A000498 Eulerian numbers (Euler's triangle: column k=4 of A008292, column k=3 of A173018).
1, 26, 302, 2416, 15619, 88234, 455192, 2203488, 10187685, 45533450, 198410786, 848090912, 3572085255, 14875399450, 61403313100, 251732291184, 1026509354985, 4168403181210, 16871482830550, 68111623139600, 274419271461131, 1103881308184906, 4434992805213952
Offset: 4
Examples
There is one permutation of 4 with exactly 3 descents (4321). There are 26 permutations of 5 with 3 descents: 15432, 21543, 25431, 31542, 32154, 32541, 35421, 41532, 42153, 42531, 43152, 43215, 43251, 43521, 45321, 51432, 52143, 52431, 53142, 53214, 53241, 53421, 54132, 54213, 54231, 54312. - Neven Juric, Jan 21 2010.
References
- L. Comtet, "Permutations by Number of Rises; Eulerian Numbers." §6.5 in Advanced Combinatorics: The Art of Finite and Infinite Expansions, rev. enl. ed. Dordrecht, Netherlands: Reidel, pp. 51 and 240-246, 1974.
- L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 243.
- F. N. David and D. E. Barton, Combinatorial Chance. Hafner, NY, 1962, p. 151.
- F. N. David, M. G. Kendall and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 260.
- J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 215.
- N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
- N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Links
- T. D. Noe, Table of n, a(n) for n = 4..200
- E. Banaian, S. Butler, C. Cox, J. Davis, J. Landgraf and S. Ponce A generalization of Eulerian numbers via rook placements, arXiv:1508.03673 [math.CO], 2015.
- L. Carlitz et al., Permutations and sequences with repetitions by number of increases, J. Combin. Theory, 1 (1966), 350-374.
- F. N. Castro, O. E. González, and L. A. Medina, The p-adic valuation of Eulerian numbers: trees and Bernoulli numbers, 2014.
- E. T. Frankel, A calculus of figurate numbers and finite differences, American Mathematical Monthly, 57 (1950), 14-25. [Annotated scanned copy]
- J. C. P. Miller, Letter to N. J. A. Sloane, Mar 26 1971
- Nagatomo Nakamura, Pseudo-Normal Random Number Generation via the Eulerian Numbers, Josai Mathematical Monographs, vol. 8, p 85-95, 2015.
- P. A. Piza, Kummer numbers, Mathematics Magazine, 21 (1947/1948), 257-260.
- P. A. Piza, Kummer numbers, Mathematics Magazine, 21 (1947/1948), 257-260. [Annotated scanned copy]
- J. Riordan, Review of Frankel (1950) [Annotated scanned copy]
- Eric Weisstein's World of Mathematics, Eulerian Number
- Robert G. Wilson v, Letter to N. J. A. Sloane, Apr. 1994
- Index entries for linear recurrences with constant coefficients, signature (20,-175,882,-2835,6072,-8777,8458,-5204,1848,-288).
Crossrefs
Programs
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Magma
[(6*4^n -6*(n+1)*3^n +3*n*(n+1)*2^n -(n-1)*n*(n+1))/6: n in [4..50]]; // G. C. Greubel, Oct 23 2017
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Magma
[EulerianNumber(n,3): n in [4..50]]; // G. C. Greubel, Dec 07 2024
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Maple
A000498:=proc(n); 4^n-(n+1)*3^n+1/2*(n)*(n+1)*2^n-1/6*(n-1)*(n)*(n+1); end:
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Mathematica
LinearRecurrence[{20, -175, 882, -2835, 6072, -8777, 8458, -5204, 1848, -288}, {1, 26, 302, 2416, 15619, 88234, 455192, 2203488, 10187685, 45533450}, 30] (* Jean-François Alcover, Feb 09 2016 *) Table[Sum[(-1)^k*Binomial[n+1,k]*(4-k)^n, {k,0,3}], {n,4,50}] (* G. C. Greubel, Oct 23 2017 *)
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PARI
for(n=4,50, print1((6*4^n -6*(n+1)*3^n +3*n*(n+1)*2^n -(n-1)*n*(n+ 1))/6, ", ")) \\ G. C. Greubel, Oct 23 2017
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SageMath
from sage.combinat.combinat import eulerian_number print([eulerian_number(n,3) for n in range(4,61)]) # G. C. Greubel, Dec 07 2024
Formula
From Mike Zabrocki, Nov 12 2004: (Start)
G.f.: x^4*(1 + 6*x - 43*x^2 + 44*x^3 + 52*x^4 - 72*x^5)/((1-x)^4 * (1-2*x)^3 * (1-3*x)^2 * (1-4*x)).
a(n) = (6*4^n - 6*(n + 1)*3^n + 3*(n)*(n + 1)*2^n - (n - 1)*(n)*(n + 1))/6. (End)
If n>3 is prime, then a(n) == 1 (mod n). A generalization: if a_t(n) denote the number of permutations of n letters with exactly t descents (column t+1 of Euler's triangle A008292), then, for prime n>t, we have a(n) == 1 (mod n). - Vladimir Shevelev, Sep 26 2010
E.g.f.: exp(x)*(exp(3*x) - (1 + 3*x)*exp(2*x) + 2*(x + 2*x^2/2!)*exp(x) - x^2/2! - x^3/3!). - Wolfdieter Lang, Apr 17 2017
Extensions
More terms from Christian G. Bower, May 12 2000
Comments