A122045 Euler (or secant) numbers E(n).
1, 0, -1, 0, 5, 0, -61, 0, 1385, 0, -50521, 0, 2702765, 0, -199360981, 0, 19391512145, 0, -2404879675441, 0, 370371188237525, 0, -69348874393137901, 0, 15514534163557086905, 0, -4087072509293123892361, 0, 1252259641403629865468285, 0, -441543893249023104553682821
Offset: 0
Keywords
Examples
G.f. = 1 - x^2 + 5*x^4 - 61*x^6 + 1385*x^8 - 50521*x^10 + 2702765*x^12 + ...
References
- Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 5, page 41.
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..200
- Fernando Barbero G., Juan Margalef-Bentabol, and Eduardo J. S. Villaseñor, A two-sided Faulhaber-like formula involving Bernoulli polynomials, arXiv:2002.00550 [math.NT], 2020.
- Marilena Barnabei, Flavio Bonetti, Niccolò Castronuovo, and Matteo Silimbani, Ascending runs in permutations and valued Dyck paths, Ars Mathematica Contemporanea (2019) Vol. 16, No. 2, 445-463.
- Hacène Belbachir and Yassine Otmani, Quadrinomial-Like Versions for Wolstenholme, Morley and Glaisher Congruences, Integers (2023) Vol. 23.
- Fatima Zohra Bensaci, Rachid Boumahdi, and Laala Khaldi, Finite Sums Involving Fibonacci and Lucas Numbers, J. Int. Seq. (2024). See p. 2.
- Michael D. Hirschhorn, Binomial Identities and Congruences for Euler Numbers, Fibonacci Quart. 53 (2015), no. 4, 319-322.
- Guodong Liu, Generating functions and generalized Euler numbers, Proc. Japan Acad. Ser. A Math. Sci., Volume 84, Number 2 (2008), 29-34. See p 32.
- F. Luca, A. Pizarro-Madariaga, and C. Pomerance, On the counting function of irregular primes, 2014.
- Emanuele Munarini, Two-Parameter Identities for q-Appell Polynomials, Journal of Integer Sequences, Vol. 26 (2023), Article 23.3.1.
- A. Niedermaier and J. Remmel, Analogues of up-down permutations for colored permutations, J. Int. Seq. 13 (2010) 10.5.6.
- Bruce E. Sagan, Generalized Euler numbers and ordered set partitions, arXiv:2501.07692 [math.NT], 2025. See p. 2.
- T. J. Stieltjes, Sur la réduction en fraction continue d'une série procédant suivant les puissances descendantes d'une variable, Ann. Fac. Sc. Toulouse 3 (1889) 1-17.
Programs
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Magma
m:=35; R
:=PowerSeriesRing(Rationals(), m); b:=Coefficients(R!( 1/Cosh(x) )); [Factorial(n-1)*b[n]: n in [1..m-1]]; // G. C. Greubel, Feb 13 2020 -
Maple
seq(euler(n) , n=0..31); # Zerinvary Lajos, Mar 15 2009 P := proc(n,x) option remember; if n = 0 then 1 else (n*x+(1/2)*(1-x))*P(n-1,x)+x*(1-x)*diff(P(n-1,x),x); expand(%) fi end: A122045 := n -> (-1)^n*subs(x=-1, P(n,x)): seq(A122045(n), n=0..30); # Peter Luschny, Mar 07 2014 ptan := proc(n) option remember; if irem(n, 2) = 1 then 0 else -add(`if`(k=0, 1, binomial(n, k)*ptan(n - k)), k = 0..n-1,2) fi end: A122045 := n -> ifelse(n = 0, 1, ptan(n)): seq(A122045(n), n = 0..30); # Peter Luschny, Jun 06 2022
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Mathematica
Table[EulerE[n], {n, 0, 30}] Range[0, 30]! CoefficientList[ Series[ Sech[x], {x, 0, 30}], x] (* Robert G. Wilson v, Aug 08 2018 *) a[0] := 1; a[n_?OddQ] := 0; a[n_?EvenQ] := a[n] = -Sum[a[k] Binomial[n, k], {k, 0, n - 1, 2}]; Map[a, Range[0, 30]] (* Oliver Seipel, May 20 2024 *)
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Maxima
a[n]:=if n<2 then 1-n else sum(-a[n-2*k]*binomial(n,2*k),k,1,floor(n/2)); makelist(a[n],n,0,50); /* Tani Akinari, Sep 15 2023 */
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PARI
x='x+O('x^66); Vec(serlaplace(1/cosh(x))) \\ Joerg Arndt, Mar 10 2014
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PARI
a(n) = 2^n*2^(n+1)*(subst(bernpol(n+1,x), x, 3/4) - subst(bernpol(n+1,x), x, 1/4))/(n+1); \\ Michel Marcus, May 20 2017
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Python
from sympy import bernoulli as B def a(n): return int(2**n*2**(n + 1)*(B(n + 1, 3/4) - B(n + 1, 1/4))/(n + 1)) print([a(n) for n in range(101)]) # Indranil Ghosh, Jun 24 2017, after PARI code by Michel Marcus
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Python
from functools import cache from math import comb as binomial @cache def ptan(n): # see also A331978 and A350972. return (0 if n % 2 == 1 else -sum(binomial(n,k) * ptan(n-k) if k > 0 else 1 for k in range(0, n-1, 2))) def A122045(n): return 1 if n == 0 else ptan(n) print([A122045(n) for n in range(31)]) # Peter Luschny, Jun 06 2022
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Sage
[euler_number(i) for i in range(31)] # Zerinvary Lajos, Mar 15 2009
Formula
E.g.f.: sech(x). - Michael Somos, Mar 11 2014
a(n) = (Sum_{k>=0} (-1)^k*(2*k+1)^n)*2. - Gottfried Helms, Mar 09 2012
From Sergei N. Gladkovskii, Oct 14 2012 - Oct 13 2013: (Start)
Continued fractions:
G.f.: 1/U(0) where U(k) = 1 - x + x*(k+1)/(1 - x*(k+1)/U(k+1)).
G.f.: 1/U(0) where U(k) = 1 - x^2 + x^2*(2*k+1)*(2*k+2)/(1 + x^2*(2*k+1)*(2*k+2)/ U(k+1)).
E.g.f.: (1-x)/U(0) where U(k) = 1 - x/(1 - x/(x - (2*k+1)*(2*k+2)/U(k+1))).
E.g.f.: 1 - x^2/U(0) where U(k) = (2*k+1)*(2*k+2) + x^2 - x^2*(2*k+1)*(2*k+2)/U(k+1).
E.g.f.: 1/U(0) where U(k) = 1 + x^2/(2*(2*k+1)*(4*k+1) - 2*x^2*(2*k+1)*(4*k+1)/(x^2 + 4*(4*k+3)*(k+1)/U(k+1))).
E.g.f.: (2 + x^4/(U(0)*(x^2-2) - 2))/(2-x^2) where U(k) = 4*k + 4 + 1/(1 + x^2/(2 - x^2 + (2*k+3)*(2*k+4)/U(k+1))).
G.f.: 1/G(0) where G(k) = 1 + x^2*(2*k+1)^2/(1 + x^2*(2*k+2)^2/G(k+1)) (due to T. J. Stieltjes).
G.f.: 1/S(0) where S(k) = 1 + x^2*(k+1)^2/S(k+1) (due to T. J. Stieltjes).
G.f.: 1 - x/(1+x) + x/(1+x)/Q(0) where Q(k) = 1 - x + x*(k+2)/(1 - x*(k+1)/Q(k+1)).
G.f.: -(1/x)/Q(0) where Q(k) = -1/x + (k+1)^2/Q(k+1) (due to T. J. Stieltjes).
G.f.: (1/(1-x))/Q(0) + 1/(1-x) where Q(k) = 1 - 1/x + (k+1)*(k+2)/Q(k+1).
G.f.: (x/(x-1))/Q(0) + 1/(1-x) where Q(k) = 1 - x + x^2*(k+1)*(k+2)/Q(k+1).
G.f.: 1 - x/(1+x) + (x/(1+x))/Q(0) where Q(k) = 1 + x + (k+1)*(k+2)*x^2/Q(k+1).
E.g.f.: 1 - T(0)*x^2/(2+x^2) where T(k) = 1 - x^2*(2*k+1)*(2*k+2)/(x^2*(2*k+1)*(2*k+2) - ((2*k+1)*(2*k+2) + x^2)*((2*k+3)*(2*k+4) + x^2)/T(k+1)).
G.f.: T(0) where T(k) = 1 - x^2*(k+1)^2/(x^2*(k+1)^2 + 1/T(k+1)). (End)
a(n) = 2^(2*n+1)*(zeta(-n,1/4) - zeta(-n,3/4)), where zeta(a, z) is the generalized Riemann zeta function. - Peter Luschny, Mar 11 2015
a(n) = 2^n*(2^(n+1)/(n+1))*(B(n+1, 3/4) - B(n+1, 1/4)) where B(n,x) is the n-th Bernoulli polynomial. See Liu link. - Michel Marcus, May 20 2017 [This is the same as: a(n) = -4^(n+1)*B(n+1, 1/4)*((n+1) mod 2)/(n+1). Peter Luschny, Oct 30 2020]
a(n) = 2*Im(PolyLog(-n, I)). - Peter Luschny, Sep 29 2020
a(4n) == 5 (mod 60) and a(4n+2) == -1 (mod 60). See Hirschhorn. - Michel Marcus, Jan 11 2022
For n > 1, a(n) = -Sum_{k=1..floor(n/2)} a(n-2*k)*binomial(n,2*k). - Tani Akinari, Sep 15 2023
Extensions
Edited by N. J. A. Sloane, Sep 17 2006
Comments