A239492 The fifth bicycle lock sequence: a(n) is the maximum value of min{x*y, (5-x)*(n-y)} over 0 <= x <= 5, 0 <= y <= n for integers x, y.

0, 0, 2, 3, 4, 6, 6, 8, 9, 10, 12, 12, 14, 15, 16, 18, 18, 20, 21, 22, 24, 24, 26, 27, 28, 30, 30, 32, 33, 34, 36, 36, 38, 39, 40, 42, 42, 44, 45, 46, 48, 48, 50, 51, 52, 54, 54, 56, 57, 58, 60, 60, 62, 63, 64, 66, 66, 68, 69, 70, 72, 72, 74, 75, 76, 78, 78, 80, 81, 82, 84, 84, 86, 87, 88, 90, 90

Offset: 0

Author

Robin Houston, Mar 23 2014

Keywords

• nonn
• easy

Comments

The minimum number of turns that always suffice to open from any starting position a bicycle lock that has n-1 dials with 5 numbers on each dial.

The minimum number of turns that always suffice to open from any starting position a bicycle lock that has 4 dials with n numbers on each dial.

(A "turn" consists of simultaneously rotating any number of adjacent dials by one place.)

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• Robin Houston, Symmetry of bicycle lock numbers.
• Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,1,-1).

Crossrefs

The fifth row of A238158.

Programs

Maple

A239492:=n->n-1+floor(n/5)+ceil((n-1)/5)-floor((n-1)/5); seq(A239492(n), n=0..50); # Wesley Ivan Hurt, Mar 29 2014

Mathematica

a[n_] :=  Max[Table[Min[x*y, (5-x)*(n-y)], {x, 0, 5}, {y, 0, n}]]
Table[n - 1 + Floor[n/5] + Ceiling[(n - 1)/5] - Floor[(n - 1)/5], {n, 0, 50}] (* Wesley Ivan Hurt, Mar 29 2014 *)
CoefficientList[Series[(2 x^5 + x^4 + x^3 + 2 x^2)/((1 - x) (1 - x^5)), {x, 0, 100}], x] (* Vincenzo Librandi, Mar 30 2014 *)

Formula

a(n) = max( min{x*y, (5-x)*(n-y)} | 0 <= x <= 5, 0 <= y <= n ).

From Ralf Stephan, Mar 29 2014: (Start)

a(n) = n + floor(n/5) - [n == 1 mod 5].

a(n) = 6*floor(n/5) + [0,0,2,3,4][n%5].

G.f.: (2*x^5 + x^4 + x^3 + 2*x^2)/((1-x)*(1-x^5)). (End)

a(n) = n - 1 + floor(n/5) + ceiling((n-1)/5) - floor((n-1)/5). - Wesley Ivan Hurt, Mar 29 2014