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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

David Barina, Sep 06 2018

See A317640 for another definition of this problem.

			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.

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

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

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;
    }
}

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

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

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)

A318489 Number of steps to reach a lower number than starting value in 7x+-1 problem, or 0 if never reached.

Original entry on oeis.org

0, 1, 12, 1, 8, 1, 4, 1, 4, 1, 42, 1, 8, 1, 4, 1, 4, 1, 23, 1, 20, 1, 4, 1, 4, 1, 12, 1, 16, 1, 4, 1, 4, 1, 282, 1, 12, 1, 4, 1, 4, 1, 229, 1, 50, 1, 4, 1, 4, 1, 8, 1, 35, 1, 4, 1, 4, 1, 8, 1, 50, 1, 4, 1, 4, 1, 46, 1, 8, 1, 4, 1, 4, 1, 225, 1, 8, 1, 4, 1, 4, 1, 35, 1, 16, 1, 4, 1, 4, 1, 46, 1, 27, 1, 4, 1, 4, 1, 16

Offset: 1

David Barina, Aug 27 2018

The least positive k for which the iterate A317640^k(n) < n.
Also called the dropping time, glide, or stopping time.
a(2n) = 1.

			a(5) = 8 because the trajectory is (5, 36, 18, 9, 64, 32, 16, 8, 4, 2, 1, ...) and the first lower number is 4. Thus 8 steps to reach the value 4 starting from the value 5.

David Barina, Table of n, a(n) for n = 1..10000
D. Barina, 7x+-1: Close Relative of Collatz Problem, arXiv:1807.00908 [math.NT], 2018.
K. Matthews, David Barina's 7x+1 conjecture.

Cf. A317640 (7x+-1 function), A102419 (3x+1 equivalent).

int a(int n0) {
        if( n0 == 1 ) return 0;
        int s = 0;
        for(int n = n0; n >= n0; s++) {
                switch(n%4) {
                        case 1: n = 7*n+1; break;
                        case 3: n = 7*n-1; break;
                        default: n = n/2;
                }
        }
        return s;
}

a7(n) = {my(m=(n+2)%4-2); if(m%2, 7*n + m, n/2)};
a(n) = if (n==1, 0, my(nb=1, m=n, nm); while((nm=a7(m)) >= n, m = nm; nb++); nb); \\ Michel Marcus, Aug 28 2018

A317753 Number of steps to reach 1 in 7x+-1 problem, or -1 if 1 is never reached.

Original entry on oeis.org

0, 1, 13, 2, 10, 14, 18, 3, 7, 11, 53, 15, 19, 19, 23, 4, 27, 8, 50, 12, 73, 54, 16, 16, 58, 20, 20, 20, 43, 24, 24, 5, 47, 28, 325, 9, 70, 51, 32, 13, 13, 74, 272, 55, 55, 17, 17, 17, 276, 59, 40, 21, 40, 21, 21, 21, 63, 44, 63

Offset: 1

David Barina, Aug 06 2018

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.
The number of steps to reach 1 is also called the total stopping time.
Also the least positive k for which the iterate A317640^k(n) = 1.

			a(5)=10 because the trajectory of 5 is (5, 36, 18, 9, 64, 32, 16, 8, 4, 2, 1).

David Barina, Table of n, a(n) for n = 1..10000
D. Barina, 7x+-1: Close Relative of Collatz Problem, arXiv:1807.00908 [math.NT], 2018.
K. Matthews, David Barina's 7x+1 conjecture.

Cf. A317640 (7x+-1 function), A006577 (3x+1 equivalent).

int a(int n) {
        int s = 0;
        while( n != 1 ) {
                switch(n%4) {
                        case 1: n = 7*n+1; break;
                        case 3: n = 7*n-1; break;
                        default: n = n/2;
                }
                s++;
        }
        return s;
}

f[n_] := Switch[Mod[n, 4], 0, n/2, 1, 7 n + 1, 2, n/2, 3, 7 n - 1]; a[n_] := Length@NestWhileList[f, n, # > 1 &] - 1; Array[a, 70] (* Robert G. Wilson v, Aug 07 2018 *)

a(n) = my(nb=0); while(n != 1, if (!((n-1)%4), n = 7*n+1, if (!((n+1)%4), n = 7*n-1, n = n/2)); nb++); nb; \\ Michel Marcus, Aug 06 2018

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.

Original entry on oeis.org

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

David Barina, Aug 02 2018

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.

			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.

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).

Cf. A006370.

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

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 *)

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

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

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)

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User: David Barina

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.

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A318489 Number of steps to reach a lower number than starting value in 7x+-1 problem, or 0 if never reached.

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A317753 Number of steps to reach 1 in 7x+-1 problem, or -1 if 1 is never reached.

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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.

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