A354463 - OEIS

%I #32 Jun 25 2022 21:44:36

%S 0,0,0,1,3,7,14,31,63,126,253,509,1021,2045,4094,8189,16380,32763,

%T 65531,131067,262140,524285,1048571,2097146,4194297,8388603,16777208,

%U 33554424,67108858,134217720,268435446,536870902,1073741816,2147483642,4294967289,8589934584,17179869176,34359738358,68719476729

%N a(n) is the number of trailing zeros in (2^n)!.

%H William Boyles, <a href="/A354463/b354463.txt">Table of n, a(n) for n = 0..650</a>

%F a(n) = A027868(A000079(n)). - _Michel Marcus_, Jun 01 2022

%F a(n) = 2^(n-2) - A055223(n) for n >= 2. - _John Keith_, Jun 06 2022

%e For n = 4, (2^4)! = 20922789888000, which has a(4) = 3 trailing zeros.

%t a[n_]:=IntegerExponent[(2^n)!]; Array[a,38,0] (* _Stefano Spezia_, Jun 01 2022 *)

%o (Haskell)

%o seq n = aux (2 ^ n) 0

%o   where

%o     aux x acc

%o       | x < 5 = acc

%o       | otherwise = aux y (acc + y)

%o       where

%o         y = x `div` 5

%o (PARI) a(n) = val(1<<n, 5)

%o val(n, p) = my(r=0); while(n, r+=n\=p); r \\ _David A. Corneth_, Jun 01 2022

%o (Python)

%o from sympy import factorial, multiplicity

%o def a(n): return multiplicity(5, factorial(2**n, evaluate=False))

%o print([a(n) for n in range(39)]) # _Michael S. Branicky_, Jun 01 2022

%o (Python)

%o def A354463(n):

%o     c, m = 0, 2**n

%o     while m >= 5:

%o         m //= 5

%o         c += m

%o     return c # _Chai Wah Wu_, Jun 02 2022

%Y Cf. A000079, A027868.

%K nonn,easy,base

%O 0,5

%A _William Boyles_, May 31 2022