G.f.: x/((1-2*x)*(1-x)).
E.g.f.: exp(2*x) - exp(x).
E.g.f. if offset 1: ((exp(x)-1)^2)/2.
a(n) = Sum_{k=0..n-1} 2^k. -
Paul Barry, May 26 2003
a(n) = a(n-1) + 2*a(n-2) + 2, a(0)=0, a(1)=1. -
Paul Barry, Jun 06 2003
Let b(n) = (-1)^(n-1)*a(n). Then b(n) = Sum_{i=1..n} i!*i*Stirling2(n,i)*(-1)^(i-1). E.g.f. of b(n): (exp(x)-1)/exp(2x). - Mario Catalani (mario.catalani(AT)unito.it), Dec 19 2003
a(n+1) = 2*a(n) + 1, a(0) = 0.
a(n) = Sum_{k=1..n} binomial(n, k).
a(n+1) = (n+1)*Sum_{k=0..n} binomial(n, k)/(k+1). -
Paul Barry, Aug 06 2004
a(n+1) = Sum_{k=0..n} binomial(n+1, k+1). -
Paul Barry, Aug 23 2004
Inverse binomial transform of
A001047. Also U sequence of Lucas sequence L(3, 2). -
Ross La Haye, Feb 07 2005
a(n) = J_n(2), where J_n is the n-th Jordan Totient function: (
A007434, is J_2).
a(n) = Sum_{d|2} d^n*mu(2/d). (End)
a(n) = det(|s(i+2,j+1)|, 1 <= i,j <= n-1), where s(n,k) are Stirling numbers of the first kind. -
Mircea Merca, Apr 06 2013
G.f.: Q(0), where Q(k) = 1 - 1/(4^k - 2*x*16^k/(2*x*4^k - 1/(1 - 1/(2*4^k - 8*x*16^k/(4*x*4^k - 1/Q(k+1)))))); (continued fraction). -
Sergei N. Gladkovskii, May 22 2013
E.g.f.: Q(0), where Q(k) = 1 - 1/(2^k - 2*x*4^k/(2*x*2^k - (k+1)/Q(k+1))); (continued fraction).
G.f.: Q(0), where Q(k) = 1 - 1/(2^k - 2*x*4^k/(2*x*2^k - 1/Q(k+1))); (continued fraction). -
Sergei N. Gladkovskii, May 23 2013
a(n) = Sum_{t_1+2*t_2+...+n*t_n=n} n*multinomial(t_1+t_2 +...+t_n,t_1,t_2,...,t_n)/(t_1+t_2 +...+t_n). -
Mircea Merca, Dec 06 2013
Sum_{n>=0} a(n)/n! =
A090142. (End)
Convolution of binomial coefficient C(n,a(k)) with itself is C(n,a(k+1)) for all k >= 3. -
Anton Zakharov, Sep 05 2016
a(n) = n + Sum_{j=1..n-1} (n-j)*2^(j-1). See a Jun 14 2017 formula for
A000918(n+1) with an interpretation. -
Wolfdieter Lang, Jun 14 2017
a(n+m) = a(n)*a(m) + a(n) + a(m). -
Yuchun Ji, Jul 27 2018
a(n+m) = a(n+1)*a(m) - 2*a(n)*a(m-1). -
Taras Goy, Dec 23 2018
a(n+1) is the determinant of n X n matrix M_(i, j) = binomial(i + j - 1, j)*2 + binomial(i+j-1, i) (empirical observation). -
Tony Foster III, May 11 2019
exp( Sum_{n >= 1} a(2*n)/a(n)*x^n/n ) = Sum_{n >= 0} a(n+1)*x^n.
The following are all examples of telescoping series:
Sum_{n >= 1} 2^n/(a(n)*a(n+1)) = 1; Sum_{n >= 1} 2^n/(a(n)*a(n+1)*a(n+2)) = 1/9.
In general, for k >= 1, Sum_{n >= 1} 2^n/(a(n)*a(n+1)*...*a(n+k)) = 1/(a(1)*a(2)*...*a(k)*a(k)).
Sum_{n >= 1} 2^n/(a(n)*a(n+2)) = 4/9, since 2^n/(a(n)*a(n+2)) = b(n) - b(n+1), where b(n) = (2/3)*(3*2^(n-1) - 1)/((2^(n+1) - 1)*(2^n - 1)).
Sum_{n >= 1} (-2)^n/(a(n)*a(n+2)) = -2/9, since (-2)^n/(a(n)*a(n+2)) = c(n) - c(n+1), where c(n) = (1/3)*(-2)^n/((2^(n+1) - 1)*(2^n - 1)).
Sum_{n >= 1} 2^n/(a(n)*a(n+4)) = 18/175, since 2^n/(a(n)*a(n+4)) = d(n) - d(n+1), where d(n) = (120*8^n - 140*4^n + 45*2^n - 4)/(15*(2^n - 1)*(2^(n+1) - 1)*(2^(n+2) - 1)*(2^(n+3) - 1)).
Sum_{n >= 1} (-2)^n/(a(n)*a(n+4)) = -26/525, since (-2)^n/(a(n)*a(n+4)) = e(n) - e(n+1), where e(n) = (-1)^n*(40*8^n - 24*4^n + 5*2^n)/(15*(2^n - 1)*(2^(n+1) - 1)*(2^(n+2) - 1)*(2^(n+3) - 1)). (End)
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