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a(1) = 1; for m > 0, a(2m) = 2m, a(2m+1) = 4m+2.
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%I #23 Feb 19 2024 03:32:46

%S 1,2,6,4,10,6,14,8,18,10,22,12,26,14,30,16,34,18,38,20,42,22,46,24,50,

%T 26,54,28,58,30,62,32,66,34,70,36,74,38,78,40,82,42,86,44,90,46,94,48,

%U 98,50,102,52,106,54,110,56,114,58,118,60,122,62,126,64,130,66,134,68

%N a(1) = 1; for m > 0, a(2m) = 2m, a(2m+1) = 4m+2.

%C Length of period of sequences r(k,n) = floor(sinh(1)*k!) - n*floor(sinh(1)*k!/n) when n is fixed. - _Benoit Cloitre_, Jun 22 2003

%H Harry J. Smith, <a href="/A066043/b066043.txt">Table of n, a(n) for n = 1..1000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0,2,0,-1).

%F O.g.f.: (x+2x^2+4x^3-x^5)/(1-x^2)^2. - _Len Smiley_, Dec 05 2001

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

%F From _Harry J. Smith_, Nov 08 2009: (Start)

%F a(n) = A109043(n), n > 1.

%F a(n) = 2*A026741(n), n > 1. (End)

%e r(k,7) is sequence 1, 2, 0, 0, 1, 6, 1, 1, 3, 2, 2, 3, 5, 0, 1, 2, 0, 0, 1, 6, 1, 1, 3, 2, 2, 3, 5, 0.... which is periodic with period (1, 2, 0, 0, 1, 6, 1, 1, 3, 2, 2, 3, 5, 0) of length 14 = a(7).

%t Join[{1}, LCM[Range[2, 100], 2]] (* _Paolo Xausa_, Feb 19 2024 *)

%o (PARI) a(n)=if(n<2,1,if(n%2,2*n,n))

%o (PARI) { for (n=1, 1000, a=if (n>1 && n%2, 2*n, n); write("b066043.txt", n, " ", a) ) } \\ _Harry J. Smith_, Nov 08 2009

%K easy,nonn

%O 1,2

%A _George E. Antoniou_, Nov 30 2001