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%I #12 Jul 17 2021 12:31:29
%S 1,1,1,-2,-1,1,-6,-5,2,1,24,14,-13,-2,1,120,94,-51,-23,3,1,-720,-444,
%T 400,87,-41,-3,1,-5040,-3828,2356,1009,-200,-62,4,1,40320,25584,
%U -22676,-5716,2609,296,-94,-4,1,362880,270576,-178500,-74120,17765,5273,-550,-130,5,1
%N Triangle read by rows: coefficients of the alternating factorial polynomial (x+1)(x-2)(x+3)(x-4)...(x+n*(-1)^(n-1)).
%C The coefficients belong to the rows of the following tree, which is built up from the following polynomials:
%C 1
%C (x+1) = x+1
%C (x+1)(x-2) = x^2-x-2
%C (x+1)(x-2)(x+3) = x^3+2x^2-5x-6
%C (x+1)(x-2)(x+3)(x-4) = x^4-2x^3-13x^2+14x+24 and so on.
%C 1
%C 1 1
%C -2 -1 1
%C -6 -5 2 1
%C 24 14 -13 -2 1
%C . . . . . .
%C Every term of the tree can be denoted by nAm as follows:
%C 1A0 1A1
%C 2A0 2A1 2A2
%C 3A0 3A1 3A2 3A3
%C 4A0 4A1 4A2 4A3 4A4
%C . . . . . .
%C nA0 nA1 nA2 nA3 nA4 . . nAn
%C Properties:
%C (1) For all positive integers n, nAn = 1
%C (2) nA0 = n!(-1)^T where T=int(n/2)
%C (3) For terms of the tree, which are not on the border of the tree, nAm = (-1)^(n+1) * n * (n-1)Am + (n-1)A(m-1). E.g., for m=1, nA1 = (-1)^(n+1) * n * (n-1)A1 + (n-1)A0 and this term has coordinates (n,m)=(n,1). The term with coordinates (3,1) = 3A1 =(-1)^2*3*2A1+2A0 =1*3*-1+-2 = -5
%C (4) All terms of the tree are integers; this follows from (1), (2) and (3).
%C (5) For n >= 2, the sum of the terms of any row is given by: Sum of terms of row n = (-1)^T*[n+(-1)^(n-1)]*Sum of terms of row (n-1).
%C (6) For all n, nA0/[(-1)^T * n! ] = 1
%C (7) As n approaches infinity, (nA2) / ((-1)^T * n!) approaches (1/2) * (log 2)^2 - (Pi^2) / 12.
%C As n approaches infinity, (nA3) / ((-1)^T * n!) approaches (1/6) * (log 2)^3 - (1/12) * (Pi^2) * (log 2) + (1/4) * zeta(3) where zeta(3) = Sum_{m >=1} 1/m^3.
%C (8) Since all terms of the tree are integers, then it follows from (7) that Zeta(3) = (1/3) * (Pi^2) * (log 2) - (2/3) * (log 2)^3 + C, where C is a rational number.
%C Remember that T equals the integer part of (n divided by 2) and n! equals (factorial n).
%D W. Dunham, Euler The Master of Us All, The Mathematical Association of America (1999), pp. 39-60.
%H Robert Israel, <a href="/A140956/b140956.txt">Table of n, a(n) for n = 1..10011</a> (first 141 rows, flattened)
%F (1) For all positive integers n, nAn = 1. (2) nA0 = n!(-1)^T where T=int(n/2). (3) For terms of the tree, which are not on the border of the tree, nAm = (-1)^(n+1) * n * (n-1)Am + (n-1)A(m-1). E.g., for m=1, nA1 = (-1)^(n+1) * n * (n-1)A1 + (n-1)A0.
%e 3A1 =(-1)^2*3*2A1+2A0 =1*3*-1+-2 = -5
%p P:= 1: A:= 1:
%p for n from 1 to 20 do
%p P:= expand(P*(x+(-1)^(n-1)*n));
%p A:= A, seq(coeff(P,x,j),j=0..n)
%p od:
%p A; # _Robert Israel_, Jul 06 2015
%K easy,sign,tabl
%O 1,4
%A Ken Grant (plkl(AT)ozemail.com.au), Jul 26 2008
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