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 A107663 a(2n) = 2*4^n-1, a(2n+1) = (2^(n+1)+1)^2; interlaces A083420 with A028400. 1

%I

%S 1,9,7,25,31,81,127,289,511,1089,2047,4225,8191,16641,32767,66049,

%T 131071,263169,524287,1050625,2097151,4198401,8388607,16785409,

%U 33554431,67125249,134217727,268468225,536870911,1073807361,2147483647

%N a(2n) = 2*4^n-1, a(2n+1) = (2^(n+1)+1)^2; interlaces A083420 with A028400.

%C a(2n) = A085903(2n) = A083420(n).

%H Colin Barker, <a href="/A107663/b107663.txt">Table of n, a(n) for n = 0..1000</a>

%H H. Bottomley, <a href="/A060919/a060919.gif">Illustration of initial terms (A028400)</a>

%H I. Strazdins, <a href="https://doi.org/10.1023/A:1005769927571">Universal affine classification of Boolean functions</a>, Acta Applic. Math. 46 (1997), 147-167.

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (1,4,-2,-4).

%F G.f.: (-1-8*x+6*x^2+16*x^3) / ((1-2*x)*(x+1)*(2*x^2-1)).

%F From _Colin Barker_, May 21 2019: (Start)

%F a(n) = a(n-1) + 4*a(n-2) - 2*a(n-3) - 4*a(n-4) for n>3.

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

%F (End)

%o Floretion Algebra Multiplication Program, FAMP Code: 4tesseq[A*B] with A = + .25'i + .25'j + .25'k + .25i' + .25j' + .25k' + .25'ii' + .25'jj' + .25'kk' + .25'ij' + .25'ik' + .25'ji' + .25'jk' + .25'ki' + .25'kj' + .25e and B = + .5'i + .5i' + 'ii' + e [Factor added to formula by _Creighton Dement_, Dec 11 2009]

%o (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

%Y Cf. A083420, A028400, A062510, A085903.

%K easy,nonn,changed

%O 0,2

%A _Creighton Dement_, May 19 2005

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Last modified May 23 19:07 EDT 2019. Contains 323528 sequences. (Running on oeis4.)