Here is a little puzzle for you: I need to figure out what the pattern is here to be able to extend this matrix further and be able to populate it pragrammatically through some kind of a loop: 1 -2 / 3 ______ 1 3 / 10 ______ -6 / 5 _____ 1-4 / 35 ______ 6 / 7 _______ -12 / 7 _______ 1 5 / 126______ -10 / 21 ______ 5 / 3 ________ -20 / 9 ________ 1-1 / 77 ______ 5 / 22 ______ -40 / 33 ________ 30 / 11 ________ -30 / 11 _________ 1There are some patterns evident: - the numerator of the first column increases by 1 ( - 1 /77 can be changed for -6 / 462) - the denominator of in the diagonal below the main diagonal increases by 2However, I still don't see the big picture. Any thoughts?

If this was a real life problem, I would try looking into each diagonal individually - they are approximating a linear function and I am positive that continuing these function is the way to go. I imported them into excel and plotted them - see below. However, "your mileage amy vary".D1 D2 D3 D4 D51 -0.6667 0.3 -0.114 0.039682541 -1.2 0.85714 -0.476 0.2272727271 -1.71414 1.66666 -0.476 1 -2.22222 2.72727 1 -2.7277 1

Oh, I should have done this also (D N where N is diagonal position from the main diagonal):D1 y = 1 (that one was obvious)D2 y = -0.5143x - 0.1631D3 y = 0.8091x - 0.6351D4 y = -0.181x + 0.0063D5 y = 0.1876x - 0.1479

HiIt looks like the first diagonal in fractional form is n(1-n)/(2n-1) for n=2,3,...and the 2nd diagonal is n(n-1)(n-2)/(8n-4) for n=3,4,...

Antoshka,it looks like the denominators in the first column are equal to C(2n+1, n+1) = 1,3,10,35,126,462,1716,6435,24310,92378... The inverse sum of the numbers in each row has the same pattern: 1,3,10,35,126... The nominators in second diagonal look like n C(2n,n) = 2, 12, 60, 280, 1260... also known as Apery numbers.The nominators in third diagonal are equal to n(2n+1)/2 C(2n,n) = 3, 30, 210, 1260...

Quoteit looks like the denominators in the first column are equal to C(2n+1, n+1) Sorry I am not familiar with this notation: C( x, x). What does this mean?

Binomial coefficient C(n,k) = n!/k!/(n-k)!

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