|
|
A354463
|
|
a(n) is the number of trailing zeros in (2^n)!.
|
|
1
|
|
|
0, 0, 0, 1, 3, 7, 14, 31, 63, 126, 253, 509, 1021, 2045, 4094, 8189, 16380, 32763, 65531, 131067, 262140, 524285, 1048571, 2097146, 4194297, 8388603, 16777208, 33554424, 67108858, 134217720, 268435446, 536870902, 1073741816, 2147483642, 4294967289, 8589934584, 17179869176, 34359738358, 68719476729
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,5
|
|
LINKS
|
|
|
FORMULA
|
|
|
EXAMPLE
|
For n = 4, (2^4)! = 20922789888000, which has a(4) = 3 trailing zeros.
|
|
MATHEMATICA
|
a[n_]:=IntegerExponent[(2^n)!]; Array[a, 38, 0] (* Stefano Spezia, Jun 01 2022 *)
|
|
PROG
|
(Haskell)
seq n = aux (2 ^ n) 0
where
aux x acc
| x < 5 = acc
| otherwise = aux y (acc + y)
where
y = x `div` 5
(PARI) a(n) = val(1<<n, 5)
(Python)
from sympy import factorial, multiplicity
def a(n): return multiplicity(5, factorial(2**n, evaluate=False))
(Python)
c, m = 0, 2**n
while m >= 5:
m //= 5
c += m
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,easy,base
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|