A107663 a(2n) = 2*4^n-1, a(2n+1) = (2^(n+1)+1)^2; interlaces A083420 with A028400.
1, 9, 7, 25, 31, 81, 127, 289, 511, 1089, 2047, 4225, 8191, 16641, 32767, 66049, 131071, 263169, 524287, 1050625, 2097151, 4198401, 8388607, 16785409, 33554431, 67125249, 134217727, 268468225, 536870911, 1073807361, 2147483647
Offset: 0
Links
- Colin Barker, Table of n, a(n) for n = 0..1000
- Henry Bottomley, Illustration of initial terms (A028400)
- I. Strazdins, Universal affine classification of Boolean functions, Acta Applic. Math. 46 (1997), 147-167.
- Index entries for linear recurrences with constant coefficients, signature (1,4,-2,-4).
Programs
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PARI
Vec((1 + 8*x - 6*x^2 - 16*x^3) / ((1 + x)*(1 - 2*x)*(1 - 2*x^2)) + O(x^35)) \\ Colin Barker, May 21 2019
Formula
G.f.: (-1-8*x+6*x^2+16*x^3) / ((1-2*x)*(x+1)*(2*x^2-1)).
From Colin Barker, May 21 2019: (Start)
a(n) = a(n-1) + 4*a(n-2) - 2*a(n-3) - 4*a(n-4) for n>3.
a(n) = ((-1)^(1+n) + 2^(1+n) + 2^((1+n)/2)*(1+(-1)^(1+n))).
(End)
Comments