A261728 -id:A261728 - cp's OEIS frontend

A261831 a(2n-1) = 2n-1; otherwise a(n) is the smallest even number not already present which is obtained from the existing terms by the rules of (3*n+1)-problem.

1, 4, 3, 2, 5, 10, 7, 16, 9, 8, 11, 22, 13, 28, 15, 14, 17, 34, 19, 40, 21, 20, 23, 46, 25, 52, 27, 26, 29, 58, 31, 64, 33, 32, 35, 70, 37, 76, 39, 38, 41, 82, 43, 88, 45, 44, 47, 94, 49, 100, 51, 50, 53, 106, 55, 112, 57, 56, 59, 118, 61, 124, 63, 62, 65, 130, 67, 136, 69, 68, 71, 142, 73, 148, 75, 74, 77, 154, 79, 160, 81, 80

Offset: 1

Vladimir Shevelev, Sep 02 2015

nonn

By the rules of the (3*n+1)-problem, an even number can appear either by the operation 3*x+1 only when x is an odd number or by the division of a number of the form 4*k by 2.
Using induction as in the proof of the Theorem in A261728, one can prove that if n == 0(mod 6), then a(n) = 2*n-2; if n == 2(mod 6), then a(n) = 2*n; if n == 4(mod 6), then a(n) = n-2.
The sequence is a permutation of the positive integers not divisible by 6 (A047253).

			Let n=28. Since 28 is of the form 6*k+4 with k=4, then a(28) = 6*4+2 = 26.

Peter J. C. Moses, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,2,0,0,0,0,0,-1).

Cf. A109732, A261690, A261728.

a[n_] := If[OddQ[n], n, Switch[Mod[n, 6], 0, 2n-2, 2, 2n, 4, n-2]]; Array[a, 81] (* Jean-François Alcover, Sep 02 2015, from given formula *)
LinearRecurrence[{0,0,0,0,0,2,0,0,0,0,0,-1},{1,4,3,2,5,10,7,16,9,8,11,22},90] (* Harvey P. Dale, Feb 24 2025 *)

a(2*n-1) = 2*n-1, for n>=1.
a(6*k) = 12*k-2, a(6*k+2) = 12*k+4 and a(6*k+4) = 6*k+2, for k>=0.
O.g.f.:(1+x (4+x (3+x (2+x (5+x (10+x (5+x (8+x (3+x (4+x (1+2 x)))))))))))/(-1+x^6)^2.

A261931 a(0)=1, a(1)=2, for n>=1, a(2n) = 6n - 2; a(2n+1) = min(((a(2n)-1)/3)^, (2a(2n))^, (2a(2n-1))^), where, instead of t, we write t^, if t has not appeared earlier in the sequence and is neither of the form 3k nor of the form 6m-2.

Original entry on oeis.org

1, 2, 4, 8, 10, 20, 16, 5, 22, 7, 28, 14, 34, 11, 40, 13, 46, 26, 52, 17, 58, 19, 64, 38, 70, 23, 76, 25, 82, 50, 88, 29, 94, 31, 100, 62, 106, 35, 112, 37, 118, 74, 124, 41, 130, 43, 136, 86, 142, 47, 148, 49, 154, 98, 160, 53, 166, 55, 172, 110, 178, 59, 184

Offset: 0

Vladimir Shevelev, Sep 06 2015

nonn

Let n>=7. If n == 1 or 3 (mod 6), then a(n) = n-2; if n == 5 (mod 6), then a(n) = 2*(n-4). This could be proved by induction similar to the theorem in A261728.
The sequence is a permutation of the positive integers not divisible by 3 which are not of the form 12*s+8, s>=2.
This sequence is connected with Collatz's (3*n+1)-conjecture. For example, if n=29, then, by the formulas, 29 = a(31) => 88 = a(30) => 11 = a(13) => 34 = a(12) => 17 = a(19) => 52 = a(18) => 13 = a(15) => 40 = a(14) => 5 = a(7) => 16 = a(6) => 1 = a(0).

			At n=3, since a(1)=2 and a(2)=6*1-2=4, a(3) should be either (4-1)/3=1 or 2*a(2)=8 or 2*a(1)=4; 1 and 4 have already appeared, so a(3)=8.
At n=5, since a(3)=8 and a(4)=6*2-2=10, a(5) should be either (10-1)/3=3 or 2*a(4)=20 or 2*a(3)=16; 3 is divisible by 3, and 16 is of the form 6*t-2, so a(5)=20.

Peter J. C. Moses, Table of n, a(n) for n = 0..9999

Cf. A261690, A261728, A261831.

For n>=1, a(2*n) = 6*n - 2.
For t>=1, a(6*t+1) = 6*t - 1; a(6*t+3) = 6*t+1; a(6*t+5) = 12*t + 2.
And from the name, for n>=3, a(2*n+1) = min(((a(2*n)-1)/3)^, (2*a(2*n-1))^).

cp's OEIS Frontend

A261831 a(2n-1) = 2n-1; otherwise a(n) is the smallest even number not already present which is obtained from the existing terms by the rules of (3*n+1)-problem.

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A261931 a(0)=1, a(1)=2, for n>=1, a(2n) = 6n - 2; a(2n+1) = min(((a(2n)-1)/3)^, (2a(2n))^, (2a(2n-1))^), where, instead of t, we write t^, if t has not appeared earlier in the sequence and is neither of the form 3k nor of the form 6m-2.

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cp's OEIS Frontend

A261831 a(2*n-1) = 2*n-1; otherwise a(n) is the smallest even number not already present which is obtained from the existing terms by the rules of (3*n+1)-problem.

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Programs

Mathematica

Formula

A261931 a(0)=1, a(1)=2, for n>=1, a(2*n) = 6*n - 2; a(2*n+1) = min(((a(2*n)-1)/3)^, (2*a(2*n))^, (2*a(2*n-1))^), where, instead of t, we write t^, if t has not appeared earlier in the sequence and is neither of the form 3*k nor of the form 6*m-2.

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A261831 a(2n-1) = 2n-1; otherwise a(n) is the smallest even number not already present which is obtained from the existing terms by the rules of (3*n+1)-problem.

A261931 a(0)=1, a(1)=2, for n>=1, a(2n) = 6n - 2; a(2n+1) = min(((a(2n)-1)/3)^, (2a(2n))^, (2a(2n-1))^), where, instead of t, we write t^, if t has not appeared earlier in the sequence and is neither of the form 3k nor of the form 6m-2.