A006577 -id:A006577 - cp's OEIS frontend

A346782 Numbers k such that A006577(k!) sets a new record.

2, 3, 4, 5, 7, 8, 9, 11, 12, 14, 17, 21, 26, 29, 33, 34, 37, 44, 45, 47, 48, 51, 52, 55, 58, 61, 70, 78, 85, 89, 91, 92, 93, 96, 97, 98, 105, 107, 113, 114, 118, 124, 129, 135, 145, 148, 149, 150, 152, 157, 161, 173, 175, 179, 184, 192, 201, 205, 206, 209, 212, 213

Offset: 1

Hugo Pfoertner, Aug 04 2021

nonn

The corresponding record numbers of steps are A346783.

Cf. A006577, A070974, A346592, A346593.

a6577(n0)={my(n=n0,k=0);while(n>1,k++;n=if(n%2,3*n+1,n/2));k};
a346782(limit)={my(msteps=0);for(k=0,limit,my(m=a6577(k!));if(m>msteps,print1(k,", ");msteps=m))};
a346782(215)

A346783 a(n) = A006577(A346782(n)), records in A070974.

Original entry on oeis.org

1, 8, 10, 20, 41, 44, 86, 147, 210, 264, 426, 500, 559, 633, 795, 950, 1069, 1171, 1218, 1327, 1330, 1342, 1402, 1551, 1672, 1974, 2274, 2528, 2725, 2730, 2911, 2928, 2994, 3117, 3315, 3443, 3495, 3524, 4061, 4347, 4651, 4703, 4818, 5049, 5322, 5447, 5449, 5756

Offset: 1

Hugo Pfoertner, Aug 04 2021

nonn

Cf. A006577, A070974, A346782.

a6577(n0)={my(n=n0, k=0); while(n>1, k++; n=if(n%2, 3*n+1, n/2)); k};
a346783(limit)={my(msteps=0);for(k=0,limit,my(m=a6577(k!));if(m>msteps,print1(m,", ");msteps=m))};
a346783(150)

A066904 Triangular numbers in A006577.

Original entry on oeis.org

1, 3, 6, 15, 15, 10, 10, 21, 21, 21, 6, 105, 105, 28, 28, 28, 28, 28, 15, 15, 15, 15, 15, 15, 36, 36, 36, 36, 10, 10, 10, 21, 21, 21, 21, 21, 21, 21, 78, 21, 91, 91, 55, 55, 55, 45, 45, 45, 45, 45, 45, 45, 45, 45, 120, 120, 120, 120, 120, 120, 120, 120, 66, 66, 136, 136

Offset: 0

K. B. Subramaniam (kb_subramaniambalu(AT)yahoo.com), Dec 20 2001

nonn

Harvey P. Dale, Table of n, a(n) for n = 0..2000

Cf. A006577.

Select[Table[Length[NestWhileList[If[EvenQ[#],#/2,3#+1]&,n,#!=1&]]-1, {n,1000}],OddQ[Sqrt[8#+1]]&] (* Harvey P. Dale, Feb 06 2015 *)

More terms from Sascha Kurz, Mar 23 2002

A066905 Squares in A006577.

Original entry on oeis.org

0, 1, 16, 9, 9, 4, 16, 16, 16, 9, 9, 9, 25, 25, 25, 25, 25, 100, 121, 36, 36, 36, 36, 49, 16, 16, 16, 16, 16, 16, 16, 16, 81, 81, 9, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 100, 100, 100, 25, 144, 25, 100, 25, 25, 144, 144, 25, 25, 144, 64, 64, 64, 64, 64, 25, 25, 64, 64

Offset: 1

K. B. Subramaniam (kb_subramaniambalu(AT)yahoo.com), Dec 20 2001

nonn

Cf. A006577, A066773.

steps[ n_ ] := For[ nn=n; ct=0, True, ct++, If[ nn==1, Return[ ct ] ]; nn=If[ EvenQ[ nn ], nn/2, 3nn+1 ] ]; Select[ steps/@Range[ 1, 1000 ], IntegerQ[ Sqrt[ # ] ]& ]

More terms from Dean Hickerson, Jan 19 2002

Offset 1 from Michel Marcus, Jul 25 2021

A213198 Number of iterations of the map n -> f(f(f(...f(n)...))) to reach the end of the cycle, where f(n) = A006577(n), the initial number n is not counted.

Original entry on oeis.org

0, 1, 5, 2, 0, 7, 4, 6, 7, 8, 11, 8, 8, 10, 10, 3, 9, 6, 6, 5, 5, 11, 11, 9, 12, 9, 13, 7, 7, 7, 10, 1, 10, 9, 9, 6, 6, 6, 10, 7, 11, 7, 8, 4, 4, 4, 10, 12, 10, 10, 10, 12, 12, 7, 7, 7, 2, 7, 2, 7, 7, 15, 15, 8, 14, 14, 14, 11, 11, 11, 14, 12, 12, 12, 11, 12

Offset: 1

Michel Lagneau, Mar 01 2013

nonn

A006577 is the number of halving and tripling steps to reach 1 in '3x+1' problem.

The end of the cycle is 1 or 5 for n = 5, 32, 57, 59, 344, 346, 348, 349, ...

			a(3) = 5 because the 5 iterations to reach 1 are A006577(3) = 7; A006577(7) = 16; A006577(16) = 4; A006577(4) = 2; A006577(2) = 1.
a(5) = 0 because A006577(5) = 5 is the end of the cycle.
a(57) = 2 because A006577(57) = 32 and A006577(32) = 5 is the end of the cycle.

Michel Lagneau, Table of n, a(n) for n = 1..10000

Cf. A006577.

for n from 1 to 200 do:
    m:=n: a:=2:
    for it from 1 to 1000
    while (a>1) do:
        jj:=0: a:=0: x:=m:
        if m=5 then
            printf(`%d, `, it-1): jj:=1:
        else
            for i from 1 to 1000
            while (x>1) do:
                if irem(x, 2)=0 then
                    x := x/2: a := a+1:
                else
                    x := 3*x+1: a := a+1:
                fi:
            od:
            m:=a:
        fi:
    od:
    if jj=0 then
        printf(`%d, `, it-1):
    fi:
od:

Collatz[n_] := NestWhileList[If[EvenQ[#], #/2, 3 # + 1] &, n, # > 1 &]; f[n_] := Length[Collatz[n]] - 1; Table[k = Rest[NestWhileList[f, n, UnsameQ, All]]; If[k[[1]] == n, 0, k = DeleteCases[k, 0]; If[Length[k] > 1 && k[[-1]] == k[[-2]], k = Most[k]]; Length[k]], {n, 100}] (* T. D. Noe, Mar 01 2013 *)

A221947 Smallest number k (different from a power of 2) such that A006577(n*k) = A006577(n) + A006577(k), or 0 if no such number exists.

Original entry on oeis.org

3, 3, 423, 3, 81, 423, 75, 3, 0, 81, 11003, 423, 155, 75, 35, 3, 239, 0, 151, 81, 23, 11003, 21, 423, 21, 155, 341, 75, 201, 35, 75, 3, 21, 239, 15, 0, 113, 151, 21, 81, 635, 23, 1131, 11003, 2017, 21, 75, 423, 1267, 21, 75, 155, 253, 341, 151, 75, 7931, 201, 75, 35, 69, 75, 213, 3, 1073, 21, 423, 239, 61, 15

Offset: 1

Michel Lagneau, Feb 25 2013

nonn

A006577 is the number of halving and tripling steps to reach 1 in the '3x+1' problem. If n is a power of 2, a(n) = 3.
If k is a power of 2, we obtain trivial results, for example A006577(n*2^m) = A006577(2^m) + A006577(n) = m + A006577(n) => the smallest k is 1.
It appears that a(n) = 0 for n of the form 9*2^a = 9, 18, 36, 72, ...

			a(3) = 423 because A006577(3*423) = A006577(1269) = 39, and A006577(3) + A006577(423) = 7 + 32 = 39.

Index entries for sequences related to 3x+1 (or Collatz) problem

Cf. A006577, A057716.

lst:={ }:C:= proc(n) a := 0 ; x := n ; while x > 1 do if type(x, 'even') then x := x/2:a:=a+1:  else x := 3*x+1 ; a := a+1 ; end if; end do; a ; end proc:
for m from 0 to 40 do:lst:=lst union {2^m}:od:for n from 1 to 73 do: ii:=0:for k from 2 to 50000 while(ii=0) do:z:=n*k : if {k} intersect lst = {} and C(z)=C(n)+C(k) then ii:=1: printf ( "%d %d \n",n,k):else fi:od: if ii=0 and {n} intersect lst = {} and {k} intersect lst = {} then printf ( "%d %d \n",n,0):else fi:od:

A277962 Least k such that A006577(k) = A006577(n) + A006577(n+1).

Original entry on oeis.org

2, 6, 12, 3, 34, 49, 9, 72, 98, 18, 25, 28, 33, 39, 36, 7, 57, 406, 65, 11, 72, 86, 98, 114, 114, 129, 913, 153, 153, 171, 27, 172, 203, 33, 39, 270, 270, 295, 270, 290, 290, 305, 361, 57, 57, 386, 73, 78, 481, 481, 78, 72, 514, 20174, 609, 641, 641, 641, 641

Offset: 1

Michel Lagneau, Nov 06 2016

nonn

A006577(n) is the number of halving and tripling steps to reach 1 in the '3x+1' problem.
The distinct squares in the sequence are 9, 25, 36, 49, 169, 361, ...
The distinct primes in the sequence are 2, 3, 7, 11, 31, 41, 47, 71, 73, 97, 103, ...

			a(5)=34 because A006577(34) = 13 = A006577(5) + A006577(6) = 5 + 8.

Michel Lagneau, Table of n, a(n) for n = 1..10000
Index entries for sequences related to 3x+1 (or Collatz) problem

Cf. A006577.

nn:=3*10^6:U:=array(1..nn):V:=array(1..nn):
for i from 1 to nn do:
m:=i:it0:=0:
   for j from 1 to nn while(m<>1) do:
    if irem(m, 2)=0
     then
     m:=m/2:it0:=it0+1:
     else
     m:=3*m+1:it0:=it0+1:
    fi:
   od:
   U[i]:=it0:
  od:
   for n from 1 to 60 do:
   ii:=0:
    for k from 1 to nn while(ii=0) do:
     if U[k]=U[n]+ U[n+1]
      then
      ii:=1:printf(`%d, `, k):
      else
     fi:
    od:
od:

f:=Table[Length[NestWhileList[If[EvenQ[#],#/2,3#+1]&,n,#!=1&]]-1,{n,3*10^6}];Do[k=1;While[f[[k]]!=f[[m]]+f[[m+1]],k++];Print[m," ",k],{m,1,60}]

A281665 Numbers m such that A006667(m)/A006577(m) = 1/3.

Original entry on oeis.org

159, 283, 377, 502, 503, 603, 615, 668, 669, 670, 799, 807, 888, 890, 892, 893, 1063, 1065, 1095, 1186, 1187, 1188, 1189, 1190, 1417, 1435, 1580, 1581, 1582, 1585, 1586, 1587, 1889, 1913, 1947, 1959, 1963, 2104, 2106, 2108, 2109, 2113, 2114, 2115, 2119, 2518

Offset: 1

Michel Lagneau, Jan 31 2017

nonn

A006667: number of tripling steps to reach 1 in '3x+1' problem.
A006577: number of halving and tripling steps to reach 1 in '3x+1' problem.
The corresponding number of iterations A006577(a(n)) is given by the sequence 54, 60, 63, 66, 66, 69, 69, 69, 69, 69, 72, 72, 72, 72, 72, 72, 75, 75, ... and the set of the distinct values of this sequence is {b(n)} = {54, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99, 102, 105, 108, 111, 114, 117, 120, ...}. We observe that {b(k)} = {54} union {60 + 3*k} for k = 1, 2, ...

			159 is in the sequence because A006667(159)/A006577(159) = 18/54 = 1/3.

Index entries for sequences related to 3x+1 (or Collatz) problem

Cf. A006577, A006666, A006667.

nn:=10000:
for n from 2 to 3000 do:
  m:=n:s1:=0:s2:=0:
   for i from 1 to nn while(m<>1) do:
    if irem(m,2)=0
     then
     s2:=s2+1:m:=m/2:
     else
     s1:=s1+1:m:=3*m+1:
    fi:
   od:
   s:=s1/(s1+s2):
    if s=1/3
     then
     printf(`%d, `,n):
     else
    fi:
od:

A058261 a(n) = n times the Collatz number of n (as given in A006577).

Original entry on oeis.org

0, 2, 21, 8, 25, 48, 112, 24, 171, 60, 154, 108, 117, 238, 255, 64, 204, 360, 380, 140, 147, 330, 345, 240, 575, 260, 2997, 504, 522, 540, 3286, 160, 858, 442, 455, 756, 777, 798, 1326, 320, 4469, 336, 1247

Offset: 0

Felix Goldberg (felixg(AT)tx.technion.ac.il), Dec 17 2000

nonn

			a(4)=8 because the Collatz number of 4 is 2 (4 -> 2 -> 1) and then 8=4*2.

a(n) = n*Collatz(n)

A174550 Run lengths of 2 or larger for consecutive prime numbers in A006577.

Original entry on oeis.org

3, 2, 2, 2, 2, 2, 4, 2, 3, 3, 2, 2, 3, 3, 4, 2, 5, 4, 3, 2, 3, 4, 2, 2, 2, 2, 3, 3, 2, 5, 2, 2, 3, 2, 3, 2, 2, 3, 8, 2, 4, 2, 2, 2, 2, 2, 3, 3, 3, 6, 3, 4, 2, 2, 3, 3, 2, 2, 4, 2, 2, 3, 6, 2, 3, 2, 2, 2, 5, 2, 2, 2, 2, 2, 2, 2, 5, 3, 2, 4, 3, 5, 3, 3, 2, 8, 2, 2, 2, 2, 3, 8, 4, 3, 3, 3, 4, 2, 3, 8, 2, 3, 3, 5, 3

Offset: 1

Michel Lagneau, Mar 22 2010

This sequence is given only for n <=5000 with max(s(n)) = 10. But we can find long sequences of primes, for example,length(s(12956))= 55, and corresponding to A006577(282984 + k), k = 0,1,...,54. We obtain a sequence of 55 consecutive prime numbers given in the example below.

			a(1) = 3 represents the run (7, 2, 5).
a(2) = 2 represents the run (3, 19).
a(3) = 2 represents the run (17, 17).
a(7) = 4 represents the run (19, 19, 107, 107).
a(12956) = 55 represents the run (83, 251, 83, 251, 127, 127, 127, 251, 83, 83, 83, 83, 83, 83, 83, 83, 83, 251, 83, 83, 83, 83, 83, 83, 101, 83, 83, 83, 83, 83, 83, 83, 83, 83, 83, 83, 83, 83, 251, 251, 83, 83, 83, 83, 83, 83, 83, 83, 83, 83, 83, 83, 83, 83, 83)

Cf. A174538, A006577, A066906.

 nn:=2000:T:=array(1..nn):for n from 1 to nn do: m:=n:for p from 0 to 1000 while (m<>1) do: if irem(m,2)=1 then m:=3*m+1:else m:=m/2:fi:od:T[n]:=p:od:ii:=1:for i from 1 to nn do:if type(T[i],prime)=true and type(T[i+1],prime)=true then ii:=ii+1:else if ii<>1 then printf(`%d, `, ii):ii:=1:else fi:fi:od:

cp's OEIS Frontend

A346782 Numbers k such that A006577(k!) sets a new record.

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PARI

A346783 a(n) = A006577(A346782(n)), records in A070974.

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A066904 Triangular numbers in A006577.

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A066905 Squares in A006577.

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A213198 Number of iterations of the map n -> f(f(f(...f(n)...))) to reach the end of the cycle, where f(n) = A006577(n), the initial number n is not counted.

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Maple

Mathematica

A221947 Smallest number k (different from a power of 2) such that A006577(n*k) = A006577(n) + A006577(k), or 0 if no such number exists.

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Maple

A277962 Least k such that A006577(k) = A006577(n) + A006577(n+1).

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Maple

Mathematica

A281665 Numbers m such that A006667(m)/A006577(m) = 1/3.

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Maple

A058261 a(n) = n times the Collatz number of n (as given in A006577).

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