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.

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A037084 Positive integers not going to 1 under iterations of the map in A001281: n->3n-1 if n odd, n->n/2 if n even.

Original entry on oeis.org

5, 7, 9, 10, 13, 14, 17, 18, 19, 20, 21, 23, 25, 26, 27, 28, 31, 33, 34, 35, 36, 37, 38, 40, 41, 42, 45, 46, 47, 49, 50, 51, 52, 54, 55, 56, 61, 62, 63, 66, 67, 68, 70, 72, 73, 74, 75, 76, 80, 81, 82, 83, 84, 89, 90, 91, 92, 93, 94, 98, 99, 100, 102
Offset: 1

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Author

Robert W. Craigen (craigen(AT)fresno.edu)

Keywords

Comments

Up to at least 100000000, every number reaches 1, 5 or 17.
Conjecture : for any x, the iterated process "x ->3x-1" if x is odd or "x ->x/2" if x is even leads to one of the following three cycles: (1, 2), (5, 14, 7, 20, 10), (41, 122, 61, 182, 91, 272, 136, 68, 34, 17, 50, 25, 74, 37, 110, 55, 164, 82). - Benoit Cloitre, May 14 2002
Complement (in N*) of A039500 ; union of A039501 and A039502 (conjectured). - M. F. Hasler, Nov 26 2007
Equivalent to the Collatz ('3n+1') problem for negative integers. - Dmitry Kamenetsky, Jan 12 2017

Examples

			Iterations of f starting at 3 are 3,8,4,2,1 - thus 3 is not in the sequence.
Iterations starting at 5 are 5,14,7,20,10,5 -periodic and 1 is not among these values, so 5 is in the sequence.
		

Crossrefs

Cf. A006370, A006577 (Collatz problem: 3n+1).

Programs

  • Mathematica
    colln[n_]:= NestWhile[If[EvenQ[#], #/2, 3#-1] &, n, FreeQ[{1, 5, 17}, #] &]; Select[Range[102], colln[#] != 1 &] (* Jayanta Basu, Jun 06 2013 *)
  • PARI
    A037084( end=999, n=0 /*starting value -1 */)={ for( i=n,end, n=i; while( n > 17 || n > 5 && n < 17, if( n%2, n=3*n-1, n>>=1)); if( n > 4, print1(i", ")))} \\ M. F. Hasler, Nov 26 2007

Extensions

More terms from Christian G. Bower, Feb 15 1999
Edited by M. F. Hasler, Nov 26 2007

A060445 "Dropping time" in 3x+1 problem starting at 2n+1 (number of steps to reach a lower number than starting value). Also called glide(2n+1).

Original entry on oeis.org

0, 6, 3, 11, 3, 8, 3, 11, 3, 6, 3, 8, 3, 96, 3, 91, 3, 6, 3, 13, 3, 8, 3, 88, 3, 6, 3, 8, 3, 11, 3, 88, 3, 6, 3, 83, 3, 8, 3, 13, 3, 6, 3, 8, 3, 73, 3, 13, 3, 6, 3, 68, 3, 8, 3, 50, 3, 6, 3, 8, 3, 13, 3, 24, 3, 6, 3, 11, 3, 8, 3, 11, 3, 6, 3, 8, 3, 65, 3, 34, 3, 6, 3, 47, 3, 8, 3, 13, 3, 6, 3, 8, 3
Offset: 0

Views

Author

N. J. A. Sloane, Apr 07 2001

Keywords

Comments

If the starting value is even then of course the next step in the trajectory is smaller (cf. A102419).
The dropping time can be made arbitrarily large: If the starting value is of form n(2^m)-1 and m > 1, the next value is 3n(2^m)-3+1. That divided by 2 is 3n(2^(m-1))-1. It is bigger than the starting value and of the same form - substitute 3n -> n and m-1 -> m, so recursively get an increasing subsequence of m odd values. The dropping time is obviously longer than that. This holds even if Collatz conjecture were refuted. For example, m=5, n=3 -> 95, 286, 143, 430, 215, 646, 323, 970, 485, 1456, 728, 364, 182, 91. So the subsequence in reduced Collatz variant is 95, 143, 215, 323, 485. - Juhani Heino, Jul 21 2017

Examples

			3 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2, taking 6 steps, so a(1) = 6.
		

Crossrefs

A060565 gives the first lower number that is reached. Cf. A060412-A060415, A217934.
See A074473, A102419 for other versions of this sequence.
Cf. A122437 (allowable dropping times), A122442 (least k having dropping time A122437(n)).
Cf. A070165.

Programs

  • Haskell
    a060445 0 = 0
    a060445 n = length $ takeWhile (>= n') $ a070165_row n'
                where n' = 2 * n + 1
    -- Reinhard Zumkeller, Mar 11 2013
    
  • Mathematica
    nxt[n_]:=If[OddQ[n],3n+1,n/2]; Join[{0},Table[Length[NestWhileList[nxt, n,#>=n&]]-1, {n,3,191,2}]]  (* Harvey P. Dale, Apr 23 2011 *)
  • Python
    def a(n):
        if n<1: return 0
        n=2*n + 1
        N=n
        x=0
        while True:
            if n%2==0: n//=2
            else: n = 3*n + 1
            x+=1
            if nIndranil Ghosh, Apr 22 2017

Extensions

More terms from Jason Earls, Apr 08 2001 and from Michel ten Voorde Apr 09 2001
Still more terms from Larry Reeves (larryr(AT)acm.org), Apr 12 2001

A208981 Number of iterations required to reach a power of 2 in the 3x+1 sequence starting at n.

Original entry on oeis.org

0, 0, 3, 0, 1, 4, 12, 0, 15, 2, 10, 5, 5, 13, 13, 0, 8, 16, 16, 3, 1, 11, 11, 6, 19, 6, 107, 14, 14, 14, 102, 0, 22, 9, 9, 17, 17, 17, 30, 4, 105, 2, 25, 12, 12, 12, 100, 7, 20, 20, 20, 7, 7, 108, 108, 15, 28, 15, 28, 15, 15, 103, 103, 0, 23, 23, 23, 10, 10, 10
Offset: 1

Views

Author

L. Edson Jeffery, Mar 04 2012

Keywords

Comments

The original name was: Number of iterations of the Collatz recursion required to reach a power of 2.
The statement that all paths must eventually reach a power of 2 is equivalent to the Collatz conjecture.
A006577(n) - a(n) gives the exponent for the first power of 2 reached in the Collatz trajectory of n. - Alonso del Arte, Mar 05 2012
Number of nonpowers of 2 in the 3x+1 sequence starting at n. - Omar E. Pol, Sep 05 2021

Examples

			a(7) = 12 because the Collatz trajectory for 7 is 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1, 4, 2, 1, ... which reached 16 = 2^4 in 12 steps.
		

Crossrefs

Row sums of A347519.
Cf. A006577 (and references therein).
Cf. A347270 (gives all 3x+1 sequences).

Programs

  • Haskell
    a208981 = length . takeWhile ((== 0) . a209229) . a070165_row
    -- Reinhard Zumkeller, Jan 02 2013
    
  • Maple
    a:= proc(n) option remember; `if`(n=2^ilog2(n), 0,
          1+a(`if`(n::odd, 3*n+1, n/2)))
        end:
    seq(a(n), n=1..100);  # Alois P. Heinz, Sep 05 2021
  • Mathematica
    Collatz[n_?OddQ] := 3*n + 1; Collatz[n_?EvenQ] := n/2; Table[-1 + Length[NestWhileList[Collatz, n, Not[IntegerQ[Log[2, #]]] &]], {n, 50}] (* Alonso del Arte, Mar 04 2012 *)
  • PARI
    ispow2(n)=n>>=valuation(n,2); n==1
    a(n)=my(s); while(!ispow2(n), n=if(n%2, 3*n+1, n/2); s++); s \\ Charles R Greathouse IV, Jul 31 2016

Formula

For x>0 an integer, define f_0(x)=x, and for r=1,2,..., f_r(x)=f_{r-1}(x)/2 if f_{r-1}(x) is even, else f_r(x)=3*f_{r-1}(x)+1. Then a(n) = min(k such that f_k(n) is equal to a power of 2).
a(n) = A006577(n) - A135282(n) (after Alonso del Arte's comment), if A006577(n) is not -1. - Omar E. Pol, Apr 10 2022

Extensions

Name clarified by Omar E. Pol, Apr 10 2022

A006584 If n mod 2 = 0 then n*(n^2-4)/12 else n*(n^2-1)/12.

Original entry on oeis.org

0, 0, 0, 2, 4, 10, 16, 28, 40, 60, 80, 110, 140, 182, 224, 280, 336, 408, 480, 570, 660, 770, 880, 1012, 1144, 1300, 1456, 1638, 1820, 2030, 2240, 2480, 2720, 2992, 3264, 3570, 3876, 4218, 4560, 4940, 5320
Offset: 0

Views

Author

Keywords

Comments

Graded dimension of L''/[L',L''] for the free Lie algebra on 2 generators. Let L be a free Lie algebra with 2 generators graded by the total degree. Set L'=[L,L] and L''=[L',L']. Then a(n) is equal to the dimension of the homogeneous subspace of degree n+2 in the quotient L''/[L',L'']. - Sergei Duzhin, Mar 15 2004
Also the 2nd Witt transform of A000027. - R. J. Mathar, Nov 08 2008
Also the number of 3-element subsets of {1..n+1} whose elements sum up to an odd integer, i.e., the third column of A159916: e.g. a(3)=2 corresponds to the two subsets {1,2,4} and {2,3,4} of {1..4}. - M. F. Hasler, May 01 2009
The set of magic numbers for an idealized harmonic oscillator nucleus with a biaxially deformed prolate ellipsoid shape and an oscillator ratio of 2:1. - Jess Tauber, May 13 2013
Quasipolynomial of order 2. - Charles R Greathouse IV, May 14 2013

References

  • W. A. Whitworth, DCC Exercises in Choice and Chance, Stechert, NY, 1945, p. 33.

Crossrefs

Partial sums of A110660.

Programs

Formula

a(n+3) = A003451(n) + A027656(n). - Yosu Yurramendi, Aug 07 2008
G.f.: 2*x^3/((1-x)^4*(1+x)^2). a(n) = 2*A006918(n-2). - R. J. Mathar, Nov 08 2008
a(n) = 2*a(n-1)+a(n-2)-4*a(n-3)+a(n-4)+2*a(n-5)-a(n-6). - Jaume Oliver Lafont, Dec 05 2008
a(n) = n*(2*n^2-5-3*(-1)^n)/24. - Luce ETIENNE, Apr 03 2015
a(n) = Sum_{i=1..n} floor(i*(n-i)/2). - Wesley Ivan Hurt, May 07 2016
E.g.f.: x*(x*(x + 3)*exp(x) - 3*sinh(x))/12. - Ilya Gutkovskiy, May 08 2016
Sum_{n>=3} 1/a(n) = 75/8 - 12*log(2). - Amiram Eldar, Sep 17 2022

A135730 Number of steps to reach the minimum of the final cycle under iterations of the map A001281: x->3x-1 if x odd, x/2 otherwise.

Original entry on oeis.org

0, 1, 4, 2, 0, 5, 3, 3, 11, 1, 6, 6, 9, 4, 9, 4, 0, 12, 7, 2, 8, 7, 3, 7, 16, 10, 5, 5, 10, 10, 6, 5, 19, 1, 13, 13, 14, 8, 13, 3, 9, 9, 8, 8, 22, 4, 16, 8, 17, 17, 11, 11, 16, 6, 12, 6, 29, 11, 11, 11, 7, 7, 19, 6, 37, 20, 20, 2, 19, 14, 19, 14, 15, 15, 9, 9, 14, 14, 14, 4
Offset: 1

Views

Author

M. F. Hasler, Nov 26 2007

Keywords

Comments

Under iterations of the map A001281, the orbit of any positive integer seems to end in one of 3 possible cycles, having 1, 5, resp. 17 as smallest element. This sequence gives the number of iterations needed to reach one of these values. Another sequence that could be considered is the number of iterations needed to reach /any/ element of the final cycle.
From N. J. A. Sloane, Sep 04 2015: (Start)
The same sequence arises as follows: Start at 2n-1 and repeatedly apply the map (see A261671): subtract 1 and divide by 2 if the result is odd, otherwise multiply by 3; a(n) is the number of steps to reach one of 1, 9, or 33.
It is conjectured that the trajectory of any odd number will eventually reach 1, 9, or 33, and so enter one of the loops (1,3), (9, 27, 13, 39, 19), or (33, 99, 49, 147, 73, 219, 109, 327, 163, 81, 243, 121, 363, 181, 543, 271, 135, 67). (End)

Crossrefs

Cf. A001281, A037084, A039500-A039505, A135727-A135729. A006370, A006577 (Collatz 3x+1 problem).
Cf. also A261671.
See A261673 and A261674 for records.

Programs

  • PARI
    A135730(n)=local(c=0);while( n>17 || n != 17 && n != 5 && n != 1, c++; if( n%2, n=3*n-1,n>>=1));c

A339991 The number of steps that n requires to reach 1 under the map: m -> m/2 if m is even, m-> m^2 - 1 if m is an odd prime, otherwise m -> m - 1. a(n) = -1 if 1 is never reached.

Original entry on oeis.org

0, 1, 4, 2, 8, 5, 9, 3, 4, 9, 15, 6, 15, 10, 11, 4, 10, 5, 22, 10, 11, 16, 11, 7, 8, 16, 17, 11, 23, 12, 18, 5, 6, 11, 12, 6, 20, 23, 24, 11, 24, 12, 21, 17, 18, 12, 19, 8, 9, 9, 10, 17, 31, 18, 19, 12, 13, 24, 27, 13, 32, 19, 20, 6, 7, 7, 21, 12, 13, 13, 27
Offset: 1

Views

Author

Ya-Ping Lu, Dec 25 2020

Keywords

Comments

Conjecture: a(n) is never equal to -1.
An even node (m) in the tree shown in Example can have up to three predecessors: 2*m, sqrt(m+1) if sqrt(m+1) is a prime, and m+1 if m+1 is a nonprime odd number. An odd node has only one predecessor: 2*m.

Examples

			The 39 starting numbers with a(n) <= 9 are given in the figure below.
10 50 7 49 96 145 288 133 264 260 258 512
  \  \ \ | /   \  /    \ /    /   /   /
   5 25 48     144     132  130  129 256
    \ | /        \       \    \   \ /
     24          72      66  65  128
       \          \       \   \  /
        12        36      33   64
          \        \       \  /
           6       18       32
             \      \      /
               3    9   16
                  \ | /
                    8
                    |
                    4
                    |
                    2
                    |
                    1
		

Crossrefs

Programs

  • Maple
    A339991 := proc(n)
        local a,x;
        x := n ;
        a := 0 ;
        while x > 1 do
            if type(x,even) then
                x := x/2 ;
            elif isprime(x) then
                x := x^2-1 ;
            else
                x := x-1 ;
            end if ;
            a := a+1 ;
        end do:
        a ;
    end proc:
    seq(A339991(n),n=1..50) ; # R. J. Mathar, Jun 27 2024
  • Mathematica
    Array[-1 + Length@ NestWhileList[Which[EvenQ@ #, #/2, PrimeQ@ #, #^2 - 1, True, # - 1] &, #, # > 1 &] &, 71] (* Michael De Vlieger, Dec 28 2020 *)
  • PARI
    f(n) = if (n%2, if (isprime(n), n^2-1, n-1), n/2);
    a(n) = my(nb=0); while (n != 1, n = f(n); nb++); nb; \\ Michel Marcus, Dec 26 2020
  • Python
    from sympy import isprime
    for n in range(1, 1001):
        ct, m = 0, n
        while m > 1:
            if m%2 == 0: m //= 2
            elif isprime(m) == 1: m = m*m - 1
            else: m -= 1
            ct += 1
        print(ct)
    

A177729 Positive integers which do not appear in a Collatz sequence starting from a smaller positive integer.

Original entry on oeis.org

1, 2, 3, 6, 7, 9, 12, 15, 18, 19, 21, 24, 25, 27, 30, 33, 36, 37, 39, 42, 43, 45, 48, 51, 54, 55, 57, 60, 63, 66, 69, 72, 73, 75, 78, 79, 81, 84, 87, 90, 93, 96, 97, 99, 102, 105, 108, 109, 111, 114, 115, 117, 120, 123, 126, 127, 129, 132, 133, 135, 138, 141
Offset: 1

Views

Author

Raul D. Miller, May 12 2010

Keywords

Comments

A variant of A061641, which is the main entry for this sequence.
The inclusion of 2 is apparently due to a non-standard definition of a Collatz sequence; A177729 assumes that the Collatz sequence ends when it reaches 1, whereas the standard definition includes the periodic 1,4,2,... from that point. The inclusion of 0 in A061641 is a bit odd, but is not actually wrong. One usually looks only at positive integers for Collatz sequences. - Franklin T. Adams-Watters, May 14 2010

Examples

			Collatz 1: 1; Collatz 2: 2,1; Collatz 3: 3,10,5,16,8,4,2,1; Collatz 6: 6,3,10,...
		

Crossrefs

Programs

  • Haskell
    a177729 = head . a192719_row  -- Reinhard Zumkeller, Jan 03 2013
  • Mathematica
    coll[n_]:=NestWhileList[If[EvenQ[#],#/2,3#+1]&,n,#>1&]; t={1}; Do[If[FreeQ[Union@@Table[coll[i],{i,n-1}],n],AppendTo[t,n]],{n,2,141}]; t (* Jayanta Basu, May 29 2013 *)

Formula

a(n) = A192719(n,1), see also A220263. - Reinhard Zumkeller, Jan 03 2013

A138753 Number of iterations of A138754 before reaching a number for the second time, when starting with n.

Original entry on oeis.org

1, 4, 5, 3, 3, 5, 3, 8, 6, 4, 21, 17, 7, 7, 5, 5, 22, 24, 20, 18, 18, 16, 8, 6, 8, 6, 29, 23, 27, 23, 23, 21, 19, 19, 17, 21, 17, 15, 7, 7, 9, 60, 9, 9, 7, 30, 28, 26, 24, 26, 24, 24, 28, 24, 22, 20, 20, 22, 20, 18, 20, 18, 20, 18, 18, 16, 14, 12, 10, 12, 10, 61, 59, 55, 12, 10, 8, 31
Offset: 1

Views

Author

M. F. Hasler, Apr 01 2008

Keywords

Comments

This is a variation of A138752, giving the number of iterations of A138754 needed to get any number for the second time, while A138752 stops counting somehow arbitrarily at 1=primepi(2) or 4=primepi(7).
The map A138754 is a variation of the Collatz map where parity of the integers is replaced by p mod 3 for the primes.
For the Collatz map, we have the only fixed point 0=f(0) and all other numbers seem to end up in the cycle 1->4->2->1.
Here the only fixed point is 1=A138754(1) and all other numbers seem to end up in the cycle 4 -> 7 -> 5 -> 4 (corresponding to primes 7 -> 17 -> 11 -> 7).
Depending on which number among primepi({2,7,11,17}) is reached first, A138753(n) = A138752(n)+1,+3,+2 resp. +1. (A138752(n) is the length of the so-called GB-sequence starting with prime(n).)

Examples

			a(1)=1 since after 1 step we find 1 again.
a(4)=3 since 4 -> 7 -> 5 -> 4 under A138754.
		

Crossrefs

Cf. A124123, A006577, A171938, A138756 (record values/indices).

Programs

Formula

a(n) = min { k>0 | A138754^k(n) = A138754^m(n) for some m>=0, m
If n is not in {1,4,5,7}, then a(n) = 1+a(A138754(n)).

A139399 Number of steps to reach a cycle in Collatz problem.

Original entry on oeis.org

0, 0, 5, 0, 3, 6, 14, 1, 17, 4, 12, 7, 7, 15, 15, 2, 10, 18, 18, 5, 5, 13, 13, 8, 21, 8, 109, 16, 16, 16, 104, 3, 24, 11, 11, 19, 19, 19, 32, 6, 107, 6, 27, 14, 14, 14, 102, 9, 22, 22, 22, 9, 9, 110, 110, 17, 30, 17, 30, 17, 17, 105, 105, 4, 25, 25, 25, 12, 12, 12, 100, 20, 113, 20
Offset: 1

Author

Reinhard Zumkeller, Apr 18 2008

Keywords

Comments

a(1)=a(2)=a(4)=0 as A006370(A006370(A006370(x)))=x for x=1,2,4 [corrected by Rémy Sigrist, Jun 28 2020];
a(n) = A006577(n) - 2 for n > 2 (if the conjecture holds).
For n>2: let L = a(n) mod 3, then A006460(n) = if L=0 then 4 else L. - Reinhard Zumkeller, Nov 17 2013

Crossrefs

Essentially the same sequence as A112695.

Programs

  • Haskell
    a139399 = f 0 where
       f k x = if x `elem` [1,2,4] then k else f (k + 1) (a006370 x)
    -- Reinhard Zumkeller, Nov 17 2013
  • Mathematica
    f[n_] := If[EvenQ[n], n/2, 3 n + 1];
    a[n_] := If[n<3, 0, Length[NestWhileList[f, n, {#1, #2, #3} != {4, 2, 1}&, 3]] - 3];
    Table[a[n], {n, 1, 100}] (* Jean-François Alcover, Aug 08 2022 *)

A238475 Rectangular array with all start numbers Me(n, k), k >= 1, for the Collatz operation ud^(2*n), n >= 1, ending in an odd number, read by antidiagonals.

Original entry on oeis.org

1, 9, 5, 17, 37, 21, 25, 69, 149, 85, 33, 101, 277, 597, 341, 41, 133, 405, 1109, 2389, 1365, 49, 165, 533, 1621, 4437, 9557, 5461, 57, 197, 661, 2133, 6485, 17749, 38229, 21845, 65, 229, 789, 2645, 8533, 25941, 70997, 152917, 87381
Offset: 1

Author

Wolfdieter Lang, Mar 10 2014

Keywords

Comments

The two operations on natural numbers m used in the Collatz 3x+1 conjecture (see the links) are here (following the M. Trümper reference) denoted by u for 'up' and d for 'down': u m = 3*m+1, if m is odd, and d m = m/2 if m is even. The present array gives all positive start numbers Me(n, k), k >= 1, for Collatz sequences following the pattern (word) ud^(2*n), for n >= 1, which end in an odd number. The end number does not depend on n and it is given by Ne(k) = 6*k - 5.
This rectangular array is Example 2.1. with x = 2*n, n >= 1, of the M. Trümper reference, pp. 4-5, written as a triangle by taking NE-SW diagonals. The case x = 2*n+1, n >= 0, for the word ud^(2*k+1) appears as array and triangle in A238476.
The first row sequences of the array Me (they become columns in the triangle Te) are A017077, A238477, A239123, ...
Note that there are also Collatz sequences starting with an odd number, following the pattern ud^(2*n) which end in an even number. For example, take n=1 and the sequence [5, 16, 8, 4]. Such sequences are here not considered.

Examples

			The rectangular array Me(n, k) begins:
n\k      1       2       3        4       5        6        7        8        9       10 ...
1:       1       9      17       25      33       41       49       57       65       73
2:       5      37      69      101     133      165      197      229      261      293
3:      21     149     277      405     533      661      789      917     1045     1173
4:      85     597    1109     1621    2133     2645     3157     3669     4181     4693
5:     341    2389    4437     6485    8533    10581    12629    14677    16725    18773
6:    1365    9557   17749    25941   34133    42325    50517    58709    66901    75093
7:    5461   38229   70997   103765  136533   169301   202069   234837   267605   300373
8:   21845  152917  283989   415061  546133   677205   808277   939349  1070421  1201493
9:   87381  611669 1135957  1660245 2184533  2708821  3233109  3757397  4281685  4805973
10: 349525 2446677 4543829  6640981 8738133 10835285 12932437 15029589 17126741 19223893
...
The triangle Te(m, n) begins (zeros are not shown):
m\n   1    2    3     4      5      6       7       8       9      10 ...
1:    1
2:    9    5
3:   17   37   21
4:   25   69  149    85
5:   33  101  277   597    341
6:   41  133  405  1109   2389   1365
7:   49  165  533  1621   4437   9557    5461
8:   57  197  661  2133   6485  17749   38229   21845
9:   65  229  789  2645   8533  25941   70997  152917   87381
10:  73  261  917  3157  10581  34133  103765  283989  611669  349525
...
----------------------------------------------------------------------------------------------
n=1, ud^2, k=1: Me(1, 1) = 1 = Te(1, 1), Ne(1) = 1 with the Collatz sequence [1, 4, 2, 1] of length 4.
n=1, ud^2, k=2: Me(1, 2) = 9 = Te(2, 1), Ne(2) = 7 with the Collatz sequence [9, 28, 14, 7] of length 4.
n=2, ud^4, k=1: Me(2, 1) = 5 = Te(2, 2), Ne(1) = 1 with the length 6 Collatz sequence [5, 16, 8, 4, 2, 1].
n=5, ud^(10), k=2: Me(5, 2) =  2389  = Te(6,5),  Ne(2) = 7 with the Collatz sequence [2389, 7168, 3584, 1792, 896, 448, 224, 112, 56, 28, 14, 7] of length 12.
		

Formula

The array: Me(n, k) = 2^(2*n+1)*k - (5*2^(2*n)+1)/3 for n >= 1 and k >= 1.
The triangle: Te(m, n) = Me(n, m-n+1) = 2*4^n*(m-n) + (4^n-1)/3 for m >= n >= 1 and 0 for m < n.
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