A088594 - OEIS

%I #41 Mar 29 2015 11:44:49

%S 1,4,44,788,18372,505156,15553372,520065572,18518471492,692900847812,

%T 26985709712524,1086313382608436,44960426477218436,

%U 1905328431907938180,82405332511166288572,3627806131038258219076,162218975410046793174404

%N Number of 3-dimensional lattice paths running from (0,0,0) to (n,n,n), lying in {(x,y,z) : 0<=x<=y<=z} and using the steps (1,0,0), (0,1,0), (0,0,1), (1,1,0), (1,0,1), (0,1,1), (1,1,1).

%C These are the 3-dimensional analogs of the large Schröder numbers, A006318.

%C R(3,n) = 4*A105124(n) for n>0, where A105124 is the three-dimensional small Schröder numbers. - _Paul D. Hanna_, Apr 19 2005

%C Number of n X 3 semi-standard Young tableaux with consecutive entries. I.e., if j is in P, and 1<=i<=j, then i is in P. - _Graham H. Hawkes_, Feb 16 2015

%H Alois P. Heinz, <a href="/A088594/b088594.txt">Table of n, a(n) for n = 0..200</a>

%H R. A. Sulanke, <a href="http://math.boisestate.edu/~sulanke/recentpapindex.html">Three-dimensional Narayana and Schröder numbers</a>

%H R. A. Sulanke, <a href="http://www.combinatorics.org/ojs/index.php/eljc/article/view/v11i1r54">Generalizing Narayana and Schroeder Numbers to Higher Dimensions</a>, Electron. J. Combin. 11 (2004), Research Paper 54, 20 pp. (see page 16).

%F For n => 1, R(3, n) := Sum[2^(k+2)*Sum[2*(-1)^(k-j)*C(3*n+1, k-j)* C(n+j, n)*C(n+j+1, n)*C(n+j+2, n)/(n+1)^2/(n+2), {j, 0, k}], {k, 0, 2*n-2}]. For n => 4, (3n-4)(n+2)(n+1)^2 R(3, n)(t) = (3n-2)(n+1)( 4(1+t+t^2) - 5(1+7t+t^2)n+3(1+7t+t^2)n^2 )R(3, n-1) - (n-2)( -12 +29n -30n^2 +9n^3)(1-t)^4 R(3, n-2) + (3n-1)(n-2)(n-3)(n-4) (1-t)^6 R(3, n-3).

%F G.f.: (1+2/x)*(1-1/x)*Int(((x-1)*(7*x^3-12*x^2+57*x+2)*hypergeom([1/3, 2/3],[1],54*x/(1-x)^3)-x*(x+5)*(x^2-8*x-11)*hypergeom([2/3, 4/3],[2],54*x/(1-x)^3))/(3*(x-1)^4*(x+2)^2),x)-(1+4*x)/(3*x). - _Mark van Hoeij_, Apr 16 2013

%F Recurrence: (n+1)^2*(n+2)*(3*n-4)*a(n) = (n+1)*(3*n-2)*(57*n^2 - 95*n + 28)*a(n-1) - (n-2)*(9*n^3 - 30*n^2 + 29*n - 12)*a(n-2) + (n-4)*(n-3)*(n-2)*(3*n-1)*a(n-3). - _Vaclav Kotesovec_, Aug 14 2013

%F a(n) ~ c*d^n/n^4, where d = 12*2^(2/3)+15*2^(1/3)+19 = 56.947628372... is the root of the equation d^3-57*d^2+3*d-1=0 and c = sqrt(4 + 10*2^(1/3)/3 + 8*2^(2/3)/3)/Pi = 1.122366540310337391196984583368763794289876... - _Vaclav Kotesovec_, Aug 14 2013, updated Mar 19 2015

%F In general, the number of SSYT of shape n X d with consecutive entries is given by:

%F [Prod_(i=1,d-1) (i/(n+i))^(d-1)] *

%F Sum_(j=0,n*(d-1)) [Prod_(i=0,d-1) (n+i+j choose n)*(Sum_(k=0,(n*(d-1)-j)) (-1)^k (n+j+k choose k)]. - _Graham H. Hawkes_, Feb 16 2015

%p 1, seq( add( add(2*(-1)^(k-j)*binomial(3*n+1, k-j)* binomial(n+j,n)*binomial(n+j+1,n)*binomial(n+j+2,n)/(n+1)^2/(n+2), j = 0 .. k) *2^(k+2), k = 0 .. 2*n-2), n = 1 ..20 );

%t Flatten[{1,Table[Sum[Sum[2*(-1)^(k-j)*Binomial[3*n+1,k-j]*Binomial[n+j,n]*Binomial[n+j+1,n]*Binomial[n+j+2,n]/(n+1)^2/(n+2),{j,0,k}]*2^(k+2),{k,0,2*n-2}],{n,1,20}]}] (* _Vaclav Kotesovec_, Aug 14 2013 *)

%o (PARI) {alias(C,binomial); R3(n)=if(n==0,1,sum(k=0,2*n-2, 2^(k+2)*sum(j=0,k, 2*(-1)^(k-j)*C(3*n+1,k-j)*C(n+j,n)*C(n+j+1,n)*C(n+j+2,n)/(n+1)^2/(n+2))))} \\ _Paul D. Hanna_, Apr 19 2005

%Y Cf. A006318, A001003, A087647.

%K nonn

%O 0,2

%A Robert A. Sulanke (sulanke(AT)math.boisestate.edu), Nov 20 2003