A100982 -id:A100982 - cp's OEIS frontend

A325904 Generator sequence for A100982.

1, 0, -3, -8, 15, -91, -54, 2531, -17021, 43035, -66258, 1958757, -24572453, 146991979, -287482322, -3148566077, 35506973089, -198639977241, 1006345648929, -8250266425561, 76832268802555, -517564939540551, 1890772860334557, 3323588929061820, -104547561696315008, 907385094824827328, -6313246535826877248

Offset: 0

Benjamin Lombardo, Sep 08 2019

sign

The name of this sequence is derived from its main purpose as a formula for A100982 (see link). Both formulas below stem from Mike Winkler's 2017 paper on the 3x+1 problem (see below), in which a recursive definition of A100982 and A076227 is created in 2-D space. These formulas redefine the sequences in terms of this 1-D recursive sequence.

Mike Winkler, The algorithmic structure of the finite stopping time behavior of the 3x + 1 function, arXiv:1709.03385 [math.GM], 2017.

Cf. A020914, A076227, A100982.

import math
numberOfTerms = 20
L6 = [1,0]
def c(n):
    return math.floor(n/(math.log2(3)-1))
def p(a,b):
    return math.factorial(a)/(math.factorial(a-b)*math.factorial(b))
def anotherTerm(newTermCount):
    global L6
    for a in range(newTermCount+1-len(L6)):
        y = len(L6)
        newElement = 0
        for k in range(y):
            newElement -= int(L6[k]*p(c(y)+y-k-2, c(y)-2))
        L6.append(newElement)
anotherTerm(numberOfTerms)
print("A325904")
for a in range(numberOfTerms+1):
    print(a, "|", L6[a])

@cached_function
def a(n):
    if n < 2: return 0^n
    A = floor(n/(log(3, 2) - 1)) - 2
    return -sum(a(k)*binomial(A + n - k, A) for k in (0..n-1))
[a(n) for n in range(100)] # Peter Luschny, Sep 10 2019

a(0)=1, a(1)=0, a(n) = -Sum_{k=0..n-1} a(k)*binomial(A325913(n)+n-k-2, A325913(n)-2) for n>1.

A056576 Highest k with 2^k <= 3^n.

Original entry on oeis.org

0, 1, 3, 4, 6, 7, 9, 11, 12, 14, 15, 17, 19, 20, 22, 23, 25, 26, 28, 30, 31, 33, 34, 36, 38, 39, 41, 42, 44, 45, 47, 49, 50, 52, 53, 55, 57, 58, 60, 61, 63, 64, 66, 68, 69, 71, 72, 74, 76, 77, 79, 80, 82, 84, 85, 87, 88, 90, 91, 93, 95, 96, 98, 99, 101, 103, 104, 106, 107

Offset: 0

Henry Bottomley, Jun 29 2000

nonn

			a(3)=4 because 3^3=27 and 2^4=16 is power of 2 immediately below 27.

Michael De Vlieger, Table of n, a(n) for n = 0..10000
Mike Winkler, The algorithmic structure of the finite stopping time behavior of the 3x+ 1 function, arXiv:1709.03385 [math.GM], 2017.

Cf. A000079 (powers of 2), A000244 (powers of 3), A020914, A022921.

Cf. A056850, A117630 (complement), A020857 (decimal expansion of log_2(3)), A076227, A100982.

a056576 = subtract 1 . a020914  -- Reinhard Zumkeller, May 17 2015

seq(ilog2(3^n), n= 0 .. 1000); # Robert Israel, Dec 11 2014

Table[Floor[Log[2, 3^n]], {n, 0, 69}] (* Robert G. Wilson v, Apr 06 2006 *)
Table[Floor[n*Log[2, 3]], {n, 0, 68}] (* L. Edson Jeffery, Dec 11 2014 *)

{a(n) = if( n<0, 0, logint(3^n, 2))}; /* Michael Somos, Dec 13 2014 */

def A056576(n): return (3**n).bit_length()-1 # Chai Wah Wu, Oct 09 2024

a(n) = floor(log_2(3^n)) = log_2(A000244(n)-A056576(n)) = a(n-1)+A022921(n-1).

a(n) = A020914(n) - 1. - L. Edson Jeffery, Dec 12 2014

A122458 "Dropping time" of the reduced Collatz iteration starting with 2n+1.

Original entry on oeis.org

0, 2, 1, 4, 1, 3, 1, 4, 1, 2, 1, 3, 1, 37, 1, 35, 1, 2, 1, 5, 1, 3, 1, 34, 1, 2, 1, 3, 1, 4, 1, 34, 1, 2, 1, 32, 1, 3, 1, 5, 1, 2, 1, 3, 1, 28, 1, 5, 1, 2, 1, 26, 1, 3, 1, 19, 1, 2, 1, 3, 1, 5, 1, 9, 1, 2, 1, 4, 1, 3, 1, 4, 1, 2, 1, 3, 1, 25, 1, 13, 1, 2, 1, 18, 1, 3, 1, 5, 1, 2, 1, 3, 1, 4, 1, 8, 1, 2, 1, 5

Offset: 0

T. D. Noe, Sep 08 2006

nonn

We count only the 3x+1 steps of the usual Collatz iteration. We stop counting when the iteration produces a number less than the initial 2n+1. For a fixed dropping time k, let N(k)=A100982(k) and P(k)=2^(A020914(k)-1). There are exactly N(k) odd numbers less than P(k) with dropping time k. Moreover, the sequence is periodic: if d is one of the N(k) odd numbers, then k=a(d)=a(d+i*P(k)) for all i>=0. This periodicity makes it easy to compute the average dropping time of the reduced Collatz iteration: Sum_{k>0} k*N(k)/P(k) = 3.492651852186... (A122791).

			a(3)=4 because, starting with 7, the iteration produces 11,17,13,5 and the last term is less than 7.
n = 13: the fr trajectory for 2*13+1 = 27 is 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 with 41 terms (without 27), hence fr^[37] = 23 < 27  and  a(13) = 37. - _Wolfdieter Lang_, Feb 20 2019

Victor Klee and Stan Wagon, Old and New Unsolved Problems in Plane Geometry and Number Theory, Mathematical Association of America (1991) pp. 225-229, 308-309. [called on p. 225 stopping time for 2n+1 and the function C(2*n+1) = A075677(n+1), n >= 0. - Wolfdieter Lang, Feb 20 2019]

T. D. Noe, Table of n, a(n) for n = 0..10000

Cf. A000265, A060445, A075677 (one step of the reduced Collatz iteration), A075680.

Cf. A087113 (indices of 1's), A017077 (indices of 2's), A122791 (limit mean).

nextOddK[n_]:=Module[{m=3n+1}, While[EvenQ[m], m=m/2]; m]; dt[n_]:=Module[{m=n, cnt=0}, If[n>1, While[m=nextOddK[m]; cnt++; m>n]]; cnt]; Table[dt[n],{n,1,301,2}]

a(n) is the least k for which fr^[k](n) < 2*n + 1, for n >= 1 and k >= 1, where fr(n) = A075677(n+1) = A000265(3*n+2). No k satisfies this for n = 0: a(0) := 0 by convention. The dropping time a(n) is finite, for n >= 1, if the Collatz conjecture is true. - Wolfdieter Lang, Feb 20 2019

a(1+i*8) = 2, for i>=0, because A100982(2) = 1 is odd, and A020914(2) = 4 gives P(2) = 2^(4-1) = 8. - Ruud H.G. van Tol, Dec 19 2021

A022921 Number of integers m such that 3^n < 2^m < 3^(n+1).

Original entry on oeis.org

1, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 1, 2, 2, 1, 2, 1

Offset: 0

Clark Kimberling

Represents increments between successive terms of allowable dropping times in the Collatz (3x+1) problem. That is, a(n) = A020914(n+1) - A020914(n). - K. Spage, Oct 23 2009

			From _Amiram Eldar_, Mar 01 2024: (Start)
a(0) = 1 because 3^0 = 1 < 2^1 = 2 < 3^1 = 3.
a(1) = 2 because 3^1 = 3 < 2^2 = 4 < 2^3 = 8 < 3^2 = 9.
a(2) = 1 because 3^2 = 9 < 2^4 = 16 < 3^3 = 27. (End)

T. D. Noe, Table of n, a(n) for n = 0..1000
Mike Winkler, The algorithmic structure of the finite stopping time behavior of the 3x + 1 function, arXiv:1709.03385 [math.GM], 2017.

Cf. A020914, A056576, A076227, A100982, A122437.

See also A020857 (decimal expansion of log_2(3)).

Digits := 100: c1 := log(3.)/log(2.): A022921 := n->floor((n+1)*c1)-floor(n*c1);
seq(ilog2(3^(n+1)) - ilog2(3^n), n=0 .. 1000); # Robert Israel, Dec 11 2014

i2 = 1; Table[p = i2; While[i2++; 2^i2 < 3^(n + 1)]; i2 - p, {n, 0, 98}] (* T. D. Noe, Feb 28 2014 *)
f[n_] := Floor[ Log2[ 3^n] + 1]; Differences@ Array[f, 106, 0] (* Robert G. Wilson v, May 25 2014 *)

a(n) = logint(3^(n+1),2) - logint(3^n,2) \\ Ruud H.G. van Tol, Dec 28 2022

Vec(matreduce([logint(2^i,3)|i<-[1..158]])[,2])[1..-2] \\ Ruud H.G. van Tol, Dec 29 2022

a(n) = floor((n+1)*log_2(3)) - floor(n*log_2(3)).
a(n) = A122437(n+2) - A122437(n+1) - 1. - K. Spage, Oct 23 2009
First differences of A020914. - Robert G. Wilson v, May 25 2014
First differences of A056576. - L. Edson Jeffery, Dec 12 2014
Asymptotic mean: lim_{m->oo} (1/m) * Sum_{k=1..m} a(k) = log_2(3) (A020857). - Amiram Eldar, Mar 01 2024

A076227 Number of surviving Collatz residues mod 2^n.

Original entry on oeis.org

1, 1, 1, 2, 3, 4, 8, 13, 19, 38, 64, 128, 226, 367, 734, 1295, 2114, 4228, 7495, 14990, 27328, 46611, 93222, 168807, 286581, 573162, 1037374, 1762293, 3524586, 6385637, 12771274, 23642078, 41347483, 82694966, 151917636, 263841377, 527682754, 967378591, 1934757182, 3611535862

Offset: 0

Labos Elemer, Oct 01 2002

nonn

Number of residue classes in which A074473(m) is not constant.
The ratio of numbers of inhomogenous r-classes versus uniform-classes enumerated here increases with n and tends to 0. For n large enough ratio < a(16)/65536 = 2114/65536 ~ 3.23%.
Theorem: a(n) can be generated for each n > 2 algorithmically in a Pascal's triangle-like manner from the two starting values 0 and 1. This result is based on the fact that the Collatz residues (mod 2^k) can be evolved according to a binary tree. There is a direct connectedness to A100982, A056576, A022921, A020915. - Mike Winkler, Sep 12 2017
Brown's criterion ensures that the sequence is complete (see formulae). - Vladimir M. Zarubin, Aug 11 2019

			n=6: Modulo 64, eight residue classes were counted: r=7, 15, 27, 31, 39, 47, 59, 63. See A075476-A075483. For other 64-8=56 r-classes u(q)=A074473(64k+q) is constant: in 32 class u(q)=2, in 16 classes u(q)=4, in 4 classes u(q)=7 and in 4 cases u(q)=9. E.g., for r=11, 23, 43, 55 A074473(64k+r)=9 independently of k.
From _Mike Winkler_, Sep 12 2017: (Start)
The next table shows how the theorem works. No entry is equal to zero.
  k =        3  4  5   6   7   8   9  10  11   12 .. | a(n)=
-----------------------------------------------------|
  n =  2  |  1                                       |    1
  n =  3  |  1  1                                    |    2
  n =  4  |     2  1                                 |    3
  n =  5  |        3   1                             |    4
  n =  6  |        3   4   1                         |    8
  n =  7  |            7   5   1                     |   13
  n =  8  |               12   6   1                 |   19
  n =  9  |               12  18   7   1             |   38
  n = 10  |                   30  25   8   1         |   64
  n = 11  |                   30  55  33   9    1    |  128
  :       |                        :   :   :    : .. |   :
-----------------------------------------------------|------
A100982(k) = 2  3  7  12  30  85 173 476 961 2652 .. |
The entries (n,k) in this table are generated by the rule (n+1,k) = (n,k) + (n,k-1). The last value of (n+1,k) is given by n+1 = A056576(k-1), or the highest value in column n is given twice only if A022921(k-2) = 2. Then a(n) is equal to the sum of the entries in row n. For k = 7 there is: 1 = 0 + 1, 5 = 1 + 4, 12 = 5 + 7, 12 = 12 + 0. It is a(9) = 12 + 18 + 7 + 1 = 38. The sum of column k is equal to A100982(k). (End)

Isaac DeJager, Madeleine Naquin, and Frank Seidl, Colored Motzkin Paths of Higher Order, VERUM 2019.
Andreas-Stephan Elsenhans, Numerical verification of the Collatz conjecture for billion digit random numbers, arXiv:2502.16743 [math.NT], 2025.
Tomás Oliveira e Silva, Computational verification of the 3x+1 conjecture.
Eric Weisstein's World of Mathematics, Brown's Criterion
M. Winkler, The algorithmic structure of the finite stopping time behavior of the 3x + 1 function, arXiv:1709.03385 [math.GM], 2017.

Cf. A006370, A074473, A075476-A075483, A100982, A056576, A022921, A020915, A243115.

/* call as follows: uint64_t s=survives(0,1,1,0,bits); */
uint64_t survives(uint64_t r, uint64_t m, uint64_t lm, int p2, int fp2)
{
    while(!(m&1) && (m>=lm)) {
        if(r&1) { r+=(r+1)>>1; m+=m>>1; }
        else { r>>=1; m>>=1; }
    }
    if(mPhil Carmody, Sep 08 2011 */

/* algorithm for the Theorem */
{limit=30; /*or limit>30*/ R=matrix(limit,limit); R[2,1]=0; R[2,2]=1; for(k=2, limit, if(k>2, print; print1("For n="k-1" in row n: ")); Kappa_k=floor(k*log(3)/log(2)); for(n=k, Kappa_k, R[n+1,k]=R[n,k]+R[n,k-1]); t=floor(1+(k-1)*log(2)/log(3)); a_n=0; for(i=t, k-1, print1(R[k,i]", "); a_n=a_n+R[k,i]); if(k>2, print; print(" and the sum is a(n)="a_n)))} \\ Mike Winkler, Sep 12 2017

a(n) = Sum_{k=A020915(n+2)..n+1} (n,k). (Theorem, cf. example) - Mike Winkler, Sep 12 2017
From Vladimir M. Zarubin, Aug 11 2019: (Start)
a(0) = 1, a(1) = 1, and for k > 0,
a(A020914(k)) = 2*a(A020914(k)-1) - A100982(k),
a(A054414(k)) = 2*a(A054414(k)-1). (End)
a(n) = 2^n - 2^n*Sum_{k=0..A156301(n)-1} A186009(k+1)/2^A020914(k). - Benjamin Lombardo, Sep 08 2019

New terms to n=39 by Phil Carmody, Sep 08 2011

A186009 Length of Collatz dropping time patterns in A186008.

Original entry on oeis.org

1, 1, 1, 2, 3, 7, 12, 30, 85, 173, 476, 961, 2652, 8045, 17637, 51033, 108950, 312455, 663535, 1900470, 5936673, 13472296, 39993895, 87986917, 257978502, 820236724, 1899474678, 5723030586, 12809477536, 38036848410, 84141805077, 248369601964

Offset: 1

T. D. Noe, Feb 09 2011

nonn

This is A100982 with 1 prepended.

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

Cf. A100982, A186008.

A177789 Irregular triangle read by rows in which row n gives the congruences (mod 2^A020914(n)) satisfied by the numbers having dropping time A122437(n+1) in the Collatz (3x+1) iteration.

Original entry on oeis.org

0, 1, 3, 11, 23, 7, 15, 59, 39, 79, 95, 123, 175, 199, 219, 287, 347, 367, 423, 507, 575, 583, 735, 815, 923, 975, 999, 231, 383, 463, 615, 879, 935, 1019, 1087, 1231, 1435, 1647, 1703, 1787, 1823, 1855, 2031, 2203, 2239, 2351, 2587, 2591, 2907, 2975, 3119

Offset: 0

T. D. Noe, May 13 2010

The dropping time is the number of Collatz iterations required to reach a lower number than starting value. Garner mentions these congruences. The first term in row n is A122442(n+1) for n > 1. The length of row n is A100982(n).
The triangle means:
  numbers 0 (mod 2) and > 0 have dropping time 1;
  numbers 1 (mod 4) and > 1 have dropping time 3;
  numbers 3 (mod 16) have dropping time 6;
  numbers 11, 23 (mod 32) have dropping time 8;
  numbers 7, 15, 59 (mod 128) have dropping time 11;
  numbers 39, 79, 95, 123, 175, 199, 219 (mod 256) have dropping time 13.
Theorem: a(n) can be evaluated using a directed rooted tree produced by a precise algorithm. Each node of this tree is given by a unique Diophantine equation whose only positive solutions are the integers with a finite stopping time. The algorithm generates (in a three step loop) the parity vectors which define the Diophantine equations. The two directions of the construction principle gives the tree a triangular form which extends ever more downwards with each column. There exist explicit arithmetic relationships between the parent and child vertices. As a consequence, a(n) can be generated algorithmically. The algorithm also generates A100982. - Mike Winkler, Sep 12 2017

			Triangle begins:
   0;
   1;
   3;
  11,  23;
   7,  15,  59;
  39,  79,  95, 123, 175, 199, 219;
  ...
From _Mike Winkler_, Sep 12 2017: (Start)
The beginning of the directed rooted tree produced by the algorithm of the Theorem. The triangular form can be seen clearly. The way the tree structure is sorting a(n), respectively the residue classes, mirrors the explicit arithmetic relationships mentioned in the Theorem.
3 (mod 2^4) -- 11 (mod 2^5) -- 59 (mod 2^7) -- 123 (mod 2^8) --
                    |                                |
                    |                          219 (mod 2^8) --
                    |
                    |
               23 (mod 2^5) --- 7 (mod 2^7) -- 199 (mod 2^8) --
                                    |                |
                                    |           39 (mod 2^8) --
                                    |
                                    |
                               15 (mod 2^7) --- 79 (mod 2^8) --
                                                     |
                                               175 (mod 2^8) --
                                                     |
                                                95 (mod 2^8) --
(End)

Michael De Vlieger, Table of n, a(n) for n = 0..12448 (rows n = 0..13, flattened)
Michael De Vlieger, Mathematica program that memoizes A060445.
Lynn E. Garner, On the Collatz 3n + 1 Algorithm, Proc. Amer. Math. Soc., Vol. 82(1981), 19-22.
Ruud H.G. van Tol, Perl code
Mike Winkler, On a stopping time algorithm of the 3n + 1 function
Mike Winkler, On the structure and the behaviour of Collatz 3n + 1 sequences - Finite subsequences and the role of the Fibonacci sequence, arXiv:1412.0519 [math.GM], 2014.
Mike Winkler, New results on the stopping time behaviour of the Collatz 3x + 1 function, arXiv:1504.00212 [math.GM], 2015.
Mike Winkler, The algorithmic structure of the finite stopping time behavior of the 3x + 1 function, arXiv:1709.03385 [math.GM], 2017.

Cf. A060445 (dropping time of odd numbers), A100982.

DroppingTime[n_] := Module[{m=n, k=0}, If[n>1, While[m>=n, k++; If[EvenQ[m], m=m/2, m=3*m+1]]]; k]; dt=Floor[1+Range[0,20]*Log[2,6]]; e=Floor[1+Range[0,20]*Log[2,3]]; Join[{0,1}, Flatten[Table[Select[Range[3,2^e[[n]],2], DroppingTime[ # ]==dt[[n]] &], {n,2,8}]]]

/* algorithm for generating the parity vectors of the Theorem, the tree structure is given by the three STEP's */
{k=3; Log32=log(3)/log(2); limit=14; /*or limit>14*/ T=matrix(limit,60000); xn=3; /*initial tuple for n=1*/ A=[]; for(i=1, 2, A=concat(A,i)); A[1]=1; A[2]=1; T[1,1]=A; for(n=2, limit, print("n="n); Sigma=floor(1+(n+1)*Log32); d=floor(n*Log32)-floor((n-1)*Log32); Kappa=floor(n*Log32); Kappa2=floor((n-1)*Log32);r=1; v=1; until(w==0, A=[]; for(i=1, Kappa2+1, A=concat(A,i)); A=T[n-1,v]; B=[]; for(i=1, Kappa+1, B=concat(B,i)); for(i=1, Kappa2+1, B[i]=A[i]); /* STEP 1 */ if(d==1, B[k]=1; T[n,r]=B; r++; v++); if(d==2, B[k]=0; B[k+1]=1; T[n,r]=B; r++; v++); /* STEP 2 */ if(B[Kappa]==0, for(j=1, Kappa-n, B[Kappa+1-j]=B[Kappa+2-j]; B[Kappa+2-j]=0; T[n,r]=B; r++; if(B[Kappa-j]==1, break(1)))); /* STEP 3 */ w=0; for(i=n+2, Kappa+1, w=w+B[i]));k=k+d; p=1; h2=3; for(i=1, r-1, h=0; B=T[n,i]; until(B[h]==0, h++); if(h>h2, p=1; h2++; print); print(T[n,i]"  "p"  "i); p++); print)} \\ Mike Winkler, Sep 12 2017

row(n) = if (n < 2, [n], my(v = vector(2^(A020914(n)-1), k, 2*k-1)); apply(x->2*x-1, Vec(select(x->(x == 1+A122437(n+1)), apply(A074473, v), 1)))); \\ Michel Marcus, Aug 15 2025

row(n)={if(n<1, [0], my(r=[1], d=[2], km=2); for(i=1, n-1, my(temp1=[], temp2=[], c=if(3^(i+1)<2^(km+1),1,2)); for(j=1, #d, temp1=concat(temp1, vector(d[j]-1, m, 3*r[j]+2^(km-d[j]+m))); temp2=concat(temp2, vector(d[j]-1, m, d[j]-m+c))); km=km+c; r=temp1; d=temp2; ); vecsort(apply(x->((-x)*lift(Mod(1/3^n, 2^km)))%2^km, r)))} \\ V. Barbera, Aug 15 2025

A374244 A Catalan-like sequence formed from the row sums of a Catalan-like triangle where row n is truncated to have ceiling((n+4)*log(3)/log(2)) - (n + 6) terms.

Original entry on oeis.org

1, 2, 5, 9, 23, 43, 113, 331, 698, 1966, 4072, 11433, 23701, 66734, 205712, 459632, 1348864, 2927822, 8499580, 26809375, 61495590, 183946295, 408179706, 1204202538, 2643267587, 7756962475, 24708004563, 57390010121, 173405214133, 389606249120, 1160606285961, 3738436950162

Offset: 1

Rob Bunce, Jul 01 2024

			Standard Catalan:
 n   Sum   Triangle terms
 1     1 = 1;
 2     2 = 1, 1;
 3     5 = 1, 2,  2;
 4    14 = 1, 3,  5; /5
 5    42 = 1, 4,  9, 14; /14
 6   132 = 1, 5, 14, 28; /42;  14
 7   429 = 1, 6, 20, 48, 90; /132; 132
...
When n=4, number of terms is restricted to 3 dropping 1 term; ceiling((4+4)*log(3)/log(2)) - (4 + 6) = 3.
When n=6, number of terms is restricted to 4 dropping 2 terms; ceiling((6+4)*log(3)/log(2)) - (6 + 6) = 4.
etc.
Truncating at the point indicated by / and summing the remaining triangle terms in the normal way results in:
n    Sum  Truncated Triangle terms
 1     1 = 1;
 2     2 = 1, 1;
 3     5 = 1, 2, 2;
 4     9 = 1, 3, 5;
 5    23 = 1, 4, 9, 9;
 6    43 = 1, 5, 14, 23;
 7   113 = 1, 6, 20, 43, 43;
 8   331 = 1, 7, 27, 70, 113, 113;
 9   698 = 1, 8, 35, 105, 218, 331;
10  1966 = 1, 9, 44, 149, 367, 698, 698;
11  4072 = 1, 10, 54, 203, 570, 1268, 1966;
12 11433 = 1, 11, 65, 268, 838, 2106, 4072, 4072;
13 23701 = 1, 12, 77, 345, 1183, 3289, 7361, 11433;
...

Andrew Howroyd, Table of n, a(n) for n = 1..1000

Cf. A009766, A000108, Half Catalan A000992.

Cf. A100982 (row sums of A368514).

lista(N) = {
  my(T=vector(N, n, vector(logint(3^(n+4), 2)-n-5)));
  for(n=1, #T
  , for(k=1, #T[n]
    , T[n][k]= if(1==k, 1, k<=#T[n-1], T[n][k-1]+T[n-1][k], T[n][k-1])
    );
  );
  vector(#T, n, vecsum(T[n]));
}

Same as for a normal Catalan triangle T(n,k), read by rows, each term is the sum of the entries above and to the left, i.e., T(n,k) = Sum_{j=0..k} T(n-1,j) where j is limited to the truncated length.

a(26) onwards from Andrew Howroyd, Oct 25 2024

A122790 Decimal expansion of Wagon's constant: the average "dropping time" of the Collatz (3x+1) iteration.

Original entry on oeis.org

9, 4, 7, 7, 9, 5, 5, 5, 5, 6, 5, 5, 9, 2, 7, 4, 7, 3, 6, 9, 1, 6, 6, 8, 1, 0, 1, 2, 5, 8, 9, 0, 4, 1, 9, 4, 2, 2, 8, 1, 8, 7, 1, 6, 0, 4, 4, 2, 7, 2, 9, 9, 7, 1, 5, 9, 2, 7, 4, 4, 2, 3, 5, 6, 6, 6, 7, 0, 1, 5, 3, 2, 2, 1, 6, 2, 4, 8, 2, 8, 1, 7, 7, 5, 6, 1, 0, 6, 8, 5, 6, 5, 5, 7, 3, 1, 0, 4, 0, 2, 6, 4, 4, 4, 3

Offset: 1

T. D. Noe, Sep 11 2006

Cf. A060445 (dropping time of the 3x+1 iteration starting at 2n+1), A100982, A122791.

Sum_{k>0} A122437(k+1)*A100982(k)/2^(A020914(k)-1) = 9.47795555655927473691668101258904...

A122791 Decimal expansion of the average "dropping time" of the reduced 3x+1 iteration.

Original entry on oeis.org

3, 4, 9, 2, 6, 5, 1, 8, 5, 2, 1, 8, 6, 4, 2, 4, 9, 1, 2, 3, 0, 5, 5, 6, 0, 3, 3, 7, 5, 2, 9, 6, 8, 0, 6, 4, 7, 4, 2, 7, 2, 9, 0, 5, 3, 4, 8, 0, 9, 0, 9, 9, 9, 0, 5, 3, 0, 9, 1, 4, 7, 4, 5, 2, 2, 2, 2, 3, 3, 8, 4, 4, 0, 7, 2, 0, 8, 2, 7, 6, 0, 5, 9, 1, 8, 7, 0, 2, 2, 8, 5, 5, 1, 9, 1, 0, 3, 4, 6, 7, 5, 4, 8, 1, 4

Offset: 1

T. D. Noe, Sep 11 2006

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

Cf. A122458 (dropping time of the reduced Collatz iteration), A100982, A122790.

Sum{k>0} k*A100982(k)/2^(A020914(k)-1) = 3.49265185218642491230556033752968...

cp's OEIS Frontend

A325904 Generator sequence for A100982.

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A056576 Highest k with 2^k <= 3^n.

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A122458 "Dropping time" of the reduced Collatz iteration starting with 2n+1.

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A022921 Number of integers m such that 3^n < 2^m < 3^(n+1).

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A076227 Number of surviving Collatz residues mod 2^n.

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A186009 Length of Collatz dropping time patterns in A186008.

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A177789 Irregular triangle read by rows in which row n gives the congruences (mod 2^A020914(n)) satisfied by the numbers having dropping time A122437(n+1) in the Collatz (3x+1) iteration.

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