OFFSET
11,1
COMMENTS
(sum of digits of n!!) / 9 is an integer for n = 9 and n > 10.
FORMULA
a(n) = A120390(n)/9 for n > 10.
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]
MATHEMATICA
Table[Sum[DigitCount[n!!][[i]]*i/9, {i, 1, 9}], {n, 11, 100}]
PROG
(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
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Michel Lagneau, Dec 23 2011
EXTENSIONS
a(37), a(41), a(45), a(46) corrected by Georg Fischer, Jul 15 2024
STATUS
approved