A173226 Partial sums of A000364.

1, 2, 7, 68, 1453, 51974, 2754739, 202115720, 19593627865, 2424473303306, 372795661540831, 69721670054678732, 15584255833611765637, 4102656765126735657998, 1256362298168756601126283, 442800255547191861154809104, 177962191835086481297819598769

Offset: 0

Author

Jonathan Vos Post, Feb 13 2010

Keywords

• easy
• nonn

Comments

Partial sums of Euler numbers. Partial sums of secant or "Zig" numbers. The subsequence of prime partial sum of Euler numbers begins 2, 7, 1453, no more through a(17). What is the next such prime?

Crossrefs

Cf. A000111, A000364, A000182, A011248, A060075, A013525, A000816, A002436, A028296, A122045.

Programs

Mathematica

Accumulate[Table[Abs[EulerE[2n]],{n,0,20}]] (* Harvey P. Dale, Aug 10 2024 *)

Python

from sympy import euler
def A173226(n): return sum(abs(euler(i)) for i in range(0,(n<<1)+1,2)) # Chai Wah Wu, Apr 16 2023

Formula

a(n) = Sum_{i=0..n} A000364(i).

G.f.: 1/U(0)/(1-x) where U(k)=1 + x - x*(2*k+1)*(2*k+2)/(1 - x*(2*k+1)*(2*k+2)/U(k+1)) ; (continued fraction, 2-step). - Sergei N. Gladkovskii, Oct 15 2012

G.f.: 1/(1-x)/Q(0), where Q(k)= 1 - x*(2*k+1)^2/(1 - x*(2*k+2)^2/Q(k+1)); (continued fraction). - Sergei N. Gladkovskii, Apr 27 2013

G.f.: Q(0)/(1-x), where Q(k) = 1 - x*(k+1)^2/( x*(k+1)^2 - 1/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 22 2013