A194222 -id:A194222 - cp's OEIS frontend

A194200 [sum{(k*e) : 1<=k<=n}], where [ ]=floor, ( )=fractional part.

0, 1, 1, 2, 2, 3, 3, 3, 4, 4, 5, 6, 6, 6, 7, 7, 7, 8, 9, 9, 9, 10, 11, 11, 12, 13, 13, 13, 14, 15, 15, 16, 16, 17, 17, 18, 18, 19, 19, 19, 20, 20, 21, 22, 22, 22, 23, 23, 23, 24, 25, 25, 25, 26, 27, 27, 28, 28, 29, 29, 30, 30, 31, 32, 32, 33, 33, 34, 34, 34, 35, 36, 37

Offset: 1

Clark Kimberling, Aug 19 2011

nonn

The defining [sum] is equivalent to
...
a(n)=[n(n+1)r/2]-sum{[k*r] : 1<=k<=n},
...
where []=floor and r=sqrt(2).  Let s(n) denote the n-th partial sum of the sequence a; then the difference sequence d defined by d(n)=s(n+1)-s(n) gives the runlengths of a.
...
Examples:
...
r...........a........s....
1/2......A002265...A001972
1/3......A002264...A001840
2/3......A002264...A001840
1/4......A194220...A194221
1/5......A194222...A118015
2/5......A057354...A011858
3/5......A194222...A118015
4/5......A057354...A011858
1/6......A194223...A194224
3/7......A057357...A194229
1/8......A194235...A194236
3/8......A194237...A194238
sqrt(2)..A194161...A194162
sqrt(3)..A194163...A194164
sqrt(5)..A194169...A194170
sqrt(6)..A194173...A194174
tau......A194165...A194166; tau=(1+sqrt(5))/2
e........A194200...A194201
2e.......A194202...A194203
e/2......A194204...A194205
pi.......A194206...A194207

			a(5)=[(e)+(2e)+(3e)+4(e)+5(e)]
    =[.718+.436+.154+.873+.591]
    =[2.77423]=2.

G. C. Greubel, Table of n, a(n) for n = 1..5000

Cf. A194201.

r = E;
a[n_] := Floor[Sum[FractionalPart[k*r], {k, 1, n}]]
Table[a[n], {n, 1, 90}]  (* A194200 *)
s[n_] := Sum[a[k], {k, 1, n}]
Table[s[n], {n, 1, 100}] (* A194201 *)

A301426 Number of steps required in the worst case for three knights to find the princess in a castle with n rooms arranged in a line (Castle and princess puzzle).

Original entry on oeis.org

1, 1, 1, 2, 2, 2, 2, 3, 4, 4, 4, 4, 5, 6, 6, 6, 6, 7, 8, 8

Offset: 1

Dmitry Kamenetsky, Mar 21 2018

The main entry for this problem is A300576. In this version there are three knights who are searching for the princess; each knight can search a different room.

Dmitry Kamenetsky, Optimal solutions

Cf. A300576, A301337.

It seems that for n >= 3:
  if n = 3 mod 5, then a(n) = (n - 3)/5*2 + 1,
  otherwise a(n) = floor((n - 4)/5)*2 + 2.
This conjecture is a(n) = A194222(n) for n>2. - R. J. Mathar, Mar 27 2025

cp's OEIS Frontend

A194200 [sum{(k*e) : 1<=k<=n}], where [ ]=floor, ( )=fractional part.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica