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A259614
Numbers congruent to {17,29} mod 36.
1
17, 29, 53, 65, 89, 101, 125, 137, 161, 173, 197, 209, 233, 245, 269, 281, 305, 317, 341, 353, 377, 389, 413, 425, 449, 461, 485, 497, 521, 533, 557, 569, 593, 605, 629, 641, 665, 677, 701, 713, 737, 749, 773, 785, 809, 821, 845, 857, 881, 893, 917, 929, 953
OFFSET
1,1
COMMENTS
Subsequence of A087445.
Let terms in this sequence be T:
Collatz sequences (C) that contain no T must terminate at 1.
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).
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.
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.
FORMULA
G.f.: x*(17+12*x+7*x^2) / ( (1+x)*(x-1)^2 ). - R. J. Mathar, Jul 26 2015
E.g.f.: 7 + (18*x - 4)*exp(x) - 3*exp(-x). - David Lovler, Sep 10 2022
MATHEMATICA
Select[Range[1000], MemberQ[{17, 29}, Mod[#, 36]] &] (* Vincenzo Librandi, Jul 01 2015 *)
PROG
(Magma) [ n : n in [1..1000] | n mod 36 in [17, 29] ] // Vincenzo Librandi Jul 01 2015
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Bob Selcoe, Jun 30 2015
STATUS
approved