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a(n) begins the first chain of 7 consecutive positive integers of h-values with symmetrical gaps about the center, where h(k) is the length of the finite sequence k, f(k), f(f(k)), ..., 1 in the Collatz (or 3x + 1) problem.
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2
943, 1377, 1494, 1495, 1680, 1681, 1682, 1991, 1992, 1993, 2358, 2359, 2987, 2988, 2989, 2990, 2991, 2992, 2993, 2994, 3288, 3289, 3360, 3542, 3543, 3982, 3983, 3984, 3985, 3986, 3987, 3988, 4193, 4481, 4482, 4722, 4723, 4724, 4725, 4897, 4936, 4937, 5313, 5314
COMMENTS
Or numbers k such that h(k) + h(k+6) = h(k+1) + h(k+5) and h(k+3) = (h(k) + h(k+6))/2, where h(k) is the length of k, f(k), f(f(k)), ..., 1 in the Collatz (or 3x + 1) problem.
The symmetry can be seen from the differences between consecutive h(k) (see the example).
The 7-tuple of consecutive h(k) are symmetric about the central value h(k+3) which are averages of both their immediate neighbors, their second neighbors and their third neighbors.
A majority of numbers of the sequence generate trivial 7-tuples {m, m, m, m, m, m, m}.
The 7-tuples {h(k)} of the form {m, p, p, p, p, p, q} are generated by the numbers of the sequence 1377, 4897, ...
The 7-tuples {h(k)} of the form {m, m, p, m, q, m, m} are generated by the numbers of the sequence 5511, 58757, ...
The 7-tuples {h(k)} of the form {m, p, m, m, m, q, p} are generated by the numbers of the sequence 9514, ...
The 7-tuples {h(k)} of the form {m, m, p, p, p, q, q} are generated by the numbers of the sequence 21442, 25666, ...
The 7-tuples {h(k)} of the form {m, m, m, p, q, q, q} are generated by the numbers of the sequence 55108, 55293, ...
EXAMPLE
In 7-tuple of consecutive {h(k)}: {h(9514),h(9515),h(9516),h(9517),h(9518),h(9519),h(9520)} = {78,52,78,78,78,104,78}, the central value is 78, and 78+78 = 52+104 = 2*78. Hence, 9514 is in the sequence.
Alternatively, the symmetry can be seen from the differences between consecutive {h(k)}. For {78,52,78,78,78,104,78}, the differences {h(k+1)-h(k)} are {-26,26,0,0,26,-26}.
MATHEMATICA
lst={}; f[n_]:=Module[{a=n, k=0}, While[a!=1, k++; If[EvenQ[a], a=a/2, a=a*3+1]]; k]; Do[If[f[m]+f[m+6]==f[m+1]+f[m+5]&&f[m+2]+f[m+4]==f[m]+f[m+6]&& f[m]+f[m+6]==f[m+2]+f[m+4]&&f[m+3]==(f[m]+f[m+6])/2, AppendTo[lst, m]], {m, 1, 6000}]; lst
a(n) begins the first chain of 9 consecutive positive integers of h-values with symmetrical gaps about the center, where h(k) is the length of the finite sequence k, f(k), f(f(k)), ...., 1 in the Collatz (or 3x + 1) problem.
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1
1680, 1991, 2987, 2988, 2989, 2990, 2991, 2992, 3982, 3983, 3984, 3985, 3986, 4722, 4723, 5313, 5314, 5315, 5316, 5317, 6576, 6577, 6578, 7083, 7084, 7085, 7086, 7087, 7088, 7089, 7090, 7091, 7794, 7795, 7976, 7977, 7978, 7979, 7980, 7981, 8769, 8770, 8771
COMMENTS
The 9-tuple of consecutive h(k) are symmetric about the central value h(k+4) which are averages of both their immediate neighbors, their second neighbors, their third neighbors and their fourth neighbors.
A majority of numbers the sequence generate trivial 9-tuples (m, m, m, m, m, m, m, m, m).
For a(n) < 200000, the following sets have been identified:
The 9-tuples {h(k)} of the form {m, p, p, p, p, p, p, p, q} are generated by the numbers of the sequence 12608, 16915, 39169, ...
The 9-tuples {h(k)} of the form {m, p, q, q, q, q, q, m, p} are generated by the numbers of the sequence 40553, ...
The 9-tuples {h(k)} of the form {m, p, p, p, q, m, m, m, p} are generated by the numbers of the sequence 55107, 124739, ...
The 9-tuples {h(k)} of the form {m, m, m, m, p, q, q, q, q} are generated by the numbers of the sequence 55292, 90396, 118109, ...
The 9-tuples {h(k)} of the form {m, m, m, p, m, q, m, m, m} are generated by the numbers of the sequence 58756, 71236, 79428, ...
The 9-tuples {h(k)} of the form {m, m, p, m, m, m, q, m, m} are generated by the numbers of the sequence 78021, ...
The 9-tuples {h(k)} of the form {m, p, m, m, m, m, m, q, m} are generated by the numbers of the sequence 93600, 124768, ...
The 9-tuples {h(k)} of the form {m, m, m, p, p, p, q, q, q} are generated by the numbers of the sequence 160705, ...
EXAMPLE
In 9-tuple of consecutive h(k): {h(55107),h(55108),...,h(55115)} = {184,60,60,60,122,184,184,184,60}, the central value is 122, and 184+60 = 2*122. Hence, 55107 is in the sequence.
Alternatively, the symmetry can be seen from the differences between consecutive h(k). For {184,60,60,60,122,184,184,184,60}, the differences h(k+1)-h(k) are (-124,0,0,62,62,0,0,-124).
MATHEMATICA
lst={}; f[n_]:=Module[{a=n, k=0}, While[a!=1, k++; If[EvenQ[a], a=a/2, a=a*3+1]]; k]; Do[If[f[m]+f[m+8]==f[m+1]+f[m+7]&&f[m+2]+f[m+6]==f[m+3]+f[m+5]&& f[m]+f[m+8]==f[m+3]+f[m+5]&&f[m+4]==(f[m]+f[m+8])/2, AppendTo[lst, m]], {m, 1, 6000}]; lst
Least k starting a chain or (2n+1)-tuple of consecutive integers {h(k+i)}, i=0,1,...,2n (excluding the trivial chain when h(k) = h(k+1) = ... = h(k+2n)) with symmetrical gaps about the center, where h(k) is the length of the finite set {k, f(k), f(f(k)),...,1} in the Collatz (or 3x + 1) problem.
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4, 507, 1377, 12608, 55291, 55290, 55289, 145645, 104455, 104454, 336734, 336733, 336732, 525907, 1960873, 1836239, 2176265, 2176264, 2176263, 2176262, 2176261, 2176260, 2176259, 2176258
COMMENTS
It is interesting to search for symmetries in the sequence A006577 (number of halving and tripling steps to reach 1 in '3x+1' problem). The symmetrical architecture is given by the following property: h(k) + h(k+2n) = h(k+1)+ h(k+2n-1)= ... = h(k+n-1)+ h(k+n+1) = 2*h(k+n) where h(k+n) is the symmetrical center. We observe two essential families of chains containing symmetries:
(i) A majority of trivial chains are obtained when a(n) begins the first chain of 2n+1 consecutive positive integers where h(k) = h(k+1) = ... = h(k+2n). This case is not considered here, but is mentioned in A078441. If this case were to be considered, it would be found that a(n) = A078441(2n+1) = 28, 98, 943, 1680, 2987, 2987, 7083, 7083, ... (ii) Chains having several distinct values. This case is more interesting, with an important question: how many distinct values can contain such a set {h(k)}? It appears that this value is equal to 3, and the sequence of the reduced corresponding sets is {4, 5, 6}, {35, 48, 61}, {96, 127, 158}, {32, 63, 94}, {91, 122, 153}, {91, 122, 153}, {91, 122, 153}, {126, 188, 250}, ... The corresponding symmetrical centers are 5, 48, 127, 63, 122, 122, 122, 188, 172, 172, 184, 184, 184, 164, 184, 202, 210, 210, 210, 210, 210, 210, 210, 210, ...
EXAMPLE
a(1) = 4 because in the first 3-tuple {h(4),h(5),h(6)} = {2, 5, 8}, the numbers are symmetric w.r.t. the central h(5)= 5 since 2+8 = 2*5. Hence 4 belongs to the sequence.
Alternatively, the symmetry can be seen from the differences between consecutive h(k). For {2,5,8}, the set of the differences is {3,3}.
a(3) = 10136 because in 7-tuple of consecutive {h(k)} = {34, 34, 34, 60, 86, 86, 86}, the numbers are symmetric w.r.t. its central h(k+3) = 60, since 34+86 = 2*60, and this is the smallest such 7-tuple. Hence 10136 belongs to the sequence.
Alternatively, the symmetry can be seen from the differences between consecutive h(k). From the set {34, 34, 34, 60, 86, 86, 86}, the set of the differences is {0,0,26,26,0,0}.
MAPLE
nn:=10^7:T:=array(1..nn):
for j from 1 to 5*10^6 do:
k:=0:m:=j:it:=0:
for i from 1 to nn while(m<>1) do:
if irem(m, 2)=0
then
m:=m/2:
else
m:=3*m+1:
fi:
it:=it+1:
od:
k:=j:T[j]:=it:
od:
for n from 3 by 2 to 50 do:
ii:=0:
for j from 1 to nn while(ii=0)do:
q:=T[j]+T[j+n-1]:
itr:=0:lst:={}:
for jj from 1 to (n-1)/2 do:
lst:=lst union {T[j+jj-1]} union {T[j+n-jj]}:
if T[j+jj-1]+T[j+n-jj]=q and T[j+(n-1)/2]=q/2
then
itr:=itr+1:
else fi:
od:
if itr=(n-1)/2 and nops(lst)>1 then ii:=1:
printf("%d %d \n", n, j):
else
fi:
od:
od:
a(n) begins the first sequence of n consecutive positive integers with the same h-value and the same d-value in the Collatz (or '3x + 1') problem.
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0
1, 54, 108, 290, 290, 386, 172146, 298200, 596400, 596400, 596400, 795201, 795201, 2849196, 2849196, 8965036, 33659819, 45529226, 52417676, 93186987, 104161282, 104161282, 104161282, 436089218, 436089218, 605581697, 934358530, 934358530, 934358530, 3826876112
COMMENTS
The Collatz function is as follows: F(x) = x/2 if x is even, otherwise F(x) = 3*x+1.
It is conjectured that starting from any number, and repeatedly applying the function on its previous result, we will always reach 1.
The d-value (or flight duration, A006577) is the number of steps needed to reach 1. The h-value (or flight height, A025586) is the maximum of the number's trajectory.
EXAMPLE
a(3) = 108 because 108, 109 and 110 have same d-value (113) and same h-value (9232).
And 108 is the smallest number starting such sequence of 3 consecutive positive integers with same d-value and same h-value.
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