A072197 a(n) = 4*a(n-1) + 1 with a(0) = 3.

3, 13, 53, 213, 853, 3413, 13653, 54613, 218453, 873813, 3495253, 13981013, 55924053, 223696213, 894784853, 3579139413, 14316557653, 57266230613, 229064922453, 916259689813, 3665038759253, 14660155037013, 58640620148053, 234562480592213, 938249922368853, 3752999689475413

Offset: 0

Author

N. Rathankar (rathankar(AT)yahoo.com), Jul 03 2002

Keywords

• nonn
• easy

Comments

Let A be the Hessenberg matrix of order n, defined by: A[1, j] = 1, A[i, i] := 2, (i > 1), A[i, i - 1] = -1, and A[i, j] = 0 otherwise. Then, for n >= 1, a(n - 1) = (-1)^n*charpoly(A, -2). - Milan Janjic, Jan 26 2010

Numbers whose binary representation is 11 together with n times 01. For example, 213 = 11010101 (2). - Omar E. Pol, Nov 22 2012

The Collatz-function starting with a(n) will terminate at 1 after 2*n + 7 steps. This is because 3*a(n) + 1 = 5*2^(2n + 1), and the Collatz-function starting with 5 terminates at 1 after 5 additional steps. So for example, a(2) = 53; Collatz sequence starting with 53 follows: 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1 (11 steps). - Bob Selcoe, Apr 03 2015

a(n) is also the sum of the numerator and denominator of the binary fractions 0.1, 0.101, 0.10101, 0.1010101... Thus 0.1 = 1/2 with 1 + 2 = 3, 0.101 = 1/2 + 1/8 = 5/8 with 5 + 8 = 13; 0.10101 = 1/2 + 1/8 + 1/32 = 21/32 with 21 + 32 = 53. - J. M. Bergot, Sep 28 2016

a(n), for n >= 2, is also the smallest odd number congruent to 5 modulo 8 for which the modified reduced Collatz map given in A324036 has n consecutive extra steps compared to the reduced Collatz map given in A075677. - Nicolas Vaillant, Philippe Delarue, Wolfdieter Lang, May 09 2019

Examples

			a(1) = 13 because a(0) = 3 and 4 * 3 + 1 = 13.
a(2) = 53 because a(1) = 13 and 4 * 13 + 1 = 53.
a(3) = 213 because a(2) = 53 and 4 * 53 + 1 = 213.
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• Petro Kosobutskyy and Volodymyr Karkulovskyy, Recurrence and structuring of sequences of transformations 3n+ 1 as arguments for confirmation of the Collatz hypothesis, Comp. Des. Sys., Theor. Practice (2023) Vol. 5, No. 1, pp. 28-33. See p. 29.
• Petro Kosobutskyy, Anastasiia Yedyharova, and Taras Slobodzyan, From Newton's binomial and Pascal's triangle to Collatz's problem, Comp. Des. Sys., Theor. Practice (2023) Vol. 5, No. 1, pp. 121-127.
• Index entries for linear recurrences with constant coefficients, signature (5,-4).
• Index entries for sequences related to 3x+1 (or Collatz) problem.

Crossrefs

Cf. A000302, A002066 (first differences), A002450, A007583, A020988, A075677, A178415, A324036, A347834.

Programs

Magma

[(10*4^n-1)/3: n in [0..30]]; // Vincenzo Librandi, Oct 31 2011

Maple

A072197:=n->(10*4^n - 1)/3: seq(A072197(n), n=0..30); # Wesley Ivan Hurt, Sep 29 2016

Mathematica

Table[(10(4^n) - 1)/3, {n, 0, 19}] (* Alonso del Arte, Nov 22 2012 *)
NestList[4#+1&,3,30] (* Harvey P. Dale, Mar 09 2019 *)

PARI

a(n)=10*4^n\3 \\ Charles R Greathouse IV, Apr 06 2016

PARI

Vec((3-2*x)/((1-x)*(1-4*x)) + O(x^30)) \\ Colin Barker, Sep 28 2016

Formula

a(n) = (10*4^n - 1)/3 = 10*A002450(n) + 3. - Henry Bottomley, Dec 02 2002

a(n) = 5*a(n-1) - 4*a(n-2), n > 1. - Vincenzo Librandi, Oct 31 2011

a(n) = 2^(2*(n + 1)) - (2^(2*n + 1) + 1)/3 = A000302(n + 1) - A007583(n). - Vladimir Pletser, Apr 12 2014

a(n) = (5*2^(2*n + 1) - 1)/3. - Bob Selcoe, Apr 03 2015

G.f.: (3-2*x)/((1-x)*(1-4*x)). - Colin Barker, Sep 28 2016

a(n) = A020988(n) + A020988(n+1) + 1 = 2*(A002450(n) + A002450(n+1)) + 1. - Yosu Yurramendi, Jan 24 2017

a(n) = A002450(n+1) + 2^(2*n+1). - Adam Michael Bere, May 13 2021

a(n) = a(n-1) + 5*2^(2*n-1), for n >= 1, with a(0) = 3. - Wolfdieter Lang, Aug 16 2021

a(n) = A178415(2,n+1) = A347834(2,n), arrays, for n >= 0. - Wolfdieter Lang, Nov 29 2021

E.g.f.: exp(x)*(10*exp(3*x) - 1)/3. - Elmo R. Oliveira, Apr 02 2025

Extensions

More terms from Henry Bottomley, Dec 02 2002