cp's OEIS Frontend

A271720 a(1)=1; for n>1, define a sequence {b(m), m >= 1} by b(1)=n b(2)=13, and b(m) = A020639(b(m-2)) + A006530(b(m-1)); then a(n) is the number of terms in that sequence before the first of the infinite string of 4s.

%I A271720 #11 Oct 01 2018 21:08:08
%S A271720 8,13,15,13,4,13,12,13,15,13,4,13,16,13,15,13,12,13,15,13,15,13,4,13,
%T A271720 4,13,15,13,8,13,12,13,15,13,4,13,12,13,15,13,4,13,8,13,15,13,12,13,
%U A271720 12,13,15,13,12,13,4,13,15,13,4,13,8,13,15,13,4,13,12,13,15,13,8,13,11,13,15,13,12,13
%N A271720 a(1)=1; for n>1, define a sequence {b(m), m >= 1} by b(1)=n b(2)=13, and b(m) = A020639(b(m-2)) + A006530(b(m-1)); then a(n) is the number of terms in that sequence before the first of the infinite string of 4s.
%C A271720 Note that the majority of the terms (every other term, initially) are equal to b(2), which is 13. This happens with several other values of b(2) less than 20. Many other values for b(2) have been tested, and it seems that for all b(2) < 100000000, a(n) < 20.
%C A271720 Records 8, 13, 15, 16, 19, 20, 24, ... occur at 1, 2, 3, 13, 349, 3919, 55633, ...
%H A271720 Antti Karttunen, <a href="/A271720/b271720.txt">Table of n, a(n) for n = 1..65537</a>
%e A271720 n = 6; the sequence is:
%e A271720 6, 13, 15, 18, 6, 5, 7, 12, 10, 7, 9, 10, 8, 4, 4, 4, ...
%e A271720 There are 13 terms before the first of the infinite 4s; a(6) = 13.
%e A271720 For n = 55633 the sequence is: 55633, 13, 55646, 27836, 6961, 6963, 7172, 166, 85, 19, 24, 22, 13, 15, 18, 6, 5, 7, 12, 10, 7, 9, 10, 8, 4, 4, 4, ... . As the first 4 comes as the 25th term, a(55633) = 24. - _Antti Karttunen_, Oct 01 2018
%t A271720 Table[
%t A271720 Clear[h];
%t A271720 h[1]=x;
%t A271720 h[2]=13;
%t A271720 h[n_]:=FactorInteger[h[n-1]][[-1,1]]+FactorInteger[h[n-2]][[1,1]];
%t A271720 Position[Array[h,100],4][[1,1]]-1,
%t A271720 {x,1,100}] (*This only works for x≠4*)
%o A271720 (PARI)
%o A271720 A006530(n) = if(n>1, vecmax(factor(n)[, 1]), 1); \\ From A006530
%o A271720 A020639(n) = if(1==n, n, factor(n)[1, 1]);
%o A271720 A271720(n) = { my(up=1001, bvec = vector(up), m=1); bvec[1] = n; bvec[2] = 13; for(n=3,oo,bvec[n] = A020639(bvec[n-2])+A006530(bvec[n-1]); if(4==bvec[n], return(n-1))); }; \\ _Antti Karttunen_, Oct 01 2018
%K A271720 nonn
%O A271720 1,1
%A A271720 _Cody M. Haderlie_, Apr 12 2016