Ryan Pythagoras Newton Critchlow

User: Ryan Pythagoras Newton Critchlow

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Ryan Pythagoras Newton Critchlow has authored 2 sequences.

A288311 Number of steps, reduced mod n, to reach 1 in the Collatz 3x+1 problem, or -1 if 1 is never reached.

0, 1, 1, 2, 0, 2, 2, 3, 1, 6, 3, 9, 9, 3, 2, 4, 12, 2, 1, 7, 7, 15, 15, 10, 23, 10, 3, 18, 18, 18, 13, 5, 26, 13, 13, 21, 21, 21, 34, 8, 27, 8, 29, 16, 16, 16, 10, 11, 24, 24, 24, 11, 11, 4, 2, 19, 32, 19, 32, 19, 19, 45, 44, 6, 27, 27, 27, 14, 14, 14, 31, 22

Offset: 1

Ryan Pythagoras Newton Critchlow, Jun 07 2017

nonn

			For n = 3, which takes 7 steps to reach 1 in the Collatz (3x+1) problem: (10, 5, 16, 8, 4, 2, 1), 7 mod 3 = 1.

Cf. A006577.

Table[Mod[-1 + Length[NestWhileList[If[EvenQ@ #, #/2, 3 # + 1] &, n, # != 1 &]], n], {n, 72}] (* Michael De Vlieger, Jun 09 2017 *)

a(n)=s=n; c=0; while(s>1, s=if(s%2, 3*s+1, s/2); c++); c % n; \\ Michel Marcus, Jun 10 2017

def stepCount(x):
    x = int(x)
    steps = 0
    while True:
        if x == 1:
            break
        elif x % 2 == 0:
            x = x/2
            steps += 1
        else:
            x = x*3 + 1
            steps += 1
    return steps
n = 1
while True:
    print(stepCount(n) % n)
    n += 1

a(n) = A006577(n) mod n.

A288023 Number of steps to reach 1 in the Collatz 3x+1 problem starting with the n-th triangular number, or -1 if 1 is never reached.

Original entry on oeis.org

0, 7, 8, 6, 17, 7, 18, 21, 16, 112, 27, 35, 92, 38, 20, 15, 36, 124, 106, 39, 127, 109, 16, 16, 24, 81, 107, 40, 27, 35, 110, 30, 43, 74, 38, 113, 170, 46, 121, 28, 103, 116, 36, 98, 124, 137, 18, 119, 132, 83, 26, 127, 26, 47, 34, 122, 91, 148, 117, 130, 37, 37, 112, 32, 76, 94, 58, 120, 120, 89, 133, 53, 115, 66

Offset: 1

Ryan Pythagoras Newton Critchlow, Jun 04 2017

nonn

			For n = 2, the 2nd triangular number is 3, which takes 7 steps to reach 1 in the Collatz (3x+1) problem: (10, 5, 16, 8, 4, 2, 1).

Ryan Pythagoras Newton Critchlow, Table of n, a(n) for n = 1..10000

Cf. A000217, A006577, A070975.

Table[Length[NestWhileList[If[EvenQ[#],#/2,3#+1]&,n,#>1&]]-1,{n,Accumulate[ Range[80]]}] (* Harvey P. Dale, Aug 17 2017 *)

num = 1
def triangleN(x):
    return x*(x+1)/2
def stepCount(x):
    x = int(x)
    steps = 0
    while True:
        if x == 1:
            break
        elif x % 2 == 0:
            x = x/2
            steps += 1
        else:
            x = x*3 + 1
            steps += 1
    return steps
while True:
    print(stepCount(triangleN(num)))
    num += 1

a(n) = A006577(A000217(n)). - Omar E. Pol, Jun 04 2017

cp's OEIS Frontend

User: Ryan Pythagoras Newton Critchlow

A288311 Number of steps, reduced mod n, to reach 1 in the Collatz 3x+1 problem, or -1 if 1 is never reached.

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