A220421 -id:A220421 - cp's OEIS frontend

A222641 Number of iterations in Collatz (3x+1) trajectory of n to reach 1 from the highest term.

0, 1, 4, 2, 4, 4, 11, 3, 11, 4, 11, 4, 8, 11, 10, 4, 11, 11, 17, 7, 6, 11, 10, 10, 17, 8, 34, 11, 17, 10, 34, 5, 25, 11, 10, 11, 20, 17, 24, 8, 34, 6, 26, 11, 15, 10, 34, 11, 23, 17, 21, 11, 10, 34, 34, 19, 26, 17, 24, 10, 18, 34, 34, 6, 26, 25, 24, 14, 13, 10

Offset: 1

Jayanta Basu, Feb 27 2013

nonn

			Collatz trajectory of 5 is 5, 16, 8, 4, 2, 1. Hence, a(5) = 4.

T. D. Noe, Table of n, a(n) for n = 1..10000

Cf. A006577, A220421.

Collatz[n_] := NestWhileList[If[EvenQ[#], #/2, 3 # + 1] &, n, # > 1 &]; Table[c = Collatz[n]; Length[c] - Position[c, Max[c]][[1, 1]], {n, 100}] (* T. D. Noe, Feb 27 2013 *)

a(n) = A006577(n) - A220421(n);

A224303 Numbers n for which number of iterations to reach the largest equals number of iterations to reach 1 from the largest in Collatz (3x+1) trajectory of n.

Original entry on oeis.org

1, 6, 120, 334, 335, 804, 1249, 2008, 2010, 2012, 2013, 6556, 6557, 6558, 6801, 6802, 6803, 7496, 7498, 7500, 7501, 7505, 10219, 22633, 25182, 25183, 27074, 27075, 27864, 27866, 27868, 31838, 31839, 32078, 36630, 36633, 36690, 36691, 36914, 39126, 39344

Offset: 1

Jayanta Basu, Apr 03 2013

nonn

For each n here we have A220421(n) = A222641(n).

			6 is in the list because Collatz trajectory of 6 is {6, 3, 10, 5, 16, 8, 4, 2, 1} and number of steps to reach largest = number of steps to reach 1 from largest = 4.

T. D. Noe, Table of n, a(n) for n = 1..1000

Cf. A220421, A222641, A224533 (odd n).

Collatz[n_] := NestWhileList[If[EvenQ[#], #/2, 3*# + 1] &, n, # > 1 &]; Select[Range[50000],Position[Collatz[#],Max[Collatz[#]]] == {{(Length[Collatz[#]] + 1)/2}} &]

cp's OEIS Frontend

A222641 Number of iterations in Collatz (3x+1) trajectory of n to reach 1 from the highest term.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

Formula