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a(n) = binomial(n^2 + 3*n, n) / (n+1)^2.
3

%I #17 Feb 09 2019 02:23:21

%S 1,1,5,51,819,18278,527085,18730855,793542167,39113958819,

%T 2201663313200,139461523272085,9824294829146550,762188806010669820,

%U 64595315110014533629,5939055918736259991759,588894813538193130767295,62651281502108852275337225

%N a(n) = binomial(n^2 + 3*n, n) / (n+1)^2.

%C a(n) = < PF_n, PF_n >, where PF_n is the parking function symmetric function and <,> denotes the usual scalar product on symmetric functions (proved). - _Richard Stanley_, Sep 24 2015

%H Seiichi Manyama, <a href="/A182316/b182316.txt">Table of n, a(n) for n = 0..339</a>

%F a(n) = [x^n] 1/(1-x)^((n+1)^2) / (n+1)^2 ; that is, a(n) equals the coefficient of x^n in 1/(1-x)^((n+1)^2) divided by (n+1)^2.

%p A182316:=n->binomial(n^2 + 3*n, n) / (n+1)^2: seq(A182316(n), n=0..20); # _Wesley Ivan Hurt_, Feb 11 2017

%o (PARI) {a(n)=binomial((n+1)^2+n-1, n)/(n+1)^2}

%o for(n=0,20,print1(a(n),","))

%Y Cf. A143669.

%K nonn,easy

%O 0,3

%A _Paul D. Hanna_, Apr 24 2012