A348407 a(n) = ((n+1)*3*2^(n+1) + 29*2^n + (-1)^n)/9.

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

Author

Paul Curtz and Thomas Scheuerle, Oct 17 2021

Keywords

• nonn
• easy

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

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

Crossrefs

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

Programs

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 *)

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(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)