A223925 a(2n+1) = 2*n-1; a(2n)= 4^n.

1, 4, 3, 16, 5, 64, 7, 256, 9, 1024, 11, 4096, 13, 16384, 15, 65536, 17, 262144, 19, 1048576, 21, 4194304, 23, 16777216, 25, 67108864, 27, 268435456, 29, 1073741824, 31

Offset: 1

Author

Paul Curtz, Mar 29 2013

Keywords

• nonn
• easy
• less

Comments

If A132050(n) has offset 1 (proposed),

A132049(n)/A132050(n) = 2, 4, 3, 16/5, 25/8, 192/61,... leads to Pi (Euler, 1735)

A132049(n)/a(n) = (2/1=2, 4/4=1, 3/3=1, 16/16=1, 25/5=5, 192/64=3,... ). The second bisection 1, 1, 3, 17, 155, begins like A110501.

Conjecture: a(2n) is always a divisor of A132049(2n).

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (0,6,0,-9,0,4).

Crossrefs

Cf. A005408, A000302, A132050, A100102.

Programs

Mathematica

Table[ If[ OddQ[n], n, 4^(n/2)], {n, 1, 31}] (* Jean-François Alcover, Apr 02 2013 *)
CoefficientList[Series[(1 + 4 x - 3 x^2 - 8 x^3 - 4 x^4 + 4 x^5) / ((1 - x)^2 (1 + x)^2 (1 - 2 x) (1 + 2 x)), {x, 0, 35}], x] (* Vincenzo Librandi, Jul 20 2013 *)
LinearRecurrence[{0,6,0,-9,0,4},{1,4,3,16,5,64},40] (* Harvey P. Dale, Jul 30 2018 *)

Formula

G.f.: x*(1+4*x-3*x^2-8*x^3-4*x^4+4*x^5)/((1-x)^2*(1+x)^2*(1-2x)*(1+2x)). - Philippe Deléham, Apr 01 2013

a(n) = 6*a(n-2) -9*a(n-4) + 4*a(n-6) with a(1) = 1, a(2) = 4, a(3) = 3, a(4) = 16, a(5) = 5, a(6) = 64. - Philippe Deléham, Apr 01 2013

Extensions

Conjecture about A132049(n)/a(n) modified by Jean-François Alcover, Apr 12 2013