V. Strehl, "Binomial identities---combinatorial and algorithmic aspects", Discrete Math. 136:1--3 (1994), 309--346.  Home/Search   Document Details and Download   Summary   Related Articles   Check

....K K There are two of them. If we want to know how they look like and what they are we could consult the manual, or display the information on the screen as shown below. In[5] Sgl2101 Do you want to set values for the equation [y n] n ( a c) a, n n Out[5] F ; 1 = 2 1 c (c) n In[6]: S2101 Summation formula (Slater, Appendix (III.4) in form of a rule. See also: SListe, SUMListe, Ers, PosListe. Let us apply S2101 (the Vandermonde summation [5, 1.7.7) Appendix (III.4) S2103 is Gauss 2 F 1 [1] evaluation [5, 1.7.6) In[6] 3 .FOrdne .S2101 Is N a nonnegative ....

.... n n Out[5] F ; 1 = 2 1 c (c) n In[6] S2101 Summation formula (Slater, Appendix (III.4) in form of a rule. See also: SListe, SUMListe, Ers, PosListe. Let us apply S2101 (the Vandermonde summation [5, 1.7.7) Appendix (III.4) S2103 is Gauss 2 F 1 [1] evaluation [5, 1.7. 6) In[6]: 3 .FOrdne .S2101 Is N a nonnegative integer [y n] y (1 M) 1 K M) N K Out[6] 1) 1 K M) K N 1.4. Manipulations of hypergeometric expressions. The result in Out[6] is not completely convincing since everybody knows that the result for the Vandermonde sum ....

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V. Strehl, Binomial identities --- Combinatorial and algorithmic aspects. 19

....K K There are two of them. If we want to know how they look like and what they are we could consult the manual, or display the information on the screen as shown below. In[5] Sgl2101 Do you want to set values for the equation [y n] n ( a c) a, n n Out[5] F ; 1 = 2 1 c (c) n In[6]: S2101 Summation formula (Slater, Appendix (III.4) in form of a rule. See also: SListe, SUMListe, Ers, PosListe. Let us apply S2101 (the Vandermonde summation [5, 1.7.7) Appendix (III.4) S2103 is Gauss 2 2 F 1 [1] evaluation [5, 1.7.6) In[6] 3 .FOrdne .S2101 Is N a nonnegative ....

....n Out[5] F ; 1 = 2 1 c (c) n In[6] S2101 Summation formula (Slater, Appendix (III.4) in form of a rule. See also: SListe, SUMListe, Ers, PosListe. Let us apply S2101 (the Vandermonde summation [5, 1.7.7) Appendix (III.4) S2103 is Gauss 2 2 F 1 [1] evaluation [5, 1.7. 6) In[6]: 3 .FOrdne .S2101 Is N a nonnegative integer [y n] y (1 M) 1 K M) N K Out[6] 1) 1 K M) K N 1.4. Manipulations of hypergeometric expressions. The result in Out[6] is not completely convincing since everybody knows that the result for the Vandermonde sum ....

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V. Strehl, Binomial identities --- Combinatorial and algorithmic aspects. 19

....Binomial[m j,m n] j, n, 2] Out[33] m n) 2 m n) SUM[n] 3 2 n) m 2 r) SUM[1 n] 2 (2 n) SUM[2 n] 0 This inversion technique can be applied in various situations. A beautiful example is provided by a transformation which was studied extensively by Strehl (1993), where also reference to its significant number theoretic relevance can be found: n X k=0 n k 2 n k k 2 = n X k=0 n k n k k k X j=0 k j 3 : As Strehl showed, in this instance another inversion pair, namely Legendre inversion, see e.g. Riordan ....

.... n k k k X j=0 k j 3 : As Strehl showed, in this instance another inversion pair, namely Legendre inversion, see e.g. Riordan (1968) can be successfully applied. In order to demonstrate the usefulness of the inversion method, we discuss another example which also stems from Strehl (1993), but was proven there by different means: n X k=0 n k 2 2k k = n X k=0 n k k X j=0 k j 3 : Because again none of the single sums finds a recurrence of order 1, we modify the problem by applying the inversion technique. Using the same inversion pair as ....

Strehl, V. (1993). Binomial identities - combinatorial and algorithmic aspects. Lehrstuhl I fuer Informatik, Interner Bericht Nr. 93 . Institut fuer Mathematische Maschinen und Datenverarbeitung der Friedrich-Alexander-Universitaet Erlangen-Nuernberg.

....to our approach and generalize as much as possible their good features. We now show using a few examples that the general approach outlined in this paper performs rather well in practice. 3.4.1. An identity between Franel and Ap ery numbers. The following identity was proved by V. Strehl [25]: n X k=0 n k 2 n k k 2 = n X k=0 n k n k k k X j=0 k j 3 : 12) Both sides of this equation satisfy the operator (n 2) 3 S 2 n Gamma Gamma (n 2) 3 (n 1) 3 4(2n 3) 3 Delta S n (n 1) 3 : 13) This operator was used by ....

Strehl, V. Binomial identities --- combinatorial and algorithmic aspects. Discrete Mathematics 136 (1994), 309--346.

....their efficiency to our approach and generalize as much as possible their good features. We now show using a few examples that the general approach outlined in this paper performs rather well in practice. 3.4.1. An identity between Franel and Ap ery numbers The following identity was proved by Strehl (1994): n X k=0 n k 2 n k k 2 = n X k=0 n k n k k k X j=0 k j 3 : 3.6) Both sides of this equation satisfy the operator (n 2) 3 S 2 n Gamma Gamma (n 2) 3 (n 1) 3 4(2n 3) 3 Delta Sn (n 1) 3 : 3.7) This operator was used by ....

Strehl, V. (1994). Binomial identities --- combinatorial and algorithmic aspects. Discrete Mathematics, 136:309--346.

....transformed the solution we got with FindRecurrence into rational function certificates, and used their denominators as input. It is interesting that they turned out to be smaller than those originally used in [WZ92a] for instance, a dramatic difference can be observed for Strehl s example ( Str94] in Section 5.2. Even with this a priori knowledge, FindRationalCertificate usually does not have a chance against FindRecurrence. The only exception is example (4) for which we found a significantly simpler recurrence with rational certificates: a recurrence of order 0 (see Section 5.5) ....

....identity X k X j n k n k k k j 3 = X k n k 2 n k k 2 ; integer n 0 (5.1) CHAPTER 5. SOME COMPUTER GENERATED PROOFS 72 originated from a number theoretical question that A. L. Schmidt had asked, and was proved in six different ways by V. Strehl in [Str94] In his proof of the irrationality of i(3) see [vdP78] Ap ery used that the famous Ap ery numbers P k Gamma n k Delta 2 Gamma n k k Delta 2 are annihilated by the recurrence operator (n 1) 3 Gamma (2n 3) 17n 2 51n 39)N (n 2) 3 N 2 : 5.2) We prove (5.1) by ....

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V. Strehl. Binomial identities - combinatorial and algorithmical aspects. Discrete Mathematics, 136:309--346, 1994.

....research as a conjecture. It was checked up to high values of n, but the colleagues of mine who came up with it were not able to prove it. I was asked if I could provide a proof, and, in fact, I was able to present quite a number of different proofs of this nontrivial result which are reported in [10]. There are now various proofs in conventional mathematical style (e.g. using results about hypergeometric functions) and there is a brute force machine based proof in the style of the proofs given above, i.e. by showing that the r.h.s. of (9) satisfies the same second order recurrence as the ....

Volker Strehl, Binomial identities - combinatorial and algorithmic aspects, Technical Report no. 93-1, Informatik-I, University of Erlangen-Nurnberg, Germany, 1993.

....now several different proofs available, none of them really trivial , and I will briefly mention below these diverse approaches that I worked out. The ordering corresponds to the chronological ordering in which these proofs were discovered. A detailed discussion of this affair will be given in [18]. 1. Identity (1) is hidden in a classical result, due to W.N. Bailey, from the theory of hypergeometric functions. In fact, it can be pulled out of Bailey s bilinear generating function for the Jacobi polynomials ( 2] 17] 19] via a kind of diagonalization. The special case where the Jacobi ....

V. Strehl. Binomial identities: combinatorial and algorithmic aspects. (in preparation).

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V. Strehl, "Binomial identities---combinatorial and algorithmic aspects", Discrete Math. 136:1--3 (1994), 309--346.

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