A299647 Positive solutions to x^2 == -2 (mod 11).
3, 8, 14, 19, 25, 30, 36, 41, 47, 52, 58, 63, 69, 74, 80, 85, 91, 96, 102, 107, 113, 118, 124, 129, 135, 140, 146, 151, 157, 162, 168, 173, 179, 184, 190, 195, 201, 206, 212, 217, 223, 228, 234, 239, 245, 250, 256, 261, 267, 272, 278, 283, 289, 294, 300, 305, 311, 316
Offset: 1
Links
- Colin Barker, Table of n, a(n) for n = 1..1000
- Index entries for linear recurrences with constant coefficients, signature (1,1,-1).
Crossrefs
Programs
-
GAP
List([1..60], n -> 5*n-2+(2*n-(-1)^n-3)/4);
-
Julia
[(11(2n-1)-(-1)^n)>>2 for n in 1:60] # Peter Luschny, Mar 07 2018
-
Magma
[5*n-2+(2*n-(-1)^n-3)/4: n in [1..60]];
-
Mathematica
Table[5 n - 2 + (2 n - (-1)^n - 3)/4, {n, 1, 60}] CoefficientList[ Series[(3 + 5x + 3x^2)/((x - 1)^2 (x + 1)), {x, 0, 57}], x] (* or *) LinearRecurrence[{1, 1, -1}, {3, 8, 14}, 58] (* Robert G. Wilson v, Mar 08 2018 *)
-
Maxima
makelist(5*n-2+(2*n-(-1)^n-3)/4, n, 1, 60);
-
PARI
vector(60, n, nn; 5*n-2+(2*n-(-1)^n-3)/4)
-
Python
[5*n-2+(2*n-(-1)**n-3)/4 for n in range(1, 60)]
-
Sage
[5*n-2+(2*n-(-1)^n-3)/4 for n in (1..60)]
Formula
O.g.f.: x*(3 + 5*x + 3*x^2)/((1 + x)*(1 - x)^2).
E.g.f.: (-1 + 12*exp(x) - 11*exp(2*x) + 22*x*exp(2*x))*exp(-x)/4.
a(n) = -a(-n+1) = a(n-1) + a(n-2) - a(n-3).
a(n) = 5*n - 2 + (2*n - (-1)^n - 3)/4.
a(n) = 4*n - 1 + floor((n - 1)/2) + floor((3*n - 1)/3).
a(n+k) - a(n) = 11*k/2 + (1 - (-1)^k)*(-1)^n/4.
a(n+k) + a(n) = 11*(2*n + k - 1)/2 - (1 + (-1)^k)*(-1)^n/4.
E.g.f.: 3 + ((22*x - 11)*exp(x) - exp(-x))/4. - David Lovler, Aug 08 2022
Sum_{n>=1} (-1)^(n+1)/a(n) = tan(5*Pi/22)*Pi/11. - Amiram Eldar, Feb 27 2023
From Amiram Eldar, Nov 23 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = cosec(3*Pi/22)/2.
Product_{n>=1} (1 + (-1)^n/a(n)) = sec(5*Pi/22)*sin(2*Pi/11). (End)
Comments