OFFSET
0,4
COMMENTS
Also, the number of bitstrings of length n containing either 101 or 111.
LINKS
David Nacin, Table of n, a(n) for n = 0..500
Index entries for linear recurrences with constant coefficients, signature (3,-2,1,-1,-2).
FORMULA
a(n) = 3*a(n-1) - 2*a(n-2) + a(n-3) - a(n-4) - 2*a(n-5), a(0)=0, a(1)=0, a(2)=0, a(3)=2, a(4)=7.
a(n) = 2^n - F(2+floor(n/2))*F(floor(2+(n+1)/2)), where F(n) are the Fibonacci numbers.
a(n) = 2^n - A006498(n+2).
G.f.: (2*x^3 + 1*x^4)/(1 - 3*x + 2*x^2 - x^3 + x^4 + 2*x^5) = x^3*(2 + x) / ((1 - 2*x)*(1 + x^2)*(1 - x - x^2)).
E.g.f.: (2*cos(x) + 5*cosh(2*x) + sin(x) + 5*sinh(2*x) - exp(x/2)*(7*cosh(sqrt(5)*x/2) + 3*sqrt(5)*sinh(sqrt(5)*x/2)))/5. - Stefano Spezia, Mar 12 2024
EXAMPLE
For n=3 the subsets containing 1 and 3 are {1,3} and {1,2,3} so a(3)=2.
MATHEMATICA
Table[2^n -Fibonacci[Floor[n/2] + 2]*Fibonacci[Floor[(n + 1)/2] + 2], {n, 0, 30}]
LinearRecurrence[{3, -2, 1, -1, -2}, {0, 0, 0, 2, 7}, 40]
CoefficientList[ Series[x^3 (x +2)/(2x^5 +x^4 -x^3 +2x^2 -3x +1), {x, 0, 33}], x] (* Robert G. Wilson v, Jan 03 2018 *)
a[n_] := Floor[ N[(2^-n ((50 - 14 Sqrt[5]) (1 - Sqrt[5])^n + ((-1 + 2I) (-2I)^n - (1 + 2I) (2I)^n + 5 4^n) (15 + 11 Sqrt[5]) - 2 (1 + Sqrt[5])^n (85 + 37 Sqrt[5])))/(150 + 110 Sqrt[5])]]; Array[a, 33] (* Robert G. Wilson v, Jan 03 2018 *)
PROG
(Python)
#Through Recurrence
def a(n, adict={0:0, 1:0, 2:0, 3:2, 4:7}):
.if n in adict:
..return adict[n]
.adict[n]=3*a(n-1)-2*a(n-2)+a(n-3)-a(n-4)-2*a(n-5)
.return adict[n]
(Python)
#Returns the actual list of valid subsets
def contains101(n):
.patterns=list()
.for start in range (1, n-1):
..s=set()
..for i in range(3):
...if (1, 0, 1)[i]:
....s.add(start+i)
..patterns.append(s)
.s=list()
.for i in range(2, n+1):
..for temptuple in comb(range(1, n+1), i):
...tempset=set(temptuple)
...for sub in patterns:
....if sub <= tempset:
.....s.append(tempset)
.....break
.return s
#Counts all such subsets using the preceding function
def countcontains101(n):
.return len(contains101(n))
(PARI) x='x+O('x^30); concat([0, 0, 0], Vec(x^3*(2+x)/((1-2*x)*(1+x^2)*(1-x-x^2)))) \\ G. C. Greubel, Jan 03 2018
(Magma) [2^n - Fibonacci(Floor(n/2) + 2)*Fibonacci(Floor((n + 1)/2) + 2): n in [0..30]]; // G. C. Greubel, Jan 03 2018
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
David Nacin, Mar 07 2012
STATUS
approved