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Let df(n,k) = Product_{i=0..k-1} (n-i) be the descending factorial and let P(m,n) = df(n-1,m-1)^2*(2*n-m)/((m-1)!*m!). Sequence gives P(6,n).
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%I #4 Jul 17 2020 00:34:51

%S 0,0,0,0,0,1,48,735,6272,37044,169344,640332,2090880,6073353,16032016,

%T 39078039,89037312,191456720,391523328,766192176,1442244096,

%U 2622518073,4623197040,7925786407,13248326784,21641442900,34616067200,54311107500,83710972800

%N Let df(n,k) = Product_{i=0..k-1} (n-i) be the descending factorial and let P(m,n) = df(n-1,m-1)^2*(2*n-m)/((m-1)!*m!). Sequence gives P(6,n).

%H <a href="/index/Rec#order_12">Index entries for linear recurrences with constant coefficients</a>, signature (12, -66, 220, -495, 792, -924, 792, -495, 220, -66, 12, -1).

%F From _Robert Israel_, Jul 16 2020: (Start)

%F a(n) = (n - 5)^2*(n - 4)^2*(n - 3)^2*(n - 2)^2*(n - 1)^2*(2*n - 6)/86400.

%F G.f.: (1 + 36*x + 225*x^2 + 400*x^3 + 225*x^4 + 36*x^5 + x^6)*x^6/(1 - x)^12. (End)

%p seq((n - 5)^2*(n - 4)^2*(n - 3)^2*(n - 2)^2*(n - 1)^2*(2*n - 6)/86400, n=1..50); # _Robert Israel_, Jul 16 2020

%Y See A132458 for further information.

%K nonn

%O 1,7

%A Ottavio D'Antona (dantona(AT)dico.unimi.it), Oct 31 2007