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.

8, 13, 15, 13, 4, 13, 12, 13, 15, 13, 4, 13, 16, 13, 15, 13, 12, 13, 15, 13, 15, 13, 4, 13, 4, 13, 15, 13, 8, 13, 12, 13, 15, 13, 4, 13, 12, 13, 15, 13, 4, 13, 8, 13, 15, 13, 12, 13, 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

Offset: 1

Author

Cody M. Haderlie, Apr 12 2016

Keywords

• nonn

Comments

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.

Records 8, 13, 15, 16, 19, 20, 24, ... occur at 1, 2, 3, 13, 349, 3919, 55633, ...

Examples

			n = 6; the sequence is:
6, 13, 15, 18, 6, 5, 7, 12, 10, 7, 9, 10, 8, 4, 4, 4, ...
There are 13 terms before the first of the infinite 4s; a(6) = 13.
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
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..65537

Programs

Mathematica

Table[
Clear[h];
h[1]=x;
h[2]=13;
h[n_]:=FactorInteger[h[n-1]][[-1,1]]+FactorInteger[h[n-2]][[1,1]];
Position[Array[h,100],4][[1,1]]-1,
{x,1,100}] (*This only works for x≠4*)

PARI

A006530(n) = if(n>1, vecmax(factor(n)[, 1]), 1); \\ From A006530
A020639(n) = if(1==n, n, factor(n)[1, 1]);
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