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Numbers that are congruent to {7,11} mod 12.
5

%I #65 Nov 24 2022 04:20:50

%S 7,11,19,23,31,35,43,47,55,59,67,71,79,83,91,95,103,107,115,119,127,

%T 131,139,143,151,155,163,167,175,179,187,191,199,203,211,215,223,227,

%U 235,239,247,251,259,263,271,275,283,287,295,299,307,311,319,323,331,335

%N Numbers that are congruent to {7,11} mod 12.

%C From _Arkadiusz Wesolowski_, Mar 16 2014: (Start)

%C Odd numbers m for which 2/m is not equal to 1/x + 1/y with x = 2*floor((m + 1)/4) + 1 and an integer y > x.

%C The primes together with 3 are in A002145. (End)

%C Odd numbers not of the form (4j+1)*3^k, {j,k>=0}. - _Bob Selcoe_, Aug 30 2015

%C Nonnegative k for which k == 3 (mod 4) and k^2 == 1 (mod 3). - _Bruno Berselli_, Apr 26 2018

%C Numbers that are not divisible by their digital root in base 5. - _Amiram Eldar_, Nov 24 2022

%H Vincenzo Librandi, <a href="/A168489/b168489.txt">Table of n, a(n) for n = 1..1000</a>

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

%F a(n) = 12*n - a(n-1) - 6 for n>1, a(1)=7.

%F From _R. J. Mathar_, Mar 21 2010: (Start)

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

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

%F G.f.: x*(7 + 4*x + x^2)/ ((1+x) * (x-1)^2). (End)

%F E.g.f.: 1 + 6*x*exp(x) - exp(-x). - _G. C. Greubel_, Aug 30 2015

%F Sum_{n>=1} (-1)^(n+1)/a(n) = sqrt(3)*Pi/12 - log(2+sqrt(3))/(2*sqrt(3)). - _Amiram Eldar_, Dec 30 2021

%t LinearRecurrence[{1, 1, -1}, {7, 11, 19}, 60] (* or *) Select[Range[350], MemberQ[{7, 11}, Mod[#, 12]]&] (* _Harvey P. Dale_, Nov 10 2011 *)

%t Rest[CoefficientList[Series[x (7 + 4 x + x^2)/((1 + x) (x - 1)^2), {x, 0, 56}], x] ] (* _Ray Chandler_, Jul 07 2015 *)

%t RecurrenceTable[{a[n] == 12 n - 6 - a[n-1], a[1]==7}, a, {n, 1, 100}] (* _G. C. Greubel_, Aug 30 2015 *)

%o (Magma) [6*n-(-1)^n: n in [1..60]]; // _Vincenzo Librandi_, Aug 10 2012

%o (PARI) x='x+O('x^100); Vec(x*(7+4*x+x^2)/((1+x)*(x-1)^2)) \\ _Altug Alkan_, Oct 22 2015

%Y Cf. A002145, A239233.

%K nonn,easy

%O 1,1

%A _Vincenzo Librandi_, Nov 27 2009