OFFSET
0,1
COMMENTS
The ratio (count of ones)/(count of zeros) in the binary expansion of a(n) is > 1/2 and <= 5 for all n > 0, this is because the division by 9 adds a repeating pattern 111000... after some binary digits.
This sequence has in its "partial binomial transform" (see formula section) no other constants than 2 and 1 despite of its more complicated looking closed form expression. This transform has a deep connection to the Grünwald-Letnikov fractional derivative if we replace the order of the derivative with the variable x: D^x*f(x).
LINKS
FORMULA
a(n) = round(((n+1)*3*2^(n+1) + 29*2^n)/9).
a(n) = 2^(n+2) + A113861(n).
a(n) = 2^(n+2) + n*2^n - A045883(n) = 2^(n+2) + n*2^n - round(((3*n+1)*2^n)/9).
a(n+1) - 2*a(n) = A001045(n+2).
A partial binomial transform in two parts:
(Partial means a diagonal in a difference table a(0), a(2)-a(1), ... . This is partial because one diagonal alone is no invertible transform.)
A001787(n+2) = Sum_{k=0..n} (-1)^(n-k)*binomial(n, k)*a(2*n-k)
= (n+2)*2^(n+1).
A052951(n+1) = Sum_{k=0..n} (-1)^(n-k)*binomial(n, k)*(a(1+2*n-k) - a(2*n-k))
= (n+2)*2^(n+1) + 2^n.
The inverse transform:
a(n+1) = Sum_{k=0..floor(n/2)} binomial(n-k, k)*(k+2)*2^(k+1)
+ Sum_{k=0..floor((n-1)/2)} binomial(n-k-1, k)*((k+2)*2^(k+1) + 2^k).
From Stefano Spezia, Oct 20 2021: (Start)
G.f.: (4 - 3*x - 6*x^2)/((1 + x)*(1 - 2*x)^2).
a(n) = 3*a(n-1) - 4*a(n-3) for n > 2. (End)
MATHEMATICA
Array[((# + 1)*3*2^(# + 1) + 29*2^# + (-1)^#)/9 &, 33, 0] (* Michael De Vlieger, Oct 19 2021 *)
LinearRecurrence[{3, 0, -4}, {4, 9, 21}, 40] (* Harvey P. Dale, Aug 12 2023 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Paul Curtz and Thomas Scheuerle, Oct 17 2021
STATUS
approved