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Product{0<=k<=floor(log_4(n)), floor(n/4^k)}, n>=1.
5

%I #2 Mar 31 2012 13:21:03

%S 1,2,3,4,5,6,7,16,18,20,22,36,39,42,45,64,68,72,76,100,105,110,115,

%T 144,150,156,162,196,203,210,217,512,528,544,560,648,666,684,702,800,

%U 820,840,860,968,990,1012,1034,1728,1764,1800,1836,2028,2067,2106,2145,2352

%N Product{0<=k<=floor(log_4(n)), floor(n/4^k)}, n>=1.

%C If n is written in base-4 as n=d(m)d(m-1)d(m-2)...d(2)d(1)d(0) (where d(k) is the digit at position k) then a(n) is also the product d(m)d(m-1)d(m-2)...d(2)d(1)d(0)*d(m)d(m-1)d(m-2)...d(2)d(1)*d(m)d(m-1)d(m-2)...d(2)*...*d(m)d(m-1)d(m-2)*d(m)d(m-1)*d(m).

%F Recurrence: a(n)=n*a(floor(n/4)); a(n*4^m)=n^m*4^(m(m+1)/2)*a(n).

%F a(k*4^m)=k^(m+1)*4^(m(m+1)/2), for 0<k<4.

%F Asymptotic behavior: a(n)=O(n^((1+log_4(n))/2)); this follows from the inequalities below.

%F a(n)<=b(n), where b(n)=n^(1+floor(log_4(n)))/4^((1+floor(log_4(n)))*floor(log_4(n))/2); equality holds for n=k*4^m, 0<k<4, m>=0. b(n) can also be written n^(1+floor(log_4(n)))/4^A000217(floor(log_4(n))).

%F Also: a(n)<=2^(1/4)*n^((1+log_4(n))/2)=1.189207...*4^A000217(log_4(n)), equality holds for n=2*4^m, m>=0.

%F a(n)>c*b(n), where c=0.4194224417951075977... (see constant A132020).

%F Also: a(n)>c*2^(1/4)*n^((1+log_4(n))/2)=0.498780...*4^A000217(log_4(n)).

%F lim inf a(n)/b(n)=0.4194224417951075977..., for n-->oo.

%F lim sup a(n)/b(n)=1, for n-->oo.

%F lim inf a(n)/n^((1+log_4(n))/2)=0.4194224417951075977...*2^(1/4), for n-->oo.

%F lim sup a(n)/n^((1+log_4(n))/2)=2^(1/4), for n-->oo.

%F lim inf a(n)/a(n+1)=0.4194224417951075977... for n-->oo (see constant A132020).

%e a(26)=floor(26/4^0)*floor(26/4^1)*floor(26/4^2)=26*6*1=156; a(34)=544 since 34=202(base-4) and so

%e a(34)=202*20*2(base-4)=34*8*2=544.

%Y Cf. A048651, A132020, A100221, A000217.

%Y For formulas regarding a general parameter p (i.e. terms floor(n/p^k)) see A132264.

%Y For the product of terms floor(n/p^k) for p=2 to p=12 see A098844(p=2), A132027(p=3)-A132033(p=9), A067080(p=10), A132263(p=11), A132264(p=12).

%Y For the products of terms 1+floor(n/p^k) see A132269-A132272, A132327, A132328.

%K nonn

%O 1,2

%A _Hieronymus Fischer_, Aug 20 2007