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
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
Programs
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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).
Comments