A209398 Number of subsets of {1,...,n} containing two elements whose difference is 2.

0, 0, 0, 2, 7, 17, 39, 88, 192, 408, 855, 1775, 3655, 7478, 15228, 30898, 62511, 126177, 254223, 511472, 1027840, 2063600, 4140015, 8300767, 16635087, 33324462, 66736764, 133615658, 267461287, 535294673, 1071191415, 2143357000, 4288290240, 8579130888

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

Cf. A006498, A209399, A209400.

Sequence in context: A154117 A173769 A067038 * A364756 A175660 A365069

Adjacent sequences: A209395 A209396 A209397 * A209399 A209400 A209401

Keyword

nonn,easy

Author

David Nacin, Mar 07 2012

Status

approved