This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
E.g.f.: A(x) = 1 + 6*x^2/2! + 330*x^4/4! + 48726*x^6/6! + 13534410*x^8/8! + 6046913046*x^10/10! + 3962771924490*x^12/12! + 3580686141374166*x^14/14! +... such that A(x) = (sin(x) + sin(6*x)) / sin(7*x). O.g.f.: F(x) = 1 + 6*x + 330*x^2 + 48726*x^3 + 13534410*x^4 + 6046913046*x^5 + 3962771924490*x^6 + 3580686141374166*x^7 +... such that the o.g.f. can be expressed as the continued fraction: F(x) = 1/(1 - 1*6*x/(1 - 7^2*x/(1 - 8*13*x/(1 - 14^2*x/(1 - 15*20*x/(1 - 21^2*x/(1 - 22*27*x/(1 - 28^2*x/(1 - 29*34*x/(1 - 35^2*x/(1 - 36*41*x/(1 - ...)))))))))))).
seq((-49)^n*euler(2*n, 1/7), n = 0..15); # Peter Luschny, Nov 26 2020
With[{nn=30},Take[CoefficientList[Series[(Sin[x]+Sin[6x])/Sin[7x],{x,0,nn}],x] Range[0,nn]!,{1,-1,2}]] (* Harvey P. Dale, Jul 08 2018 *)
{a(n) = my(A=1, X=x+x*O(x^(2*n+1))); (2*n)! * polcoeff( (sin(1*X) + sin(6*X))/sin(7*X), 2*n)}
for(n=0, 20, print1(a(n), ", "))
{a(n) = my(A=1, X=x+x*O(x^(2*n+1))); (2*n)! * polcoeff( (cos(1*X) + cos(6*X))/(1 + cos(7*X)), 2*n)}
for(n=0, 20, print1(a(n), ", "))
{a(n) = my(A=1, X=x+x*O(x^(2*n+1))); (2*n)! * polcoeff( (exp(1*I*X) + exp(6*I*X))/(1 + exp(7*I*X)), 2*n)}
for(n=0, 20, print1(a(n), ", "))
The array of the general case starts: [k] [1] 1, 1, 0, -1, 0, 1, 0, -17, 0, ... [A198631] [2] 1, 0, -1, 0, 5, 0, -61, 0, 1385, ... [A122045] [3] 1, -1, -2, 13, 22, -121, -602, 18581, 30742, ... [A156179] [4] 1, -2, -3, 44, 57, -722, -2763, 196888, 250737, ... [this sequence] [5] 1, -3, -4, 99, 116, -2523, -8764, 1074243, 1242356, ... [A156182] ...
a := n -> 4^n*euler(n, 1/4)*2^padic[ordp](n+1, 2): seq(a(n), n=0..9);
Array[4^#*EulerE[#, 1/4]*2^IntegerExponent[# + 1, 2] &, 22, 0] (* Michael De Vlieger, Mar 15 2022 *)
def euler_sum(n):
return (-1)^n*sum(2^k*binomial(n, k)*euler_number(k) for k in (0..n))
def a(n): return euler_sum(n) << valuation(n + 1, 2)
print([a(n) for n in range(22)])