A268253
a(n) begins the first chain of 5 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
98, 130, 290, 354, 386, 387, 418, 507, 514, 610, 628, 802, 840, 841, 866, 943, 944, 945, 1003, 1121, 1122, 1154, 1172, 1186, 1272, 1314, 1378, 1442, 1494, 1495, 1496, 1497, 1538, 1634, 1680, 1681, 1682, 1683, 1684, 1698, 1699, 1826, 1890, 1922, 1923, 1991, 1992
Offset: 1
In 5-tuple of consecutive {h(k)}: {h(1272),h(1273),h(1274),h(1275),h(1276)} = {57,31,57,83,57}, the central value is 57, and 57+57 = 31+83 = 2*57. Hence, 1272 is in the sequence.
Alternatively, the symmetry can be seen from the differences between consecutive {h(k)}. For {57,31,57,83,57}, the differences {h(k+1)-h(k)} are {-26,26,26,-26}.
-
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+4]==f[m+1]+f[m+3]&&f[m+2]==(f[m]+f[m+4])/2,AppendTo[lst,m]],{m,1,4000}];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}.
-
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:
Showing 1-2 of 2 results.
Comments