Testwiki:Reference desk/Archives/Mathematics/2015 April 1
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Contents
April 1
Adding numbers
If you have columns of numbers:
# | k1 | k2 | k3 | k4 | k5 | k6 | # 0| 1 | 1 | 1 | 1 | 1 | 1 | # 1| 1 | 2 | 2 | 2 | 2 | 2 | # 2| 1 | 3 | 4 | 4 | 4 | 4 | # 3| 1 | 4 | 7 | 8 | 8 | 8 | # 4| 1 | 5 | 11 | 15 | 16 | 16 | # 5| 1 | 6 | 16 | 26 | 31 | 32 | # 6| 1 | 7 | 22 | 42 | 57 | 63 | # 7| 1 | 8 | 29 | 64 | 99 | 120| # 8| 1 | 9 | 37 | 93 | 163| 219| # 9| 1 | 10 | 46 | 130| 256| 382| n| n^0| n |(n^2+n+2)/2| (n^3+5n+6)/6| ? | ? |
Where a number equals to the number below it plus the one on its bottom right. I heard somewhere you add the equations or something. Can you come up with a formula where you input n to get the number for each column? Is there a generalization for all columns? K as the column and n as the row. But most importantly, how do you create equations like this? — Preceding unsigned comment added by Someone with a Question (talk • contribs) 12:48, 1 April 2015 (UTC)
- First of all, your description seems to be upside-down: each cell in your diagram is the sum of cell above it and the cell on its top left; e.g. the 16 in the fifth row is 5 + 11. This sort of thing is called a recurrence relation, though that article has very little to say on multi-variable RRs, which is what your problem is. I found this question on StackExchange which discusses a method of generating a formula for any cell in the grid (the example there is essentially Pascal's triangle, where the rule is that the value of a cell is the sum of the two cells above and to the left of it. (Googling "recurrence relation two variables" found this, and lots of similar hits.) AndrewWTaylor (talk) 14:30, 1 April 2015 (UTC)
- The value in column k of row n is the coefficient of x^(k-1) in the formal power series expansion of
- (assuming column numbering starts at 1). Alternatively, it is the sum of the first k values in the nth row of Pascal's triangle if k <= n, or 2^n otherwise. Gandalf61 (talk) 14:47, 1 April 2015 (UTC)
- I believe this is Template:Oeis. --RDBury (talk) 16:11, 1 April 2015 (UTC)