A360464 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.

1, 1, 1, 2, 3, 5, 7, 10, 17, 21, 29, 34, 43, 49, 59, 66, 77, 85, 97, 106, 119, 129, 143, 154, 169, 193, 209, 234, 251, 277, 295, 322, 341, 369, 389, 418, 439, 469, 491, 522, 545, 577, 601, 634, 659, 693, 719, 754, 781, 817, 845, 882, 911, 949, 979, 1018, 1049

Offset: 1

Author

Jack Braxton, Feb 08 2023

Keywords

• nonn
• easy

Comments

All terms beyond a(458) are divisible by 5. - Jack Braxton, Feb 14 2023

From Robert Israel, Feb 15 2023: (Start)

a(n) is divisible by 25 for n >= 8857.

a(n) is divisible by 125 for n >= 8861.

a(n) is divisible by 625 for n >= 8945.

a(n) is divisible by 1875 for n >= 9060.

a(n) is divisible by 5625 for n >= 9064.

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)

From Pontus von Brömssen, Feb 17 2023: (Start)

(Answer to the question above.) Yes:

a(n) has an additional factor 5 for n >= 64423404 (so a(n) is divisible by 28125);

a(n) has an additional factor 5 for n >= 64423410;

a(n) has an additional factor 3 for n >= 64424073;

a(n) has an additional factor 21 for n >= 64424144;

a(n) has an additional factor 3 for n >= 64428745;

a(n) has an additional factor 7 for n >= 64428748;

a(n) has an additional factor 3 for n >= 64428756;

a(n) has an additional factor 3 for n >= 64428821;

a(n) has an additional factor 3 for n >= 64514757;

a(n) has an additional factor 5 for n >= 64514783;

a(n) has an additional factor 3 for n >= 797299454;

a(n) has an additional factor 3 for n >= 797299480;

a(n) has an additional factor 5 for n >= 797299487;

a(n) has an additional factor 3 for n >= 797299490;

a(n) has an additional factor 5 for n >= 797299652;

a(n) has an additional factor 3 for n >= 797299667;

a(n) has an additional factor 7 for n >= 797299846;

a(n) has an additional factor 3 for n >= 797299933.

The index for which the next additional factor occurs (if it exists) is larger than 2*10^10.

(End)

Examples

			a(5) = 2 + 1 - 1 + gcd(2, 1) = 3.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A083658, A248098.

Programs

Maple

A:= Vector(200):
A[1]:= 1: A[2]:= 1: A[3]:= 1:
for n from 4 to 200 do
  A[n]:= A[n-1] + A[n-2] - A[n-3] + igcd(A[n-1],A[n-3])
od:
convert(A,list); # Robert Israel, Feb 15 2023

Mathematica

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 *)

Python

from math import gcd
a = [0, 1, 1, 1]
[a.append(a[n-1]+a[n-2]-a[n-3]+gcd(a[n-1], a[n-3])) for n in range(4, 58)]
print(a[1:]) # Michael S. Branicky, Feb 09 2023

Formula

a(n) = a(n-1) + a(n-2) - a(n-3) + gcd(a(n-1), a(n-3)).