|
| |
|
|
A347838
|
|
Positive numbers that are congruent to 2, 5, or 11 modulo 12.
|
|
1
|
|
|
|
2, 5, 11, 14, 17, 23, 26, 29, 35, 38, 41, 47, 50, 53, 59, 62, 65, 71, 74, 77, 83, 86, 89, 95, 98, 101, 107, 110, 113, 119, 122, 125, 131, 134, 137, 143, 146, 149, 155, 158, 161, 167, 170, 173, 179, 182, 185, 191, 194, 197, 203, 206, 209, 215, 218, 221, 227, 230, 233, 239
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,1
|
|
|
COMMENTS
|
This sequence follows from the first column sequence of the array A347834, namely A047529 ({1,3,7} (mod 8)), as given in the formula below.
Together with A017617, the positive integers congruent to 8 modulo 12, one obtains A016789, the positive integers congruent to 2 modulo 3. See the array A347839.
|
|
|
LINKS
|
|
|
|
FORMULA
|
Trisection: a(3*k+1) = 2 + 12*k, a(3*k+2) = 5 + 12*k, a(3*k+3) = 11 + 12*k, or with a(3*k) = -1 + 12*k for k >= 0.
O.g.f. with a(0) =-1: G(x) = (-1 + 3*x + 3*x^2 + 7*x^3)/((1 - x)*(1 - x^3)) = -6/(1-x) + 4/(1-x)^2 + (1 + x)/(1 + x + x^2). Note that (1 - x)*(1 - x^3) = (1-x)^2*(1 + x + x^2) = 1 - x - x^3 + x^4.
a(n) = a(n-1) + a(n-3) - a(n-4), for n >= 4, given a(n) for 0..3, with a(0) = -1.
a(n) = 2*b(n) + 3*b(n-1) + 6*b(n-2) + b(n-3), with b(n) = floor((n+2)/3) = A002264(n+2).
a(n) = -1 + 3*n + 3*floor(n/3) (from the partial fraction decomposition of G).
E.g.f.: 1 + 2*exp(x)*(2*x - 1) + exp(-x/2)*(3*cos(sqrt(3)*x/2) + sqrt(3)*sin(sqrt(3)*x/2))/3. - Stefano Spezia, Dec 08 2021
Sum_{n>=1} (-1)^(n+1)/a(n) = ((sqrt(2)+1)*Pi + sqrt(3)*log(sqrt(3)+2) + sqrt(6)*log(5-2*sqrt(6)))/12. - Amiram Eldar, Dec 30 2021
|
|
|
MATHEMATICA
|
Map[(3 # + 1)/2 &, LinearRecurrence[{1, 0, 1, -1}, {1, 3, 7, 9}, 60]] (* Michael De Vlieger, Oct 21 2021 *)
|
|
|
CROSSREFS
|
Cf. A002264, A016921, A016789, A017617, A017653, A040117(k+1), A047261, A047529, A102283, A347834, A347839.
|
|
|
KEYWORD
|
nonn,easy
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
