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%I #14 Oct 06 2020 02:09:48
%S 6,30,78,246,582,1830,4758,14358,41118,122430,360438,1079742,3213222,
%T 9624990,28798926,86319894,258673542,775718310,2326095798,6976966422,
%U 20927018814,62775429150,188311523478,564911643486,1694677889958
%N a(n) = Sum_{d|n} (2^d*3^(n/d)).
%e For n = 3, the divisors are 1 and 3, the sum is 2^1 * 3^3 + 2^3 * 3^1 = 54 + 24 = 78.
%t Table[Total[2^Divisors[n] 3^(n/Divisors[n])],{n,30}] (* _Harvey P. Dale_, Nov 08 2017 *)
%o (PARI) a(n) = sumdiv(n, d, 2^d*3^(n/d)); \\ _Michel Marcus_, Oct 05 2020
%Y Equals 6*A034735(n).
%K nonn,easy
%O 1,1
%A _Christian G. Bower_
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