A268268
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.
Original entry on oeis.org
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
Offset: 1
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}.
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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
A268288
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.
Original entry on oeis.org
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
Offset: 1
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).
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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
A268468
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.
Original entry on oeis.org
4, 507, 1377, 12608, 55291, 55290, 55289, 145645, 104455, 104454, 336734, 336733, 336732, 525907, 1960873, 1836239, 2176265, 2176264, 2176263, 2176262, 2176261, 2176260, 2176259, 2176258
Offset: 1
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}.
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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:
A341362
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.
Original entry on oeis.org
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
Offset: 1
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.
Showing 1-4 of 4 results.
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