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%I #15 Jul 23 2018 02:49:08
%S 1,-2,1,7,-4,1,-36,21,-6,1,246,-144,42,-8,1,-2100,1230,-360,70,-10,1,
%T 21510,-12600,3690,-720,105,-12,1,-257040,150570,-44100,8610,-1260,
%U 147,-14,1,3510360,-2056320,602280,-117600,17220,-2016,196,-16,1,-53933040
%N Triangle T(n,k) read by rows: matrix inverse of A106246.
%C It appears that any square root can be calculated as a limiting ratio from the first column of the matrix inverse of a triangle similar to A106246. Replace the 2:s call them "x", in A167194 with an integer and multiply the A167194-like-triangle elementwise with the Pascal triangle A007318. Then divide the elements below the main diagonal in the resulting triangle with another integer.
%C For example x=3 so that the sequence in the columns is [1,3,1,0,0,0,0...]. The limiting ratio of the form n*a(n-1)/a(n) taken on the numbers in the first column of the matrix inverse will then converge to a square root of an integer - x. That is the square root = x+n*a(n-1)/a(n) where a(n) is the first column in the matrix inverse. The multiplying integer is related to the squares. The pattern of the dividing integer seems be every second square root downwards beginning minus two steps below the square root of the multiplying integer. The higher the dividing integer the lower the square root.
%C If instead as dividing number a fraction of the form [0.5; 1.5; 2.5; 3.5; ...] i.e., "a half, one and a half, two and a half ...", is chosen then the other every second square roots can be calculated. There is also a recursion such that if you instead of the 2:s in A167194 put a value of the previous every second square roots (either one of the every second square roots) you will get a square root that is two steps lower.
%C From _Mats Granvik_, Oct 31 2009: (Start)
%C To calculate the square root of x consider the triangle A:
%C 1,
%C x,1,
%C 1,x,1,
%C 0,1,x,1,
%C 0,0,1,x,1,
%C 0,0,0,1,x,1,
%C 0,0,0,0,1,x,1,
%C 0,0,0,0,0,1,x,1
%C Define y=(x-1)*x/2 and consider the triangle B:
%C 1,
%C y,1,
%C y,y,1,
%C y,y,y,1,
%C y,y,y,y,1,
%C y,y,y,y,y,1,
%C y,y,y,y,y,y,1,
%C y,y,y,y,y,y,y,1
%C Consider the Pascal triangle:
%C 1
%C 1..1
%C 1..2..1
%C 1..3..3..1
%C 1..4..6..4..1
%C 1..5.10.10..5..1
%C 1..6.15.20.15..6..1
%C 1..7.21.35.35.21..7..1
%C Multiply the elements in the Pascal triangle with the elements in triangle A and divide with the elements in triangle B so that you get:
%C 1*1/1
%C 1*x/y...1*1/1
%C 1*1/y...2*x/y...1*1/1
%C ....0...3*1/y...3*x/y.....1*1/1
%C ....0.......0...6*1/y.....4*x/y...1*1/1
%C ....0.......0.......0....10*1/y...5*x/y...1*1/1
%C ....0.......0.......0.........0..15*1/y...6*x/y...1*1/1
%C ....0.......0.......0.........0.......0..21*1/y...7*x/y...1*1/1
%C which when simplified becomes triangle D:
%C ..1
%C x/y.......1
%C 1/y...2*x/y.......1
%C ..0.....3/y...3*x/y.........1
%C ..0.......0.....6/y.....4*x/y.......1
%C ..0.......0.......0......10/y...5*x/y.......1
%C ..0.......0.......0.........0....15/y...6*x/y.......1
%C ..0.......0.......0.........0.......0....21/y...7*x/y.......1
%C Calculate the matrix inverse of triangle D. In the case of x=2 the matrix inverse of triangle D is this triangle A167196. The ratio of the form: x+n*a(n-1)/a(n) appears to converge to sqrt(x) as n-->infinity. See A167199 for example. (End)
%e From _Mats Granvik_, Oct 31 2009: (Start)
%e Table begins:
%e ......1
%e .....-2.......1
%e ......7......-4.......1
%e ....-36......21......-6.......1
%e ....246....-144......42......-8.......1
%e ..-2100....1230....-360......70.....-10.......1
%e ..21510..-12600....3690....-720.....105.....-12.......1
%e -257040..150570..-44100....8610...-1260.....147.....-14.......1
%e (End)
%o (PARI) tabl(nn) = {m = matrix(nn, nn, n, k, binomial(n-1,k-1)*binomial(2,n-k))^(-1); for (n=1, nn, for (k=1, n, print1(m[n,k], ", ");); print;);} \\ _Michel Marcus_, Dec 23 2017
%Y Cf. A106246, A167199.
%K sign,tabl
%O 1,2
%A _Mats Granvik_, _Roger L. Bagula_, _Gary W. Adamson_, _Paul Barry_, Oct 30 2009
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