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 A223925 a(2n+1) = 2*n-1; a(2n)= 4^n. 2
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 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). 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 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 *) CROSSREFS Cf. A005408, A000302, A132050, A100102. Sequence in context: A322144 A272580 A065679 * A062776 A269512 A270128 Adjacent sequences: A223922 A223923 A223924 * A223926 A223927 A223928 KEYWORD nonn,easy,less AUTHOR Paul Curtz, Mar 29 2013 EXTENSIONS Conjecture about A132049(n)/a(n) modified by Jean-François Alcover, Apr 12 2013 STATUS approved

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Last modified February 8 11:17 EST 2023. Contains 360138 sequences. (Running on oeis4.)