A116640 -id:A116640 - cp's OEIS frontend

A274048 a(n) = A116640(A018900(n)) = A116623(A059893(A018900(n))).

19, 23, 29, 31, 37, 49, 47, 53, 65, 89, 79, 85, 97, 121, 169, 143, 149, 161, 185, 233, 329, 271, 277, 289, 313, 361, 457, 649, 527, 533, 545, 569, 617, 713, 905, 1289, 1039, 1045, 1057, 1081, 1129, 1225, 1417, 1801

Offset: 1

Joe Slater, Jun 07 2016

A subset of A116640 containing all terms A116640(m) such that m has binary weight of 2. This sequence is related to the Collatz and Terras trajectories; specifically those trajectories that include three odd numbers besides 1.

			The first three terms of A018900 are 3,5,6. Taking these terms from A116640 gives 19,23,29, which are the first three terms of this sequence. The sequence is generated from the regular triangle
1;
1,2;
1,2,3;
etc., so the first three terms are
2^(1+1) + 3*(3+2^1) = 19;
2^(2+1) + 3*(3+2^1) = 23;
2^(2+1) + 3*(3+2^2)= 29.

Subsequence of A116640.

Cf. A116623.

a(n) = 2^(Row(n)+1) + 3*(3+2^Col(n)) where Row(n) = A002024(n) = the row position of n when the sequence is viewed as a regular triangle; and Col(n) = A002260(n) = the column position of n when the sequence is viewed as a regular triangle.

A116623 a(0)=1, a(2n) = a(n)+A000079(A000523(2n)), a(2n+1) = 3*a(n) + A000079(A000523(2n+1)+1).

Original entry on oeis.org

1, 5, 7, 19, 11, 29, 23, 65, 19, 49, 37, 103, 31, 85, 73, 211, 35, 89, 65, 179, 53, 143, 119, 341, 47, 125, 101, 287, 89, 251, 227, 665, 67, 169, 121, 331, 97, 259, 211, 601, 85, 223, 175, 493, 151, 421, 373, 1087, 79, 205, 157, 439, 133, 367, 319, 925, 121

Offset: 0

Antti Karttunen, Feb 20 2006. Proposed by Pierre Lamothe (plamothe(AT)aei.ca), May 21 2004

Viewed as a binary tree, this is (1); 5; 7,19; 11,29,23,65; ... Related to the parity vectors of Collatz and Terras trajectories.

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

Cf. a(n) = A116640(A059893(n)). a(A000225(n)) = A001047(n+1). For n>= 1 a(A000079(n)) = A062709(n+1). A116641 gives the terms in ascending order and without duplicates.

A116623 := proc(n)
    option remember;
    if n = 0 then
        1;
    elif type(n,'even') then
        procname(n/2)+2^A000523(n) ;
    else
        3*procname(floor(n/2))+2^(1+A000523(n)) ;
    end if;
end proc: # R. J. Mathar, Nov 28 2016

a[n_] := a[n] = Which[n == 0, 1, EvenQ[n], a[n/2] + 2^Floor@Log2[n], True, 3a[Floor[n/2]] + 2^(1 + Floor@Log2[n])];
Table[a[n], {n, 0, 56}] (* Jean-François Alcover, Sep 01 2023 *)

A119733 Offsets of the terms of the nodes of the reverse Collatz function.

Original entry on oeis.org

0, 1, 2, 5, 4, 7, 10, 19, 8, 11, 14, 23, 20, 29, 38, 65, 16, 19, 22, 31, 28, 37, 46, 73, 40, 49, 58, 85, 76, 103, 130, 211, 32, 35, 38, 47, 44, 53, 62, 89, 56, 65, 74, 101, 92, 119, 146, 227, 80, 89, 98, 125, 116, 143, 170, 251, 152, 179, 206, 287, 260, 341, 422, 665, 64, 67

Offset: 0

William Entriken, Jun 14 2006

nonn

Create a binary tree starting with x. To follow 0 from the root, apply f(x)=2x. To follow 1, apply g(x)=(2x-1)/3. For example, starting with x, the string 010 {also known as f(g(f(x)))}, you would get (8x-2)/3. These expressions represent the reverse Collatz function and will provide numbers whose Collatz path may include x. These expressions will all be of the form (2^a*x-b)/3^c. This sequence concerns b. What makes b interesting is that if you draw the tree, each level of the tree will have the same sequence of values for b. The root of the tree x, can be written as (2^0*x-0)/3^0, which has the first value for b. Each subsequent level contains twice as many values of b.
This sequence is 0 followed by a permutation of A213539, and therefore consists of 0 plus the elements of A116640 multiplied by 2^k, where k >= 0. E.g., 1, 5, 7, 19 becomes 0, 2^0*1, 2^1*1, 2^0*5, 2^2*1, 2^0*7, 2^1*5, 2^0*19, ... - Joe Slater, Dec 19 2016
When this sequence is arranged as an irregular triangle the sum of each row a(2^k)...a(2^(k+1)-1) equals A081039(2^k). The cumulative sum from a(0) to a(2^k-1) equals A002697(k). - Joe Slater, Apr 12 2018

			a(1) = 1 = 2 * 0 + 3^0 since 0 written in binary contains no 1's.

Alois P. Heinz, Table of n, a(n) for n = 0..16383
Víctor Martín Chabrera, An algebraic fractal approach to Collatz Conjecture, Bachelor tesis, Universitat Politècnica de Catalunya (Barcelona, 2019), see lemma 6.1 with a(n) = r(A005836(n)).

Cf. A002697, A081039, A116623, A116640, A116641, A213539, A226383.

a:= proc(n) `if`(n=0, 0, `if`(irem(n, 2, 'r')=0, 0,
      3^add(i, i=convert(r, base, 2)))+2*a(r))
    end:
seq(a(n), n=0..127);  # Alois P. Heinz, Aug 13 2017

a[0] := 0; a[n_?OddQ] := 2a[(n - 1)/2] + 3^Plus@@IntegerDigits[(n - 1)/2, 2]; a[n_?EvenQ] := 2a[n/2]; Table[a[n], {n, 0, 65}] (* Alonso del Arte, Apr 21 2011 *)

a(n) = my(ret=0); if(n, for(i=0,logint(n,2), if(bittest(n,i), ret=3*ret+1<Kevin Ryde, Oct 22 2021

# call with n to get 2^n values
$depth=shift; sub funct { my ($i, $b, $c) = @_; if ($i < $depth) { funct($i+1, $b*2, $c); funct($i+1, 2*$b+$c, $c*3); } else { print "$b, "; } } funct(0, 0, 1); print " ";

from sympy.core.cache import cacheit
@cacheit
def a(n): return 0 if n==0 else 2*a((n - 1)//2) + 3**bin((n - 1)//2).count('1') if n%2 else 2*a(n//2)
print([a(n) for n in range(131)]) # Indranil Ghosh, Aug 13 2017

a(0) = 0, a(2*n + 1) = 2*a(n) + 3^wt(n) = 2*a(n) + A048883(n), a(2*n) = 2*a(n), where wt(n) = A000120(n) = the number 1's in the binary representation of n.
a(k) = [z^k] 1 + (1/(1-z)) * Sum_{s=0..n-1} 2^s*z^(2^s)*(1 - z^(2^s)) * Product_{r=s+1..n-1} (1 + 3*z^(2^r)), for 0 < k <= 2^n-1. - Wolfgang Hintze, Jul 28 2017
a(n) = Sum_{i=0..k} 2^e[i] * 3^i where binary expansion n = 2^e[0] + 2^e[1] + ... + 2^e[k] with descending e[0] > e[1] > ... > e[k] (A272011). [Martín Chabrera lemma 6.1, adapting index i] - Kevin Ryde, Oct 22 2021

A213539 Variant of numbers for which there is at least one 3-smooth representation that is special of level k.

Original entry on oeis.org

1, 2, 4, 5, 7, 8, 10, 11, 14, 16, 19, 20, 22, 23, 28, 29, 31, 32, 35, 37, 38, 40, 44, 46, 47, 49, 53, 56, 58, 62, 64, 65, 67, 70, 73, 74, 76, 79, 80, 85, 88, 89, 92, 94, 97, 98, 101, 103, 106, 112, 116, 119, 121, 124, 125, 128, 130, 131, 133, 134, 140, 143, 146

Offset: 0

Kenneth Vollmar, Mar 03 2013

nonn

These numbers are of the form 3^k*2^{a_0} + 3^{k-1}*2^{a_1} + ... + 3^1*2^{a_{k-1}} + 3^0*2^{a_k} in which every power 3^i appears, 0 <= i <= k, and where a_i satisfies 0 <= a_0 < a_1 < ... < a_k.

These values are those of sequence A116640 in addition to any multiple of two of elements of this sequence. - Kenneth Vollmar, Jun 05 2013

			n=19 has two 3-smooth representations that are special of level k. At k=1, 19 = 3^1*2^0 + 3^0*2^4. At k=2, 19 = 3^2*2^0 + 3^1*2^1 + 3^0*2^2.

Kenneth Vollmar, Recursive calculation of 3-smooth representations special of level k, To be submitted mid-2013.

R. Blecksmith, M. McCallum and J. L. Selfridge, 3-smooth representations of integers, Amer. Math. Monthly, 105 (1998), 529-543.

Cf. A116623, A116640, A116641, A119733, A226383, A003586.

Corrected a reference to another sequence and added cross references - Joe Slater, Dec 19 2016

A292044 Wiener index of the n-halved cube graph.

Original entry on oeis.org

0, 1, 6, 32, 160, 768, 3584, 16384, 73728, 327680, 1441792, 6291456, 27262976, 117440512, 503316480, 2147483648, 9126805504, 38654705664, 163208757248, 687194767360, 2886218022912, 12094627905536, 50577534877696, 211106232532992, 879609302220800, 3659174697238528

Offset: 1

Eric W. Weisstein, Sep 08 2017

a(n) is the sum of the first 2^(n-1) entries of A116640. - Joe Slater, Apr 11 2018

Eric Weisstein's World of Mathematics, Halved Cube Graph.
Eric Weisstein's World of Mathematics, Wiener Index.
Index entries for linear recurrences with constant coefficients, signature (8,-16).

Cf. A116640.

Table[If[n == 1, 0, 2^(2 n - 5) n], {n, 40}]
Join[{0}, LinearRecurrence[{8, -16}, {1, 6}, 20]]
CoefficientList[Series[((1 - 2 x) x)/(1 - 4 x)^2, {x, 0, 20}], x]

a(n) = if(n<2, n-1, 2^(2*n-5)*n); \\ Altug Alkan, Apr 12 2018

a(n) = 2^(2*n-5)*n for n > 1.
a(n) = 8*a(n-1) - 16*a(n-2) for n > 3.
G.f.: ((1 - 2 x) x^2)/(1 - 4 x)^2.
a(n) = 4*a(n-1) + 2^(2*n-5) for n > 2. - Joe Slater, Apr 11 2018
From Amiram Eldar, Apr 16 2022: (Start)
Sum_{n>=2} 1/a(n) = 32*log(4/3) - 8.
Sum_{n>=2} (-1)^n/a(n) = 8 - 32*log(5/4). (End)
From Stefano Spezia, Aug 04 2022: (Start)
E.g.f.: (exp(4*x) - 1)*x/8.
a(n) = (-1)^n*det(M(n-1))/2 for n > 1, where M(n) is the n X n symmetric Toeplitz matrix whose first row consists of 2, 4, ..., 2*n. (End)

cp's OEIS Frontend

A274048 a(n) = A116640(A018900(n)) = A116623(A059893(A018900(n))).

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A116623 a(0)=1, a(2n) = a(n)+A000079(A000523(2n)), a(2n+1) = 3*a(n) + A000079(A000523(2n+1)+1).

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A119733 Offsets of the terms of the nodes of the reverse Collatz function.

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A213539 Variant of numbers for which there is at least one 3-smooth representation that is special of level k.

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A292044 Wiener index of the n-halved cube graph.

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Mathematica

PARI

Formula