|
%I #37 Nov 09 2023 07:30:21
%S 1,3,9,7,17,11,25,15,33,19,41,23,49,27,57,31,65,35,73,39,81,43,89,47,
%T 97,51,105,55,113,59,121,63,129,67,137,71,145,75,153,79,161,83,169,87,
%U 177,91,185,95,193,99,201,103,209,107,217,111,225,115,233,119,241,123,249
%N Least odd number in the Collatz (3x+1) preimage of odd numbers not a multiple of 3.
%C The odd non-multiples of 3 are 1, 5, 7, 11,... (A007310). The odd multiples of 3 have no odd numbers their Collatz pre-image. The next odd number in the Collatz iteration of a(2n) is 6n-1. The next odd number in the Collatz iteration of a(2n+1) is 6n+1. For each non-multiple of 3, there are an infinite number of odd numbers in its Collatz pre-image. For example:
%C Odd pre-images of 1: 1, 5, 21, 85, 341,... (A002450)
%C Odd pre-images of 5: 3, 13, 53, 213, 853,... (A072197)
%C Odd pre-images of 7: 9, 37, 149, 597, 2389,...
%C Odd pre-images of 11: 7, 29, 117, 469, 1877,...(A072261)
%C In each case, the pre-image sequence is t(k+1) = 4*t(k) + 1 with t(0)=a(n). The array of pre-images is in A178415.
%C a(n) = A047529(P(n)), with the permutation P(n) = A006368(n-1) + 1, for n >= 1. This shows that this sequence gives the numbers {1, 3, 7} (mod 8) uniquely. - _Wolfdieter Lang_, Sep 21 2021
%H Matthew House, <a href="/A178414/b178414.txt">Table of n, a(n) for n = 1..10000</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0,2,0,-1).
%F a(n) = (n - 1)*(3 - (-1)^n) + 1. [_Bogart B. Strauss_, Sep 20 2013, adapted to the offset by _Matthew House_, Feb 14 2017]
%F From _Matthew House_, Feb 14 2017: (Start)
%F G.f.: x*(1 + 3*x + 7*x^2 + x^3)/((1 - x)^2*(1 + x)^2).
%F a(n) = 2*a(n-2) - a(n-4). (End)
%F From _Philippe Deléham_, Nov 06 2023: (Start)
%F a(2*n) = 4*n-1, a(2*n+1) = 8*n+1.
%F a(n) = 2*A022998(n-1)+1.
%F a(n) = 2*A114752(n)-1. (End)
%t Riffle[1+8*Range[0,50], 3+4*Range[0,50]]
%Y Cf. A002450, A006368, A007310, A022998, A047529, A072197, A072261, A114752.
%K nonn,easy
%O 1,2
%A _T. D. Noe_, May 28 2010
|
