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%I #23 Aug 06 2018 14:04:00
%S 0,8,1,20,2,36,3,48,4,64,5,76,6,92,7,104,8,120,9,132,10,148,11,160,12,
%T 176,13,188,14,204,15,216,16,232,17,244,18,260,19,272,20,288,21,300,
%U 22,316,23,328,24,344,25,356,26,372,27,384,28,400,29,412,30,428,31,440,32,456,33
%N The 7x+-1 function: a(n) = 7n+1 if n == +1 (mod 4), a(n) = 7n-1 if n == -1 (mod 4), otherwise a(n) = n/2.
%C The 7x+-1 problem is as follows. Start with any natural number n. If 4 divides n-1, multiply it by 7 and add 1; if 4 divides n+1, multiply it by 7 and subtract 1; otherwise divide it by 2. The 7x+-1 problem concerns the question whether we always reach 1.
%H Colin Barker, <a href="/A317640/b317640.txt">Table of n, a(n) for n = 0..1000</a>
%H D. Barina, <a href="https://arxiv.org/abs/1807.00908">7x+-1: Close Relative of Collatz Problem</a>, arXiv:1807.00908 [math.NT], 2018.
%H K. Matthews, <a href="http://www.numbertheory.org/php/barina.html">David Barina's 7x+1 conjecture</a>.
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0,1,0,1,0,-1).
%F a(n) = a(a(2*n)).
%F From _Colin Barker_, Aug 03 2018: (Start)
%F G.f.: x*(8 + x + 12*x^2 + x^3 + 8*x^4) / ((1 - x)^2*(1 + x)^2*(1 + x^2)).
%F a(n) = a(n-2) + a(n-4) - a(n-6) for n>5.
%F (End)
%e a(3)=20 because 3 == -1 (mod 4), and thus 7*3 - 1 results in 20.
%e a(5)=36 because 5 == +1 (mod 4), and thus 7*5 + 1 results in 36.
%t Array[Which[#2 == 1, 7 #1 + 1, #2 == 3, 7 #1 - 1, True, #1/2] & @@ {#, Mod[#, 4]} &, 67, 0] (* _Michael De Vlieger_, Aug 02 2018 *)
%o (C)
%o int a(int n) {
%o ....switch(n%4) {
%o ........case 1: return 7*n+1;
%o ........case 3: return 7*n-1;
%o ........default: return n/2;
%o ....}
%o }
%o (PARI) a(n)={my(m=(n+2)%4-2); if(m%2, 7*n + m, n/2)} \\ _Andrew Howroyd_, Aug 02 2018
%o (PARI) concat(0, Vec(x*(8 + x + 12*x^2 + x^3 + 8*x^4) / ((1 - x)^2*(1 + x)^2*(1 + x^2)) + O(x^70))) \\ _Colin Barker_, Aug 03 2018
%Y Cf. A006370.
%K nonn,easy
%O 0,2
%A _David Barina_, Aug 02 2018
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