This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A261723 #30 Sep 08 2022 08:46:13
%S A261723 4,3,6,5,10,9,18,17,34,33,66,65,130,129,258,257,514,513,1026,1025,
%T A261723 2050,2049,4098,4097,8194,8193,16386,16385,32770,32769,65538,65537,
%U A261723 131074,131073,262146,262145,524290,524289,1048578,1048577,2097154,2097153,4194306,4194305,8388610,8388609
%N A261723 Interleave 2^n + 2 and 2^n + 1.
%C A261723 There is only one fundamental difference between the x + 1 problem and the 3x + 1 problem, that being one of definition: if x is odd, then the next value is x + 1 rather than 3x + 1. But there are also important differences between the two problems, such as that with the x + 1 problem it is easy to prove that every positive integer x reaches 1 after a finite number of iterations.
%C A261723 As with the 3x + 1 problem, the minimum necessary number of iterations is obvious: given an arbitrary positive n, it will take at least floor(log_2(n)) iterations to reach 1, especially if n is a power of 2 to begin with.
%C A261723 But with the x + 1 problem, it is almost as easy to determine the maximum number of iterations needed to reach 1: 2*ceiling(log_2(n)). This is the case when n is one of the odd numbers in this sequence, in which case there are no consecutive halving steps prior to reaching 4.
%C A261723 If an initial 1 and 2 are added to this sequence, the complete x + 1 trajectory of any number in this sequence can be obtained by reading backwards from that number.
%C A261723 A061313(n) = number of steps to reach 1 when starting with n. - _Reinhard Zumkeller_, Sep 05 2015
%H A261723 Colin Barker, <a href="/A261723/b261723.txt">Table of n, a(n) for n = 1..1000</a>
%H A261723 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0,3,0,-2).
%F A261723 a(2n - 1) = 2^n + 2, a(2n) = 2^n + 1.
%F A261723 a(1) = 4, a(2n) = 2a(n - 1), a(2n + 1) = a(n - 1) - 1.
%F A261723 a(1) = 4, a(n) = a(n - 1) - 1 if a(n - 1) is even, a(n) = 2a(n - 1) if a(n - 1) is odd.
%F A261723 a(n) = 3*a(n-2) - 2*a(n-4) for n > 4. - _Colin Barker_, Aug 31 2015
%F A261723 G.f.: -x*(4*x^3+6*x^2-3*x-4) / ((x-1)*(x+1)*(2*x^2-1)). - _Colin Barker_, Aug 31 2015
%F A261723 a(n) = (2^((n+1)/2)+2)*(1-(-1)^n)/2+(2^(n/2)+1)*(1+(-1)^n)/2. - _Wesley Ivan Hurt_, Sep 06 2015
%F A261723 E.g.f.: sqrt(2)*sinh(sqrt(2)*x) + cosh(sqrt(2)*x) + 2*sinh(x) + cosh(x) - 2. - _Robert Israel_, Sep 06 2015
%p A261723 A261723:=n->(2^((n+1)/2)+2)*(1-(-1)^n)/2+(2^(n/2)+1)*(1+(-1)^n)/2: seq(A261723(n), n=1..60); # _Wesley Ivan Hurt_, Sep 06 2015
%t A261723 Flatten[Table[{2^n + 2, 2^n + 1}, {n, 25}]]
%o A261723 (Magma) &cat[[2^n+2, 2^n+1]: n in [1..30]]; // _Vincenzo Librandi_, Aug 31 2015
%o A261723 (PARI) Vec(-x*(4*x^3+6*x^2-3*x-4)/((x-1)*(x+1)*(2*x^2-1)) + O(x^100)) \\ _Colin Barker_, Aug 31 2015
%o A261723 (Haskell)
%o A261723 a261723 n = a261723_list !! (n-1)
%o A261723 a261723_list = concat $ transpose [tail a052548_list, tail a000051_list]
%o A261723 -- _Reinhard Zumkeller_, Sep 05 2015
%Y A261723 Cf. A000051, A052548, A061313.
%K A261723 easy,nonn
%O A261723 1,1
%A A261723 _Alonso del Arte_, Aug 29 2015
%E A261723 Two incorrect terms corrected by _Colin Barker_, Aug 31 2015