A208903 The sum over all bitstrings b of length n with at least two runs of the number of runs in b not immediately followed by a longer run.

0, 4, 12, 32, 76, 180, 412, 940, 2108, 4700, 10364, 22716, 49404, 106876, 229884, 492284, 1049596, 2229756, 4720636, 9964540, 20975612, 44046332, 92282876, 192950268, 402669564, 838885372, 1744863228, 3623927804, 7516258300, 15569354748, 32212385788

Offset: 1

Author

David Nacin, Mar 03 2012

Keywords

• nonn
• easy

Comments

A run is a maximal subsequence of (possibly just one) identical bits.

Examples

			n=3: 101, 010 each have 3; 100, 011 each have 1; 001, 110 each have 2. (000, 111 do not have at least two runs so they do not contribute.) Summing these gives 6+2+4 = 12 so a(3) = 12.
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..1000
• Aruna Gabhe, Problem 11623, Am. Math. Monthly 119 (2012) 161.
• Index entries for linear recurrences with constant coefficients, signature (5,-6,-6,16,-8).

Crossrefs

Cf. A056453, A208900, A208901, A208902.

Programs

Mathematica

Table[2^n*(2 + (n-1)/2 - (1/2)^(n-1) - 2*(1 - (1/2)^Floor[n/2]) + (1/2)^(Floor[n/2] + 1) (1 + (-1)^n)) - 2, {n, 1, 40}]
LinearRecurrence[{5, -6, -6, 16, -8}, {0, 4, 12, 32, 76}, 40]

Formula

a(n) = 2^n * (2 + (n - 1)/2 - (1/2)^(n - 1) - 2 (1 - (1/2)^floor(n/2)) + (1/2)^(floor(n/2) + 1) (1 + (-1)^n)) - 2.

a(n) = A208902(n) - 2.

a(n) = 5*a(n-1) - 6*a(n-2) - 6*a(n-3) + 16*a(n-4) - 8*a(n-5), a(1) = 0, a(2) = 4, a(3) = 12, a(4) = 32, a(5) = 76.

G.f.: (4*x - 8*x^2 - 4*x^3 + 12*x^4)/(1 - 5*x + 6*x^2 + 6*x^3 - 16*x^4 +

8*x^5).