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Re: List:{1}, {2,3},{4,5,6}....
Posted:
Jan 14, 2013 12:01 AM


> to get the following result: > {{1}, {2, 3}, {4, 5, 6}, {7, 8, 9, 10}, {11, 12, 13, 14, 15}}
Hi. If you don't actually need to generate a large list, here's another interesting idea:
Series[(12 (1+x) x)/(1+x)^3,{x,0,10}] //Normal
1+x+2 x^2+4 x^3+7 x^4+11 x^5+16 x^6+22 x^7+29 x^8+37 x^9+46 x^10
If you just need the fifth item in the list, then with 11 x^5, that list starts with 11, and has a length of 5. ie. {11, 12, 13, 14, 15}
Another variation might be just the list of starting values, with the length based on it's position.
z=CoefficientList[Series[(1+xx^2)/(1+x)^3,{x,0,10}],x]
{1,2,4,7,11,16,22,29,37,46,56}
(ie 6th item starts with 16, and has length 6. ie 16=8521)
The 1000th item starts with 499501.
Coefficient[Series[(12 (1+x) x)/(1+x)^3,{x,0,1000}],x,1000] // Timing
{0.004363, 499501}
Just thought it interesting:
= = = = = = = = = = HTH :>) Dana DeLouis Mac & Mathematica 8 = = = = = = = = = =
On Jan 10, 9:38 pm, hknetman <hknet...@hotmail.com> wrote: > I use the command > list = Table[i + 1, {n, 1, 5}, {i, n (n  1)/2, n (n  1)/2 + n  1}] > to get the following result: > {{1}, {2, 3}, {4, 5, 6}, {7, 8, 9, 10}, {11, 12, 13, 14, 15}} > > Is there any better method? > > Thanks a lot.
Re: List:{1}, {2,3},{4,5,6}....



