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A249916 a(n) = 4*(n - 1) - a(n-3), n >= 3, a(0) = a(1) = 1, a(2) = 5.

%I A249916 #44 May 10 2020 09:17:26
%S A249916 1,1,5,7,11,11,13,13,17,19,23,23,25,25,29,31,35,35,37,37,41,43,47,47,
%T A249916 49,49,53,55,59,59,61,61,65,67,71,71,73,73,77,79,83,83,85,85,89,91,95,
%U A249916 95,97,97,101,103,107,107,109,109,113,115,119,119,121,121,125,127
%N A249916 a(n) = 4*(n - 1) - a(n-3), n >= 3, a(0) = a(1) = 1, a(2) = 5.
%C A249916 Conjecture: These are the natural numbers of the form 6*j +- 1 in which those of the form 12*k +- 1 are repeated.
%C A249916 From _Jianing Song_, Jan 28 2019: (Start)
%C A249916 The second conjecture in Formula section is correct.
%C A249916 We can see this from the recurrence: a(n) = 4*(n - 1) - a(n-3)            (1)
%C A249916 Replace n with n+1:  a(n+1) = 4*n - a(n-2)                                (2)
%C A249916 Subtract (1) from (2): a(n+1) = 4 + a(n) - a(n-2) + a(n-3)                (3)
%C A249916 Replace n with n+1:  a(n+2) = 4 + a(n+1) - a(n-1) + a(n-2)                (4)
%C A249916 Subtract (3) from (4): a(n+2) = 2*a(n+1)-a(n)-a(n-1)+2*a(n-2)-a(n-3)      (5)
%C A249916 This also confirms the conjecture in Comment section and the conjecture on the g.f. in Formula section. (End)
%H A249916 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (2,-1,-1,2,-1).
%F A249916 G.f.: (1 - x + 4*x^2 - x^3 + x^4)/((1 - x)^2*(1 + x^3)).  [Confirmed, see _Jianing Song_ in Comment section.]
%F A249916 Recurrence: a(n) = 2*a(n-1) - a(n-2) - a(n-3) + 2*a(n-4) - a(n-5) for n > 4, a(0)=a(1)=1, a(2)=5, a(3)=7, a(4)=11. [Confirmed, see _Jianing Song_ in Comment section.]
%t A249916 a[0] = a[1] = 1; a[2] = 5; a[n_] := 4*(n - 1) - a[n - 3]; Table[a[n], {n, 0, 63}]
%t A249916 RecurrenceTable[{a[0]==a[1]==1,a[2]==5,a[n]==4(n-1)-a[n-3]},a,{n,70}] (* _Harvey P. Dale_, Jan 26 2019 *)
%Y A249916 Cf. A007310, A091998.
%K A249916 nonn,easy
%O A249916 0,3
%A A249916 _L. Edson Jeffery_, Jan 14 2015