0,2

The triangle of numbers of n-sequences of 0,1 with k subsequences of consecutive 01 is A034867 because this number is C(n+1,2*k+1). I have not yet found a formula for subsequences 00.

The problem is equivalent to one encountered by David W. Wilson, Dept of Geography, University of Southampton, UK, in his work on Markov models for rainfall disaggregation. He asked for the number of ways in which there can be k instances of adjacent rainy days in a period of n consecutive days. Representing a rainy day by 0 and a fine day by 1, the problem is equivalent to that solved by this sequence. - E. Keith Lloyd (ekl(AT)soton.ac.uk), Nov 29 2004

Row n (n>=1) contains n terms.

L. Carlitz and R. Scoville, Zero-one sequences and Fibonacci numbers, Fibonacci Quarterly, 15 (1977), 246-254.

Recurrence: a(n, k) = (a(n-1, k) + a(n-2, k))+(a(n-3, k-1)+a(n-4, k-2)+...+a(n-k-2, 0)). Special values: a(n, 0)=Fibonacci(n+1); a(n, n-1) = 1 for n >= 2; a(n, n-2)=2 for n>=3; a(n, n-3)=n+1 for n>=4, etc.

a(n, n-4)=3n-5 for n>=5, a(n, n-5)=(n^2+5n-26)/2 for n>=6, a(n, n-6)=2n^2-8n-4, for n>=7 etc. Recurrence relation: a(n+1, k)=a(n, k)+a(n-1, k)+a(n, k-1)-a(n-1, k-1) for k>=1, n>=1. Generating function: a(n, k) is coefficient of x^n in {[x^(k+1)]x[(1-x)^(k-1)]}/[(1-x-x^2)^(k+1)] for k>=1. - E. Keith Lloyd (ekl(AT)soton.ac.uk), Nov 29 2004

G.f.=G(t,x)=[1+(1-t)x]/[1-(1+t)x-(1-t)x^2]. [Carlitz-Scoville] - Emeric Deutsch (deutsch(AT)duke.poly.edu), May 19 2006

a(5, 2)=6 because the binary sequences of length 5 with 2 subsequences 00 are 10001, 11000, 01000, 00100, 00010, 00011.

Triangle begins

1;

2;

3,1;

5,2,1;

8,5,2,1;

13,10,6,2,1;

Cf. a(n, 1)=A001629, a(n, 2)=A055243.

Sequence in context: A053815 A066909 A095195 this_sequence A075014 A086686 A021816

Adjacent sequences: A076788 A076789 A076790 this_sequence A076792 A076793 A076794

nonn,tabf

Roger Cuculiere (cuculier(AT)imaginet.fr), Nov 16 2002

More terms from E. Keith Lloyd (ekl(AT)soton.ac.uk), Nov 29 2004