A372446 -id:A372446 - cp's OEIS frontend

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

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)).

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)));

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];

A372453 a(n) = A372443(n) - A086893(1+A372447(n)).

Original entry on oeis.org

6, -12, 10, -6, -14, 22, -52, 36, 6, -76, 18, -58, 20, -38, -78, 54, -260, 104, -46, 38, 36, -58, 84, -22, 138, -134, -286, 254, -984, 58, 2, -1362, -336, -276, 92, -16, 8, 2, -18, 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 05 2024

sign

These are the differences obtained when the term of A086893 that has the same binary length as A372443(n) is subtracted from the latter. Here A372443(n) gives the n-th iterate of 27 with Reduced Collatz-function R, where R(n) = A000265(3*n+1).

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.

			The term of A086893 that has same binary length as A372443(0) = 27 is 21 [as 21 = 10101_2 in binary, and 27 = 11011_2 in binary], therefore a(0) = 27-21 = 6.
The term of A086893 that has same binary length as A372443(1) = 41 is 53, therefore a(1) = 41-53 = -12.

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

Cf. A000265, A000523, A086893, A372443.

Cf. also A372446.

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);
A372443(n) = { my(x=27); while(n, x=A000265(3*x+1); n--); (x); };
A372453(n) = { my(x=A372443(n)); (x - A086893(1+A000523(x))); };

a(n) = A372443(n) - A086893(1+A000523(A372443(n))).

cp's OEIS Frontend

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

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

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

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A372453 a(n) = A372443(n) - A086893(1+A372447(n)).

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