A317640 The 7x+-1 function: a(n) = 7n+1 if n == +1 (mod 4), a(n) = 7n-1 if n == -1 (mod 4), otherwise a(n) = n/2.

0, 8, 1, 20, 2, 36, 3, 48, 4, 64, 5, 76, 6, 92, 7, 104, 8, 120, 9, 132, 10, 148, 11, 160, 12, 176, 13, 188, 14, 204, 15, 216, 16, 232, 17, 244, 18, 260, 19, 272, 20, 288, 21, 300, 22, 316, 23, 328, 24, 344, 25, 356, 26, 372, 27, 384, 28, 400, 29, 412, 30, 428, 31, 440, 32, 456, 33

Offset: 0

Author

David Barina, Aug 02 2018

Keywords

• nonn
• easy

Comments

The 7x+-1 problem is as follows. Start with any natural number n. If 4 divides n-1, multiply it by 7 and add 1; if 4 divides n+1, multiply it by 7 and subtract 1; otherwise divide it by 2. The 7x+-1 problem concerns the question whether we always reach 1.

Examples

			a(3)=20 because 3 == -1 (mod 4), and thus 7*3 - 1 results in 20.
a(5)=36 because 5 == +1 (mod 4), and thus 7*5 + 1 results in 36.
		

Links

• Colin Barker, Table of n, a(n) for n = 0..1000
• D. 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).

Crossrefs

Cf. A006370.

Programs

C

int a(int n) {
    switch(n%4) {
        case 1: return 7*n+1;
        case 3: return 7*n-1;
        default: return n/2;
    }
}

Mathematica

Array[Which[#2 == 1, 7 #1 + 1, #2 == 3, 7 #1 - 1, True, #1/2] & @@ {#, Mod[#, 4]} &, 67, 0] (* Michael De Vlieger, Aug 02 2018 *)

PARI

a(n)={my(m=(n+2)%4-2); if(m%2, 7*n + m, n/2)} \\ Andrew Howroyd, Aug 02 2018

PARI

concat(0, Vec(x*(8 + x + 12*x^2 + x^3 + 8*x^4) / ((1 - x)^2*(1 + x)^2*(1 + x^2)) + O(x^70))) \\ Colin Barker, Aug 03 2018

Formula

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

From Colin Barker, Aug 03 2018: (Start)

G.f.: x*(8 + x + 12*x^2 + x^3 + 8*x^4) / ((1 - x)^2*(1 + x)^2*(1 + x^2)).

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

(End)