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%I #52 Sep 10 2021 17:49:34
%S 1,-1,6,1,-54,36,-1,210,-264,216,1,-450,792,-1284,1296,-1,540,-1188,
%T 2940,-6204,7776,1,-252,792,-3000,10692,-29724,46656,-1,-252,0,960,
%U -7128,37860,-140844,279936,1,540,0,168,792,-15600,129492,-657564,1679616
%N Square array T(n, k) read by ascending antidiagonals, T(n, k) = Sum_{j=0..n} (-1)^(n + j)*(6 - n + j)^k * binomial(12, n - j) if k > 0 and (-1)^n otherwise. T(n, k) for 0 <= n, k <= 11.
%C T(n, k) are the numerators of the coefficients in the standardized probability distribution where the variable is the sum of 12 uniformly distributed random variables.
%C Given that each of the twelve random variables in the sum is uniformly distributed in [-1/2..1/2], the expected value for the sum is 0, and its variance is 1, so that the distribution, whose graph is a bell curve, results "naturally standardized" [this only happens if we add 12 random variables, since the variance of the single variable is 1/12].
%C The probability distribution is a piecewise-defined function; it consists of twelve 11th-degree polynomials for x in ranges [k-6, k-5], k=0..11.
%C In each interval the distribution has the polynomial closed form
%C P(n, x)= Sum_{k=0..11} R(n, k) * x^(11 - k), powers of x in decreasing order: x^11, x^10, x^9, ..., x, 1.
%C The numerators of R(n, k) are the elements of the square array T(n, k); row n contains the numerators of the coefficients for the n-th interval [n-6, n-5].
%C The denominators of R(n, k) are k!*(11 - k)! if k > 0 and n!*(11 - n)! otherwise.
%D S. Brandt, "Data Analysis: Statistical and Computational Methods for Scientists and Engineers", Springer, 3rd edition (1998), 128-129.
%D F. Martinelli, "Somma di variabili aleatorie distribuite uniformemente: probabilità in forma chiusa", Atti della Fondazione Giorgio Ronchi, Anno LXV n. 1 Gen-Feb 2010, 115-132. http://ronchi.isti.cnr.it/index.php/atti-della-fondazione
%H Franco Martinelli, <a href="/A347211/b347211.txt">Antidiagonals for n = 0..11, flattened.</a>
%F Simplified expressions: T(0, k) = 6^k; for k > 0, T(1, k) = 6^k - 12*5^k.
%F The square array is essentially symmetric:
%F T(11 - n, k) = T(n, k) for odd k, T(11 - n, k) = -T(n, k) for even k.
%e Upper left corner of the table (which has 12 rows and 12 columns):
%e n/k | 0 1 2 3 4 5 6
%e ==============================================================
%e 0 | 1 6 36 216 1296 7776 46656
%e 1 | -1 -54 -264 -1284 -6204 -29724 -140844
%e 2 | 1 210 792 2940 10692 37860 129492
%e 3 | -1 -450 -1188 -3000 -7128 -15600 -30888
%e 4 | 1 540 792 960 792 240 792
%e 5 | -1 -252 0 168 0 -552 0
%e 6 | 1 -252 0 168 0 -552 0
%e .
%e Let R(n, k) = T(n, k) / S(n, k), where S(n, k) = k!*(11 - k)! if k > 0 and n!*(11 - n)! otherwise.
%e T(3, 2) = -1188, the distribution in its 4th interval (n = 3) is a polynomial whose 9th-degree term coefficient R(3, 2) is -1188/725760 = -11/6720.
%e For x in range [-3..-2]: P(3, x) = -(1/241920)*x^11 - (1/8064)*x^10 - (11/6720)*x^9 - (25/2016)*x^8 - (33/560)*x^7 - (13/72)*x^6 - (143/400)*x^5 - (239/504)*x^4 - (583/1120)*x^3 - (1619/3024)*x^2 - (781/5600)*x + 61297/166320.
%e Row n = 5 for 0 <= k <= 11 is [-1, -252, 0, 168, 0, -552, 0, 5208, 0, -135912, 0, 15724248]. It stands for the polynomial P(5, x) = Sum_{k=0..11} R(5, k) * x^(11 - k). Explicitly this becomes P(5, x) = -(1/86400)*x^11 - (1/14400)*x^10 + (1/1440)*x^8 - (23/3600)*x^6 + (31/720)*x^4 - (809/4320)*x^2 + 655177/1663200 for x in range (-1,0).
%e The list of coefficients begins: 1/39916800, 1/604800, 1/20160, 1/1120, 3/280, 9/100, 27/50, 81/35, 243/35, 486/35, 2916/175, 17496/1925, -1/3628800, -1/67200, -11/30240, -107/20160, -517/10080, ....
%p T := (n, k) -> `if`(k = 0, (-1)^n, add((-1)^(n+j)*(6-n+j)^k*binomial(12, n-j), j=0..n)): for n from 0 to 10 do seq(T(n-k, k), k=0..n) od; for n from 0 to 11 do seq(T(11+n-k, k), k=n..11) od;
%Y Cf. A000400 (powers of 6).
%K frac,sign,tabf,fini
%O 0,3
%A _Franco Martinelli_, Aug 23 2021