

A360464


a(n) = a(n1) + a(n2)  a(n3) + gcd(a(n1), a(n3)), with a(1) = a(2) = a(3) = 1.


2



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
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OFFSET

1,4


COMMENTS

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


LINKS



FORMULA

a(n) = a(n1) + a(n2)  a(n3) + gcd(a(n1), a(n3)).


EXAMPLE

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


MAPLE

A:= Vector(200):
A[1]:= 1: A[2]:= 1: A[3]:= 1:
for n from 4 to 200 do
A[n]:= A[n1] + A[n2]  A[n3] + igcd(A[n1], A[n3])
od:


MATHEMATICA

a[1] = a[2] = a[3] = 1; a[n_] := a[n] = a[n1] + a[n2]  a[n3] + GCD[a[n1], a[n3]]; Array[a, 100] (* Amiram Eldar, Feb 08 2023 *)


PROG

(Python)
from math import gcd
a = [0, 1, 1, 1]
[a.append(a[n1]+a[n2]a[n3]+gcd(a[n1], a[n3])) for n in range(4, 58)]


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



STATUS

approved



