|
|
A182137
|
|
Size of the set of b for numbers of the form 2^n*x + b that cannot be the smallest element of a set giving a duration of infinite flight in the Collatz problem.
|
|
1
|
|
|
1, 3, 6, 13, 28, 56, 115, 237, 474, 960, 1920, 3870, 7825, 15650, 31473, 63422, 126844, 254649, 509298, 1021248, 2050541, 4101082, 8219801, 16490635, 32981270, 66071490, 132455435, 264910870, 530485275, 1060970550, 2123841570, 4253619813, 8507239626, 17027951548, 34095896991, 68191793982, 136471574881, 272943149762, 546144278026, 1093108792776, 2186217585552
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,2
|
|
COMMENTS
|
In the Collatz Problem A014682, it is possible to apply the algorithm to first degree polynomials like 2^n*x+b, where n is integer and 0 <= b < 2^n. The iteration terminates by two cases:
1) a*x+b where a < 2^n: the polynomial is "minimized"
2) a*x+b where a is odd and a > 2^n, parity cannot be found. The polynomial cannot be minimized.
The sequence counts how many first degree polynomials end like first case for each n > 0.
The interest of this sequence is that every number that can be described by a minimized polynomial cannot be the smallest element of a set of value of T(n) = infinity.
|
|
LINKS
|
|
|
FORMULA
|
|
|
EXAMPLE
|
Example with 4x+b (0 <= b < 4):
4x is even, thus gives 2x, 2 < 4 (first case).
4x+1, is odd thus 3(4x+1)+1 = 12x+4 is even, thus (12x+4)/2/2=3x+1 3 < 4, first case.
4x+2 is even, (4x+2)/2=2x+1, 2 < 4, first case.
4x+3 with same way gives 9x+8. 9 is odd and 9 > 4, second case.
That explains why the second (n=2) term in sequence is 3.
|
|
MATHEMATICA
|
a[n_] := Module[{b, p0, p1, minimized = 0}, For[b = 1, b <= 2^n, b++, {p0, p1} = {b, 2^n}; While[Mod[p1, 2] == 0 && p1 >= 2^n, {p0, p1} = If[Mod[p0, 2] == 0, {p0/2, p1/2}, {3*p0+1, 3*p1}]; If[p1<2^n, minimized += 1]]]; minimized]; Table[Print[an = a[n]]; an, {n, 1, 40}] (* Jean-François Alcover, Feb 12 2014, translated from D. S. McNeil's Sage code *)
|
|
PROG
|
(Sage)
minimized = 0
for b in range(2**n):
p = [b, 2**n]
while p[1] % 2 == 0 and p[1] >= 2**n:
p[0], p[1] = [p[0]/2, p[1]/2] if p[0] % 2 == 0 else [3*p[0]+1, 3*p[1]]
if p[1] < 2**n: minimized += 1
(PARI) upto(P=18)= my(r=Vec([1, 1], P)); forstep(x=3, 2^P, 4, my(s=x, p=0); until(s<=x, s= if(s%2, 3*s+1, s)/2; if(p++ > P, next(2))); if((2^p>x), r[p]++)); for(i=2, #r, r[i]+= 2*r[i-1]); print(r); \\ Ruud H.G. van Tol, Mar 13 2023
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|