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A087445 Numbers that are congruent to 1 or 5 mod 12. 6

%I #99 Nov 08 2022 01:40:42

%S 1,5,13,17,25,29,37,41,49,53,61,65,73,77,85,89,97,101,109,113,121,125,

%T 133,137,145,149,157,161,169,173,181,185,193,197,205,209,217,221,229,

%U 233,241,245,253,257,265,269,277,281,289,293,301,305,313,317,325,329

%N Numbers that are congruent to 1 or 5 mod 12.

%C From _Bob Selcoe_, Jun 03 2015: (Start)

%C For k >= 1: all numbers congruent to A002450(k) mod 2^(2k+1) and A072197(k) mod 4^(k+1) not congruent to 0 mod 3. Equivalently, for k >= 3: all numbers congruent to A096773(k) mod 2^k not congruent to 0 mod 3.

%C Conjecture: at least one number in this sequence must appear in all Collatz sequences.

%C (End)

%C The sequence is composed of all numbers in congruence classes T(n,1) mod 2^(n+k) in A259663 (i.e., T"(1) in array T259663(n,k)) not congruent to 0 mod 3. Therefore the conjecture above is true (see A259663 for additional explanation). - _Bob Selcoe_, Jul 15 2017

%C Closure of {1} under the map (x,y)->2x+3y [Klarner-Rado, see Lagarias (2016), p. 755]. - _N. J. A. Sloane_, Oct 06 2016

%C The above conjecture is true: this is because even numbers and odd numbers divisible by 3 will lead to the set of odd numbers not divisible by 3. Odd numbers of the form 4k - 1 can also be ignored, as this consists of odd numbers that grow between themselves and the next odd term through Collatz iteration. No infinite sequence of growth between consecutive odd terms is possible, so all numbers of the form 4k - 1 will lead to an odd number that shrinks between itself and the next odd number. All numbers 4k - 1 will lead to a number in 4k - 3, the odd numbers that shrink between themselves and the following odd term. What we are left after that elimination is this sequence. - _Aidan Simmons_, Feb 25 2019

%H Jeffrey C. Lagarias, <a href="http://www.jstor.org/stable/10.4169/amer.math.monthly.123.8.753">Erdős, Klarner and the 3x+ 1 Problem</a>, Amer. Math. Monthly, Vol. 123, No. 8 (2016), pp. 753-776.

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

%F G.f.: x*(1+4*x+7*x^2)/((1+x)*(1-x)^2).

%F E.g.f.: 6*(x-1)*exp(x) + 7 - exp(-x). - corrected by _Robert Israel_, Jun 10 2015

%F a(n) = 6*(n-1) - (-1)^n. - _Rolf Pleisch_, Aug 04 2009

%F a(n) = 12*n - a(n-1) - 18 (with a(1)=1). - _Vincenzo Librandi_, Aug 08 2010

%F a(n) = a(n-1) + a(n-2) - a(n-3), with a(0)=1, a(1)=5, a(2)=13. - _Harvey P. Dale_, Jun 13 2011

%F Sum_{n>=1} (-1)^(n+1)/a(n) = sqrt(3)*Pi/12 + log(2 + sqrt(3))/(2*sqrt(3)). - _Amiram Eldar_, Dec 28 2021

%p seq(6*(n-1)-(-1)^n,n=1..100); # _Robert Israel_, Jun 10 2015

%t LinearRecurrence[{1,1,-1},{1,5,13},70] (* or *) Rest[CoefficientList[ Series[x (1+4x+7x^2)/((1+x)(1-x)^2),{x,0,70}],x]] (* _Harvey P. Dale_, Jun 13 2011 *)

%o (PARI) a(n)=(n-1)\2*12 + [5,1][n%2+1] \\ _Charles R Greathouse IV_, Jun 03 2015

%o (Magma) [k:k in [1..330]| k mod 12 in [1,5]]; // _Marius A. Burtea_, Feb 08 2020

%Y Cf. A001651, A002450, A047241, A087444, A087446, A096773, A259663.

%K easy,nonn

%O 1,2

%A _Paul Barry_, Sep 04 2003

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