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Sum of terms in simple continued fraction for n!!/(n-1)!!, where n!! is a double factorial, n(n-2)(n-4)....
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%I #11 Mar 28 2015 15:36:34

%S 1,2,3,5,9,8,10,17,19,23,15,18,28,19,43,33,59,34,65,39,104,46,44,46,

%T 148,61,64,62,250,167,219,79,63,122,221,106,78,120,247,152,94,99,133,

%U 97,134,98,637,126,216,660,2159,195,150,209,283,701,138,164,290,218

%N Sum of terms in simple continued fraction for n!!/(n-1)!!, where n!! is a double factorial, n(n-2)(n-4)....

%e 4!!/3!! = 2*4/(1*3) = 8/3 = 2 + 1/(1 + 1/2), so the 4th term is 2 + 1 + 2 = 5.

%K easy,nonn

%O 1,2

%A _Leroy Quet_, Dec 10 2000