login
Numbers congruent to {17,29} mod 36.
1

%I #28 Sep 10 2022 07:36:12

%S 17,29,53,65,89,101,125,137,161,173,197,209,233,245,269,281,305,317,

%T 341,353,377,389,413,425,449,461,485,497,521,533,557,569,593,605,629,

%U 641,665,677,701,713,737,749,773,785,809,821,845,857,881,893,917,929,953

%N Numbers congruent to {17,29} mod 36.

%C Subsequence of A087445.

%C Let terms in this sequence be T:

%C Collatz sequences (C) that contain no T must terminate at 1.

%C Define C containing at least one T as C(T), and let T(i) {i=1..z} be T in order of appearance in C(T).

%C All T(i) i>=2 have odd preimages congruent to either {1,5} mod 12 or {11,19} mod 24. Preimages of the second type (P2) are congruent to B mod 2^m (m>=4), where B is a set of numbers with a predictable recurrence pattern (a bit cumbersome to describe here) starting with A259663(n,2), i.e., {11, 19, 3, 35, 99, 483, ...}. All P2 lead to T(i) == A002450((m-2)/2) mod 2^(m-1) when m is even, and T(i) == A072197((m-3)/2) mod 2^(m-1) when m is odd. So, for example, T(i) == 1 mod 8 when P2 == 11 mod 16; T(i) == 13 mod 16 when P2 == 19 mod 32; T(i) == 5 mod 32 when P2 == 3 mod 64; T(i) == 53 mod 64 when P2 == 35 mod 128; etc.

%C If the Collatz conjecture is true (i.e., all C terminate at 1), then all C(T) contain T(z) after which all subsequent odd terms decrease and are congruent to {1,5} mod 12 that are not congruent to {17,29} mod 36. The first few T(z) are {17, 53, 341, 1109, 1205, ...}. So, for example, the trajectory of odd terms in C with initial term 950 is [475, 713, 535, 803, 1205, 113, 85, 1], where T(1) = 713 and T(2) = T(z) = 1205. In this example, P2 = 803 because 803 == 11 mod 24.

%H G. C. Greubel, <a href="/A259614/b259614.txt">Table of n, a(n) for n = 1..2000</a>

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

%F G.f.: x*(17+12*x+7*x^2) / ( (1+x)*(x-1)^2 ). - _R. J. Mathar_, Jul 26 2015

%F E.g.f.: 7 + (18*x - 4)*exp(x) - 3*exp(-x). - _David Lovler_, Sep 10 2022

%t Select[Range[1000], MemberQ[{17, 29}, Mod[#, 36]] &] (* _Vincenzo Librandi_, Jul 01 2015 *)

%o (Magma) [ n : n in [1..1000] | n mod 36 in [17, 29] ] // _Vincenzo Librandi_ Jul 01 2015

%Y Cf. A002450, A072197, A087445, A259663.

%K nonn,easy

%O 1,1

%A _Bob Selcoe_, Jun 30 2015