cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A277365 a(n) is the smallest number k such that f(k) = f(n) + f(n+1) and g(k) = g(n) + g(n+1), where f(n) (resp. g(n)) is the number of halving (resp. tripling) steps to reach 1 in the Collatz ('3x+1') problem.

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%I A277365 #13 Sep 14 2017 03:59:14
%S A277365 2,6,12,20,34,49,56,72,98,112,144,176,196,228,224,272,344,406,392,384,
%T A277365 448,520,576,688,688,772,913,912,912,1028,992,1040,1220,1152,1376,
%U A277365 1624,1624,1708,1624,1728,1728,1824,2160,2080,2080,2215,2559,2752,2884,2884,2752
%N A277365 a(n) is the smallest number k such that f(k) = f(n) + f(n+1) and g(k) = g(n) + g(n+1), where f(n) (resp. g(n)) is the number of halving (resp. tripling) steps to reach 1 in the Collatz ('3x+1') problem.
%C A277365 A006666: Number of halving steps to reach 1 in '3x+1' problem.
%C A277365 A006667: Number of tripling steps to reach 1 in '3x+1' problem.
%C A277365 We observe an interesting property: the subsequence {b(i)} of perfect squares is {49, 144, 196, 576, 3844, 12544, 15376, 51529, 61504, 246016, ...} with the property that b(3) = 4*b(1), b(4) = 4*b(2), b(7) = 4*b(5), b(9) = 4*b(7), b(10) = 4*b(9), ...
%C A277365 The primes of the sequence are 2, 2953, 3739, 9931, 38303, 44641, ...
%e A277365 a(3) = 12 because (A006666(3), A006667(3)) = (f(3), g(3)) = (5, 2) => f(12) = f(3) + f(4) = 5 + 2 = 7 and g(12) = g(3) + g(4) = 2 + 0 = 2.
%p A277365 nn:=10^6:U:=array(1..nn):V:=array(1..nn):
%p A277365 for i from 1 to nn do:
%p A277365 m:=i:it0:=0:it1:=0:
%p A277365    for j from 1 to nn while(m<>1) do:
%p A277365     if irem(m,2)=0
%p A277365      then
%p A277365      m:=m/2:it0:=it0+1:
%p A277365      else
%p A277365      m:=3*m+1:it1:=it1+1:
%p A277365     fi:
%p A277365    od:
%p A277365    U[i]:=it0:V[i]:=it1:
%p A277365   od:
%p A277365    for n from 1 to 100 do:
%p A277365    ii:=0:
%p A277365     for k from 1 to nn while(ii=0) do:
%p A277365      if U[k]=U[n]+U[n+1] and V[k]=V[n]+V[n+1]
%p A277365       then
%p A277365       ii:=1:printf(`%d, `,k):
%p A277365       else
%p A277365      fi:
%p A277365     od:
%p A277365 od:
%Y A277365 Cf. A006666, A006667.
%K A277365 nonn
%O A277365 1,1
%A A277365 _Michel Lagneau_, Oct 11 2016
%E A277365 Name edited by _Michel Marcus_, Sep 13 2017