cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 10 results.

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

Views

Author

T. D. Noe, Sep 25 2002

Keywords

Comments

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)

Examples

			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)
		

References

  • 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.

Crossrefs

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.

Programs

  • Haskell
    a075677 = a000265 . subtract 2 . (* 6) -- Reinhard Zumkeller, Jan 08 2014
    
  • Maple
    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
  • Mathematica
    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 *)
  • PARI
    a(n)=n+=2*n-1;n>>valuation(n,2) \\ Charles R Greathouse IV, Jul 05 2013
    
  • Python
    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

Formula

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

Views

Author

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

Keywords

Comments

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.

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
		

Crossrefs

Cf. A016921, A372351 (even and odd bisection), A372290 (numbers occurring in the latter).
Cf. also A302338.

Programs

  • Mathematica
    Array[#2*(2^#3) + ((4^#3) - 1)/3 & @@ {#1, #2, IntegerExponent[#2, 2]} & @@ {#, 3 #1 + 1} &, 67, 0] (* Michael De Vlieger, Apr 19 2024 *)
  • PARI
    A371094(n) = { my(m=1+3*n, e=valuation(m,2)); ((m*(2^e)) + (((4^e)-1)/3)); };
    
  • Python
    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

Formula

a(n) = A372289(A016777(n)).
a(2n) = A016777(2n) = A016921(n).

A257852 Array A read by upward antidiagonals in which the entry A(n,k) in row n and column k is defined by A(n,k) = (2^n*(6*k - 3 - 2*(-1)^n) - 1)/3, n,k >= 1.

Original entry on oeis.org

3, 1, 7, 13, 9, 11, 5, 29, 17, 15, 53, 37, 45, 25, 19, 21, 117, 69, 61, 33, 23, 213, 149, 181, 101, 77, 41, 27, 85, 469, 277, 245, 133, 93, 49, 31, 853, 597, 725, 405, 309, 165, 109, 57, 35, 341, 1877, 1109, 981, 533, 373, 197, 125, 65, 39
Offset: 1

Views

Author

L. Edson Jeffery, Jul 12 2015

Keywords

Comments

Sequence is a permutation of the odd natural numbers.
Let N_1 denote the set of odd natural numbers, and let |y|_2 denote 2-adic valuation of y. Define the map F : N_1 -> N_1 by F(x) = (3*x + 1)/2^|3*x+1|_2 (cf. A075677). Then row n of A is the set of all x in N_1 for which |3*x + 1|_2 = A371093(x) = n. Hence F(A(n,k)) = 6*k - 3 - 2*(-1)^n.

Examples

			From _Ruud H.G. van Tol_, Oct 17 2023, corrected and extended by _Antti Karttunen_, Apr 18 2024: (Start)
Array A begins:
n\k|   1|   2|   3|   4|   5|   6|   7|   8| ...
---+---------------------------------------------
1  |   3,   7,  11,  15,  19,  23,  27,  31, ...
2  |   1,   9,  17,  25,  33,  41,  49,  57, ...
3  |  13,  29,  45,  61,  77,  93, 109, 125, ...
4  |   5,  37,  69, 101, 133, 165, 197, 229, ...
5  |  53, 117, 181, 245, 309, 373, 437, 501, ...
6  |  21, 149, 277, 405, 533, 661, 789, 917, ...
... (End)
		

Crossrefs

Cf. A006370, A075677, A096773 (after its initial 0, column 1 of this array).
Cf. A004767, A017077, A082285, A238477 (rows 1-4).
Cf. A371092, A371093 (column and row indices for odd numbers).

Programs

  • Mathematica
    (* Array: *)
    Grid[Table[(2^n*(6*k - 3 - 2*(-1)^n) - 1)/3, {n, 10}, {k, 10}]]
    (* Array antidiagonals flattened: *)
    Flatten[Table[(2^(n - k + 1)*(6*k - 3 - 2*(-1)^(n - k + 1)) - 1)/ 3, {n, 10}, {k, n}]]
  • PARI
    up_to = 105;
    A257852sq(n,k) = ((2^n * (6*k - 3 - 2*(-1)^n) - 1)/3);
    A257852list(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] = A257852sq((a-(col-1)),col))); (v); };
    v257852 = A257852list(up_to);
    A257852(n) = v257852[n]; \\ Antti Karttunen, Apr 18 2024

Formula

From Ruud H.G. van Tol, Oct 17 2023: (Start)
A(n,k+1) = A(n,k) + 2^(n+1).
A(n+2,k) = A(n,k)*4 + 1.
A(1,k) = A004767(k-1).
A(2,k) = A017077(k-1).
A(3,k) = A082285(k-1).
A(4,k) = A238477(k). (End)
For all odd positive numbers n, A(A371093(n), A371092(n)) = n. - Antti Karttunen, Apr 24 2024

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

Views

Author

Antti Karttunen, May 01 2024

Keywords

Crossrefs

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:
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.

Programs

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

Formula

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)).

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

Views

Author

Antti Karttunen, Apr 28 2024

Keywords

Comments

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

Examples

			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,
		

Crossrefs

Cf. A005408 (row 1), A075677 (row 2), A372443 (column 14).
Arrays derived from this one or related to:
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.

Programs

  • Mathematica
    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 *)
  • PARI
    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];

Formula

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].

A371092 a(n) = floor((A000265(3*n+1)+5)/6), where A000265 gives the odd part of its argument.

Original entry on oeis.org

1, 1, 2, 1, 3, 1, 4, 2, 5, 2, 6, 3, 7, 1, 8, 4, 9, 3, 10, 5, 11, 1, 12, 6, 13, 4, 14, 7, 15, 2, 16, 8, 17, 5, 18, 9, 19, 2, 20, 10, 21, 6, 22, 11, 23, 3, 24, 12, 25, 7, 26, 13, 27, 1, 28, 14, 29, 8, 30, 15, 31, 4, 32, 16, 33, 9, 34, 17, 35, 3, 36, 18, 37, 10, 38, 19, 39, 5, 40, 20, 41, 11, 42, 21, 43, 1, 44, 22, 45, 12
Offset: 0

Views

Author

Antti Karttunen, Apr 19 2024

Keywords

Comments

When a(n) is applied to square array A257852 we obtain square array A002260, or in other words, a(n) applied to any odd number gives the index of the column where it is located in array A257852.

Crossrefs

Programs

  • Maple
    with(padic): A37109 := n -> floor(1/6*(3*n + 1)/2^ordp(3*n + 1, 2) + 5/6); seq(A37109(n), n = 0 .. 89); # Miles Wilson, Oct 10 2024
  • Mathematica
    A371092[n_] := With[{k = 3*n + 1}, Floor[(k/2^IntegerExponent[k, 2] + 5)/6]];
    Array[A371092, 100, 0] (* Paolo Xausa, Apr 20 2024 *)
  • PARI
    A000265(n) = (n>>valuation(n,2));
    A371092(n) = floor((A000265(1+(3*n))+5)/6);

Formula

a(n) = a(4*n + 1) = a(16*n + 5) = a(64*n + 21) = ... = a(4^k * n + ((4^k)-1)/3).

A372449 a(n) = A000523(A372444(n)); One less than the length of binary expansion of the n-th iterate of 27 with A371094.

Original entry on oeis.org

4, 7, 12, 23, 44, 84, 165, 326, 650, 1297, 2590, 5177, 10349, 20695, 41386, 82766, 165527, 331048, 662093, 1324181, 2648358, 5296712, 10593418, 21186832, 42373658, 84747311, 169494616, 338989224, 677978441, 1355956875, 2711913744, 5423827481, 10847654953, 21695309901, 43390619796, 86781239588, 173562479173, 347124958346
Offset: 0

Views

Author

Antti Karttunen, May 04 2024

Keywords

Crossrefs

Programs

Formula

a(n) = A000523(A372444(n)).
a(0) = A372447(0) = 4, and for n > 0, a(n) = A372447(n) + 2*A372448(n-1).

A372448 a(n) is the 2-adic valuation of 1 + 3*{the n-th iterate of 27 with A371094}.

Original entry on oeis.org

1, 4, 9, 19, 39, 79, 160, 322, 645, 1292, 2585, 5171, 10344, 20689, 41379, 82759, 165520, 331043, 662087, 1324175, 2648352, 5296705, 10593412, 21186825, 42373651, 84747303, 169494607, 338989215, 677978433, 1355956867, 2711913735, 5423827471, 10847654946, 21695309894, 43390619790, 86781239584, 173562479171, 347124958343
Offset: 0

Views

Author

Antti Karttunen, May 04 2024

Keywords

Crossrefs

Programs

Formula

a(n) = A371093(A372444(n)).
a(0) = 1, and for n > 0, a(n) = 2*a(n-1) + A371093(A372443(n)).

A372362 a(n) is the 2-adic valuation of 1 + 3*{the n-th iterate of 27 with Reduced Collatz-function R}.

Original entry on oeis.org

1, 2, 1, 1, 1, 1, 2, 2, 1, 2, 1, 1, 2, 1, 1, 1, 2, 3, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 3, 1, 1, 1, 4, 2, 2, 4, 3, 1, 1, 5, 4, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2
Offset: 0

Views

Author

Antti Karttunen, May 06 2024

Keywords

Crossrefs

The first 41 terms form the row 13 of A351122.

Programs

Formula

a(n) = A371093(A372443(n)).

A375961 2-adic valuation of 6*n + 2.

Original entry on oeis.org

1, 3, 1, 2, 1, 5, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 7, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 5, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 6, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 5, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 9, 1
Offset: 0

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Author

Ruud H.G. van Tol, Sep 04 2024

Keywords

Comments

6*i+2 is the first (3*x+1)/2 successor of 4*i+1, with i >= 0.
The first occurrence of odd t is before that of t-1.

Examples

			a(21) = A007814(6*21+2) = 7.
		

Crossrefs

Programs

  • Mathematica
    a[n_] := IntegerExponent[6*n + 2, 2]; Array[a, 100, 0] (* Amiram Eldar, Sep 04 2024 *)
  • PARI
    a(n) = valuation(6*n+2, 2);
    
  • Python
    def A375961(n): return (~(3*n+1)&3*n).bit_length()+1 # Chai Wah Wu, Sep 27 2024

Formula

a(n) = A007814(6*n + 2).
a(n) = A371093(n) + 1.
a(n) = A087229(n) - 1.
a(n) = k for n == A096773(k) (mod 2^k), k >= 1.
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 2. - Amiram Eldar, Oct 01 2024
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