A055203 Number of different relations between n intervals on a line.
1, 1, 13, 409, 23917, 2244361, 308682013, 58514835289, 14623910308237, 4659168491711401, 1843200116875263613, 886470355671907534969, 509366445167037318008557, 344630301458257894126724041, 271188703889907190388528763613, 245570692377888837925941696215449
Offset: 0
Examples
In case n = 2 this is the Delannoy number a(2) = D(2,2) = 13. a(2) = 13 because if you have two intervals [a1,a2] and [b1,b2], using a for a1 or a2 and b for b1 or b2 and writing c if an a is at the same place as a b, we get the following possibilities: aabb, acb, abab, cab, abc, baab, abba, cc, bac, cba, baba, bca, bbaa.
References
- S. R. Schwer, Dépendances temporelles: les mots pour le dire, Journées Intelligence Artificielle, 1998.
- S. R. Schwer, Enumerating and generating Allen's algebra, in preparation.
Links
- T. D. Noe, Table of n, a(n) for n = 0..100
- Tim Fernando and Carl Vogel, Prior Probabilities of Allen Interval Relations over Finite Orders, 11th International Conference on Agents and Artificial Intelligence (NLPinAI 2019) Prague.
- IBM Ponder This, Jan 01 2001
- J. B. Slowinski, The Number of Multiple Alignments, Molecular Phylogenetics and Evolution 10:2 (1998), 264-266. doi:10.1006/mpev.1998.0522
- Peter Stockman, Upper Bounds on the Time Complexity of Temporal CSPs, Linköping University | Department of Computer science, Master's thesis, 30 ECTS | Datateknik 2016 | LIU-IDA/LITH-EX-A--16/022--SE.
- David Adam Woods, Strings for Temporal Annotation and Semantic Representation of Events, Ph. D. Dissertation, Univ. of Dublin, Trinity College (Ireland, 2022).
Programs
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Maple
lambda := proc(p,n) option remember; if n = 1 then if p = 2 then RETURN(1) else RETURN(0) fi; else RETURN((p*(p-1)/2)*(lambda(p,n-1)+2*lambda(p-1,n-1)+lambda(p-2,n-1))) fi; end; A055203 := n->add(lambda(i,n),i=2..2*n); A055203 := proc(n) local k; add(A078739(n,k)*k!,k=0..2*n)/2^n end: seq(A055203(n),n=0..15); # Peter Luschny, Mar 25 2011 # second Maple program: b:= proc(n) option remember; `if`(n=0, 1, add(b(n-j)*binomial(n, j), j=1..n)) end: a:= n-> ceil(add(b(n+k)*binomial(n, k), k=0..n)/2^(n+1)): seq(a(n), n=0..20); # Alois P. Heinz, Jul 10 2018
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Mathematica
a[n_] := Sum[((m-1)*m)^n / 2^(m+n+1), {m, 0, Infinity}]; Table[a[n], {n, 0, 15}] (* Jean-François Alcover, Oct 10 2011, after Vladeta Jovovic *) With[{r = 2}, Flatten[{1, Table[Sum[Sum[(-1)^i*Binomial[j, i]*Binomial[j - i, r]^k, {i, 0, j}], {j, 0, k*r}], {k, 1, 15}]}]] (* Vaclav Kotesovec, Mar 22 2016 *)
Formula
a(n) = Sum_{i=2..2n} lambda(i, n), with lambda(p, 1) = 1 if p = 2, otherwise 0; lambda(p, n) = (p*(p-1)/2)*(lambda(p, n-1) + 2*lambda(p-1, n-1) + lambda(p-2, n-1)).
lambda(p, n) = Sum_k[( - 1)^(p + k) * C(p, k) * ((k - 1)*k/2)^n]. So if T(m, 0), T(m, 1), ..., T(m, m) is any row of A035317 with m >= 2n - 1 then a(n) = Sum_j[(-1)^j * T(m, j) * ((m - j + 1)*(m - j)/2)^n]; e.g., a(2) = 13 = 1*6^2 - 3*3^2 + 4*1^2 - 2*0^2 = 1*10^2 - 4*6^2 + 7*3^2 - 6*1^2 + 3*0^2 = 1*15^2 - 5*10^2 + 11*6^2 - 13*3^2 + 9*1^2 - 3*0^2 etc. while a(3) = 409 = 1*15^3 - 5*10^3 + 11*6^3 - 13*3^3 + 9*1^3 - 3*0^3 etc. - Henry Bottomley, Jan 03 2001
Row sums of A122193. - Vladeta Jovovic, Aug 24 2006
a(n) = Sum_{k=0..n} k!*Stirling2(n,k)*A121251(k). - Vladeta Jovovic, Aug 25 2006
E.g.f.: Sum_{m>=0} exp(x*binomial(m,2))/2^(m+1). - Vladeta Jovovic, Sep 24 2006
a(n) = Sum_{m>=0} binomial(m,2)^n/2^(m+1). - Vladeta Jovovic, Aug 17 2006
a(n) = (1/2^n)*Sum_{k=0..n} (-1)^(n-k)*binomial(n,k)*A000670(n+k). - Vladeta Jovovic, Aug 17 2006
a(n) ~ n! * n^n * 2^(n-1) / (exp(n) * (log(2))^(2*n+1)). - Vaclav Kotesovec, Mar 15 2014
From Peter Bala, Jan 30 2018: (Start)
a(n) = Sum_{k = 2..2*n} Sum_{i = 0..k} (-1)^(k-i)*binomial(k,i)*(i*(i-1)/2)^n.
a(n) = (1/2^(n+1))*Sum_{k = 0..n} binomial(n,k)*A000670(n+k) for n >= 1. (End)
Extensions
More terms from Larry Reeves (larryr(AT)acm.org), Oct 04 2000
More terms from N. J. A. Sloane, Jan 03 2001
Comments