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%I #21 Jul 15 2024 05:04:28
%S 2,2,2,2,2,2,4,5,5,4,5,5,4,6,7,5,8,5,8,8,10,10,10,9,12,9,12,9,14,10,
%T 12,12,16,9,16,10,15,13,16,14,17,12,20,15,20,15,19,20,20,19,22,20,22,
%U 18,21,21,24,25,27,19,27,23,26,23,28,27,25,23,27,23,29,31
%N (Sum of digits of n!!) / 9.
%C (sum of digits of n!!) / 9 is an integer for n = 9 and n > 10.
%F a(n) = A120390(n)/9 for n > 10.
%F a(n) << n log n. Presumably a(n) ~ n log n but proving this requires showing that not too many digits are 0. (The trailing 0's in even terms are not a problem, being only about n/8.) The expected constant is 1 / (4 log 10) = 0.10857.... [_Charles R Greathouse IV_, Dec 23 2011]
%t Table[Sum[DigitCount[n!!][[i]]*i/9,{i,1,9}],{n,11,100}]
%o (PARI) a(n)=my(v=eval(Vec(Str(prod(k=1,n\2,2*k+n%2)))));sum(i=1,#v,v[i])/9 \\ _Charles R Greathouse IV_, Dec 23 2011
%Y Cf. A120390.
%K nonn,base
%O 11,1
%A _Michel Lagneau_, Dec 23 2011
%E a(37), a(41), a(45), a(46) corrected by _Georg Fischer_, Jul 15 2024