (A082633). - Amiram Eldar, Oct 03 2024
Comments