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May 4[edit]

Explicit equations for these three sequences[edit]

I've looked at the pages for the up/down numbers, the Bernoulli numbers, and the Euler numbers linked to from the page of trigonometric functions, but I don't see the functions produce them. What are they? --Melab±1 ☎ 03:49, 4 May 2013 (UTC)[reply]

Obviously there may be a function, but these numbers are described and the way the values are generated are by counting the instances. For n=0, there is only one permutation {}. For n=1 there is only 1 permutation {1}. For n=2, we must have c1<c2, so there is only one permutation {1,2}. For n=3, with c1<c2>c3 there are two possible permutations: 1,3,2 and 2,3,1 so An=2. For n=4, see the list on the Alternating permutation page. That is where 1,1,1,2,5,... come from. -- SGBailey (talk) 06:16, 4 May 2013 (UTC)[reply]

Jack: It isn't bad spelling, just bad typing, there's a difference! sgb

That's OK. I accept your spelling is not as bad as your typography would indicate. -- Jack of Oz ^[Talk] 01:15, 7 May 2013 (UTC) [reply]

There are explicit formula for the last two Bernoulli number#Definitions, Euler number#Explicit formulas and Up/down number can be easily derived from those. Indeed

A_{n}=i^{n+1}\sum _{k=1}^{n+1}\sum _{j=0}^{k}{k \choose {j}}{\frac {(-1)^{j}(k-2j)^{n+1}}{2^{k}i^{k}k}}

this might not be the best way to generate the numbers.--Salix (talk): 06:51, 4 May 2013 (UTC)[reply]