A295340 Numbers congruent to 11 or 13 mod 15.
11, 13, 26, 28, 41, 43, 56, 58, 71, 73, 86, 88, 101, 103, 116, 118, 131, 133, 146, 148, 161, 163, 176, 178, 191, 193, 206, 208, 221, 223, 236, 238, 251, 253, 266, 268, 281, 283, 296, 298, 311, 313, 326, 328, 341, 343, 356, 358, 371, 373, 386, 388, 401, 403, 416, 418, 431, 433
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).
Programs
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Magma
[n: n in [1..500] | n mod 15 in [11, 13]]; // Vincenzo Librandi, Sep 06 2018
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Mathematica
ParallelMap[11 * Ceiling[#/2] + 2 * # - 2 &, Range@ 10^3] CoefficientList[ Series[(2x^2 + 2x + 11)/((1 + x) (x - 1)^2), {x, 0, 60}], x] (* or *) LinearRecurrence[{1, 1, -1}, {11, 13, 26}, 60] (* Robert G. Wilson v, Jan 09 2018 *) Select[Range[500], MemberQ[{11, 13}, Mod[#, 15]] &] (* Vincenzo Librandi, Sep 06 2018 *) 11/2 * Mod[#, 2] + 15 * #/2 - 2 &/@ Range@ 500 (* Mikk Heidemaa, Sep 08 2018 *) -
PARI
Vec(x*(11 + 2*x + 2*x^2) / ((1 - x)^2*(1 + x)) + O(x^40)) \\ Colin Barker, Dec 07 2017
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PARI
a(n) = if(n%2, (15*n+7)/2, (15*n-4)/2); \\ Altug Alkan, Sep 06 2018
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PARI
a(n) = [11, -2][(n - 1)%2 + 1] + 15*(n \ 2) \\ David A. Corneth, Sep 06 2018
Formula
a(n) = (1/4)*(-1)^n*(3*(-1)^n*(10*n + 1) - 11) for n > 0.
From Colin Barker, Dec 07 2017: (Start)
G.f.: x*(11 + 2*x + 2*x^2) / ((1 - x)^2*(1 + x)).
a(n) = (15*n - 4) / 2 for n even.
a(n) = (15*n + 7) / 2 for n odd.
a(n) = a(n-1) + a(n-2) - a(n-3) for n > 3.
(End)
a(n) = ceiling(15*n/2) + 5*(n mod 2) - 2 for n > 0. - Mikk Heidemaa, Sep 06 2018
a(n + 2) = a(n) + 15. - David A. Corneth, Sep 06 2018
a(n) = (11/2)*(n mod 2) + 15*n/2 - 2 for n > 0. - Mikk Heidemaa, Sep 08 2018
f(n) = 15*n - ((13*n + 17) mod 26) for n > 0 yields odd terms. - Mikk Heidemaa, Oct 28 2019
a(n) = 11*ceiling(1/2*n) + 2*n - 2 for n > 0. - Mikk Heidemaa, Nov 04 2019
E.g.f.: 2 + ((30*x + 3)*exp(x) - 11*exp(-x))/4. - David Lovler, Sep 08 2022
Extensions
Name simplified by David A. Corneth, Sep 06 2018
Comments