A186107
Numerator of the frequency of the n-th dropping time in the Collatz iteration.
Original entry on oeis.org
1, 1, 1, 1, 3, 7, 3, 15, 85, 173, 119, 961, 663, 8045, 17637, 51033, 54475, 312455, 663535, 950235, 5936673, 1684037, 39993895, 87986917, 128989251, 205059181, 949737339, 2861515293, 400296173, 19018424205
Offset: 1
The frequencies are 1/2, 1/4, 1/16, 1/16, 3/128, 7/256, 3/256, 15/2048, 85/8192,....
A186110
Denominator of the cumulative frequency of the dropping time in the Collatz iteration.
Original entry on oeis.org
2, 4, 16, 8, 128, 256, 16, 2048, 8192, 32768, 32768, 262144, 16384, 2097152, 8388608, 16777216, 33554432, 134217728, 536870912, 1073741824, 4294967296, 4294967296, 34359738368, 137438953472, 274877906944, 137438953472, 2199023255552, 4398046511104, 4398046511104, 35184372088832
Offset: 1
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.
Original entry on oeis.org
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
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.
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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 *)
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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
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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
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