A372283 -id:A372283 - cp's OEIS frontend

A372287 Array read by upward antidiagonals: A(n, k) = A371092(A372283(n, k)), n,k >= 1.

1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 1, 3, 2, 3, 1, 1, 1, 1, 3, 3, 1, 1, 1, 1, 1, 3, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 6, 3, 1, 1, 1, 1, 1, 1, 1, 9, 1, 5, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 6, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 9, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 7

Offset: 1

Antti Karttunen, Apr 28 2024

A(n, k) gives the column index of A372282(n, k) [or equally, of A372283(n, k)] in array A257852.

Collatz conjecture is equal to the claim that in each column 1 will eventually appear.

			Array begins:
n\k| 1  2  3  4  5  6  7  8  9 10 11 12 13  14 15  16 17 18 19  20
---+---------------------------------------------------------------
1  | 1, 1, 1, 2, 2, 3, 1, 4, 3, 5, 1, 6, 4,  7, 2,  8, 5, 9, 2, 10,
2  | 1, 1, 1, 3, 2, 3, 1, 6, 1, 2, 1, 9, 5,  6, 3, 12, 4, 1, 2, 15,
3  | 1, 1, 1, 3, 3, 1, 1, 9, 1, 3, 1, 1, 2,  8, 3, 18, 5, 1, 3, 12,
4  | 1, 1, 1, 1, 3, 1, 1, 1, 1, 3, 1, 1, 3, 12, 1, 27, 2, 1, 3, 17,
5  | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 18, 1, 21, 3, 1, 1,  4,
6  | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 27, 1, 16, 3, 1, 1,  5,
7  | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 21, 1, 23, 1, 1, 1,  2,
8  | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 16, 1, 18, 1, 1, 1,  3,
9  | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 23, 1, 26, 1, 1, 1,  3,
10 | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 18, 1, 39, 1, 1, 1,  1,
11 | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 26, 1, 30, 1, 1, 1,  1,
12 | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 39, 1, 44, 1, 1, 1,  1,
13 | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 30, 1, 66, 1, 1, 1,  1,
14 | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 44, 1, 99, 1, 1, 1,  1,
15 | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 66, 1, 75, 1, 1, 1,  1,
16 | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 99, 1, 28, 1, 1, 1,  1,
17 | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 75, 1, 42, 1, 1, 1,  1,
18 | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 28, 1, 63, 1, 1, 1,  1,
19 | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 42, 1, 48, 1, 1, 1,  1,
20 | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 63, 1, 71, 1, 1, 1,  1,

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

Cf. A257852, A371092, A371094, A372282, A372283, A372288.

Cf. also A371097 (array giving every fourth column, 1, 5, 9, ...), A371103 (array giving every even numbered column), also array A371101.

up_to = 105;
A000265(n) = (n>>valuation(n,2));
A371092(n) = floor((A000265(1+(3*n))+5)/6);
R(n) = { n = 1+3*n; n>>valuation(n, 2); };
A372283sq(n,k) = if(1==n,2*k-1,R(A372283sq(n-1,k)));
A372287sq(n,k) = A371092(A372283sq(n,k));
A372287list(up_to) = { my(v = vector(up_to), i=0); for(a=1,oo, for(col=1,a, i++; if(i > up_to, return(v)); v[i] = A372287sq((a-(col-1)),col))); (v); };
v372287 = A372287list(up_to);
A372287(n) = v372287[n];

A(n, k) = A371092(A372282(n,k)) = A371092(A372283(n,k)).

A372361 Array read by upward antidiagonals: A(n, k) = A372358(A372283(n, k)), n,k >= 1.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 6, 4, 0, 0, 0, 4, 2, 6, 0, 0, 0, 0, 6, 4, 0, 0, 0, 0, 0, 4, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 2, 4, 0, 0, 0, 0, 0, 0, 0, 22, 0, 6, 0, 0, 0, 0, 0, 0, 0, 0, 0, 8, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 6, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0, 22, 12, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 6, 14

Offset: 1

Antti Karttunen, May 01 2024

			Array begins:
n\k| 1  2  3  4  5  6  7   8  9 10 11  12  13   14 15   16  17  18  19  20
---+------------------------------------------------------------------------
1  | 0, 0, 0, 2, 4, 6, 0,  2, 4, 6, 0,  2, 12,  14, 8,  10, 20, 22, 16, 18,
2  | 0, 0, 0, 6, 2, 4, 0,  2, 0, 8, 0, 22,  6,  28, 6,  26, 12,  0,  2, 14,
3  | 0, 0, 0, 4, 6, 0, 0, 22, 0, 6, 0,  0,  8,  10, 4,  18,  6,  0,  6, 12,
4  | 0, 0, 0, 0, 4, 0, 0,  0, 0, 4, 0,  0,  6,  26, 0,  62,  8,  0,  4, 22,
5  | 0, 0, 0, 0, 0, 0, 0,  0, 0, 0, 0,  0,  4,  18, 0, 116,  6,  0,  0, 48,
6  | 0, 0, 0, 0, 0, 0, 0,  0, 0, 0, 0,  0,  0,  62, 0,  44,  4,  0,  0,  6,
7  | 0, 0, 0, 0, 0, 0, 0,  0, 0, 0, 0,  0,  0, 116, 0,  14,  0,  0,  0,  8,
8  | 0, 0, 0, 0, 0, 0, 0,  0, 0, 0, 0,  0,  0,  44, 0,  92,  0,  0,  0,  6,
9  | 0, 0, 0, 0, 0, 0, 0,  0, 0, 0, 0,  0,  0,  14, 0,  50,  0,  0,  0,  4,
10 | 0, 0, 0, 0, 0, 0, 0,  0, 0, 0, 0,  0,  0,  92, 0,  78,  0,  0,  0,  0,
11 | 0, 0, 0, 0, 0, 0, 0,  0, 0, 0, 0,  0,  0,  50, 0,  60,  0,  0,  0,  0,
12 | 0, 0, 0, 0, 0, 0, 0,  0, 0, 0, 0,  0,  0,  78, 0, 122,  0,  0,  0,  0,
13 | 0, 0, 0, 0, 0, 0, 0,  0, 0, 0, 0,  0,  0,  60, 0,  82,  0,  0,  0,  0,
14 | 0, 0, 0, 0, 0, 0, 0,  0, 0, 0, 0,  0,  0, 122, 0, 222,  0,  0,  0,  0,
15 | 0, 0, 0, 0, 0, 0, 0,  0, 0, 0, 0,  0,  0,  82, 0, 260,  0,  0,  0,  0,
16 | 0, 0, 0, 0, 0, 0, 0,  0, 0, 0, 0,  0,  0, 222, 0, 232,  0,  0,  0,  0,
17 | 0, 0, 0, 0, 0, 0, 0,  0, 0, 0, 0,  0,  0, 260, 0, 114,  0,  0,  0,  0,
18 | 0, 0, 0, 0, 0, 0, 0,  0, 0, 0, 0,  0,  0, 232, 0,  46,  0,  0,  0,  0,
19 | 0, 0, 0, 0, 0, 0, 0,  0, 0, 0, 0,  0,  0, 114, 0,  44,  0,  0,  0,  0,
20 | 0, 0, 0, 0, 0, 0, 0,  0, 0, 0, 0,  0,  0,  46, 0,  78,  0,  0,  0,  0,
21 | 0, 0, 0, 0, 0, 0, 0,  0, 0, 0, 0,  0,  0,  44, 0, 252,  0,  0,  0,  0,

Cf. A075677, A086893, A372283, A372358, A372360 (binary weights), A372446 (column 14).

Cf. also A372359.

up_to = 105;
R(n) = { n = 1+3*n; n>>valuation(n, 2); };
A372283sq(n,k) = if(1==n,2*k-1,R(A372283sq(n-1,k)));
A000523(n) = logint(n,2);
A086893(n) = (if(n%2, 2^(n+1), 2^(n+1)+2^(n-1))\3); \\ From A086893
A372358(n) = bitxor(A086893(1+A000523(n)),n);
A372361sq(n,k) = A372358(A372283sq(n,k));
A372361list(up_to) = { my(v = vector(up_to), i=0); for(a=1,oo, for(col=1,a, i++; if(i > up_to, return(v)); v[i] = A372361sq((a-(col-1)),col))); (v); };
v372361 = A372361list(up_to);
A372361(n) = v372361[n];

A075677 Reduced Collatz function R applied to the odd integers: a(n) = R(2n-1), where R(k) = (3k+1)/2^r, with r as large as possible.

Original entry on oeis.org

1, 5, 1, 11, 7, 17, 5, 23, 13, 29, 1, 35, 19, 41, 11, 47, 25, 53, 7, 59, 31, 65, 17, 71, 37, 77, 5, 83, 43, 89, 23, 95, 49, 101, 13, 107, 55, 113, 29, 119, 61, 125, 1, 131, 67, 137, 35, 143, 73, 149, 19, 155, 79, 161, 41, 167, 85, 173, 11, 179, 91, 185, 47, 191, 97, 197

Offset: 1

T. D. Noe, Sep 25 2002

The even-indexed terms a(2i+2) = 6i+5 = A016969(i), i >= 0 [Comment corrected by Bob Selcoe, Apr 06 2015]. The odd-indexed terms are the same as A067745. Note that this sequence is A016789 with all factors of 2 removed from each term. Also note that a(4i-1) = a(i). No multiple of 3 is in this sequence. See A075680 for the number of iterations of R required to yield 1.
From Bob Selcoe, Apr 06 2015: (Start)
All numbers in this sequence appear infinitely often.
From Eq. 1 and Eq. 2 in Formulas: Eq. 1 is used with 1/3 of the numbers in this sequence, Eq. 2 is used with 2/3 of the numbers.
(End)
Empirical: For arbitrary m, Sum_{n=2..A007583(m)} (a(n) - a(n-1)) = 0. - Fred Daniel Kline, Nov 23 2015
From Wolfdieter Lang, Dec 07 2021: (Start)
Only positive numbers congruent to 1 or 5 modulo 6 appear.
i) For the sequence entry with value A016921(m), for m >= 0, that is, a value from {1, 7, 13, ...}, the indices n are given by the row of array A178415(2*m+1, k), for k >= 1.
ii) For the sequence entry with value A007528(m), for m >= 1, that is, a value from {5, 11, 17, ...}, the indices n are given by the row of array A178415(2*m, k), for k >= 1.
See also the array A347834 with permuted row numbers and columns k >= 0. (End)

			a(11) = 1 because 21 is the 11th odd number and R(21) = 64/64 = 1.
From _Wolfdieter Lang_, Dec 07 2021: (Start)
i) 1 (mod 6) entry 1 = A016921(0) appears for n = A178415(1, k) = A347834(1, k-1) (the arrays), for k >= 1, that is, for {1, 5, 21, ..} = A002450.
ii) 5 (mod 6) entry 11 = A007528(2) appears for n = A178415(4, k) = A347835(3, k-1) (the arrays), for k >= 1, that is, for {7, 29, 117, ..} = A072261. (End)

Richard K. Guy, Unsolved Problems in Number Theory, 3rd Edition, Springer, 2004, Section E16, pp. 330-336.
Victor Klee and Stan Wagon, Old and new unsolved problems in plane geometry and number theory, The Mathematical Association of America, 1991, p. 225, C(2n+1) = a(n+1), n >= 0.
Jeffrey C. Lagarias, ed., The Ultimate Challenge: The 3x+1 Problem, Amer. Math. Soc., 2010; see p. 57, also (90-9), p. 306.

T. D. Noe, Table of n, a(n) for n = 1..1000
Marc Chamberland, Una actualizacio del problema 3x + 1, Butl. Soc. Catalana Mat. (18), 19-45, 2003.
Jeffrey C. Lagarias, The 3x+1 problem and its generalizations, Amer. Math. Monthly, 92 (1985), 3-23.
Fabian S. Reid, The Visual Pattern in the Collatz Conjecture and Proof of No Non-Trivial Cycles, arXiv:2105.07955 [math.GM], 2021.
Eric Weisstein's World of Mathematics, Collatz Problem.
Index entries for sequences related to 3x+1 (or Collatz) problem.

Cf. A000265, A000302, A002450, A007528, A016789, A016921, A016969, A065677, A072197, A072261, A075680, A081294, A178415, A191669, A329480, A347834.
Cf. A006370, A014682 (for non-reduced Collatz maps), A087230 (A371093), A371094.
Odd bisection of A139391.
Even bisection of A067745, which is also the odd bisection of this sequence.
After the initial 1, the second leftmost column of A256598.
Row 2 of A372283.

a075677 = a000265 . subtract 2 . (* 6) -- Reinhard Zumkeller, Jan 08 2014

f:=proc(n) local t1;
if n=1 then RETURN(1) else
t1:=3*n+1;
while t1 mod 2 = 0 do t1:=t1/2; od;
RETURN(t1); fi;
end;
# N. J. A. Sloane, Jan 21 2011

nextOddK[n_] := Module[{m=3n+1}, While[EvenQ[m], m=m/2]; m]; (* assumes odd n *) Table[nextOddK[n], {n, 1, 200, 2}]
v[x_] := IntegerExponent[x, 2]; f[x_] := (3*x + 1)/2^v[3*x + 1]; Table[f[2*n - 1], {n, 66}] (* L. Edson Jeffery, May 06 2015 *)

a(n)=n+=2*n-1;n>>valuation(n,2) \\ Charles R Greathouse IV, Jul 05 2013

from sympy import divisors
def a(n):
    return max(d for d in divisors(n) if d % 2)
print([a(6*n - 2) for n in range(1, 101)]) # Indranil Ghosh, Apr 15 2017, after formula by Reinhard Zumkeller

a(n) = A000265(6*n-2) = A000265(3*n-1). - Reinhard Zumkeller, Jan 08 2014
From Bob Selcoe, Apr 05 2015: (Start)
For all n>=1 and for every k, there exists j>=0 dependent upon n and k such that either:
Eq. 1: a(n) = (3n-1)/2^(2j+1) when k = ((4^(j+1)-1)/3) mod 2^(2j+3). Alternatively: a(n) = A016789(n-1)/A081294(j+1) when k = A002450(j+1) mod A081294(j+2). Example: n=51; k=101 == 5 mod 32, j=1. a(51) = 152/8 = 19.
  or
Eq. 2: a(n) = (3n-1)/4^j when k = (5*2^(2j+1) - 1)/3 mod 4^(j+1). Alternatively: a(n) = A016789(n-1)/A000302(j) when k = A072197(j) mod A000302(j+1). Example: n=91; k=181 == 53 mod 64, j=2. a(91) = 272/16 = 17.
(End) [Definition corrected by William S. Hilton, Jul 29 2017]
a(n) = a(n + g*2^r) - 6*g, n > -g*2^r. Examples: n=59; a(59)=11, r=5. g=-1: 11 = a(27) = 5 - (-1)*6; g=1: 11 = a(91) = 17 - 1*6; g=2: 11 = a(123) = 23 - 2*6; g=3: 11 = a(155) = 29 - 3*6; etc. - Bob Selcoe, Apr 06 2015
a(n) = a((1 + (3*n - 1)*4^(k-1))/3), k>=1 (cf. A191669). - L. Edson Jeffery, Oct 05 2015
a(n) = a(4n-1). - Bob Selcoe, Aug 03 2017
a(n) = A139391(2n-1). - Antti Karttunen, May 06 2024
Sum_{k=1..n} a(k) ~ n^2. - Amiram Eldar, Aug 26 2024
G.f.: Sum_{k>=1} ((3 + 2*(-1)^k)*x^(3*2^(k - 1) - (-2)^k/3 + 1/3) + (3 - 2*(-1)^k)*x^(2^(k - 1) - (-2)^k/3 + 1/3))/(x^(2^k) - 1)^2. - Miles Wilson, Oct 26 2024

A371094 a(n) = m*(2^e) + ((4^e)-1)/3, where m = 3n+1, and e is the 2-adic valuation of m.

Original entry on oeis.org

1, 21, 7, 21, 13, 341, 19, 45, 25, 117, 31, 69, 37, 341, 43, 93, 49, 213, 55, 117, 61, 5461, 67, 141, 73, 309, 79, 165, 85, 725, 91, 189, 97, 405, 103, 213, 109, 1877, 115, 237, 121, 501, 127, 261, 133, 1109, 139, 285, 145, 597, 151, 309, 157, 5461, 163, 333, 169, 693, 175, 357, 181, 1493, 187, 381, 193, 789, 199

Offset: 0

Antti Karttunen (proposed by Ali Sada), Apr 19 2024

Construction: take the binary expansion of 3n+1 (A016777(n)), and substitute "01" for all trailing 0-bits that follow after its odd part (= A067745(1+n)), of which there are A371093(n) in total. See the examples.

			For n=1, 3*n+1 = 4, "100" in binary, when we substitute 01's for the two trailing 0's, we obtain 21, "10101" in binary, therefore a(1) = 21.
For n=6, 3*6+1 = 19, "10011" in binary, and there are no trailing 0's, and no changes, therefore a(6) = 19.
For n=7, 3*7+1 = 22, "10110" in binary, with one trailing 0, which when replaced with 01 gives us 45, "101101" in binary, therefore a(7) = 45.
For n=229, there are e=4 trailing bit expansions 0 -> 01,
  3n+1 = binary  101011  0 0 0 0
  a(n) = binary  101011 01010101

Antti Karttunen, Table of n, a(n) for n = 0..16384
Michael De Vlieger, Plot binary expansion of a(n) arranged with smallest bit at bottom and largest at top, n increasing from left to right for n = 0..3071. Black indicates 1, white indicates 0.
Index entries for sequences related to 3x+1 (or Collatz) problem
Index entries for sequences related to binary expansion of n

Cf. A016777, A067745, A075677, A086893, A096773, A372289, A371093.
Cf. A016921, A372351 (even and odd bisection), A372290 (numbers occurring in the latter).
Cf. arrays A372282, A372283 and A371096, A371100, A371102.
Cf. also A302338.

Array[#2*(2^#3) + ((4^#3) - 1)/3 & @@ {#1, #2, IntegerExponent[#2, 2]} & @@ {#, 3 #1 + 1} &, 67, 0] (* Michael De Vlieger, Apr 19 2024 *)

A371094(n) = { my(m=1+3*n, e=valuation(m,2)); ((m*(2^e)) + (((4^e)-1)/3)); };

def A371094(n): return ((m:=3*n+1)<<(e:=(~m & m-1).bit_length()))+((1<<(e<<1))-1)//3 # Chai Wah Wu, Apr 28 2024

a(n) = A372289(A016777(n)).

a(2n) = A016777(2n) = A016921(n).

A372282 Array read by upward antidiagonals: A(n, k) = A371094(A(n-1, k)) for n > 1, k >= 1; A(1, k) = 2*k-1.

Original entry on oeis.org

1, 21, 3, 5461, 21, 5, 357913941, 5461, 341, 7, 1537228672809129301, 357913941, 1398101, 45, 9, 28356863910078205288614550619314017621, 1537228672809129301, 23456248059221, 1109, 117, 11, 9649340769776349618630915417390658987772498722136713669954798667326094136661, 28356863910078205288614550619314017621, 6602346876188694799461995861, 873813, 11605, 69, 13

Offset: 1

Antti Karttunen, Apr 28 2024

			Array begins:
n\k|    1     2        3     4      5     6        7     8      9     10
---+----------------------------------------------------------------------
1  |    1,    3,       5,    7,     9,   11,      13,   15,    17,    19,
2  |   21,   21,     341,   45,   117,   69,     341,   93,   213,   117,
3  | 5461, 5461, 1398101, 1109, 11605, 3413, 1398101, 2261, 87381, 11605,

Cf. A005408 (row 1), A372351 (row 2, bisection of A371094), A372444 (column 14).
Arrays derived from this one:
  A372283,
  A372285 the number of terms of A086893 in the interval [A(n, k), A(1+n, k)],
  A372287 the column index of A(n, k) in array A257852,
  A372288 the sum of digits of A(n, k) in "Jacobsthal greedy base",
  A372353 differences between A(n,k) and the largest term of A086893 <= A(n,k),
  A372354 floor(log_2(.)) of terms, A372356 (and their columnwise first differences),
  A372359 terms xored with binary words of the same length, either of the form 10101...0101 or 110101...0101, depending on whether the binary length is odd or even.
Cf. also arrays A371096, A371102 that give subsets of columns of this array, and array A371100 that gives the terms of the row 2 in different order.

up_to = 28;
A371094(n) = { my(m=1+3*n, e=valuation(m,2)); ((m*(2^e)) + (((4^e)-1)/3)); };
A372282sq(n,k) = if(1==n,2*k-1,A371094(A372282sq(n-1,k)));
A372282list(up_to) = { my(v = vector(up_to), i=0); for(a=1,oo, for(col=1,a, i++; if(i > up_to, return(v)); v[i] = A372282sq((a-(col-1)),col))); (v); };
v372282 = A372282list(up_to);
A372282(n) = v372282[n];

A372443 The n-th iterate of 27 with Reduced Collatz-function R, which gives the odd part of 3n+1.

Original entry on oeis.org

27, 41, 31, 47, 71, 107, 161, 121, 91, 137, 103, 155, 233, 175, 263, 395, 593, 445, 167, 251, 377, 283, 425, 319, 479, 719, 1079, 1619, 2429, 911, 1367, 2051, 3077, 577, 433, 325, 61, 23, 35, 53, 5, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1

Offset: 0

Antti Karttunen, May 01 2024

nonn

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

Cf. A000265, A075677, A371093.
Column 14 of A372283, Row 13 of A256598 (but only up to the first 1).
Row 1 of A372560.
From term 47 to the first 1 same as A088593.
Sequences derived from this one or related to:
  A372444,
  A372445 column index of a(n) in array A257852,
  A372362 the 2-adic valuation of 1 + 3*a(n), equal to row index of a(n) in array A257852,
  A372447 binary lengths minus 1,
  A372446 a(n) xored with the term of A086893 having the same binary length,
  A372453 a(n) minus the term of A086893 having the same binary length.

R(n) = { n = 1+3*n; n>>valuation(n, 2); };
A372443(n) = { my(x=27); while(n, x=R(x); n--); (x); };

a(0) = 27; for n > 0, a(n) = R(a(n-1)), where R(n) = (3*n+1)/2^A371093(n) = A000265(3*n+1).

For n > 0, a(n) = R(A372444(n-1)) = A000265(1+3*A372444(n-1)).

A256598 Irregular triangle where row n contains the odd terms in the Collatz sequence beginning with 2n+1.

Original entry on oeis.org

1, 3, 5, 1, 5, 1, 7, 11, 17, 13, 5, 1, 9, 7, 11, 17, 13, 5, 1, 11, 17, 13, 5, 1, 13, 5, 1, 15, 23, 35, 53, 5, 1, 17, 13, 5, 1, 19, 29, 11, 17, 13, 5, 1, 21, 1, 23, 35, 53, 5, 1, 25, 19, 29, 11, 17, 13, 5, 1, 27, 41, 31, 47, 71, 107, 161, 121, 91, 137, 103, 155

Offset: 0

Bob Selcoe, Apr 03 2015

The Collatz function is an integer-valued function given by n/2 if n is even and 3n+1 if n is odd. We build a Collatz sequence by beginning with a natural number and iterating the function indefinitely. It is conjectured that all such sequences terminate at 1.
In this triangle, row n is made up of the odd terms of the Collatz sequence beginning with 2n+1. Therefore, it is conjectured that this sequence is well-defined, i.e., that all rows terminate at 1.
The last index k in row n gives the number of iterations required for the Collatz sequence to terminate if even terms are omitted.
T(n,k)/T(n,k+1) is of form: ceiling(T(n,k)*3/2^j) for some j>=1. Therefore, the coefficients in each row may be read as a series of iterated ceilings, where j may vary. For example, row 3 has initial term 7, which is followed by ceiling(7*3/2), ceiling(ceiling(7*3/2)*3/2), ceiling(ceiling(ceiling(7*3/2)*3/2)*3/4), ceiling(ceiling(ceiling(ceiling(7*3/2)*3/2)*3/4)*3/8), ceiling(ceiling(ceiling(ceiling(ceiling(7*3/2)*3/2)*3/4)*3/8)*3/16).
The length of row n is A258145(n) (set to 0 if 1 is not reached after a finite number of steps). - Wolfdieter Lang, Aug 11 2021

			Triangle starts T(0,0):
n\k   0   1   2   3   4    5   6   7   8   9  10 ...
0:    1
1:    3   5   1
2:    5   1
3:    7   11  17  13  5    1
4:    9   7   11  17  13   5   1
5:    11  17  13  5   1
6:    13  5   1
7:    15  23  35  53  5    1
8:    17  13  5   1
9:    19  29  11  17  13   5   1
10:   21  1
11:   23  35  53  5    1
12:   25  19  29  11  17  13   5   1
...
n=13 starts with 27 and takes 41 steps: (27), 41, 31, 47, 71, 107,... 53, 5, 1, (see A372443).
Row 8 is [17, 13, 5, 1] because it is the subsequence of odd terms for the Collatz sequence starting with 17: [17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1].

Eric Weisstein's World of Mathematics, Collatz Problem
Wikipedia, Collatz conjecture
Index entries for sequences related to 3x+1 (or Collatz) problem

Cf. A006370, A070165, A075677, A258145.
Cf. A372443 (row 13 up to its first 1).
Cf. also array A372283 showing the same terms in different orientation.

f[n_] := NestWhileList[(3*# + 1)/2^IntegerExponent[3*# + 1, 2] &, 2*n + 1, # > 1 &]; Grid[Table[f[n], {n, 0, 12}]] (* L. Edson Jeffery, Apr 25 2015 *)

row(n) = {my(oddn = 2*n+1, vl = List(oddn), x); while (oddn != 1, x = 3*oddn+1; oddn = x >> valuation(x, 2); listput(vl, oddn)); Vec(vl);}
tabf(nn) = {for (n=0, nn, my(rown = row(n)); for (k=1, #rown, print1(rown[k], ", ")); print;);} \\ Michel Marcus, Oct 04 2019

def Collatz(n):
    A = [n]
    b = A[-1]
    while b != 1:
        if is_even(b):
            A.append(b//2)
        else:
            A.append(3*b+1)
    return A
[y for sublist in [[x for x in Collatz(2*n+1) if is_odd(x)] for n in [0..15]] for y in sublist] # Tom Edgar, Apr 04 2015

T(n,0) = 2n+1 and T(n,k) = A000265(3*T(n,k-1)+1) for k>0. - Tom Edgar, Apr 04 2015

A372288 Array read by upward antidiagonals: A(n, k) = A265745(A372282(n, k)), n,k >= 1, where A265745(n) is the sum of digits in "Jacobsthal greedy base".

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 3, 3, 1, 1, 1, 3, 3, 1, 1, 1, 1, 3, 3, 3, 3, 1, 1, 1, 1, 3, 3, 1, 3, 1, 1, 1, 1, 3, 1, 1, 3, 3, 1, 1, 1, 1, 1, 1, 1, 5, 3, 3, 1, 1, 1, 1, 1, 1, 1, 3, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 5, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 3, 5, 3

Offset: 1

Antti Karttunen, Apr 28 2024

Collatz conjecture is equal to the claim that each column will eventually settle to constant 1's, somewhere under its topmost row. This works as only the bisection A002450 of Jacobsthal numbers (A001045) contains numbers of the form 4k+1, while the other bisection contains only numbers of the form 4k+3, which do not occur among the range of A372351. See also the comments in A371094.

			Array begins:
n\k| 1  2  3  4  5  6  7  8  9 10 11 12 13     14 15    16 17 18 19 20    21 22
---+----------------------------------------------------------------------------
1  | 1, 1, 1, 3, 3, 1, 3, 3, 3, 3, 1, 3, 3,     3, 3,    3, 3, 3, 3, 5,    5, 1,
2  | 1, 1, 1, 3, 3, 3, 1, 3, 3, 3, 1, 5, 5,     5, 3,    5, 3, 3, 3, 5,    5, 3,
3  | 1, 1, 1, 3, 3, 3, 1, 5, 1, 3, 1, 3, 3,     5, 3,    5, 5, 1, 3, 3,    5, 3,
4  | 1, 1, 1, 3, 3, 1, 1, 3, 1, 3, 1, 1, 3,     5, 3,    3, 3, 1, 3, 5,    5, 3,
5  | 1, 1, 1, 1, 3, 1, 1, 1, 1, 3, 1, 1, 3,     5, 1,    5, 3, 1, 3, 3,    3, 3,
6  | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3,     3, 1,    5, 3, 1, 1, 5,    5, 3,
7  | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,     5, 1,    3, 3, 1, 1, 3,    5, 3,
8  | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,     5, 1,    5, 1, 1, 1, 3,    3, 3,
9  | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,     3, 1,    5, 1, 1, 1, 3,    5, 1,
10 | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,     5, 1,    5, 1, 1, 1, 3,    5, 1,
11 | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,     5, 1, 2155, 1, 1, 1, 1,    5, 1,
12 | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,     5, 1,    5, 1, 1, 1, 1, 6251, 1,
13 | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 10347, 1,    5, 1, 1, 1, 1,    5, 1,
14 | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,     5, 1,    5, 1, 1, 1, 1,    5, 1,
15 | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,     5, 1,    7, 1, 1, 1, 1,    5, 1,
16 | 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,     5, 1,    5, 1, 1, 1, 1,    7, 1,

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

Cf. A001045, A002450, A265745, A265747, A371094, A372351, A372282, A372283, A372287.

Cf. also array A372561 (formed by columns whose indices in this array are given by A372443).

up_to = 105;
A130249(n) = (#binary(3*n+1)-1);
A001045(n) = (2^n - (-1)^n) / 3;
A265745(n) = { my(s=0); while(n,s++; n -= A001045(A130249(n))); (s); };
A371094(n) = { my(m=1+3*n, e=valuation(m,2)); ((m*(2^e)) + (((4^e)-1)/3)); };
A372282sq(n,k) = if(1==n,2*k-1,A371094(A372282sq(n-1,k)));
A372288sq(n,k) = A265745(A372282sq(n,k));
A372288list(up_to) = { my(v = vector(up_to), i=0); for(a=1,oo, for(col=1,a, i++; if(i > up_to, return(v)); v[i] = A372288sq((a-(col-1)),col))); (v); };
v372288 = A372288list(up_to);
A372288(n) = v372288[n];

A372360 Array read by upward antidiagonals: A(n, k) = A000120(A372361(n, k)), n,k >= 1; Binary weights of terms of arrays A372359 and A372361.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 2, 1, 0, 0, 0, 1, 1, 2, 0, 0, 0, 0, 2, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 3, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 3, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 3

Offset: 1

Antti Karttunen, May 01 2024

Entry A(n, k) at row n and column k tells how many bits needs to be flipped in the binary expansion of the (n-1)-th iterate of Reduced Collatz function R, when started from 2*k-1, to obtain the unique term of A086893 with the same binary length as that (n-1)-th iterate. That is, A(n, k) gives the Hamming distance between A372283(n, k) and A086893(1+A000523(A372283(n, k))).

Zeros occur in the same locations as where they occur in A372359, etc.

			Array begins:
n\k| 1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
---+-------------------------------------------------------------------------
1  | 0, 0, 0, 1, 1, 2, 0, 1, 1, 2, 0, 1, 2, 3, 1, 2, 2, 3, 1, 2, 3, 4, 2, 3,
2  | 0, 0, 0, 2, 1, 1, 0, 1, 0, 1, 0, 3, 2, 3, 2, 3, 2, 0, 1, 3, 2, 2, 1, 2,
3  | 0, 0, 0, 1, 2, 0, 0, 3, 0, 2, 0, 0, 1, 2, 1, 2, 2, 0, 2, 2, 3, 1, 0, 5,
4  | 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 2, 3, 0, 5, 1, 0, 1, 3, 2, 1, 0, 4,
5  | 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 0, 4, 2, 0, 0, 2, 5, 1, 0, 3,
6  | 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 5, 0, 3, 1, 0, 0, 2, 4, 2, 0, 3,
7  | 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0, 3, 0, 0, 0, 1, 3, 1, 0, 4,
8  | 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 4, 0, 0, 0, 2, 3, 0, 0, 3,
9  | 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 3, 0, 0, 0, 1, 4, 0, 0, 4,
10 | 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0, 4, 0, 0, 0, 0, 3, 0, 0, 4,
11 | 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 4, 0, 0, 0, 0, 4, 0, 0, 5,
12 | 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0, 5, 0, 0, 0, 0, 4, 0, 0, 3,
13 | 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0, 3, 0, 0, 0, 0, 5, 0, 0, 6,
14 | 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 5, 0, 6, 0, 0, 0, 0, 3, 0, 0, 2,
15 | 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 2, 0, 0, 0, 0, 6, 0, 0, 4,
16 | 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 6, 0, 4, 0, 0, 0, 0, 2, 0, 0, 4,
17 | 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 4, 0, 0, 0, 0, 4, 0, 0, 4,
18 | 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0, 4, 0, 0, 0, 0, 4, 0, 0, 3,
19 | 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0, 3, 0, 0, 0, 0, 4, 0, 0, 4,
20 | 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0, 4, 0, 0, 0, 0, 3, 0, 0, 6,
21 | 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 6, 0, 0, 0, 0, 4, 0, 0, 4,
We have A372283(5, 14) = 71, and when we compare the binary expansion of 71 = 1000111_2 with the term of A086893 that has a binary expansion of the same length, which in this case is 85 = 1010101_2, we see that only the bits at positions 1 and 4 (indexed from the right hand end, with 0 being the least significant bit position at right) need to be toggled to obtain the 71 from 85 or vice versa, therefore A(5, 14) = 2.
We have A372283(6, 14) = 107 = 1101011_2, and when xored with A086893(7) = 85 = 1010101_2, we obtain A372361(6, 14) = 62 = 111110_2, with five 1-bits, therefore A(6, 14) = 5. I.e., five bits (all except the least and the most significant bit) need to be flipped to change 85 to 107 or vice versa.

Cf. A000120, A003987, A086893, A372282, A372283, A372287, A372354, A372358, A372360.
Binary weights of A372359 and A372361.
Cf. also A372288.

up_to = 105;
R(n) = { n = 1+3*n; n>>valuation(n, 2); };
A372283sq(n,k) = if(1==n,2*k-1,R(A372283sq(n-1,k)));
A000523(n) = logint(n,2);
A086893(n) = (if(n%2, 2^(n+1), 2^(n+1)+2^(n-1))\3);
A372358(n) = bitxor(A086893(1+A000523(n)),n);
A372361sq(n,k) = A372358(A372283sq(n,k));
A372361list(up_to) = { my(v = vector(up_to), i=0); for(a=1,oo, for(col=1,a, i++; if(i > up_to, return(v)); v[i] = A372361sq((a-(col-1)),col))); (v); };
v372361 = A372361list(up_to);
A372361(n) = v372361[n];

A(n, k) = A000120(A372361(n, k)) = A000120(A372358(A372283(n, k))).

A(n, k) = A000120(A372359(n, k)) = A000120(A372358(A372282(n, k))).

cp's OEIS Frontend

A372287 Array read by upward antidiagonals: A(n, k) = A371092(A372283(n, k)), n,k >= 1.

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A372361 Array read by upward antidiagonals: A(n, k) = A372358(A372283(n, k)), n,k >= 1.

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A075677 Reduced Collatz function R applied to the odd integers: a(n) = R(2n-1), where R(k) = (3k+1)/2^r, with r as large as possible.

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Maple

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A371094 a(n) = m*(2^e) + ((4^e)-1)/3, where m = 3n+1, and e is the 2-adic valuation of m.

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A372282 Array read by upward antidiagonals: A(n, k) = A371094(A(n-1, k)) for n > 1, k >= 1; A(1, k) = 2*k-1.

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A372443 The n-th iterate of 27 with Reduced Collatz-function R, which gives the odd part of 3n+1.

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A256598 Irregular triangle where row n contains the odd terms in the Collatz sequence beginning with 2n+1.

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A372288 Array read by upward antidiagonals: A(n, k) = A265745(A372282(n, k)), n,k >= 1, where A265745(n) is the sum of digits in "Jacobsthal greedy base".