Timeline for Recurrent sequences and Bernoulli-like numbers
Current License: CC BY-SA 2.5
8 events
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| Dec 11, 2010 at 9:29 | comment | added | Johann Cigler | Sorry, it should be $$M(a(2n,s)) = 2((2n + 1)B_{2n} - (2n + 3)B_{2n + 2} ).$$ | |
| Dec 9, 2010 at 11:29 | comment | added | Johann Cigler | Define the functional $L$ by $ L(a(2n,s)) = \delta_ {n,0}. $ Then $$L(a(2n+1,s)) = (-1)^n ( G_{2n + 2} + G_{2n + 4} ),$$ where ${(G_{2n})} = (1,1,3,17,155, \cdots)$ is the sequence of unsigned Genocchi numbers OEIS A110501. This gives the sequence $(2, - 4,20, - 172, \cdots )$. If we define the linear functional $M$ by $M(a(2n + 1,s)) = 2\delta_{n,0} ,$ then $$M(a(2n,s)) = (2n + 1)B_{2n} - (2n + 3)B_{2n + 2} .$$ | |
| Dec 9, 2010 at 11:28 | comment | added | Johann Cigler | Thank you for your last remarks. This goes into the direction I am looking for. Your example is in fact a linear combination of Fibonacci polynomials. Your sequence is $a{(n,2s)}$ with $a(n,s) = F_n (1,s) + F_{n + 1} (1,s)$. The linear functionals give of course the same numbers in both cases. So it suffices to consider the sequence $a(n,s).$ | |
| Dec 8, 2010 at 21:07 | history | edited | Aaron Meyerowitz | CC BY-SA 2.5 |
added 1096 characters in body
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| Dec 8, 2010 at 19:48 | comment | added | Johann Cigler | @Aaron: Concerning your later remarks: My question is not about abstract generalities which are routine. I am interested in concrete cases where the functionals or the corresponding Seidel matrices give "famous" numbers as you call them or where the generating functions of the odd resp. even subsequences are connected via "simple" functions as in the examples I have given. | |
| Dec 8, 2010 at 19:04 | history | edited | Aaron Meyerowitz | CC BY-SA 2.5 |
typos
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| Dec 8, 2010 at 10:05 | comment | added | Johann Cigler | Thank you for your interesting comments. The numbers $M(s^n)$ are the so called mean Genocchi numbers. For the other results consider the more general Fibonacci polynomials defined by $F_n(s)=xF_{n-1}(s)+sF_{n-2}(s)$. They satisfy the identity $$(1+e^{x z}){\sum{\frac{F_{2n}(s)}{(2n)!} z^{2n}\}}= (e^{x z}-1) {\sum{\frac{F_{2n+1}(s)}{(2n+1)!}z^{2n+1}\}}$$ which is easily proved using the Binet formulas (cf. uk.arxiv.org/ftp/arxiv/papers/0908/0908.1219.pdf ). | |
| Dec 8, 2010 at 7:46 | history | answered | Aaron Meyerowitz | CC BY-SA 2.5 |