a(n) = (2*n^2-1+(-1)^n)/8. -
Paul Barry, May 27 2003
G.f.: x^2/((1-x)^2*(1-x^2)) = x^2 / ( (1+x)*(1-x)^3 ). -
Simon Plouffe in his 1992 dissertation, leading zeros dropped
E.g.f.: exp(x)*(2*x^2+2*x-1)/8 + exp(-x)/8.
a(-n) = a(n) for all n in Z.
a(n) = a(n-1) + a(n-2) - a(n-3) + 1 [with a(-1) = a(0) = a(1) = 0], a(2k) = k^2, a(2k-1) = k(k-1). -
Henry Bottomley, Mar 08 2000
0*0, 0*1, 1*1, 1*2, 2*2, 2*3, 3*3, 3*4, ... with an obvious pattern.
a(n) = Sum_{k=1..n} floor(k/2). - Yong Kong (ykong(AT)curagen.com), Mar 10 2001
a(n) = n*floor((n-1)/2) - floor((n-1)/2)*(floor((n-1)/2)+ 1); a(n) = a(n-2) + n-2 with a(1) = 0, a(2) = 0. -
Santi Spadaro, Jul 13 2001
Also: a(n) = binomial(n, 2) - a(n-1) =
A000217(n-1) - a(n-1) with a(0) = 0. -
Labos Elemer, Apr 26 2003
a(n) = Sum_{k=0..n} (-1)^(n-k)*C(k, 2). -
Paul Barry, Jul 01 2003
a(n) = (-1)^n * partial sum of alternating triangular numbers. -
Jon Perry, Dec 30 2003
a(n) = a(n-2) + n - 1, n > 1. -
Paul Barry, Jul 14 2004
a(n+1) = Sum_{i=0..n} min(i, n-i). -
Marc LeBrun, Feb 15 2005
a(n+1) = Sum_{k = 0..floor((n-1)/2)} n-2k; a(n+1) = Sum_{k=0..n} k*(1-(-1)^(n+k-1))/2. -
Paul Barry, Apr 16 2005
1 + 1/(1 + 2/(1 + 4/(1 + 6/(1 + 9/(1 + 12/(1 + 16/(1 + ...))))))) = 6/(Pi^2 - 6) = 1.550546096730... -
Philippe Deléham, Jun 20 2005
Sequence starting (2, 2, 4, 6, 9, ...) =
A128174 (as an infinite lower triangular matrix) * vector [1, 2, 3, ...]; where
A128174 = (1; 0,1; 1,0,1; 0,1,0,1; ...). -
Gary W. Adamson, Jul 27 2007
a(n) = Sum_{i=k..n} P(i, k) where P(i, k) is the number of partitions of i into k parts. -
Thomas Wieder, Sep 01 2007
a(n) = round((2*n^2-1)/8) = round(n^2/4) = ceiling((n^2-1)/4). -
Mircea Merca, Nov 29 2010
n*a(n+2) = 2*a(n+1) + (n+2)*a(n). Holonomic Ansatz with smallest order of recurrence. -
Thotsaporn Thanatipanonda, Dec 12 2010
a(n) = floor(b(n)) with b(n) = b(n-1) + n/(1+e^(1/n)) and b(0)= 0. -
Richard R. Forberg, Jun 08 2013
a(n) = floor((n+2)/2 - 1)*(floor((n+2)/2)-1 + (n+2) mod 2). -
Wesley Ivan Hurt, Jun 09 2013
0 = a(n)*a(n+2) + a(n+1)*(-2*a(n+2) + a(n+3)) for all integers n. -
Michael Somos, Nov 22 2014
a(n) = Sum_{j=1..n} Sum_{i=1..n} ceiling((i+j-n-1)/2). -
Wesley Ivan Hurt, Mar 12 2015
a(n) = a(n-3) + floor(3*n/2) - 2. -
Yuchun Ji, Aug 14 2020
Sum_{n>=2} (-1)^n/a(n) = Pi^2/6 - 1. -
Amiram Eldar, Mar 10 2022
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