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A006877
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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)
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27
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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
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OFFSET
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1,2
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COMMENTS
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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
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REFERENCES
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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).
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LINKS
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G. T. Leavens and M. Vermeulen, 3x+1 search programs, Computers and Mathematics with Applications, 24 (1992), 79-99. (Annotated scanned copy)
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MAPLE
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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;
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MATHEMATICA
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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 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn,nice
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AUTHOR
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STATUS
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approved
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