OFFSET
1,1
COMMENTS
For any n, either a(n+1)-a(n)=0 or a(n+1)-a(n)=1.
a(n)/b(n) tends to 1 - 2/Pi as n tends to infinity, where b(n) is the n-th term of A159061.
REFERENCES
M. Griffiths, The Backbone of Pascal's Triangle, United Kingdom Mathematics Trust, 2008, pp. 68-72.
LINKS
M. Griffiths, How many children?, Math. Gaz., 90 (2006), 146-149.
FORMULA
a(n) is the nearest integer to 2*n*(1+binomial(2*n,n)/(2^(2*n)))-((n*binomial(2*n,n))/(2^(2*n-1)))^2.
MATHEMATICA
f[n_] := Round[2^(1 - 4 n) n (16^n + Binomial[2 n, n] (4^n - 2 n Binomial[2 n, n]))]; Array[f, 72]
PROG
(PARI) a(n) = round(2*n*(1+binomial(2*n, n)/(2^(2*n)))-((n*binomial(2*n, n))/(2^(2*n-1)))^2) \\ Felix Fröhlich, Jan 23 2019
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Martin Griffiths, Apr 04 2009
EXTENSIONS
More terms from Robert G. Wilson v, Apr 05 2009
Formula clarified by the author, Apr 06 2009
STATUS
approved