

A167196


Triangle T(n,k) read by rows: matrix inverse of A106246.


2



1, 2, 1, 7, 4, 1, 36, 21, 6, 1, 246, 144, 42, 8, 1, 2100, 1230, 360, 70, 10, 1, 21510, 12600, 3690, 720, 105, 12, 1, 257040, 150570, 44100, 8610, 1260, 147, 14, 1, 3510360, 2056320, 602280, 117600, 17220, 2016, 196, 16, 1, 53933040
(list;
table;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,2


COMMENTS

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 A167194liketriangle elementwise with the Pascal triangle A007318. Then divide the elements below the main diagonal in the resulting triangle with another integer.
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(n1)/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(n1)/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 squareroot of the multiplying integer. The higher the dividing integer the lower the square root.
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.
From Mats Granvik, Oct 31 2009: (Start)
To calculate the square root of x consider the triangle A:
1,
x,1,
1,x,1,
0,1,x,1,
0,0,1,x,1,
0,0,0,1,x,1,
0,0,0,0,1,x,1,
0,0,0,0,0,1,x,1
Define y=(x1)*x/2 and consider the triangle B:
1,
y,1,
y,y,1,
y,y,y,1,
y,y,y,y,1,
y,y,y,y,y,1,
y,y,y,y,y,y,1,
y,y,y,y,y,y,y,1
Consider the Pascal triangle:
1
1..1
1..2..1
1..3..3..1
1..4..6..4..1
1..5.10.10..5..1
1..6.15.20.15..6..1
1..7.21.35.35.21..7..1
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:
1*1/1
1*x/y...1*1/1
1*1/y...2*x/y...1*1/1
....0...3*1/y...3*x/y.....1*1/1
....0.......0...6*1/y.....4*x/y...1*1/1
....0.......0.......0....10*1/y...5*x/y...1*1/1
....0.......0.......0.........0..15*1/y...6*x/y...1*1/1
....0.......0.......0.........0.......0..21*1/y...7*x/y...1*1/1
which when simplified becomes triangle D:
..1
x/y.......1
1/y...2*x/y.......1
..0.....3/y...3*x/y.........1
..0.......0.....6/y.....4*x/y.......1
..0.......0.......0......10/y...5*x/y.......1
..0.......0.......0.........0....15/y...6*x/y.......1
..0.......0.......0.........0.......0....21/y...7*x/y.......1
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(n1)/a(n) appears to converge to sqrt(x) as n>infinity. See A167199 for example. (End)


LINKS

Table of n, a(n) for n=1..46.


EXAMPLE

From Mats Granvik, Oct 31 2009: (Start)
Table begins:
......1
.....2.......1
......7......4.......1
....36......21......6.......1
....246....144......42......8.......1
..2100....1230....360......70.....10.......1
..21510..12600....3690....720.....105.....12.......1
257040..150570..44100....8610...1260.....147.....14.......1
(End)


PROG

(PARI) tabl(nn) = {m = matrix(nn, nn, n, k, binomial(n1, k1)*binomial(2, nk))^(1); for (n=1, nn, for (k=1, n, print1(m[n, k], ", "); ); print; ); } \\ Michel Marcus, Dec 23 2017


CROSSREFS

Cf. A106246, A167199.
Sequence in context: A092276 A011274 A122843 * A241881 A107865 A089225
Adjacent sequences: A167193 A167194 A167195 * A167197 A167198 A167199


KEYWORD

sign,tabl


AUTHOR

Mats Granvik, Roger L. Bagula, Gary W. Adamson, Paul Barry, Oct 30 2009


STATUS

approved



