A348405 -id:A348405 - cp's OEIS frontend

4, 9, 21, 47, 105, 231, 505, 1095, 2361, 5063, 10809, 22983, 48697, 102855, 216633, 455111, 953913, 1995207, 4165177, 8679879, 18058809, 37515719, 77827641, 161247687, 333680185, 689729991, 1424199225, 2937876935, 6054710841, 12467335623, 25650499129, 52732654023, 108328619577

Offset: 0

Paul Curtz and Thomas Scheuerle, Oct 17 2021

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).

Index entries for linear recurrences with constant coefficients, signature (3,0,-4).

Cf. A001045, A001787, A034007, A045883, A052951, A113861, A348405.

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 *)

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(n) = A034007(n+3) + A045883(n-1) for n > 0.
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)

cp's OEIS Frontend

A348407 a(n) = ((n+1)32^(n+1) + 29*2^n + (-1)^n)/9.

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