A214551 - OEIS

%I #83 Mar 28 2022 11:20:20

%S 1,1,1,2,3,4,3,2,3,2,2,5,7,9,14,3,4,9,4,2,11,15,17,28,43,60,22,65,25,

%T 47,112,137,184,37,174,179,216,65,244,115,36,70,37,73,143,180,253,36,

%U 6,259,295,301,80,75,376,57,44,105,54,49,22,38,87,109,147

%N Reed Kelly's sequence: a(n) = (a(n-1) + a(n-3))/gcd(a(n-1), a(n-3)) with a(0) = a(1) = a(2) = 1.

%C Like Narayana's Cows sequence A000930, except that the sums are divided by the greatest common divisor (gcd) of the prior terms.

%C It is a strong conjecture that 8 and 10 are missing from this sequence, but it would be nice to have a proof! See A214321 for the conjectured values. [I have often referred to this as "Reed Kelly's sequence" in talks.] - _N. J. A. Sloane_, Feb 18 2017

%H T. D. Noe and N. J. A. Sloane, <a href="/A214551/b214551.txt">Table of n, a(n) for n = 0..10000</a>

%H Benoit Cloitre, <a href="/A214551/a214551.png">Graph of a(n)^(1/n) for n=1 up to 381817</a>

%H N. J. A. Sloane, <a href="https://www.youtube.com/watch?v=9ogbsh8KuEM">Exciting Number Sequences</a> (video of talk), Mar 05 2021.

%F It appears that, very roughly, a(n) ~ constant*exp(0.123...*n). - _N. J. A. Sloane_, Sep 07 2012. See next comment for more precise estimate.

%F If a(n)^(1/n) converges the limit should be near 1.126 (see link). - _Benoit Cloitre_, Nov 08 2015

%F _Robert G. Wilson v_ reports that at around 10^7 terms a(n)^(1/n) is about exp(1/8.4). - _N. J. A. Sloane_, May 05 2021

%e a(14)=9, a(16)=3, therefore a(17)=(9+3)/gcd(9,3) = 12/3 = 4.

%e a(24)=28, a(26)=60, therefore a(27)=(28+60)/gcd(28,60) = 88/4 = 22.

%p a:= proc(n) a(n):= `if`(n<3, 1, (a(n-1)+a(n-3))/igcd(a(n-1), a(n-3))) end:

%p seq(a(n), n=0..100); # _Alois P. Heinz_, Oct 18 2012

%t t = {1, 1, 1}; Do[AppendTo[t, (t[[-1]] + t[[-3]])/GCD[t[[-1]], t[[-3]]]], {100}]

%t f[l_List] := Append[l, (l[[-1]] + l[[-3]])/GCD[l[[-1]], l[[-3]]]]; Nest[f, {1, 1, 1}, 62] (* _Robert G. Wilson v_, Jul 23 2012 *)

%t RecurrenceTable[{a[0]==a[1]==a[2]==1,a[n]==(a[n-1]+a[n-3])/GCD[ a[n-1], a[n-3]]},a,{n,70}] (* _Harvey P. Dale_, May 06 2014 *)

%o (Perl)

%o use bignum;

%o my @seq = (1, 1, 1);

%o print "1 1\n2 1\n3 1\n";

%o for ( my $i = 3; $i < 400; $i++ )

%o {

%o     my $next = ( $seq[$i-1] + $seq[$i-3] ) /

%o         gcd( $seq[$i-1], $seq[$i-3] );

%o     my $ind = $i+1;

%o     print "$ind $next\n";

%o     push( @seq, $next );

%o }

%o sub gcd {

%o     my ($x, $y) = @_;

%o     ($x, $y) = ($y, $x % $y) while $y;

%o     return $x;

%o }

%o (Haskell)

%o a214551 n = a214551_list !! n

%o a214551_list = 1 : 1 : 1 : zipWith f a214551_list (drop 2 a214551_list)

%o    where f u v = (u + v) `div` gcd u v

%o -- _Reinhard Zumkeller_, Jul 23 2012

%o (Sage)

%o def A214551Rec():

%o     x, y, z = 1, 1, 1

%o     yield x

%o     while True:

%o         x, y, z =  y, z, (z + x)//gcd(z, x)

%o         yield x

%o A214551 = A214551Rec();

%o print([next(A214551) for _ in range(65)])  # _Peter Luschny_, Oct 18 2012

%o (PARI) first(n)=my(v=vector(n+1)); for(i=1,min(n,3),v[i]=1); for(i=4,#v, v[i]=(v[i-1]+v[i-3])/gcd(v[n-1],v[i-3])); v \\ _Charles R Greathouse IV_, Jun 21 2017

%o (Python)

%o from math import gcd

%o def aupton(nn):

%o     alst = [1, 1, 1]

%o     for n in range(3, nn+1):

%o         alst.append((alst[n-1] + alst[n-3])//gcd(alst[n-1], alst[n-3]))

%o     return alst

%o print(aupton(64)) # _Michael S. Branicky_, Mar 28 2022

%Y Similar to A000930. Cf. A341312, A341313, which are also similar.

%Y Cf. also A214320, A214321, A214322, A214323 (gcd's), A219898 (records), A214324, A214325, A214330, A214331, A214809, A227836, A227837.

%Y Starting with a(2) = 3 gives A214626. - _Reinhard Zumkeller_, Jul 23 2012

%K nonn,nice

%O 0,4

%A _Reed Kelly_, Jul 20 2012