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Numbers that are congruent to {4, 20} mod 24.
7

%I #53 Sep 06 2022 14:38:05

%S 4,20,28,44,52,68,76,92,100,116,124,140,148,164,172,188,196,212,220,

%T 236,244,260,268,284,292,308,316,332,340,356,364,380,388,404,412,428,

%U 436,452,460,476,484,500,508,524,532,548,556,572,580,596,604,620,628

%N Numbers that are congruent to {4, 20} mod 24.

%H Danny Rorabaugh, <a href="/A259755/b259755.txt">Table of n, a(n) for n = 1..10000</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1,1,-1).

%F a(n) = 2*(6*n + (-1)^n - 3).

%F A259748(a(n))/a(n) = 3/4.

%F a(n) = 4*A007310(n). - _Michel Marcus_, Sep 22 2015

%F G.f.: 4*x*(1 + 4*x + x^2) / ((1 + x)*(1 - x)^2). - _Bruno Berselli_, Oct 23 2015

%F Sum_{n>=1} (-1)^(n+1)/a(n) = sqrt(3)*Pi/24. - _Amiram Eldar_, Dec 31 2021

%F E.g.f.: 2*(2 + (6*x - 3)*exp(x) + exp(-x)). - _David Lovler_, Sep 05 2022

%t A[n_] := A[n] = Sum[a b, {a, 1,n}, {b, a + 1, n}]; Select[Range[200], Mod[A[#], #]/# == 3/4 &]

%t Table[2 (6 n + (-1)^n - 3), {n, 1, 60}] (* _Bruno Berselli_, Oct 23 2015 *)

%t LinearRecurrence[{1,1,-1},{4,20,28},60] (* _Harvey P. Dale_, Jul 19 2016 *)

%o (Magma) [2*(6*n+(-1)^n-3): n in [1..60]]; // _Vincenzo Librandi_, Aug 27 2015

%o (PARI) vector(100, n, 2*(6*n+(-1)^n-3)) \\ _Altug Alkan_, Oct 23 2015

%Y Cf. A000914, A007310.

%Y Other sequences of numbers k such that A259748(k)/k equals a constant: A008606, A073762, A259749, A259750, A259751, A259752, A259754.

%K nonn,easy

%O 1,1

%A _José María Grau Ribas_, Jul 18 2015

%E Better name from _Danny Rorabaugh_, Oct 22 2015