

A140956


Triangle read by rows: coefficients of the alternating factorial polynomial (x+1)(x2)(x+3)(x4)...(x+n*(1)^(n1)).


1



1, 1, 1, 2, 1, 1, 6, 5, 2, 1, 24, 14, 13, 2, 1, 120, 94, 51, 23, 3, 1, 720, 444, 400, 87, 41, 3, 1, 5040, 3828, 2356, 1009, 200, 62, 4, 1, 40320, 25584, 22676, 5716, 2609, 296, 94, 4, 1, 362880, 270576, 178500, 74120, 17765, 5273, 550, 130, 5, 1
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OFFSET

1,4


COMMENTS

The coefficients belong to the rows of the following tree, which is built up from the following polynomials:
1
(x+1) = x+1
(x+1)(x2) = x^2x2
(x+1)(x2)(x+3) = x^3+2x^25x6
(x+1)(x2)(x+3)(x4) = x^42x^313x^2+14x+24 and so on.
1
1 1
2 1 1
6 5 2 1
24 14 13 2 1
. . . . . .
Every term of the tree can be denoted by nAm as follows:
1A0 1A1
2A0 2A1 2A2
3A0 3A1 3A2 3A3
4A0 4A1 4A2 4A3 4A4
. . . . . .
nA0 nA1 nA2 nA3 nA4 . . nAn
Properties:
(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 * (n1)Am + (n1)A(m1). E.g., for m=1, nA1 = (1)^(n+1) * n * (n1)A1 + (n1)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
(4) All terms of the tree are integers; this follows from (1), (2) and (3).
(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)^(n1)]*Sum of terms of row (n1).
(6) For all n, nA0/[(1)^T * n! ] = 1
(7) As n approaches infinity, (nA2) / ((1)^T * n!) approaches (1/2) * (log 2)^2  (Pi^2) / 12.
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.
(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.
Remember that T equals the integer part of (n divided by 2) and n! equals (factorial n).


REFERENCES

W. Dunham, Euler The Master of Us All, The Mathematical Association of America (1999), pp. 3960.


LINKS

Robert Israel, Table of n, a(n) for n = 1..10011 (first 141 rows, flattened)


FORMULA

(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 * (n1)Am + (n1)A(m1). E.g., for m=1, nA1 = (1)^(n+1) * n * (n1)A1 + (n1)A0.


EXAMPLE

3A1 =(1)^2*3*2A1+2A0 =1*3*1+2 = 5


MAPLE

P:= 1: A:= 1:
for n from 1 to 20 do
P:= expand(P*(x+(1)^(n1)*n));
A:= A, seq(coeff(P, x, j), j=0..n)
od:
A; # Robert Israel, Jul 06 2015


CROSSREFS

Sequence in context: A190782 A330490 A199063 * A166919 A338874 A338876
Adjacent sequences: A140953 A140954 A140955 * A140957 A140958 A140959


KEYWORD

easy,sign,tabl,changed


AUTHOR

Ken Grant (plkl(AT)ozemail.com.au), Jul 26 2008


STATUS

approved



