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A186187 Period 8 sequence [ 2, 2, 1, 2, 4, 2, 1, 2, ...] except a(0) = 1. 0

%I #36 Jul 21 2023 06:01:15

%S 1,2,1,2,4,2,1,2,2,2,1,2,4,2,1,2,2,2,1,2,4,2,1,2,2,2,1,2,4,2,1,2,2,2,

%T 1,2,4,2,1,2,2,2,1,2,4,2,1,2,2,2,1,2,4,2,1,2,2,2,1,2,4,2,1,2,2,2,1,2,

%U 4,2,1,2,2,2,1,2,4,2,1,2,2,2,1,2,4,2,1,2,2,2,1,2,4,2,1,2,2,2,1,2,4,2,1,2,2

%N Period 8 sequence [ 2, 2, 1, 2, 4, 2, 1, 2, ...] except a(0) = 1.

%C Also continued fraction expansion of sqrt(2717)/38. - _Bruno Berselli_, Mar 07 2011

%H Michael Somos, <a href="http://cis.csuohio.edu/~somos/rfmc.txt">Rational Function Multiplicative Coefficients</a>

%H <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (1,-1,1,-1,1,-1,1).

%F Euler transform of length 8 sequence [ 2, -2, 2, 0, 0, -2, 0, 1].

%F Moebius transform is length 8 sequence [ 2, -1, 0, 3, 0, 0, 0, -2].

%F a(n) = 2 * b(n) where b() is multiplicative with b(2) = 1/2, b(4) = 2, b(2^e) = 1 if e>2, b(p^e) = 1 if p>2.

%F G.f.: (1 + x)^4 * (1 - x + x^2)^2 / (1 - x^8) = (1-x+x^2)^2*(1+x)^3 / ((1-x) *(1+x^2) *(1+x^4)). a(-n) = a(n). a(2*n + 1) = 2, a(4*n + 2) = 1, a(8*n + 4) = 4, a(8*n) = 2 except a(0) = 1.

%F a(n) = A056594(n)-A014017(n)+2 for n>0. - Bruno Berselli, Feb 15 2011

%e 1 + 2*x + x^2 + 2*x^3 + 4*x^4 + 2*x^5 + x^6 + 2*x^7 + 2*x^8 + 2*x^9 + ...

%t PadRight[{1},108,{2,2,1,2,4,2,1,2}] (* _Harvey P. Dale_, Mar 22 2012 *)

%o (PARI) {a(n) = - (n==0) + [ 2, 2, 1, 2, 4, 2, 1, 2] [n%8 + 1]}

%o (PARI) {a(n) = polcoeff( (1 + x)^4 * (1 - x + x^2)^2 / (1 - x^8) + x * O(x^abs(n)), abs(n))}

%o (Magma) [1] cat &cat[ [2, 1, 2, 4, 2, 1, 2, 2]: n in [1..13]]; // _Bruno Berselli_, Mar 07 2011

%K nonn,easy

%O 0,2

%A _Michael Somos_, Feb 14 2011

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Last modified May 3 21:49 EDT 2024. Contains 372225 sequences. (Running on oeis4.)