Nico Brown - cp's OEIS frontend

A248573 An irregular triangle giving the Collatz-Terras tree.

1, 2, 4, 8, 5, 16, 3, 10, 32, 6, 20, 21, 64, 12, 13, 40, 42, 128, 24, 26, 80, 84, 85, 256, 48, 17, 52, 53, 160, 168, 170, 512, 96, 11, 34, 104, 35, 106, 320, 336, 113, 340, 341, 1024, 192, 7, 22, 68, 69, 208, 23, 70, 212, 213, 640, 672, 75, 226, 680, 227, 682, 2048

Offset: 0

Nico Brown, Oct 08 2014

From Wolfdieter Lang, Oct 31 2014: (Start)
(old name corrected)
Irregular triangle CT(l, m) such that the first three rows l = 0, 1 and 2 are 1, 2, 4, respectively, and for l >= 3 the row entries CT(l, m) are obtained from replacing the numbers of row l-1 by (2*x-1)/3, 2*x if they are 2 (mod 3) and by 2*x otherwise.
The modified Collatz (or Collatz-Terras) map sends a positive number x to x/2 if it is even and to (3*x+1)/2 if it is odd (see A060322). The present tree (without the complete tree originating at CT(2,1) = 1) can be considered as an incomplete binary tree, with nodes (vertices) of out-degree 2 if they are 2 (mod 3) and out-degree 1 otherwise. In the example below, the edges (branches) could be labeled L (left) or V (vertical).
The row length sequence is A060322(l+1), l>=0. (End)
The Collatz conjecture is true if and only if all odd numbers appear in this sequence.
This sequence is similar to A127824.

			The irregular triangle CT(l,m) begins:
l\m   1   2  3   4   5   6   7   8   9  10  11   12  13   14   15  16  17   18   19  20  21   22   23   24 ...
0:    1
1:    2
2:    4  here the 1, which would generate the complete tree again, is omitted
3:    8
4:    5  16
5:    3  10 32
6:    6  20 21  64
7:   12  13 40  42 128
8:   24  26 80  84  85 256
9:   48  17 52  53 160 168 170 512
10:  96  11 34 104  35 106 320 336 113 340 341 1024
11: 192   7 22  68  69 208  23  70 212 213 640  672  75  226  680 227 682 2048
12: 384  14 44  45 136 138 416  15  46 140 141  424 426 1280 1344 150 452  453 1360 151 454 1364 1365 4096
... reformatted, and extended - _Wolfdieter Lang_, Oct 31 2014
--------------------------------------------------------------------------------------------------------------
From _Wolfdieter Lang_, Oct 31 2014: (Start)
The Collatz-Terras tree starting with 4 looks like (numbers x == 2 (mod 3) are marked with a left bar, and the left branch ends then in (2*x-1)/3 and the vertical one in 2*x)
l=2:                                                                                        4
l=3:                                                                                       |8
l=4:                                                    |5                                 16
l=5:    3                                               10                                |32
l=6:    6                                              |20   21                            64
l=7:   12                     13                        40   42                          |128
l=8:   24                    |26                       |80   84            85             256
l=9:   48           |17       52              |53      160  168          |170            |512
l=10:  96     |11    34     |104        |35   106      320  336     |113  340      |341  1024
l=11: 192   7  22   |68  69  208   23|   70   212  213 640  672  75  226  680  227  682  2048
...
E.g., x = 7 = CT(11, 2) leads back to 4 via 7, 11, 17, 26, 13, 20, 10, 5, 8, 4, and from there back to 2, 1.
(End)
--------------------------------------------------------------------------------------------------------------

Sebastian Karlsson, Rows l = 0..35, flattened
Riho Terras, A stopping time problem on the positive integers, Acta Arith. 30 (1976) 241-252.
Eric Weisstein's World of Mathematics, Collatz Problem.

Cf. A127824, A060322, A088975.

Join[{{1}, {2}}, NestList[Flatten[Map[If[Mod[#, 3] == 2, {(2*#-1)/3, 2*#}, 2*#]&, #]]&, {4}, 10]] (* Paolo Xausa, Jan 25 2024 *)

rows(N) = my(r=List(),x); for(i=0, min(2, N), listput(r, x=[2^i])); for(n=3, N, my(w=List()); for(i=1, #x, my(q=2*x[i]); if(1==q%3, listput(w, (q-1)/3)); listput(w, q)); listput(r, x=Vec(w))); Vec(r); \\ Ruud H.G. van Tol, Jan 25 2024

Edited. New name (old corrected name as comment). - Wolfdieter Lang, Oct 31 2014

A246758 Prime numbers of the form (2^(mn)-1)/((2^m-1)(2^n-1)).

Original entry on oeis.org

3, 11, 43, 151, 683, 2731, 43691, 174763, 599479, 2796203, 715827883, 2932031007403, 10052678938039, 145295143558111, 581283643249112959, 658812288653553079, 768614336404564651, 9520972806333758431, 201487636602438195784363

Offset: 1

Nico Brown, Sep 02 2014

nonn

The sequence contains A000979 as a subsequence.

Both m and n must be prime.

			For m=3 and n=5, (2^15-1)/((2^3-1)(2^5-1))=151 is prime, so 151 is a member of the sequence.

Robert Israel, Table of n, a(n) for n = 1..39

Primes in A140803.

N:= 200: # to use all (p, q) with p*q < N
Primes:= select(isprime, [$2..floor(N/2)]):
A:= {}:
for i from 1 to nops(Primes) do
  p:= Primes[i];
  Qs:= select(q -> q < N/p, [seq(Primes[j], j=1..i-1)]);
  A:= A union {seq((2^(p*q)-1)/(2^p-1)/(2^q-1), q=Qs)};
od:
# in Maple 12 and up
select(isprime, A);
# or in earlier Maple versions
sort([select(isprime, , A); # _)[]])[];
# Robert Israel, Sep 02 2014

A225784 Denominators of the sum of the reciprocals of the Collatz (3x+1) sequence beginning at n.

Original entry on oeis.org

1, 2, 240, 4, 80, 80, 272272, 8, 350064, 80, 38896, 240, 208, 272272, 4095840, 16, 3536, 116688, 21431696, 80, 1344, 38896, 1365280, 80, 535792400, 208, 44841486948146266934850832405421294927083491752830032389039800908293040266400, 38896, 1127984, 1365280

Offset: 1

Nico Brown, May 15 2013

nonn

If the sum of the reciprocals of a Collatz sequence is bounded, there are no Collatz cycles other than 4,2,1,4,2,1,...

a(n) = denominator of Sum_{k = 1..A006577(n)} 1/A070165(n,k). - Reinhard Zumkeller, May 16 2013

			For n=9 the Collatz sequence is {9, 28, 14, 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 4, 2, 1}.  So the sum of the reciprocals is 1/9 + 1/28 + 1/14 + 1/7 + 1/22 + 1/11 + ... + 1/4 + 1/2 + 1/1 = 1061683/350064, whose denominator is 350064.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Index entries for sequences related to 3x+1 (or Collatz) problem

Cf. A225761 (numerators), A087226.

Cf. A225843.

import Data.Ratio (denominator)
a225784 = denominator . sum . map (recip . fromIntegral) . a070165_row
-- Reinhard Zumkeller, May 16 2013

Collatz[n_] := NestWhileList[If[EvenQ[#], #/2, 3 # + 1] &, n, # > 1 &]; Table[Denominator[Total[1/Collatz[n]]], {n, 40}] (* T. D. Noe, May 15 2013 *)

Extended by T. D. Noe, May 15 2013

A225761 Numerators of the sums of reciprocals of the Collatz (3x+1) sequence beginning with n and stopping at 1.

Original entry on oeis.org

1, 3, 617, 7, 171, 219, 766329, 15, 1061683, 179, 102151, 677, 497, 785777, 10380059, 31, 8861, 360377, 60226515, 183, 2731, 103919, 3339321, 229, 1548244271, 505, 129481899470258402665619129356105706380861444925035330406812603986229803685477, 113643

Offset: 1

Nico Brown, May 14 2013

nonn

If the sum of the reciprocals of a Collatz sequence is bounded, there are no cycles other than 4,2,1.

a(n) = numerator of sum (1/A070165(n,k): k = 1..A006577(n)). - Reinhard Zumkeller, May 16 2013

			For n=9 the Collatz sequence is {9, 28, 14, 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 4, 2, 1}.  So the sum of the reciprocals is 1/9 + 1/28 + 1/14 + 1/7 + 1/22 + 1/11 + ... + 1/4 + 1/2 + 1/1 = 1061683/350064, whose numerator is 1061683.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Index entries for sequences related to 3x+1 (or Collatz) problem

Cf. A087226, A225784 (denominators).

Cf. A225843.

import Data.Ratio (numerator)
a225761 = numerator . sum . map (recip . fromIntegral) . a070165_row
-- Reinhard Zumkeller, May 16 2013

Table[Numerator[Total[1/Collatz[n]]], {n, 40}] (* T. D. Noe, May 15 2013 *)

Extended by T. D. Noe, May 15 2013

A218253 The number of steps it takes, by taking the smallest number so that it minus the number of ones in its binary representation is the previous number, to reach a number which has no others that satisfy this property.

Original entry on oeis.org

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

Offset: 1

Nico Brown, Oct 24 2012

A218252 is divided into groups with length the numbers with odd position in this sequence.

			For n=7, the numbers you get are 7,8,10,12, and because this takes 3 steps, the 7th term is 3.

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Cf. A218252, A218254.

f(n)=for(k=n+1,n+log(n)\log(2)+1,if(k-hammingweight(k)==n,return(k)))
a(n)=my(k);while(n=f(n),k++);k \\ Charles R Greathouse IV, Oct 29 2012

A218254 Irregular table, where row n (n >= 0) starts with n, the next term is n-A000120(n), and the successive terms are obtained by repeatedly subtracting the number of 1's in the previous term's binary expansion, until zero is reached, after which the next row starts with one larger n.

Original entry on oeis.org

0, 1, 0, 2, 1, 0, 3, 1, 0, 4, 3, 1, 0, 5, 3, 1, 0, 6, 4, 3, 1, 0, 7, 4, 3, 1, 0, 8, 7, 4, 3, 1, 0, 9, 7, 4, 3, 1, 0, 10, 8, 7, 4, 3, 1, 0, 11, 8, 7, 4, 3, 1, 0, 12, 10, 8, 7, 4, 3, 1, 0, 13, 10, 8, 7, 4, 3, 1, 0, 14, 11, 8, 7, 4, 3, 1, 0, 15, 11, 8, 7, 4, 3, 1, 0

Offset: 0

Nico Brown, Oct 24 2012

			The n-th row (starting indexing from zero) in this irregular table consists of block of length A071542(n)+1: 1,2,3,3,4,4,5,5,... which always ends with zero, as:
0
1,0
2,1,0
3,1,0
4,3,1,0
5,3,1,0
6,4,3,1,0
7,4,3,1,0
The 17th term is 6, which in binary is 110. The 18th term is then 6-2=4.

Antti Karttunen, Rows 0..255, flattened

Cf. A218252, A218253. A213707 gives the positions of zeros (i.e. the ending index of each row). A071542, A000120.

The reversed tails of the rows converge towards A179016.

for(n=0,9,k=n;while(k, print1(k", "); k-=hammingweight(k)); print1("0, ")) \\ Charles R Greathouse IV, Oct 30 2012

A218271 a(n) = a(n-1)^a(n-2) + a(n-3) with a(0) = a(1) = a(2) = 1.

Original entry on oeis.org

1, 1, 1, 2, 3, 10, 1002, 1020180963368077455371525121027

Offset: 0

Nico Brown, Oct 24 2012

nonn

a(8) has over 30000 digits.

			a(3) = a(2)^a(1) + a(0) = 1^1+1 = 1+1 = 2.

Cf. A184163, A218270, A218122.

t = {1, 1, 1}; Do[AppendTo[t, t[[-1]]^t[[-2]] + t[[-3]]], {5}]; t (* T. D. Noe, Oct 29 2012 *)

A218252 Start with 1. For each term m, the next term is the smallest positive integer k such that k - (sum of base 2 digits of k) = m. If no such k exists, use the smallest natural number not already in the sequence.

Original entry on oeis.org

1, 2, 3, 4, 6, 5, 7, 8, 10, 12, 9, 11, 14, 13, 15, 16, 18, 20, 17, 19, 22, 24, 21, 23, 26, 30, 25, 28, 27, 29, 31, 32, 34, 36, 33, 35, 38, 40, 37, 39, 42, 46, 48, 41, 44, 43, 45, 47, 50, 54, 58, 49, 52, 51, 53, 56, 60, 55, 57, 62, 59, 61, 63, 64, 66, 68, 65, 67, 70, 72, 69

Offset: 1

Nico Brown, Oct 24 2012

nonn

The sequence is a permutation of the positive integers.

			To obtain the 2nd term, take the first, 1.  What is the smallest integer k so that k - the number of 1's in k's binary representation is 1? The answer, obviously, is 2. [A213723(1) = 2.]
There is no number that is 2 more than its binary weight [as A213723(2) = 0], therefore we just take 3 as the next term.
Following 3 we can choose either 4 or 5, but 4 is smaller, and is thus the next term of the sequence. [A213723(3) = 4.]

Antti Karttunen, Table of n, a(n) for n = 1..8192
Index entries for sequences that are permutations of the natural numbers

Inverse permutation: A257683.
Cf. A000120, A005187, A011371, A213723, A218253, A218254.
Cf. also A257676.

Name slightly edited and links to A213723 in examples added by Antti Karttunen, May 04 2015

A218270 a(0)=1, a(1)=1, a(n) = a(n-2)^a(n-1)+1.

Original entry on oeis.org

1, 1, 2, 2, 5, 33, 116415321826934814453126

Offset: 0

Nico Brown, Oct 24 2012

nonn

The next term has roughly 1.7*10^23 digits.

			a(2) = a(0)^a(1)+1 = 1+1 = 2.

Cf. A184163, A218271, A218122.

nxt[{a_,b_}]:={b,a^b+1}; Join[{1},NestList[nxt,{1,1},5][[All,2]]] (* Harvey P. Dale, Nov 25 2016 *)

a[0]:1$
a[1]:1$
a[n]:=a[n-2]^a[n-1]+1$
makelist(a[n],n,1,6); /* Martin Ettl, Oct 29 2012 */

A218219 Define a(x,y) to be 1 if x is 0 or 1 and y*a(x-1,y)-a(x-2,y) otherwise. Then the n-th term of the sequence is a(n,n).

Original entry on oeis.org

1, 1, 1, 5, 41, 436, 5741, 90481, 1663585, 34988311, 828931049, 21851881930, 634556225161, 20129592507025, 692665874901013, 25699370092119569, 1022757988416562049, 43461563755791470416, 1964206882303435582865, 94076863910519354420761, 4760163616791818063701801

Offset: 0

Nico Brown, Oct 23 2012

nonn

The sequence grows about as fast as x!*e^x/p(x), where p(x) is a quadratic.

			For 3, by definition, a(0,3)=1 and a(1,3)=1, so a(2,3)=3*a(1,3)-a(0,3)=3-1=2 and a(3,3)=3*a(2,3)-a(1,3)=6-1=5, so the third element is 5

Cf. A218220.

cp's OEIS Frontend

User: Nico Brown

A248573 An irregular triangle giving the Collatz-Terras tree.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Extensions

A246758 Prime numbers of the form (2^(m*n)-1)/((2^m-1)*(2^n-1)).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

A225784 Denominators of the sum of the reciprocals of the Collatz (3x+1) sequence beginning at n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Haskell

Mathematica

Extensions

A225761 Numerators of the sums of reciprocals of the Collatz (3x+1) sequence beginning with n and stopping at 1.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Haskell

Mathematica

Extensions

A218253 The number of steps it takes, by taking the smallest number so that it minus the number of ones in its binary representation is the previous number, to reach a number which has no others that satisfy this property.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

A218254 Irregular table, where row n (n >= 0) starts with n, the next term is n-A000120(n), and the successive terms are obtained by repeatedly subtracting the number of 1's in the previous term's binary expansion, until zero is reached, after which the next row starts with one larger n.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

PARI

A218271 a(n) = a(n-1)^a(n-2) + a(n-3) with a(0) = a(1) = a(2) = 1.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

A218252 Start with 1. For each term m, the next term is the smallest positive integer k such that k - (sum of base 2 digits of k) = m. If no such k exists, use the smallest natural number not already in the sequence.

Views

Author

Keywords

A246758 Prime numbers of the form (2^(mn)-1)/((2^m-1)(2^n-1)).