A239124 a(n) = 64*n - 11 for n >= 1. Third column of triangle A238476.
53, 117, 181, 245, 309, 373, 437, 501, 565, 629, 693, 757, 821, 885, 949, 1013, 1077, 1141, 1205, 1269, 1333, 1397, 1461, 1525, 1589, 1653, 1717, 1781, 1845, 1909, 1973, 2037, 2101, 2165, 2229, 2293, 2357, 2421, 2485, 2549, 2613, 2677, 2741, 2805, 2869, 2933, 2997
Offset: 1
Examples
a(1) = 53 because the Collatz sequence of length 7 following the pattern uddddd, ending in an odd number is [53, 160, 80, 40, 20, 10, 5]. The end number is 6*1 - 1 = 5.
Links
- Vincenzo Librandi, Table of n, a(n) for n = 1..1000
- Wolfdieter Lang, On Collatz' Words, Sequences, and Trees, J. of Integer Sequences, Vol. 17 (2014), Article 14.11.7.
- Manfred Trümper, The Collatz Problem in the Light of an Infinite Free Semigroup, Chinese Journal of Mathematics, Vol. 2014, Article ID 756917, 21 pages.
- Index entries for linear recurrences with constant coefficients, signature (2,-1).
Programs
-
Magma
[64*n-11: n in [1..50]]; // Vincenzo Librandi, Mar 13 2014
-
Mathematica
CoefficientList[Series[(53 + 11 x)/(1 - x)^2, {x, 0, 50}], x] (* Vincenzo Librandi, Mar 13 2014 *) 64 Range[40]-11 (* Harvey P. Dale, Nov 21 2018 *)
Formula
O.g.f.: x*(53+11*x)/(1-x)^2.
From Elmo R. Oliveira, Apr 04 2025: (Start)
E.g.f.: 11 + exp(x)*(64*x - 11).
a(n) = 2*a(n-1) - a(n-2) for n > 2. (End)
Comments