STATUS
reviewed
approved
The OEIS is supported by the many generous donors to the OEIS Foundation.
reviewed
approved
proposed
reviewed
editing
proposed
a(n) = least integer j >= 0 such that n =Floor[ floor((5^j)/(3^k)] ) for some integer k >= 0.
1 = floor(5^0 / 3^0),
2 = floor(5^2 / 3^2),
3 = floor(5^7 / 3^9),
4 = floor(5^3 / 3^3), ...,
1=[5^0/3^0], 2=[5^2/3^2], 3=[5^7/3^9], 4=[5^3/3^3],..., so j-sequence = (0,2,7,3,...); k-sequence = (0,2,9,3,...).
proposed
editing
editing
proposed
V:=Vector(N, -1): count:= 0:
if m <= N and V[m] = 0 -1 then V[m]:= j; count:= count+1 fi
approved
editing
Robert Israel, <a href="/A124910/b124910_1.txt">Table of n, a(n) for n = 1..10000</a>
proposed
approved
editing
proposed
Robert Israel, <a href="/A124910/b124910_1.txt">Table of n, a(n) for n = 1..10000</a>
N:= 100: # for a(1) .. a(N)
V:=Vector(N): count:= 0:
for j from 0 while count < N do
x:= 5^j;
k0:= max(0, floor(log[3](x/N)));
x:= x/3^(k0-1);
for k from k0 do
x:= x/3;
if x < 1 then break fi;
m:= floor(x);
if m <= N and V[m] = 0 then V[m]:= j; count:= count+1 fi
od od:
convert(V, list); # Robert Israel, Mar 08 2024
approved
editing