1967
DOI: 10.2307/2005010
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Computation of Tangent, Euler, and Bernoulli Numbers

Abstract: Abstract. Some elementary methods are described which may be used to calculate tangent numbers, Euler numbers, and Bernoulli numbers much more easily and rapidly on electronic computers than the traditional recurrence relations which have been used for over a century. These methods have been used to prepare an accompanying table which extends the existing tables of these numbers. Some theorems about the periodicity of the tangent numbers, which were suggested by the tables, are also proved.

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Cited by 32 publications
(49 citation statements)
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“…However for large values of k more efficient algorithms are needed to calculate tangent numbers. One of them can be found in [26]. A connection with Bernoulli numbers is established by the well-known formula (see, for example, [26]), valid for n > 1,…”
Section: Zeta Function Values At Positive Even Integersmentioning
confidence: 99%
“…However for large values of k more efficient algorithms are needed to calculate tangent numbers. One of them can be found in [26]. A connection with Bernoulli numbers is established by the well-known formula (see, for example, [26]), valid for n > 1,…”
Section: Zeta Function Values At Positive Even Integersmentioning
confidence: 99%
“…Let k be a positive integer. According to Euler-MacLaurin formula [15], P n (k) = 1 n + 2 n + · · · + k n .…”
Section: Hook Lengths and D-polynomialsmentioning
confidence: 99%
“…It is known that the tangent number T 2n+1 is equal to the number of all alternating permutations of length 2n + 1 (see [1,9,13,15]). Also, T 2n+1 counts the number of increasing labelled complete binary trees with 2n + 1 vertices.…”
Section: Introductionmentioning
confidence: 99%
“…Evaluate m n := M k 2 n−kn+k (mod k) for k = p t , where p is a prime number and t ≥ 3. It seems that the sequence (m n ) n≥0 is always periodic for any p and t. Computer calculation has provided the initial values: (m n ) n≥0 = (1, 1, 5, 5, 1, 1, 5, 5, · · · ) for k = 2 3 , (m n ) n≥0 = (1, 1, 10, 1, 1, 10, 1, 1, 10 · · ·) for k = 3 3 , (m n ) n≥0 = (1, 1, 126, 376, 126, 1, 1, 126, 376, 126, · · ·) for k = 5 4 , (m n ) n≥0 = (1,1,13,5,9,9,5,13,1,1,13,5,9,9,5,13, · · ·) for k = 2 4 .…”
Section: Introductionmentioning
confidence: 99%