OFFSET
0,4
COMMENTS
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
LINKS
Tomás Oliveira e Silva, Computational verification of the 3x+1 conjecture.
Isaac DeJager, Madeleine Naquin, and Frank Seidl, Colored Motzkin Paths of Higher Order, VERUM 2019.
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.
FORMULA
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(n) = 2^n - 2^n*Sum_{k=0..A156301(n)-1} A186009(k+1)/2^A020914(k). - Benjamin Lombardo, Sep 08 2019
EXAMPLE
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)
PROG
(C) /* 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(m<lm) { return 0; }
if(p2==fp2) { return 1; }
return survives(r, m<<1, lm<<1, p2+1, fp2)
+ survives(r+m, m<<1, lm<<1, p2+1, fp2);
} /* Phil Carmody, Sep 08 2011 */
(PARI) /* 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
CROSSREFS
KEYWORD
nonn
AUTHOR
Labos Elemer, Oct 01 2002
EXTENSIONS
New terms to n=39 by Phil Carmody, Sep 08 2011
STATUS
approved