The 2-adic Valuation of Stirling Numbers

Amdeberhan, Tewodros; Manna, Dante; Moll, Victor H.

doi:10.1080/10586458.2008.10129026

Cited by 38 publications

(51 citation statements)

References 10 publications

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“…This gives an alternative proof of the Amdeberhan conjecture [5] which was proved in [8]. Thus ν(S(n + 1, k + 1)) = σ(k) − σ(n) iff S(n, k) is a MZC, in which case ν(S(n + 1, k + 1)) = ν(S(n, k)).…”

Section: Basic Properties Of the Estimates And Cases Some Examples Amentioning

confidence: 80%

The p-adic analysis of Stirling numbers via higher order Bernoulli numbers

Adelberg

2018

Int. J. Number Theory

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Several new estimates for the 2-adic valuations of Stirling numbers of the second kind are proved. These estimates, together with criteria for when they are sharp, lead to improvements in several known theorems and their proofs, as well as to new theorems. The estimates and criteria all depend on our previous analysis of powers of 2 in the denominators of coefficients of higher order Bernoulli polynomials. The corresponding estimates for Stirling numbers of the first kind are also proved.Some attention is given to asymptotic cases, which will be further explored in subsequent publications.Keywords: Stirling numbers of the second kind, 2-adic analysis, higher order Bernoulli numbers and polynomials, estimates and exact values, Newton polygons. MSC[2010] : 11B68, 11B73, 05A10, 11S05.

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Section: Basic Properties Of the Estimates And Cases Some Examples Amentioning

confidence: 80%

The p-adic analysis of Stirling numbers via higher order Bernoulli numbers

Adelberg

2018

Int. J. Number Theory

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“…Furthermore, they found that the 2-adic valuation of the Stirling numbers of second kind of order 5 is the first non-trivial case. In fact, they showed that v 2 (S(4n + 1, 5)) = v 2 (S(4n + 2, 5)) = 0 for all natural numbers n. It was also observed in [1] that v 2 (S(4n, 5)) = v 2 (S(4n + 3, 5)) for most indices. Consequently, Amdeberhan, Manna and Moll proposed a conjecture describing those indices n such that v 2 (S(4n, 5)) = v 2 (S(4n + 3, 5)).…”

Section: Introductionmentioning

confidence: 93%

“…Amdeberhan, Manna and Moll [1] studied the 2-adic valuations of Stirling numbers of second kind. Actually, they computed the 2-adic valuation v 2 (S(n, k)) for k ≤ 4.…”

Section: Introductionmentioning

confidence: 99%

The 2-Adic Valuations of Stirling Numbers of the Second Kind

Hong

Zhao

Zhao³

2012

Int. J. Number Theory

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In this paper, we investigate the 2-adic valuations of the Stirling numbers S(n, k) of the second kind. We show that v 2 (S(4i, 5)) = v 2 (S(4i + 3, 5)) if and only if i ≡ 7 (mod 32). This confirms a conjecture of Amdeberhan, Manna and Moll raised in 2008. We show also that v 2 (S(2 n + 1, k + 1)) = s 2 (n) − 1 for any positive integer n, where s 2 (n) is the sum of binary digits of n. It proves another conjecture of Amdeberhan, Manna and Moll.

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“…The number that counts this process of division of a set in disjoint sets that go from 2 to k is the Stirling number of the second type. In this way, the number of configurations of the space of solutions of CaRS is at least O(2 n ) greater than the space of solutions of the Traveling Salesman Problem, since that for k = 2 the associated number of Stirling is O(2 n ) (Amdeberhan et al 2008). For a general case of CaRS the dimension of the space of solutions is still greater once there is not a theoretical limit for the number of different cars to be considered in the problem.…”

Section: The Difficulty Of Solving Carsmentioning

confidence: 99%