A223911 Number of tiered orders on n nodes (corrected version of A006860).
1, 1, 3, 13, 111, 1381, 25623, 678133, 26169951, 1447456261, 114973232583, 13034314621813, 2103826463800911, 481932523110975301, 156356753093586913143, 71729530379673590609653, 46471511649877647638694591, 42487759521494442057018000901, 54781291469300608901184153800103
Offset: 0
Keywords
Links
- Joerg Arndt and Alois P. Heinz, Table of n, a(n) for n = 0..100 (terms n = 1..22 from Joerg Arndt)
- D. Klarner, The number of tiered posets modulo six, Discrete Math., 62 (1986), 295-297.
Programs
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Maple
f:= (i, j)-> add((-1)^(i-k)*binomial(i, k) *(2^k-1)^j, k=0..i): b:= proc(n, i) option remember; `if`(n=0, 1, add(b(n-j, j)/j!*f(i, j), j=1..n)) end: a:= n-> n!*b(n, 1): seq(a(n), n=0..20); # Alois P. Heinz, Jul 23 2013 -
Mathematica
f[i_, j_] := Sum[(-1)^(i-k)*Binomial[i, k]*(2^k-1)^j, {k, 0, i}]; b[n_, i_] := b[n, i] = If[n==0, 1, Sum[b[n-j, j]/j!*f[i, j], {j, 1, n}]]; a[n_] := n!*b[n, 1]; Table[a[n], {n, 0, 20}] (* Jean-François Alcover, Apr 08 2015, after Alois P. Heinz *) -
PARI
f(m,n) = sum(k=0, m, (-1)^(m-k) * binomial(m,k) * (2^k-1)^n ); mn(n,V,m) = n! / prod(k=1, m, V[k]! ); /* multinomial of V[1..m] */ A223911(n)= { my(m=n, C=vector(n,j,1), z, t, ret); while ( 1, /* for all compositions C[1..m] of n */ t = mn(n,C,m) * prod(j=1, m-1, f(C[j],C[j+1]) ); ret += t; if ( m<=1, break() ); /* last composition? */ /* create next composition: */ C[m-1] += 1; z = C[m]; C[m] = 1; m += z - 2; ); return(ret); }
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PARI
\\ here f(m,n) is A218695. f(m,n) = {sum(k=0, m, (-1)^(m-k) * binomial(m, k) * (2^k-1)^n )} seq(n)={my(N=matrix(n,n,i,j, f(i,j)), T=vector(n), v=vector(n+1)); v[1]=1; for(r=1, n, T[r]=vector(r, k, (r==k) + binomial(r,k)*sum(i=1, r-k, T[r-k][i]*N[i,k])); v[1+r]=vecsum(T[r])); v} \\ Andrew Howroyd, Mar 29 2023
Formula
a(n) = sum(all composition C of n, M(C) * prod(j=1..m-1, f(C[j]*C[j+1]) ) ) where m is the number of parts of the current composition P, f(i,j) = sum(k=0..i, (-1)^(i-k) * binomial(i,k) * (2^k-1)^j ), and M(C) is the multinomial coefficient n!/prod(j=1..m, C[j]! ); see Pari code.
Klarner incorrectly gives prod(j=1..m-1, f(C[j]*C[m]) ) in the formula for a(n).
Conjecture: a(n) ~ c * 2^(n^2/4 + 3*n/2) / sqrt(n), where c = EllipticTheta[3, 0, 1/2^(1/4)] / (sqrt(Pi) * 2^(1/4)) = 2.020039... (based on the numerical analysis of 600 terms). - Vaclav Kotesovec, Apr 10 2015
Extensions
Added a(0) = 1 by Alois P. Heinz, Jul 23 2013
Comments