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In the '3x+1' problem, these values for the starting value set new records for number of steps to reach 1.
(Formerly M0748)
+10
30
1, 2, 3, 6, 7, 9, 18, 25, 27, 54, 73, 97, 129, 171, 231, 313, 327, 649, 703, 871, 1161, 2223, 2463, 2919, 3711, 6171, 10971, 13255, 17647, 23529, 26623, 34239, 35655, 52527, 77031, 106239, 142587, 156159, 216367, 230631, 410011, 511935, 626331, 837799
COMMENTS
Both the 3x+1 steps and the halving steps are counted.
This sequence without a(2) = 2 specifies where records occur in A208981. - Omar E. Pol, Apr 14 2022
REFERENCES
D. R. Hofstadter, Goedel, Escher, Bach: an Eternal Golden Braid, Random House, 1980, p. 400.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
LINKS
G. T. Leavens and M. Vermeulen, 3x+1 search programs, Computers and Mathematics with Applications, 24 (1992), 79-99. (Annotated scanned copy)
MAPLE
A006877 := proc(n) local a, L; L := 0; a := n; while a <> 1 do if a mod 2 = 0 then a := a/2; else a := 3*a+1; fi; L := L+1; od: RETURN(L); end;
MATHEMATICA
numberOfSteps[x0_] := Block[{x = x0, nos = 0}, While [x != 1 , If[Mod[x, 2] == 0 , x = x/2, x = 3*x + 1]; nos++]; nos]; a[1] = 1; a[n_] := a[n] = Block[{x = a[n-1] + 1}, record = numberOfSteps[x - 1]; While[ numberOfSteps[x] <= record, x++]; x]; A006877 = Table[ Print[a[n]]; a[n], {n, 1, 44}](* Jean-François Alcover, Feb 14 2012 *) DeleteDuplicates[Table[{n, Length[NestWhileList[If[EvenQ[#], #/2, 3#+1]&, n, #>1&]]}, {n, 838000}], GreaterEqual[#1[[2]], #2[[2]]]&][[All, 1]] (* Harvey P. Dale, May 13 2022 *)
PROG
(PARI) A006577(n)=my(s); while(n>1, n=if(n%2, 3*n+1, n/2); s++); s step(n, r)=my(t); forstep(k=bitor(n, 1), 2*n, 2, t= A006577(k); if(t>r, return([k, t]))); [2*n, r+1] (Python)
c1 = lambda x: (3*x+1 if (x%2) else x>>1)
r = -1
for n in range(1, 10**5):
a=0 ; n1=n
while n>1: n=c1(n); a+=1;
if a > r: print(n1, end = ', '); r=a
In the '3x+1' problem, these values for the starting value set new records for highest point of trajectory before reaching 1.
(Formerly M0843)
+10
25
1, 2, 3, 7, 15, 27, 255, 447, 639, 703, 1819, 4255, 4591, 9663, 20895, 26623, 31911, 60975, 77671, 113383, 138367, 159487, 270271, 665215, 704511, 1042431, 1212415, 1441407, 1875711, 1988859, 2643183, 2684647, 3041127, 3873535, 4637979, 5656191
COMMENTS
Both the 3x+1 steps and the halving steps are counted.
REFERENCES
R. B. Banks, Slicing Pizzas, Racing Turtles and Further Adventures in Applied Mathematics, Princeton Univ. Press, 1999. See p. 96.
D. R. Hofstadter, Goedel, Escher, Bach: an Eternal Golden Braid, Random House, 1980, p. 400.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
LINKS
G. T. Leavens and M. Vermeulen, 3x+1 search problems, Computers and Mathematics with Applications, 24 (1992), 79-99. G. T. Leavens and M. Vermeulen, 3x+1 search programs, Computers and Mathematics with Applications, 24 (1992), 79-99. (Annotated scanned copy) Tomás Oliveira e Silva, Tables (gives many more terms).
MATHEMATICA
mcoll[n_]:=Max@@NestWhileList[If[EvenQ[#], #/2, 3#+1]&, n, #>1&]; t={1, max=2}; Do[If[(y=mcoll[n])>max, max=y; AppendTo[t, n]], {n, 3, 705000, 4}]; t (* Jayanta Basu, May 28 2013 *) DeleteDuplicates[Parallelize[Table[{n, Max[NestWhileList[If[EvenQ[#], #/2, 3#+1]&, n, #>1&]]}, {n, 57*10^5}]], GreaterEqual[#1[[2]], #2[[2]]]&][[;; , 1]] (* Harvey P. Dale, Apr 23 2023 *)
PROG
(Haskell)
a006884 n = a006884_list !! (n-1)
a006884_list = f 1 0 a025586_list where
f i r (x:xs) = if x > r then i : f (i + 1) x xs else f (i + 1) r xs
(PARI) A025586(n)=my(r=n); while(n>2, if(n%2, n=3*n+1; if(n>r, r=n)); n>>=1); r
CROSSREFS
A060409 gives associated "dropping times", A060410 the maximal values and A060411 the steps at which the maxima occur.
Record highest point of trajectory before reaching 1 in '3x+1' problem, corresponding to starting values in A006884.
(Formerly M2086)
+10
14
1, 2, 16, 52, 160, 9232, 13120, 39364, 41524, 250504, 1276936, 6810136, 8153620, 27114424, 50143264, 106358020, 121012864, 593279152, 1570824736, 2482111348, 2798323360, 17202377752, 24648077896, 52483285312, 56991483520, 90239155648, 139646736808
COMMENTS
Both the 3x+1 steps and the halving steps are counted.
In an email of Aug 06 2023, Guy Chouraqui observes that the digital root of a(n) appears to be 7 for all n > 2. - N. J. A. Sloane, Aug 11 2023
REFERENCES
R. B. Banks, Slicing Pizzas, Racing Turtles and Further Adventures in Applied Mathematics, Princeton Univ. Press, 1999. See p. 96.
D. R. Hofstadter, Goedel, Escher, Bach: an Eternal Golden Braid, Random House, 1980, p. 400.
G. T. Leavens and M. Vermeulen, 3x+1 search problems, Computers and Mathematics with Applications, 24 (1992), 79-99.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
LINKS
G. T. Leavens and M. Vermeulen, 3x+1 search programs, Computers and Mathematics with Applications, 24 (1992), 79-99. (Annotated scanned copy)
MATHEMATICA
mcoll[n_]:=Max@@NestWhileList[If[EvenQ[#], #/2, 3#+1]&, n, #>=n&]; t={1, max=2}; Do[If[(y=mcoll[n])>max, AppendTo[t, max=y]], {n, 3, 10^6, 4}]; t (* Jayanta Basu, May 28 2013 *)
Record number of steps to reach 1 in '3x+1' problem, corresponding to starting values in A006877.
(Formerly M4335)
+10
11
0, 1, 7, 8, 16, 19, 20, 23, 111, 112, 115, 118, 121, 124, 127, 130, 143, 144, 170, 178, 181, 182, 208, 216, 237, 261, 267, 275, 278, 281, 307, 310, 323, 339, 350, 353, 374, 382, 385, 442, 448, 469, 508, 524, 527, 530, 556, 559, 562, 583, 596, 612, 664, 685, 688, 691, 704
COMMENTS
Both the 3x+1 steps and the halving steps are counted.
REFERENCES
D. R. Hofstadter, Goedel, Escher, Bach: an Eternal Golden Braid, Random House, 1980, p. 400.
G. T. Leavens and M. Vermeulen, 3x+1 search problems, Computers and Mathematics with Applications, 24 (1992), 79-99.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
LINKS
G. T. Leavens and M. Vermeulen, 3x+1 search programs, Computers and Mathematics with Applications, 24 (1992), 79-99. (Annotated scanned copy)
MAPLE
f := proc(n) local a, L; L := 0; a := n; while a <> 1 do if a mod 2 = 0 then a := a/2; else a := 3*a+1; fi; L := L+1; od: RETURN(L); end;
MATHEMATICA
numberOfSteps[x0_] := Block[{x = x0, nos = 0}, While[x != 1, If[Mod[x, 2] == 0, x = x/2, x = 3*x+1]; nos++]; nos]; A006878 = numberOfSteps /@ A006877 (* Jean-François Alcover, Feb 22 2012 *) DeleteDuplicates[Table[Length[NestWhileList[If[EvenQ[#], #/2, 3#+1]&, n, #>1&]], {n, 0, 10^6}], GreaterEqual]-1 (* The program generates the first 44 terms of the sequence, derived from all starting values from 1 up to and including 1 million. *) (* Harvey P. Dale, Nov 26 2022 *)
In the '3x+1' problem, these values for the starting value set new records for number of steps to reach 1.
+10
5
1, 3, 7, 9, 25, 27, 73, 97, 129, 171, 231, 313, 327, 703, 871, 1161, 2463, 2919, 3711, 6171, 10971, 13255, 17647, 23529, 26623, 34239, 35655, 52527, 77031, 106239, 142587, 156159, 216367, 230631, 410011, 511935, 626331, 837799, 1117065, 1501353, 1723519, 2298025, 3064033
COMMENTS
Only the 3x+1 steps, not the halving steps, are counted.
REFERENCES
D. R. Hofstadter, Goedel, Escher, Bach: an Eternal Golden Braid, Random House, 1980, p. 400.
G. T. Leavens and M. Vermeulen, 3x+1 search problems, Computers and Mathematics with Applications, 24 (1992), 79-99.
MATHEMATICA
f[ nn_ ] := Module[ {c, n}, c = 0; n = nn; While[ n != 1, If[ Mod[ n, 2 ] == 0, n /= 2, n = 3*n + 1; c++ ] ]; Return[ c ] ] maxx = -1; For[ n = 1, n <= 10^8, n++, Module[ {val}, val = f[ n ]; If[ val > maxx, maxx = val; Print[ n, " ", val ] ] ] ] (* Winston C. Yang (winston(AT)cs.wisc.edu), Aug 27 2000 *)
PROG
(Haskell)
a033958 n = a033958_list !! (n-1)
-- For definition of a033958_list: see A033959.
EXTENSIONS
Corrected with Mathematica code by Winston C. Yang (winston(AT)cs.wisc.edu), Aug 27 2000
Record number of steps to reach 1 in '3x+1' problem, corresponding to starting values in A033958.
+10
4
0, 2, 5, 6, 7, 41, 42, 43, 44, 45, 46, 47, 52, 62, 65, 66, 76, 79, 87, 96, 98, 101, 102, 103, 113, 114, 119, 125, 129, 130, 138, 141, 142, 164, 166, 174, 189, 195, 196, 197, 207, 208, 209, 217, 222, 228, 248, 256, 257, 258, 263, 278, 357, 358, 359, 362, 370
COMMENTS
Only the 3x+1 steps, not the halving steps, are counted.
REFERENCES
D. R. Hofstadter, Goedel, Escher, Bach: an Eternal Golden Braid, Random House, 1980, p. 400.
G. T. Leavens and M. Vermeulen, 3x+1 search problems, Computers and Mathematics with Applications, 24 (1992), 79-99.
MAPLE
A033959 := proc(n) local a, L; L := 0; a := n; while a <> 1 do if a mod 2 = 0 then a := a/2; else a := 3*a+1; L := L+1; fi; od: RETURN(L); end;
MATHEMATICA
f[ nn_ ] := Module[ {c, n}, c = 0; n = nn; While[ n != 1, If[ Mod[ n, 2 ] == 0, n /= 2, n = 3*n + 1; c++ ] ]; Return[ c ] ] maxx = -1; For[ n = 1, n <= 10^8, n++, Module[ {val}, val = f[ n ]; If[ val > maxx, maxx = val; Print[ n, " ", val ] ] ] ]
PROG
(Haskell)
a033959 n = a033959_list !! (n-1)
(a033959_list, a033958_list) = unzip $ (0, 1) : f 1 1 where
f i x | y > x = (y, 2 * i - 1) : f (i + 1) y
| otherwise = f (i + 1) x
where y = a075680 i
EXTENSIONS
More terms from Winston C. Yang (winston(AT)cs.wisc.edu), Aug 27 2000
More terms from Larry Reeves (larryr(AT)acm.org), Sep 27 2000
Maximal number of halving and tripling steps to reach 1 in '3x+1' problem for range (1, ..., n).
+10
0
1, 2, 8, 8, 8, 9, 17, 17, 20, 20, 20, 20, 20, 20, 20, 20, 20, 21, 21, 21, 21, 21, 21, 21, 24, 24, 112, 112, 112, 112, 112, 112, 112, 112, 112, 112, 112, 112, 112, 112, 112, 112, 112, 112, 112, 112, 112, 112, 112, 112, 112, 112, 112, 113, 113, 113, 113, 113
MATHEMATICA
nst[n_]:=Length[NestWhileList[If[EvenQ[#], #/2, 3#+1]&, n, #>1&]]; nn=60; With[ {stps= Array[nst, nn]}, Table[Max[Take[stps, n]], {n, nn}]] (* Harvey P. Dale, Apr 17 2014 *)
AUTHOR
Benjamin Frost (benjamin.frost(AT)students.adelaide.edu.au), Sep 02 2008
Inputs n that yield a record-breaking value of A008908(n)/(log_2(n)+1) for the Collatz conjecture.
+10
0
1, 3, 7, 9, 27, 26623, 35655, 52527, 142587, 156159, 230631, 626331, 837799, 1723519, 3542887, 3732423, 5649499, 6649279, 8400511, 63728127, 3743559068799, 100759293214567, 104899295810901231
COMMENTS
The metric A008908(n)/(log_2(n)+1) is always equal to 1 for any power of 2 (where 1 is the smallest possible value).
EXAMPLE
a(1) = 1, which is trivial, because the first element in any sequence is record setting.
a(5) = 27, because A008908(n)/(log_2(n)+1) yields a maximum value at n=27 among the first 27 elements, and there are 4 record-breaking elements beforehand.
PROG
(Python)
import math
oeis_ A006877 = [1, 2, 3, 6, 7, 9, 18, 25, 27, 54, 73, 97, 129, 171, 231, 313, 327, 649, 703, 871, 1161, 2223, 2463, 2919, 3711, 6171, 10971, 13255, 17647, 23529, 26623, 34239, 35655, 52527, 77031, 10623, 142587, 156159, 216367, 230631, 410011, 511935, 626331, 837799, 1117065, 1501353, 1723519, 2298025, 3064033, 3542887, 3732423, 5649499, 6649279, 8400511, 11200681, 14934241, 15733191, 31466382, 36791535, 63728127] def stopping_time(n):
time = 1
while n>1:
n = 3*n + 1 if n & 1 else n//2
time += 1
return time
def stopping_time_metric(n):
time = stopping_time(n)
logarithmic_distance = (math.log(n, 2)+1)
return float(time/logarithmic_distance)
result = []
record_stopping_time_metric = stopping_time_metric(record_input)
result.append(record_input)
for n in range(1, len(oeis_ A006877)): current_stopping_time_metric = stopping_time_metric(current_input)
if current_stopping_time_metric > record_stopping_time_metric:
record_input = current_input
record_stopping_time_metric = current_stopping_time_metric
result.append(record_input)
for n in range(len(result)):
print(result[n], end=", ")
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