A318972 The 7x+-1 function ("shortcut" definition): a(n) = (7n+1)/4 if n == +1 (mod 4), a(n) = (7n-1)/4 if n == -1 (mod 4), otherwise a(n) = n/2.

0, 2, 1, 5, 2, 9, 3, 12, 4, 16, 5, 19, 6, 23, 7, 26, 8, 30, 9, 33, 10, 37, 11, 40, 12, 44, 13, 47, 14, 51, 15, 54, 16, 58, 17, 61, 18, 65, 19, 68, 20, 72, 21, 75, 22, 79, 23, 82, 24, 86, 25, 89, 26, 93, 27, 96, 28, 100, 29, 103, 30, 107, 31, 110, 32, 114, 33, 117, 34, 121

Offset

0,2

Comments

See A317640 for another definition of this problem.

Links

Harvey P. Dale, Table of n, a(n) for n = 0..1000

David Barina, 7x+-1: Close Relative of Collatz Problem, arXiv:1807.00908 [math.NT], 2018.

K. Matthews, David Barina's 7x+1 conjecture.

Index entries for linear recurrences with constant coefficients, signature (0, 1, 0, 1, 0, -1).

Formula

a(n) = a(a(2*n))

From Chai Wah Wu, Nov 09 2018: (Start)

a(n) = a(n-2) + a(n-4) - a(n-6) for n > 5.

G.f.: x*(2*x^4 + x^3 + 3*x^2 + x + 2)/(x^6 - x^4 - x^2 + 1). (End)

Example

a(3) = 5 because 3 == -1 (mod 4), and thus (7*3 - 1)/4 results in 5.
a(5) = 9 because 5 == +1 (mod 4), and thus (7*5 + 1)/4 results in 9.

Mathematica

Table[Which[Mod[n, 4]==1, (7n+1)/4, Mod[n, 4]==3, (7n-1)/4, True, n/2], {n, 0, 70}] (* Harvey P. Dale, Oct 25 2025 *)
(* Alternative: *)
LinearRecurrence[{0, 1, 0, 1, 0, -1}, {0, 2, 1, 5, 2, 9}, 80] (* Harvey P. Dale, Oct 25 2025 *)

Prog

(C)
int a(int n) {
    switch(n%4) {
        case 1: return (7*n+1)/4;
        case 3: return (7*n-1)/4;
        default: return n/2;
    }
}
(PARI) a(n) = my(m=n%4); if (m==1, (7*n+1)/4, if (m==3, (7*n-1)/4, n/2)); \\ Michel Marcus, Sep 06 2018
(Python)
from __future__ import division
def A318972(n):
    return (7*n+1)//4 if n % 4 == 1 else (7*n-1)//4 if n % 4 == 3 else n//2 # Chai Wah Wu, Nov 09 2018

Crossrefs

Cf. A014682 (3x+1 equivalent), A317640.

Sequence in context: A257971 A205377 A082010 * A341495 A384952 A337943

Adjacent sequences: A318969 A318970 A318971 * A318973 A318974 A318975

Keyword

nonn,easy

Author

David Barina, Sep 06 2018

Status

approved