A372361 -id:A372361 - cp's OEIS frontend

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.

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))).

A372283 Array read by upward antidiagonals: A(n, k) = R(A(n-1, k)) for n > 1, k >= 1; A(1, k) = 2*k-1, where Reduced Collatz function R(n) gives the odd part of 3n+1.

Original entry on oeis.org

1, 1, 3, 1, 5, 5, 1, 1, 1, 7, 1, 1, 1, 11, 9, 1, 1, 1, 17, 7, 11, 1, 1, 1, 13, 11, 17, 13, 1, 1, 1, 5, 17, 13, 5, 15, 1, 1, 1, 1, 13, 5, 1, 23, 17, 1, 1, 1, 1, 5, 1, 1, 35, 13, 19, 1, 1, 1, 1, 1, 1, 1, 53, 5, 29, 21, 1, 1, 1, 1, 1, 1, 1, 5, 1, 11, 1, 23, 1, 1, 1, 1, 1, 1, 1, 1, 1, 17, 1, 35, 25

Offset: 1

Antti Karttunen, Apr 28 2024

Collatz conjecture is equal to the claim that in each column 1 will eventually appear. See also arrays A372287 and A372288.

			Array begins:
n\k| 1  2  3   4   5   6   7   8   9  10  11  12  13   14  15   16  17  18
---+-----------------------------------------------------------------------
1  | 1, 3, 5,  7,  9, 11, 13, 15, 17, 19, 21, 23, 25,  27, 29,  31, 33, 35,
2  | 1, 5, 1, 11,  7, 17,  5, 23, 13, 29,  1, 35, 19,  41, 11,  47, 25, 53,
3  | 1, 1, 1, 17, 11, 13,  1, 35,  5, 11,  1, 53, 29,  31, 17,  71, 19,  5,
4  | 1, 1, 1, 13, 17,  5,  1, 53,  1, 17,  1,  5, 11,  47, 13, 107, 29,  1,
5  | 1, 1, 1,  5, 13,  1,  1,  5,  1, 13,  1,  1, 17,  71,  5, 161, 11,  1,
6  | 1, 1, 1,  1,  5,  1,  1,  1,  1,  5,  1,  1, 13, 107,  1, 121, 17,  1,
7  | 1, 1, 1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  5, 161,  1,  91, 13,  1,
8  | 1, 1, 1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  1, 121,  1, 137,  5,  1,
9  | 1, 1, 1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  91,  1, 103,  1,  1,
10 | 1, 1, 1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  1, 137,  1, 155,  1,  1,
11 | 1, 1, 1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  1, 103,  1, 233,  1,  1,
12 | 1, 1, 1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  1, 155,  1, 175,  1,  1,
13 | 1, 1, 1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  1, 233,  1, 263,  1,  1,
14 | 1, 1, 1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  1, 175,  1, 395,  1,  1,
15 | 1, 1, 1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  1, 263,  1, 593,  1,  1,
16 | 1, 1, 1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  1, 395,  1, 445,  1,  1,

Cf. A005408 (row 1), A075677 (row 2), A372443 (column 14).
Cf. A258145 (A075680), A371093.
Arrays derived from this one or related to:
  A372282,
  A372287 the column index of A(n, k) in array A257852,
  A372361 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,
  A372360 binary weights of A372361.
Cf. also array A371095 (giving every fourth column, 1, 5, 9, ...) and irregular array A256598 which gives the terms of each column, but only down to the first 1.

With[{dmax = 15}, Table[#[[k, n-k+1]], {n, dmax}, {k, n}] & [Array[NestList[(3*# + 1)/2^IntegerExponent[3*# + 1, 2] &, 2*# - 1, dmax - #] &, dmax]]] (* Paolo Xausa, Apr 29 2024 *)

up_to = 91;
R(n) = { n = 1+3*n; n>>valuation(n, 2); };
A372283sq(n,k) = if(1==n,2*k-1,R(A372283sq(n-1,k)));
A372283list(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] = A372283sq((a-(col-1)),col))); (v); };
v372283 = A372283list(up_to);
A372283(n) = v372283[n];

For n > 1, A(n, k) = R(A372282(n-1, k)), where R(n) = (3*n+1)/2^A371093(n).

For all k >= 1, A(A258145(k-1), k) = 1 [which is the topmost 1 in each column].

A372358 a(n) = n XOR A086893(1+A000523(n)), where XOR is a bitwise-XOR, A003987.

Original entry on oeis.org

0, 1, 0, 1, 0, 3, 2, 5, 4, 7, 6, 1, 0, 3, 2, 5, 4, 7, 6, 1, 0, 3, 2, 13, 12, 15, 14, 9, 8, 11, 10, 21, 20, 23, 22, 17, 16, 19, 18, 29, 28, 31, 30, 25, 24, 27, 26, 5, 4, 7, 6, 1, 0, 3, 2, 13, 12, 15, 14, 9, 8, 11, 10, 21, 20, 23, 22, 17, 16, 19, 18, 29, 28, 31, 30, 25, 24, 27, 26, 5, 4, 7, 6, 1, 0, 3, 2, 13, 12, 15

Offset: 1

Antti Karttunen, May 01 2024

a(n) gives n xored with the unique term of A086893 that has the same binary length as n itself. The binary expansions of the terms of A086893 are of the form 10101...0101 (i.e., alternating 1's and 0's starting and ending with 1) when the binary length is odd, and of the form 110101...0101 (i.e., 1 followed by alternating 1's and 0's, and ending with 1) when the binary length is even. In other words, a(n) is n with its all its even-positioned bits (indexing starts from 0 which stands for the least significant bit) inverted, and additionally also the odd-positioned most significant bit inverted if the number of significant bits is even (i.e., n is a nonzero term of A053754).

			25 in binary is 11001_2, and inverting all the even-positioned bits gives 01100_2, and as A007088(12) = 1100, a(25) = 12.
46 in binary is 101110_2, so we flip all the even-positioned bits (starting from the rightmost, with position 0), and because there are even number of bits in the binary expansion, we flip also the most significant bit, thus we obtain 011011_2, and as A007088(27) = 11011, a(46) = 27.

Antti Karttunen, Table of n, a(n) for n = 1..16384
Index entries for sequences related to binary expansion of n

Cf. A000523, A003987, A007088, A053754, A086893 (positions of 0's), A372359, A372360, A372361, A372446.

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(n, A086893(1+A000523(n)));

A372446 a(n) = A372358(A372443(n)).

Original entry on oeis.org

14, 28, 10, 26, 18, 62, 116, 44, 14, 92, 50, 78, 60, 122, 82, 222, 260, 232, 114, 46, 44, 78, 252, 106, 138, 410, 354, 774, 1064, 218, 2, 1366, 336, 276, 228, 16, 8, 2, 22, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0

Offset: 0

Antti Karttunen, May 03 2024

nonn

These are the bitmasks (or symmetric differences) obtained when the n-th iterate of 27 with Reduced Collatz-function R [= A372443(n), where R(n) = A000265(3*n+1)] is xored with that term of A086893 that has the same binary length. The binary expansions of the terms of A086893 are always of the form 10101...0101 (i.e., alternating 1's and 0's starting and ending with 1) when the binary length is odd, and of the form 110101...0101 (i.e., 1 followed by alternating 1's and 0's, and ending with 1) when n is even. Note that for all n >= 1, R(A086893(2n-1)) = 1, and R(A086893(2n)) = 5 (with R(5) = 1), so the first zero here, a(39) = 0 indicates that the iteration will soon have reached the terminal 1, and indeed, A372443(41) = 1.

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

Column 14 of A372361.

Cf. A000265, A075677, A086893, A372358, A372443.

A000265(n) = (n>>valuation(n,2));
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);
A372443(n) = { my(x=27); while(n, x=A000265(3*x+1); n--); (x); };
A372446(n) = A372358(A372443(n));

cp's OEIS Frontend

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.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

PARI

Formula

A372283 Array read by upward antidiagonals: A(n, k) = R(A(n-1, k)) for n > 1, k >= 1; A(1, k) = 2*k-1, where Reduced Collatz function R(n) gives the odd part of 3n+1.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A372358 a(n) = n XOR A086893(1+A000523(n)), where XOR is a bitwise-XOR, A003987.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

A372446 a(n) = A372358(A372443(n)).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

PARI