|
|
A091998
|
|
Numbers that are congruent to {1, 11} mod 12.
|
|
17
|
|
|
1, 11, 13, 23, 25, 35, 37, 47, 49, 59, 61, 71, 73, 83, 85, 95, 97, 107, 109, 119, 121, 131, 133, 143, 145, 155, 157, 167, 169, 179, 181, 191, 193, 203, 205, 215, 217, 227, 229, 239, 241, 251, 253, 263, 265, 275, 277, 287, 289, 299, 301, 311, 313, 323, 325, 335
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,2
|
|
COMMENTS
|
Cf. property described by Gary Detlefs in A113801: more generally, these numbers are of the form (2*h*n + (h-4)*(-1)^n-h)/4 (h and n in A000027), then ((2*h*n + (h-4)*(-1)^n - h)/4)^2 - 1 == 0 (mod h); in our case, a(n)^2 - 1 == 0 (mod 12). Also a(n)^2 - 1 == 0 (mod 24).
|
|
LINKS
|
|
|
FORMULA
|
a(n) = 6*n + 2*(-1)^n - 3.
G.f.: x*(1+10*x+x^2)/((1+x)*(1-x)^2).
a(n) - a(n-1) - a(n-2) + a(n-3) = 0 for n > 3.
a(n) = a(n-2) + 12 for n > 2.
a(n) = 12*A000217(n-1) + 1 - 2*Sum_{i=1..n-1} a(i) for n > 1.
Sum_{n>=1} (-1)^(n+1)/a(n) = (2 + sqrt(3))*Pi/12. - Amiram Eldar, Dec 04 2021
E.g.f.: 1 + (6*x - 3)*exp(x) + 2*exp(-x). - David Lovler, Sep 04 2022
|
|
MATHEMATICA
|
LinearRecurrence[{1, 1, -1}, {1, 11, 13}, 100] (* Harvey P. Dale, Jul 26 2017 *)
|
|
PROG
|
(Magma) [ n: n in [1..350] | n mod 12 eq 1 or n mod 12 eq 11 ];
(Haskell)
a091998 n = a091998_list !! (n-1)
a091998_list = 1 : 11 : map (+ 12) a091998_list
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
|
|
EXTENSIONS
|
Formulae and comment added by Bruno Berselli, Nov 17 2010 - Nov 18 2010
|
|
STATUS
|
approved
|
|
|
|