A281500 Reduced denominators of f(n) = (n+1)/(2^(2+n)-2) with A026741(n+1) as numerators.

2, 3, 14, 15, 62, 63, 254, 255, 1022, 1023, 4094, 4095, 16382, 16383, 65534, 65535, 262142, 262143, 1048574, 1048575, 4194302, 4194303, 16777214, 16777215, 67108862, 67108863, 268435454, 268435455, 1073741822, 1073741823, 4294967294, 4294967295, 17179869182, 17179869183

Offset: 0

Author

Paul Curtz, Jan 23 2017

Keywords

• nonn
• easy

Comments

f(n) = (n+1)/A000918(n+2) = 1/2, 2/6, 3/14, 4/30, 5/62, 6/126, 7/254, 8/510, 9/1022, 10/2046, 11/4094, 12/8190, ... .

Partial reduction: 1/2, 1/3, 3/14, 2/15, 5/62, 3/63, 7/254, 4/255, 9/1022, 5/1023, 11/4094, 6/4095, ... = A026741(n+1)/a(n).

Full reduction: 1/2, 1/3, 3/14, 2/15, 5/62, 1/21, 7/254, ... = A111701(n+1)/(2, 3, 14, 15, 62, 21, ... )

A164555(n+1)/A027642(n) = 1/2, 1/6, 0, -1/30, 0, 1/42, ... = f(n) * A198631(n)/A006519(n+1) = 1, 1/2, 0, -1/4, 0, 1/2, ... .).

Via f(n), we go from the second fractional Euler numbers to the second Bernoulli numbers.

a(n) mod 10: periodic sequence of length 4: repeat [2, 3, 4, 5].

a(n) differences table:

. 2, 3, 14, 15, 62, 63, 254, 255, ...

. 1, 11, 1, 47, 1, 191, 1, 767, ... see A198693

. 10, -10, 46, -46, 190, -190, 766, -766, ... see A096045, from Bernoulli(2n).

Extension of a(n): a(-2) = -1, a(-1) = 0.

Links

• Colin Barker, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (0,5,0,-4).

Crossrefs

Cf. A000027, A000918, A001477, A006519, A026741, A027642, A096045, A111701, A117856, A164555, A198631, A198693, A209308, A267921.

Programs

Mathematica

a[n_] := (3+(-1)^n)*(2^(n+1)-1)/2; (* or *) a[n_] := If[EvenQ[n], 4^(n/2+1)-2, 4^((n+1)/2)-1]; Table[a[n], {n, 0, 35}] (* Jean-François Alcover, Jan 24 2017 *)

PARI

Vec((2 + 3*x + 4*x^2) / ((1 - x)*(1 + x)*(1 - 2*x)*(1 + 2*x)) + O(x^50)) \\ Colin Barker, Jan 24 2017

Formula

From Colin Barker, Jan 24 2017: (Start)

G.f.: (2 + 3*x + 4*x^2) / ((1 - x)*(1 + x)*(1 - 2*x)*(1 + 2*x)).

a(n) = 5*a(n-2) - 4*a(n-4) for n>3. (End)

From Jean-François Alcover, Jan 24 2017: (Start)

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

a(n) = 4^((n + 1 + ((n + 1) mod 2))/2) - 1 - ((n + 1) mod 2). (End)

a(n) = a(n-2) + A117856(n+1) for n>1.

a(2*k) = 4^(k + 1) - 2, a(2*k+1) = a(2*k) + 1 = 4^(k+1) - 1.

a(2*k) + a(2*k+1) = A267921(k+1).