OFFSET
1,2
COMMENTS
If n is written in base-5 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).
LINKS
Harvey P. Dale, Table of n, a(n) for n = 1..1000
FORMULA
Recurrence: a(n)=n*a(floor(n/5)); a(n*5^m)=n^m*5^(m(m+1)/2)*a(n).
a(k*5^m)=k^(m+1)*5^(m(m+1)/2), for 0<k<5.
Asymptotic behavior: a(n)=O(n^((1+log_5(n))/2)); this follows from the inequalities below.
a(n)<=b(n), where b(n)=n^(1+floor(log_5(n)))/5^((1+floor(log_5(n)))*floor(log_5(n))/2); equality holds for n=k*5^m, 0<k<5, m>=0. b(n) can also be written n^(1+floor(log_5(n)))/5^A000217(floor(log_5(n))).
Also: a(n)<=2^((1-log_5(2))/2)*n^((1+log_5(n))/2)=1.2181246...*5^A000217(log_5(n)), equality holds for n=2*5^m, m>=0.
a(n)>c*b(n), where c=0.438796837203638531... (see constant A132021).
Also: a(n)>c*(sqrt(2)/2^log_5(sqrt(2)))*n^((1+log_5(n))/2)=0.534509224...*5^A000217(log_5(n)).
lim inf a(n)/b(n)=0.438796837203638531..., for n-->oo.
lim sup a(n)/b(n)=1, for n-->oo.
lim inf a(n)/n^((1+log_5(n))/2)=0.438796837203638531...*sqrt(2)/2^log_5(sqrt(2)), for n-->oo.
lim sup a(n)/n^((1+log_5(n))/2)=sqrt(2)/2^log_5(sqrt(2))=1.2181246..., for n-->oo.
lim inf a(n)/a(n+1)=0.438796837203638531... for n-->oo (see constant A132021).
EXAMPLE
a(26)=floor(26/5^0)*floor(26/5^1)*floor(26/5^2)=26*5*1=130; a(34)=204 since 34=114(base-5) and so a(34)=114*11*1(base-5)=34*6*1=204.
MATHEMATICA
Table[Product[Floor[n/5^k], {k, 0, Floor[Log[5, n]]}], {n, 60}] (* Harvey P. Dale, Oct 16 2019 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Hieronymus Fischer, Aug 20 2007
STATUS
approved