[go: up one dir, main page]

login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

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
OFFSET
0,5
LINKS
FORMULA
a(n) = A027868(A000079(n)). - Michel Marcus, Jun 01 2022
a(n) = 2^(n-2) - A055223(n) for n >= 2. - John Keith, Jun 06 2022
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)
val(n, p) = my(r=0); while(n, r+=n\=p); r \\ David A. Corneth, Jun 01 2022
(Python)
from sympy import factorial, multiplicity
def a(n): return multiplicity(5, factorial(2**n, evaluate=False))
print([a(n) for n in range(39)]) # Michael S. Branicky, Jun 01 2022
(Python)
def A354463(n):
c, m = 0, 2**n
while m >= 5:
m //= 5
c += m
return c # Chai Wah Wu, Jun 02 2022
CROSSREFS
Sequence in context: A123707 A011947 A129629 * A223136 A354603 A089526
KEYWORD
nonn,easy,base
AUTHOR
William Boyles, May 31 2022
STATUS
approved