a(n) = a(n-1) + 2*n, a(0) = 0.
Sum_{n >= 1} a(n) = n*(n+1)*(n+2)/3 (cf.
A007290, partial sums).
Sum_{n >= 1} 1/a(n) = 1. (Cf. Tijdeman)
Sum_{n >= 1} (-1)^(n+1)/a(n) = log(4) - 1 =
A016627 - 1 [Jolley eq (235)].
1 = 1/2 + Sum_{n >= 1} 1/(2*a(n)) = 1/2 + 1/4 + 1/12 + 1/24 + 1/40 + 1/60 + ... with partial sums: 1/2, 3/4, 5/6, 7/8, 9/10, 11/12, 13/14, ... -
Gary W. Adamson, Jun 16 2003
a(n)*a(n+1) = a(n*(n+2)); e.g., a(3)*a(4) = 12*20 = 240 = a(3*5). -
Charlie Marion, Dec 29 2003
Log 2 = Sum_{n >= 0} 1/a(2n+1) = 1/2 + 1/12 + 1/30 + 1/56 + 1/90 + ... = (1 - 1/2) + (1/3 - 1/4) + (1/5 - 1/6) + (1/7 - 1/8) + ... = Sum_{n >= 0} (-1)^n/(n+1) =
A002162. -
Gary W. Adamson, Jun 22 2003
(2, 6, 12, 20, 30, ...) = binomial transform of (2, 4, 2). -
Gary W. Adamson, Nov 28 2007
a(0) = 0, a(n) = a(n-1) + 1 + floor(x), where x is the minimal positive solution to fract(sqrt(a(n-1) + 1 + x)) = 1/2. -
Hieronymus Fischer, Dec 31 2008
E.g.f.: ((-x+1)*log(-x+1)+x)/x^2 also Integral_{x = 0..1} ((-x+1)*log(-x+1) + x)/x^2 = zeta(2) - 1. -
Stephen Crowley, Jul 11 2009
a(n-1) = floor(n^5/(n^3 + n^2 + 1)). -
Gary Detlefs, Feb 11 2010
For n > 0 a(n) = 1/(Integral_{x=0..Pi/2} 2*(sin(x))^(2*n-1)*(cos(x))^3). -
Francesco Daddi, Aug 02 2011
Sum_{n >= 1} 1/(a(n))^(2s) = Sum_{t = 1..2*s} binomial(4*s - t - 1, 2*s - 1) * ( (1 + (-1)^t)*zeta(t) - 1). See Arxiv:1301.6293. -
R. J. Mathar, Feb 03 2013
a(n) = floor(n^2 * e^(1/n)) and a(n-1) = floor(n^2 / e^(1/n)). -
Richard R. Forberg, Jun 22 2013
Binomial transform of [0, 2, 2, 0, 0, 0, ...]. -
Alois P. Heinz, Mar 10 2015
For n > 0, a(n) = 1/(Integral_{x=0..1} (x^(n-1) - x^n) dx). -
Rick L. Shepherd, Oct 26 2015
For n > 0, a(n) = lim_{m -> oo} (1/m)*1/(Sum_{i=m*n..m*(n+1)} 1/i^2), with error of ~1/m. -
Richard R. Forberg, Jul 27 2016
Dirichlet g.f.: zeta(s-2) + zeta(s-1).
Convolution of nonnegative integers (
A001477) and constant sequence (
A007395).
Sum_{n >= 0} a(n)/n! = 3*exp(1). (End)
a(n)*a(n+2k-1) + (n+k)^2 = ((2n+1)*k + n^2)^2.
a(n)*a(n+2k) + k^2 = ((2n+1)*k + a(n))^2. (End)
Product_{n>=1} (1 + 1/a(n)) = cosh(sqrt(3)*Pi/2)/Pi. -
Amiram Eldar, Jan 20 2021
A generalization of the Dec 29 2003 formula, a(n)*a(n+1) = a(n*(n+2)), follows. a(n)*a(n+k) = a(n*(n+k+1)) + (k-1)*n*(n+k+1). -
Charlie Marion, Jan 02 2023
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