A245663 The first number k such that the sum of the base n digits of k! does not divide k!.

10, 43, 86, 87, 188, 156, 291, 364, 432, 410, 7, 510, 4, 4, 4, 813, 4, 1079, 4, 1900, 6, 10, 6, 2330, 2147, 5, 3463, 2401, 7, 2522, 5, 3884, 5, 5, 8316, 3621, 5, 8, 8, 4866, 5, 5, 5, 5, 6302, 5, 5, 8616, 5

Offset

2,1

Comments

a(n)! > n. - Robert Israel, Aug 17 2014

Links

Table of n, a(n) for n=2..50.

Example

The sum of the base-2 digits of 10! is 1+1+0+1+1+1+0+1+0+1+1+1+1+1+0+0+0+0+0+0+0+0=11, which does not divide 10!.  Since the sum of the base-2 digits of k! divides k! for 0 <= k <= 9, a(2) = 10.
The sum of the base-3 digits of 43! is 106, which does not divide 43!.  Since the sum of the base-3 digits of k! divides k! for 0 <= k <= 42, a(3) = 43.

Maple

f:= proc(n)
  local f, k;
  for k from 1 do
    f:= k!;
    if f mod convert(convert(f, base, n), `+`) <> 0 then return k fi;
  od
end proc:
seq(f(n), n=2..30); # Robert Israel, Aug 10 2014

Mathematica

a245663[n_Integer] := Module[{f = 2, k = 2}, While[Divisible[f, Total[IntegerDigits[f, n]]] == True, k++; f = k!]; k]; a245663 /@ Range[2, 50] (* Michael De Vlieger, Aug 15 2014 *)

Prog

(Haskell)
fac :: Integer -> Integer
fac 0 = 1
fac n = foldl (*) 1 [2..n]
base 0 b = []
base a b = (a `mod` b) : base ((a-(a `mod` b)) `div` b) b
bAse a b = reverse (base a b)
sigbAse a b = foldl (+) 0 (bAse a b)
f n = [k | k <- [1..], not ((fac k) `mod` (sigbAse (fac k) n) == 0)] !! 0
main = print (map f [2..20]) -- generates values for n = 2 through 20. May be slow for values over 30.
(PARI) sumd(k, n) = my(d = digits(k, n)); sum(j=1, #d, d[j]);
a(n) = {k = 2; fk = k!; while (fk % sumd(fk, n) == 0, k++; fk = k!); k; } \\ Michel Marcus, Aug 10 2014

Crossrefs

Sum of the base n digits of k for n = 2, 3 and 10 respectively: A000120, A053735, A007953

Sequence in context: A077541 A084036 A092117 * A244802 A336288 A211070

Adjacent sequences: A245660 A245661 A245662 * A245664 A245665 A245666

Keyword

nonn,base

Author

G. H. Faust, Jul 28 2014

Status

approved