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A269026 a(1)=1; for n>1, define a sequence {b(m), m >= 1} by b(1)=a(n-1), b(2)=n, 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. 1

%I #49 Apr 15 2016 13:55:51

%S 1,9,12,1,4,10,5,6,8,6,5,3,15,7,12,2,17,7,4,6,13,11,8,10,9,3,12,9,11,

%T 3,12,2,16,6,12,10,5,11,12,6,9,7,12,14,13,11,16,10,5,7,12,14,8,10,5,

%U 11,4,10,17,15,15,7,8,2,5,3,15,7,4,9,12,10,5,10,13,3,11,11,11

%N a(1)=1; for n>1, define a sequence {b(m), m >= 1} by b(1)=a(n-1), b(2)=n, 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.

%H Michel Marcus, <a href="/A269026/b269026.txt">Table of n, a(n) for n = 1..10000</a>

%e n = 3:

%e a(n-1) = a(2) = 9;

%e b(1) = 9, b(2) = 3;

%e the sequence generated is: 9, 3, 6, 6, 5, 7, 12, 10, 7, 9, 10, 8, 4, 4, 4, ...

%e There are 12 terms before the first of the infinite 4s, so a(3) = 12.

%o (PARI) spf(n) = if (n==1, 1, vecmin(factor(n)[,1]));

%o gpf(n) = if (n==1, 1, vecmax(factor(n)[,1]));

%o nbt(a, n) = {x = a; y = n; nb = 0; while (!((x==4) && (y==4)), z = spf(x) + gpf(y); x = y; y = z; nb++;); nb;}

%o lista(nn) = { print1(a=1, ", "); for (n=2, nn, na = nbt(a, n); print1(na, ", "); a = na;);} \\ _Michel Marcus_, Apr 12 2016

%Y Cf. A006530, A020639, A271621.

%K nonn

%O 1,2

%A _Cody M. Haderlie_, Apr 11 2016

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