A376908 - OEIS

%I #14 Oct 16 2024 21:35:42

%S 0,0,2,2,4,4,0,3,2,6,0,4,12,12,0,5,4,16,4,0,6,12,0,7,3,6,18,4,24,3,9,

%T 6,24,0,8,2,8,16,32,24,0,9,18,36,8,24,0,10,0,11,4,28,24,0,12,48,0,13,

%U 12,0,14,0,15,5,10,20,6,30,9,27,4,32,64,10,30,60,14

%N a(0) = 0. For n >= 0, if a(n-1) is a novel term, a(n) = a(a(n-1)). If a(n-1) > 0 has been seen k(>1) times, a(n) = k*a(n-1), unless k*a(n-1) > n, in which case a(n) = 0. If a(n-1) = 0 has been seen k(>1) times, a(n) = k.

%C It seems that a(n) <= n with equality (in first 2^24 terms) only for n = 0,2,4,12,200, 216, 360, 2646.

%C a(n-1) = 0 is followed by k, the number of zeros seen so far, (rather than k*a(n-1) as for nonzero terms), otherwise the sequence would contain only zeros. The a(n) = 0 clause is necessary to ensure continuation of the sequence at points, since without it there would be occasions where a(n) = a(a(m)) but a(m) is not yet defined. The first such instance follows a(5) = 4, the second time 4 appears, upon which we would have a(6) = 2*4 = 8, a novel term so a(7) = a(a(8)), but a(8) is not yet defined. By setting a(6) = 0 the sequence can continue (see Example).

%H Michael De Vlieger, <a href="/A376908/b376908.txt">Table of n, a(n) for n = 0..10000</a>

%H Michael De Vlieger, <a href="/A376908/a376908.png">Log log scatterplot of a(n)</a> n = 0..2^20, showing a(n) = 0 instead as a(n) = 1/2 in red for visibility.

%e a(0) = 0 a novel term, implies a(1) = a(a(0)) = a(0) = 0.

%e Now we have a(0) = a(1) = 0, so k = 2, which from the definition means a(2) = 2.

%e a(2) = 2 is a novel term so a(3) = a(a(2)) = a(2) = 2.

%e Because 2 has occurred twice we have a(4) = 2*2 = 4.

%e a(4) = 4 is a novel term so a(5) = a(a(4)) = a(4) = 4, the second occurrence of 4.

%e Since 2*4 = 8 exceeds 6, the index of the next term we invoke the 0 clause so a(6) = 0.

%e Now 0 has occurred three times, so a(7) = 3, a novel term making a(8) = a(a(3)) = 2.

%t nn = 120; c[_] := 0; j = a[0] = 0;

%t Do[If[c[j] == 0,

%t   k = a[j],

%t   If[j == 0,

%t     k = c[0] + 1,

%t     If[# > n, k = 0, k = #] &[j*(c[j] + 1)] ] ];

%t   c[j]++; Set[{a[n], j}, {k, k}], {n, nn}];

%t Array[a, nn, 0] (* _Michael De Vlieger_, Oct 10 2024 *)

%Y Cf. A377027.

%K nonn

%O 0,3

%A _David James Sycamore_, Oct 08 2024

%E More terms from _Michael De Vlieger_, Oct 10 2024