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%I #58 Feb 26 2023 20:09:54
%S 1,1,1,2,3,5,7,10,17,21,29,34,43,49,59,66,77,85,97,106,119,129,143,
%T 154,169,193,209,234,251,277,295,322,341,369,389,418,439,469,491,522,
%U 545,577,601,634,659,693,719,754,781,817,845,882,911,949,979,1018,1049
%N a(n) = a(n-1) + a(n-2) - a(n-3) + gcd(a(n-1), a(n-3)), with a(1) = a(2) = a(3) = 1.
%C All terms beyond a(458) are divisible by 5. - _Jack Braxton_, Feb 14 2023
%C From _Robert Israel_, Feb 15 2023: (Start)
%C a(n) is divisible by 25 for n >= 8857.
%C a(n) is divisible by 125 for n >= 8861.
%C a(n) is divisible by 625 for n >= 8945.
%C a(n) is divisible by 1875 for n >= 9060.
%C a(n) is divisible by 5625 for n >= 9064.
%C Do there exist N > 9064 and m > 5625 such that a(n) is divisible by m for n >= N? If so, N >= 2*10^7. (End)
%C From _Pontus von Brömssen_, Feb 17 2023: (Start)
%C (Answer to the question above.) Yes:
%C a(n) has an additional factor 5 for n >= 64423404 (so a(n) is divisible by 28125);
%C a(n) has an additional factor 5 for n >= 64423410;
%C a(n) has an additional factor 3 for n >= 64424073;
%C a(n) has an additional factor 21 for n >= 64424144;
%C a(n) has an additional factor 3 for n >= 64428745;
%C a(n) has an additional factor 7 for n >= 64428748;
%C a(n) has an additional factor 3 for n >= 64428756;
%C a(n) has an additional factor 3 for n >= 64428821;
%C a(n) has an additional factor 3 for n >= 64514757;
%C a(n) has an additional factor 5 for n >= 64514783;
%C a(n) has an additional factor 3 for n >= 797299454;
%C a(n) has an additional factor 3 for n >= 797299480;
%C a(n) has an additional factor 5 for n >= 797299487;
%C a(n) has an additional factor 3 for n >= 797299490;
%C a(n) has an additional factor 5 for n >= 797299652;
%C a(n) has an additional factor 3 for n >= 797299667;
%C a(n) has an additional factor 7 for n >= 797299846;
%C a(n) has an additional factor 3 for n >= 797299933.
%C The index for which the next additional factor occurs (if it exists) is larger than 2*10^10.
%C (End)
%H Robert Israel, <a href="/A360464/b360464.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n) = a(n-1) + a(n-2) - a(n-3) + gcd(a(n-1), a(n-3)).
%e a(5) = 2 + 1 - 1 + gcd(2, 1) = 3.
%p A:= Vector(200):
%p A[1]:= 1: A[2]:= 1: A[3]:= 1:
%p for n from 4 to 200 do
%p A[n]:= A[n-1] + A[n-2] - A[n-3] + igcd(A[n-1],A[n-3])
%p od:
%p convert(A,list); # _Robert Israel_, Feb 15 2023
%t a[1] = a[2] = a[3] = 1; a[n_] := a[n] = a[n-1] + a[n-2] - a[n-3] + GCD[a[n-1], a[n-3]]; Array[a, 100] (* _Amiram Eldar_, Feb 08 2023 *)
%o (Python)
%o from math import gcd
%o a = [0, 1, 1, 1]
%o [a.append(a[n-1]+a[n-2]-a[n-3]+gcd(a[n-1], a[n-3])) for n in range(4, 58)]
%o print(a[1:]) # _Michael S. Branicky_, Feb 09 2023
%Y Cf. A083658, A248098.
%K nonn,easy
%O 1,4
%A _Jack Braxton_, Feb 08 2023