A220071 -id:A220071 - cp's OEIS frontend

A213678 Number of terms k such that difference between halving and tripling steps in Collatz (3x+1) trajectory of k is n.

1, 1, 1, 3, 3, 5, 8, 14, 20, 29, 40, 59, 87, 130, 196, 294, 439, 658, 985, 1459, 2203, 3328, 5001, 7482, 11205, 16805, 25220, 37850, 56713, 85108, 127728, 191635

Offset: 0

Jayanta Basu, Mar 04 2013

nonn

			a(5) = 5 since there are only five numbers 12, 13, 20, 21, 32 such that difference between number of halving and tripling steps is 5.

Cf. A220071, A222599 (lists of numbers).

Collatz[n_] := NestWhileList[If[EvenQ[#], #/2, 3 # + 1] &, n, # > 1 &]; nn = 15; t = Table[0, {nn}]; Do[c = Collatz[n]; e = Select[c, EvenQ]; diff = 2*Length[e] - Length[c]; If[diff < nn - 1, t[[diff + 2]]++], {n, 2^(nn - 1)}]; t (* T. D. Noe, Mar 04 2013 *)

Corrected and extended by T. D. Noe, Mar 06 2013

A304715 For any n > 0, if A006666(n) >= 0, then a(n) = Sum_{i = 0..A006666(n)-1} 2^i * [T^i(n) == 0 (mod 2)] (where [] is an Iverson bracket and T^i denotes the i-th iterate of the Collatz function A014682); otherwise a(n) = -1.

Original entry on oeis.org

0, 1, 28, 3, 14, 57, 1896, 7, 7586, 29, 948, 115, 118, 3793, 3824, 15, 474, 15173, 15180, 59, 62, 1897, 1912, 231, 60722, 237, 1102691417057682138372, 7587, 7590, 7649, 137836427132210267296, 31, 242890, 949, 956, 30347, 30350, 30361, 7772616, 119

Offset: 1

Rémy Sigrist, May 17 2018

In other words, when a(n) >= 0, the binary representation of a(n) encodes the tripling and halvings steps of the Collatz compressed trajectory of n up to the first occurrence of the number 1 (where zeros and ones respectively denote tripling and halving steps).

			The first terms, alongside the binary representation of a(n) and the Collatz compressed trajectory of a(n) up to the first 1 in reverse order, are:
  n    a(n)       bin(a(n))  rev(traj(n))
  --   ----       ---------  ------------
   1      0               0  (1)
   2      1               1  (1, 2)
   3     28           11100  (1, 2, 4, 8, 5, 3)
   4      3              11  (1, 2, 4)
   5     14            1110  (1, 2, 4, 8, 5)
   6     57          111001  (1, 2, 4, 8, 5, 3, 6)
   7   1896     11101101000  (1, 2, 4, 8, 5, 10, 20, 13, 26, 17, 11, 7)
   8      7             111  (1, 2, 4, 8)
   9   7586   1110110100010  (1, 2, 4, 8, 5, 10, 20, 13, 26, 17, 11, 7, 14, 9)
  10     29           11101  (1, 2, 4, 8, 5, 10)
  11    948      1110110100  (1, 2, 4, 8, 5, 10, 20, 13, 26, 17, 11)
  12    115         1110011  (1, 2, 4, 8, 5, 3, 6, 12)
  13    118         1110110  (1, 2, 4, 8, 5, 10, 20, 13)
  14   3793    111011010001  (1, 2, 4, 8, 5, 10, 20, 13, 26, 17, 11, 7, 14)
  15   3824    111011110000  (1, 2, 4, 8, 5, 10, 20, 40, 80, 53, 35, 23, 15)
  16     15            1111  (1, 2, 4, 8, 16)
  17    474       111011010  (1, 2, 4, 8, 5, 10, 20, 13, 26, 17)
  18  15173  11101101000101  (1, 2, 4, 8, 5, 10, 20, 13, 26, 17, 11, 7, 14, 9, 18)

Cf. A000120, A006666, A014682, A029837, A220071, A248573.

a(n) = my (v=0); for (k=0, oo, if (n==1, return (v), n%2, n = (3*n+1)/2, n = n/2; v += 2^k))

a(2^k) = 2^k - 1 for any k >= 0.
a(2*n) = 2*a(n) + 1.
A029837(a(n)+1) = A006666(n).
A000120(a(n)) = A220071(n).
a(A248573(n)) < a(A248573(n+1)) for any n >= 0. - Rémy Sigrist, Nov 09 2018

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A213678 Number of terms k such that difference between halving and tripling steps in Collatz (3x+1) trajectory of k is n.

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A304715 For any n > 0, if A006666(n) >= 0, then a(n) = Sum_{i = 0..A006666(n)-1} 2^i * [T^i(n) == 0 (mod 2)] (where [] is an Iverson bracket and T^i denotes the i-th iterate of the Collatz function A014682); otherwise a(n) = -1.

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