cp's OEIS Frontend

A379757 a(n) = a(n-1) + 1 with two exceptions: if a(n-1) is prime, a(n) = a(n-2) + a(n-1), or if a(n-1) is a power, a(n) = a(n-1) / (root factor), with initial three terms are 0, 1, 2.

%I A379757 #27 Apr 13 2025 16:24:18
%S A379757 0,1,2,3,5,8,4,2,6,7,13,20,21,22,23,45,46,47,93,94,95,96,97,193,290,
%T A379757 291,292,293,585,586,587,1173,1174,1175,1176,1177,1178,1179,1180,1181,
%U A379757 2361,2362,2363,2364,2365,2366,2367,2368,2369,2370,2371,4741,4742,4743
%N A379757 a(n) = a(n-1) + 1 with two exceptions: if a(n-1) is prime, a(n) = a(n-2) + a(n-1), or if a(n-1) is a power, a(n) = a(n-1) / (root factor), with initial three terms are 0, 1, 2.
%C A379757 The construction rules are very basic, but lead to somewhat surprising results. Terms that are perfect powers are extremely rare (only n=6,7 so far). Additionally, the sequence is nearly all composites. Comparing to A000045, eight early distinct terms are in common, but it is unclear when another intersection is seen.
%F A379757 Conjecture: log(a(n)) ~ k*sqrt(n).
%e A379757 We know a(1)=0, a(2)=1, a(3)=2. Since a(3) is prime, a(4)=a(2)+a(3)=3. Since a(4) is prime, a(5)=a(3)+a(4)=5. Similarly, a(6)=a(4)+a(5)=8. Since a(6) is a perfect power, a(7) = a(6)/2 since 8=2^3. Since a(7)=4 is another perfect power, a(8)=4/2=2. Since a(8) is prime, a(9)=a(7)+a(8)=6. For clarity, if a(n-1) = r^k, then a(n) = a(n-1)/r.
%t A379757 a[n_] := a[n] = If[n < 4, n-1, If[PrimeQ[a[n-1]], a[n-1] + a[n-2], If[(g = GCD @@ FactorInteger[a[n-1]][[;; , 2]]) > 1, a[n-1]^(1 - 1/g), a[n-1] + 1]]]; Array[a, 54] (* _Amiram Eldar_, Apr 10 2025 *)
%Y A379757 Cf. A000045, A001597, A089580.
%K A379757 nonn,easy
%O A379757 1,3
%A A379757 _Bill McEachen_, Jan 02 2025