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 1-4 of 4 results.

A100982 Number of admissible sequences of order j; related to 3x+1 problem and Wagon's constant.

Original entry on oeis.org

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

Views

Author

Steven Finch, Jan 13 2005

Keywords

Comments

Eric Roosendaal counted all admissible sequences up to order j=1000 (2005). Note: there is a typo in both Wagon and Chamberland in the definition of Wagon's constant 9.477955... The expression floor(1 + 2*i + i*log_2(3)) should be replaced by floor(1 + i + i*log_2(3)).
The length of all admissible sequences of order j is A020914(j). - T. D. Noe, Sep 11 2006
Conjecture: a(n) is given for each n > 3 by a formula using a(2)..a(n-1). This allows us to create an iterative algorithm which generates a(n) for each n > 6. This has been proved for each n <= 53. For higher values of n the algorithm must be slightly modified. - Mike Winkler, Jan 03 2018
Theorem 1: a(k) is given for each k > 1 by a formula using a(1)..a(k-1). Namely, a(1)=1 and a(k+1) = Sum_{m=1..k} (-1)^(m-1)*binomial(floor((k-m+1)*(log(3)/log(2))) + m - 1, m)*a(k-m+1) for k >= 1. - Vladimir M. Zarubin, Sep 25 2015
Theorem 2: 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 connection with A076227, A056576 and A022921. - Mike Winkler, Sep 12 2017
A177789 shows another theorem and algorithm for generating a(n). - Mike Winkler, Sep 12 2017

Examples

			The unique admissible sequence of order 1 is 3/2, 1/2.
The unique admissible sequence of order 2 is 3/2, 3/2, 1/2, 1/2.
The two admissible sequences of order 3 are 3/2, 3/2, 3/2, 1/2, 1/2 and 3/2, 3/2, 1/2, 3/2, 1/2.
a(13) = 8045 = binomial(floor(5*(13-2)/3), 13-2)
- Sum_{i=2..6} binomial(floor((3*(13-i)+0)/2), 13-i)*a(i)
- Sum_{i=7..11} binomial(floor((3*(13-i)-1)/2), 13-i)*a(i)
- Sum_{i=12..12} binomial(floor((3*(13-i)-2)/2), 13-i)*a(i)
= 31824 - 4368*1 - 3003*2 - 715*3 - 495*7 - 120*12 - 28*30 - 21*85 - 5*173 - 4*476 - 1*961 - 0*2652. (Conjecture)
From _Mike Winkler_, Sep 12 2017: (Start)
The next table shows how Theorem 2 works. No entry is equal to zero.
n =       3  4  5   6   7   8   9  10  11   12 .. |A076227(k)=
--------------------------------------------------|
k =  2 |  1                                       |     1
k =  3 |  1  1                                    |     2
k =  4 |     2  1                                 |     3
k =  5 |        3   1                             |     4
k =  6 |        3   4   1                         |     8
k =  7 |            7   5   1                     |    13
k =  8 |               12   6   1                 |    19
k =  9 |               12  18   7   1             |    38
k = 10 |                   30  25   8   1         |    64
k = 11 |                   30  55  33   9    1    |   128
:      |                        :   :   :    : .. |    :
--------------------------------------------------|---------
a(n) =    2  3  7  12  30  85 173 476 961 2652 .. |
The entries (k,n) in this table are generated by the rule (k+1,n) = (k,n) + (k,n-1). The last value of (k+1,n) is given by k+1 = A056576(n-1), or the highest value in column n is given twice only if A022921(n-2) = 2. Then a(n) is equal to the sum of the entries in column n. For n = 7 there is 1 = 0 + 1, 5 = 1 + 4, 12 = 5 + 7, 12 = 12 + 0. Therefore a(7) = 1 + 5 + 12 + 12 = 30. The sum of row k is equal to A076227(k). (End)
From _Ruud H.G. van Tol_, Dec 04 2023: (Start)
A tree view.
n-tree--A098294--ids-----paths-----------------a(n)
0 ._          0  0       0                       -
1 |_          1  1       10                      1
2 |_._        2  2       1100                    1
3 |_|_        2  3-4     11010     -   11100     2
4 |_|_._      3  5-7     1101100   -  1111000    3
5 |_|_|_      3  8-14    11011010  - 11111000    7
6 |_|_|_._    4  15-26   1101101100-1111110000  12
7 |_|_|_|_._  5  27-56   ...                    30
8 |_|_|_|_|_  5  57-141  ...                    85
...
For n>=1, the endpoints are at A098294(n) to the right.
(End)

Links

Crossrefs

Cf. A122790 (Wagon's constant), A076227, A056576, A022921, A098294, A177789.
Cf. A060941, A293946.

Programs

  • Mathematica
    (* based on Eric Roosendaal's algorithm *) nn=100; Clear[x,y]; Do[x[i]=0, {i,0,nn+1}]; x[1]=1; t=Table[Do[y[cnt]=x[cnt]+x[cnt-1], {cnt,p+1}]; Do[x[cnt]=y[cnt], {cnt,p+1}]; admis=0; Do[If[(p+1-cnt)*Log[3]T. D. Noe, Sep 11 2006 *)
  • PARI
    /* translation of the above code from T. D. Noe */
{limit=100; n=1; x=y=vector(limit+1); x[1]=1; for(b=2, limit, for(c=2, b+1, y[c]=x[c]+x[c-1]); for(c=2, b+1, x[c]=y[c]); a_n=0; for(c=1, b+1, if((b+1-c)*log(3)Mike Winkler, Feb 28 2015
  • PARI
    /* algorithm for the Conjecture */
{limit=53; zn=vector(limit); zn[2]=1; zn[3]=2; zn[4]=3; zn[5]=7; zn[6]=12; f=1; e1=-1; e2=-2; for(n=7, limit, m=floor((n-1)*log(3)/log(2))-(n-1); j=(m+n-2)!/(m!*(n-2)!); if(n>6*f, if(frac(n/2)==0, e=e1, e=e2)); if(frac((n-6 )/12)==0, f++; e1=e1+2); if(frac((n-12)/12)==0, f++; e2=e2+2); Sum=a=b=0; c=1; d=5; until(c>=n-1, for(i=2+a*5+b, 1+d+a*5, if(i>11 && frac((i+2)/6)==0, b++); delta=e-a; Sum=Sum+binomial(floor((3*(n-i)+delta)/2),n-i)*zn[i]; c++); a++; for(k=3, 50, if(n>=k*6 && a==k-1, d=k+3))); zn[n]=j-Sum; print(n" "zn[n]))} \\ Mike Winkler, Jan 03 2018
  • PARI
    /* cf. code for Theorem 2 */
{limit=100; /*or limit>100*/ p=q=vector(limit); c=2; w=log(3)/log(2); for(n=3, limit, p[1]=Sum=1; for(i=2, c, p[i]=p[i-1]+q[i]; Sum=Sum+p[i]); a_n=Sum; print(n" "a_n); for(i=1, c, q[i]=p[i]); d=floor(n*w)-floor((n-1)*w); if(d==2, c++)); } \\ Mike Winkler, Apr 14 2015
  • PARI
    /* algorithm for Theorem 1 */
n=20; a=vector(n); log32=log(3)/log(2);
{a[1]=1; for ( k=1, n-1, a[k+1]=sum( m=1,k,(-1)^(m-1)*binomial( floor( (k-m+1)*log32)+m-1,m)*a[k-m+1] ); print(k" "a[k]) );
} \\ Vladimir M. Zarubin, Sep 25 2015
  • PARI
    /* algorithm for Theorem 2 */
{limit=30; /*or limit>30*/ R=matrix(limit,limit); R[2,1]=0; R[2,2]=1; for(n=2, limit, print; print1("For n="n" in column n: "); Kappa_n=floor(n*log(3)/log(2)); a_n=0; for(k=n, Kappa_n, R[k+1,n]=R[k,n]+R[k,n-1]; print1(R[k+1,n]", "); a_n=a_n+R[k+1,n]); print; print(" and the sum is a(n)="a_n))} \\ Mike Winkler, Sep 12 2017

Formula

A sequence s(k), where k=1, 2, ..., n, is *admissible* if it satisfies s(k)=3/2 exactly j times, s(k)=1/2 exactly n-j times, s(1)*s(2)*...*s(n) < 1 but s(1)*s(2)*...*s(m) > 1 for all 1 < m < n.
a(n) = (m+n-2)!/(m!*(n-2)!) - Sum_{i=2..n-1} binomial(floor((3*(n-i)+b)/2), n-i)*a(i), where m = floor((n-1)*log_2(3))-(n-1) and b assumes different integer values within the sum at intervals of 5 or 6 terms. (Conjecture)
a(n) = Sum_{k=n-1..A056576(n-1)} (k,n). (Theorem 2, cf. example)
a(k) = 2*A076227(A020914(k)-1) - A076227(A020914(k)), for k > 0. - Vladimir M. Zarubin, Sep 29 2019
a(1)=1, a(n) = Sum_{k=0..A020914(n-1)-n-2} A325904(k)*binomial(A020914(n-1)-k-2, n-2) for n>1. - Benjamin Lombardo, Oct 18 2019

Extensions

Two more terms from Jules Renucci (jules.renucci(AT)wanadoo.fr), Nov 02 2005
More terms from T. D. Noe, Sep 11 2006

A186008 Irregular triangle T(n,k) read by rows, in which row n has the pattern of conjectured dropping times in the Collatz iteration.

Original entry on oeis.org

2, 4, 16, 12, 32, 8, 52, 128, 40, 56, 84, 136, 160, 180, 256, 60, 80, 136, 220, 288, 296, 448, 528, 636, 688, 712, 1024, 152, 232, 384, 648, 704, 788, 856, 1000, 1204, 1416, 1472, 1556, 1592, 1624, 1800, 1972, 2008, 2120, 2356, 2360, 2676, 2744, 2888, 2912, 3064, 3328, 3444, 3680, 3832, 4096
Offset: 1

Views

Author

T. D. Noe, Feb 09 2011

Keywords

Comments

Consider A126241, the sequence of dropping times in the Collatz iteration. Only zero and the numbers in A020914 can be dropping times. The dropping times in A126241 have a definite pattern. For example, 1 appears at positions n = 2 + 2*i, for i=0,1,2,3,... Similarly, 2 appears at positions n = 5 + 4*i; 4 appears at n = 3 + 16*i; 5 appears at n = 11 + {12,32}*i; and 7 appears at 7 + {8, 52, 128}*i. In general, if we let s=A020914(r) be the r-th possible stopping time, then A126241(n) = s for n = A122442(r) + T(r)*i, where T(r) is the r-th row of this triangle. The length of row n is A186009(n). The n-th row ends with 2^A020914(n).
The frequency of the r-th dropping time s=A020914(r) can be computed as A186009(r)/2^s. The first few frequencies are 1/2, 1/4, 1/16, 1/16, 3/128, 7/256, 3/256, 15/2048, and 85/8192.
The term "stopping time" is sometimes used instead of "dropping time", but the former usually refers to A006666.
This sequence is closely related to A177789.

Examples

			The triangle begins
2
4
16
12, 32
8, 52, 128
40, 56, 84, 136, 160, 180, 256
60, 80, 136, 220, 288, 296, 448, 528, 636, 688, 712, 1024

References

  • J. C. Lagarias, ed., The Ultimate Challenge: The 3x+1 Problem, Amer. Math. Soc., 2010. See pp. 33, 35ff.

Links

Crossrefs

Cf. A014682, A126241, A006666, A186009.

A243115 Starting values of the reduced Collatz function (A014682) where 2 to the power of the "dropping time" is greater than the starting value.

Original entry on oeis.org

3, 7, 11, 15, 23, 27, 31, 39, 47, 59, 63, 71, 79, 91, 95, 103, 111, 123, 127, 155, 159, 167, 175, 191, 199, 207, 219, 223, 231, 239, 251, 255, 283, 287, 303, 319, 327, 347, 359, 367, 383, 411, 415, 423, 447, 463, 479, 487, 495, 507, 511, 539, 543, 559, 575
Offset: 1

Views

Author

K. Spage, Aug 20 2014

Keywords

Comments

a(n) is the lowest positive starting value of the reduced Collatz function such that all starting values (>1) that are congruent to a(n) (mod 2^d) have the same dropping time (d). The dropping time here counts the (3x+1)/2 and the x/2 steps as listed in A126241. A number is included in this sequence if 2^A126241(a(n)) > a(n).
Starting values that produce new record dropping times as listed in A060412 are necessarily a subset of this sequence.
If at least one iteration is carried out before checking that the absolute iterated value has become less than or equal to the absolute starting value, then a(n) is the lowest positive starting value such that all starting values (positive, zero or negative) that are congruent to a(n) (mod 2^d) have the same dropping time (d). Defined like this, the sequence would start with 0, 1, 3, 7.
For k>0, A076227(k) is the number of terms between 2^k and 2^(k+1)-1. - Ruud H.G. van Tol, Dec 18 2022
All terms are congruent to 3 (mod 4) since any 1 (mod 4) has dropping time A126241(4k+1) = 2, for k>=1. - Ruud H.G. van Tol, Jan 11 2023

Examples

			3 is in this sequence because the dropping time starting with 3 is A126241(3) = 4 and 2^4 > 3.

Crossrefs

Cf. A014682, A060412, A076227, A126241, A177789, A239126.

Programs

  • PARI
    is(t)= if(t<3||3!=t%4,0,my(x=t, d=0); until(x<=t, if(x%2, x=(x*3+1)/2, x/=2); d++); 2^d>t); \\ updated by Ruud H.G. van Tol, Jan 10 2023

Extensions

Offset 1 from Ruud H.G. van Tol, Jan 10 2023

A260591 a(n) is the number of odd numbers k < 2^n such that A260590(k) = n.

Original entry on oeis.org

0, 1, 0, 1, 2, 0, 3, 7, 0, 12, 0, 30, 85, 0, 173, 476, 0, 961, 0, 2652, 8045, 0, 17637, 51033, 0, 108950, 312455, 0, 663535, 0, 1900470, 5936673, 0, 13472296, 39993895, 0, 87986917, 0, 257978502, 820236724, 0, 1899474678, 5723030586, 0, 12809477536, 38036848410, 0, 84141805077, 0, 248369601964
Offset: 1

Views

Author

Joseph K. Horn, O. Praem, and Robert G. Wilson v, Jul 29 2015

Keywords

Comments

a(n) is either 0 or about c^(n-1) with c = log(3)/log(2).
Nonzero values give A100982. - Ruud H.G. van Tol, Nov 25 2021
A close variant of this sequence, that starts at offset 0, but with a(0)=0 and a(1)=1, maps it to the count of dropping patterns of 2^n+c(2^n), with the c(2^n) as mentioned with A177789. The positions of the zeros of that variant sequence might be a close variant of A054414, again with a(0)=0 (not properly checked yet). - Ruud H.G. van Tol, Nov 28 2021
It appears that the proportion of zeros is 1-log(2)/log(3) = 36.907...%. - Jesse Randall, Oct 10 2024

Examples

			a(1) = 0 since there exists no odd number whose msa is 1;
a(2) = 1 since there is only one odd number, 5 with k=2 2k+1, with k less than 2^2 whose msa is 2;
a(3) = 0 since there exists no odd number whose msa is 3;
a(4) = 1 since there is only one number, 1, less than 2^(4+1) whose msa is 4;
a(5) = 2 since there are two numbers, 11 & 23, less than 2^(4+1) whose msa is 4; etc.

Crossrefs

Cf. A260590, A054414, A100982, A186009, A186008, A177789, A346640.

Programs

  • Mathematica
    msa[n_] := If[ OddQ@ n, (3n + 1)/2, n/2]; f[n_] := Block[{k = 2n + 1}, Length@ NestWhileList[ msa@# &, k, # >= k &] - 1]; g[n_] := Length@ Select[ Range[ 2^(n - 1)], f@# == n &]; Array[ g, 20]

Extensions

a(31) onwards from Jesse Randall, Sep 09 2024
Showing 1-4 of 4 results.

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