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%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
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