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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
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.
FORMULA
a(n) = 2^n - A076227(n) for n >= 2. - Ruud H.G. van Tol, Mar 13 2023
For n not in A020914, a(n) = 2*a(n-1). - Ruud H.G. van Tol, Apr 12 2023
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)
def A182137(n):
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
return minimized # D. S. McNeil, Apr 14 2012
(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
Jérôme STORTI, Apr 14 2012
EXTENSIONS
More terms from D. S. McNeil, Apr 14 2012
a(31) from Jérôme STORTI, Apr 22 2012
a(32)-a(38) from Jérôme STORTI, Jul 21 2012
a(39) from Jérôme STORTI, Jul 26 2012
a(40) from Jérôme STORTI, Feb 08 2014
a(37) and a(39) corrected by Jérôme STORTI, Dec 29 2021
STATUS
approved