

A351871


a(1) = 1, a(2) = 2; a(n) = gcd(a(n1), a(n2)) + (a(n1) + a(n2))/gcd(a(n1), a(n2)).


6



1, 2, 4, 5, 10, 8, 11, 20, 32, 17, 50, 68, 61, 130, 192, 163, 356, 520, 223, 744, 968, 222, 597, 276, 294, 101, 396, 498, 155, 654, 810, 250, 116, 185, 302, 488, 397, 886, 1284, 1087, 2372, 3460, 1462, 2463, 3926, 6390, 5160, 415, 1120, 312, 187, 500, 688, 301, 66, 368, 219, 588, 272, 219, 492, 240, 73, 314, 388
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,2


COMMENTS

After the first 277 terms, the sequence values repeat periodically with a period of 2901. The maximum value of a(n) is 2269429312765395470820, whose first occurrence appears at n = 2006.
Changing the initial terms a(1) and a(2) generates other periodic sequences. The periods found empirically are 3, 9, 155, 2901. It is not known whether the number of possible periods is finite.
Let a(1), a(2) be the first two (positive) integers, and for n>2 define a(n)=g+(a(n1)+(a2))/g, where g=gcd(a(n1),a(n2)).
If a(1) and a(2) are odd then it is easy to see that all numbers in the sequence are odd.
If a(1) or a(2) is even, then by induction out of every two consecutive numbers in the sequence at least one of them is even.
This partitions the sequences into two groups.
Conjecture: In the first group the sequence always goes to infinity (as in A355898), and in the second group it always goes to a cycle (as in the present sequence).
Here are three more cycle lengths:
For a(1)=52, a(2)=378 the sequence starts with: 52, 378, 217, 92, 310, 203, 514, 718, 618, 670, 646, 660, 655, 268, 924, 302, 615, 918, 514, 718, ... and has a cycle length of 12, starting at 514.
For a(1)=264, a(2)=1037 the sequence starts with
264, 1037, 1302, 2340, 613, 2954, 3568, 3263, 6832, 10096, 1074, 5587, 6662, 12250, 9458, 10856, 10159, 21016, 31176, 6532, 9431, 15964, 25396, 10344, 8939, 19284, 28224, 3971, 32196, 36168, 5709, 1302, 2340, ...
and has a cycle length of 29, starting at 1302.
for a(1)=542, a(2)=6017 the cycle has length 802 and the maximum term is 557981456058.
(End)


LINKS



FORMULA

For n >= 278, a(2901 + n) = a(n).


EXAMPLE

a(3) = gcd(1,2) + (1+2)/gcd(1,2) = 1 + 3/1 = 4.
a(4) = gcd(2,4) + (2+4)/gcd(2,4) = 2 + 6/2 = 5.
a(5) = gcd(4,5) + (4+5)/gcd(4,5) = 1 + 9/1 = 10.
a(6) = gcd(5,10) + (5+10)/gcd(5,10) = 5 + 15/5 = 8.
...
a(3179) = a(2901 + 278) = a(278) = 40.


MAPLE

A351871 := proc(u, v, M) local n, r, s, g, t, a;
a:=[u, v]; r:=u; s:=v;
for n from 1 to M do g:=gcd(r, s); t:=g+(r+s)/g; a:=[op(a), t];
r:=s; s:=t; od;
a;
end proc;


MATHEMATICA

a[1] = 1; a[2] = 2; a[n_] := a[n] = GCD[a[n  1], a[n  2]] + (a[n  1] + a[n  2])/GCD[a[n  1], a[n  2]]; Array[a, 50] (* Amiram Eldar, Feb 24 2022 *)


PROG

(Python)
from math import gcd
a, terms = [1, 2], 65
[a.append(gcd(a[1], a[2]) + (a[1] + a[2])//gcd(a[1], a[2])) for n in range(3, terms+1)]
(PARI) {a351871(N=65, A1=1, A2=2)= my(a=vector(N)); a[1]=A1; a[2]=A2; for(n=1, N, if(n>2, my(g=gcd(a[n1], a[n2])); a[n]=g+(a[n1]+a[n2])/g); print1(a[n], ", ")) } \\ Ruud H.G. van Tol, Sep 19 2022


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



STATUS

approved



