OFFSET
1,2
LINKS
Hieronymus Fischer, Table of n, a(n) for n = 1..1000
FORMULA
a(2^n) = 2^(n*(n+1)/2) = A006125(n+1).
From Hieronymus Fischer, Aug 13 2007: (Start)
a(n) = Product_{k=0..floor(log_2(n))} floor(n/2^k), n>=1.
Recurrence:
a(n*2^m) = n^m*2^(m(m+1)/2)*a(n).
a(n) <= n^((1+log_2(n))/2) = 2^A000217(log_2(n)); equality iff n is a power of 2.
a(n) >= c(n)*(n+1)^((1 + log_2(n+1))/2) for n != 2,
where c(n) = Product_{k=1..floor(log_2(n))} (1 - 1/2^k); equality holds iff n+1 is a power of 2.
a(n) > c*(n+1)^((1 + log_2(n+1))/2)
where c = 0.288788095086602421... (see constant A048651).
lim inf a(n)/n^((1+log_2(n))/2)=0.288788095086602421... for n-->oo.
lim sup a(n)/n^((1+log_2(n))/2) = 1 for n-->oo.
lim inf a(n)/a(n+1) = 0.288788095086602421... for n-->oo (see constant A048651).
a(n) = O(n^((1+log_2(n))/2)). (End)
EXAMPLE
a(10) = floor(10/2^0)*floor(10/2^1)*floor(10/2^2)*floor(10/2^3) = 10*5*2*1 = 100;
a(17) = 1088 since 17 = 10001(base 2) and so a(17) = 10001*1000*100*10*1(base 2) = 17*8*4*2*1 = 1088.
MATHEMATICA
lst={}; Do[p=n; s=1; While[p>1, p=IntegerPart[p/2]; s*=p; ]; AppendTo[lst, s], {n, 1, 6!, 2}]; lst (* Vladimir Joseph Stephan Orlovsky, Jul 28 2009 *)
PROG
(PARI) a(n)=if(n<2, 1, n*a(floor(n/2)))
(Python)
from math import prod
def A098844(n): return n*prod(n//2**k for k in range(1, n.bit_length()-1)) # Chai Wah Wu, Jun 07 2022
CROSSREFS
KEYWORD
nonn
AUTHOR
Benoit Cloitre, Nov 03 2004
EXTENSIONS
Formula section edited by Hieronymus Fischer, Jun 13 2012
STATUS
approved