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 A214551 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. 32
 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, 47, 112, 137, 184, 37, 174, 179, 216, 65, 244, 115, 36, 70, 37, 73, 143, 180, 253, 36, 6, 259, 295, 301, 80, 75, 376, 57, 44, 105, 54, 49, 22, 38, 87, 109, 147 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,4 COMMENTS Like Narayana's Cows sequence A000930, except that the sums are divided by the greatest common divisor (gcd) of the prior terms. 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 LINKS T. D. Noe and N. J. A. Sloane, Table of n, a(n) for n = 0..10000 Benoit Cloitre, Graph of a(n)^(1/n) for n=1 up to 381817 N. J. A. Sloane, Exciting Number Sequences (video of talk), Mar 05 2021. FORMULA 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. If a(n)^(1/n) converges the limit should be near 1.126 (see link). - Benoit Cloitre, Nov 08 2015 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 EXAMPLE a(14)=9, a(16)=3, therefore a(17)=(9+3)/gcd(9,3) = 12/3 = 4. a(24)=28, a(26)=60, therefore a(27)=(28+60)/gcd(28,60) = 88/4 = 22. MAPLE a:= proc(n) a(n):= `if`(n<3, 1, (a(n-1)+a(n-3))/igcd(a(n-1), a(n-3))) end: seq(a(n), n=0..100); # Alois P. Heinz, Oct 18 2012 MATHEMATICA t = {1, 1, 1}; Do[AppendTo[t, (t[[-1]] + t[[-3]])/GCD[t[[-1]], t[[-3]]]], {100}] 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 *) 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 *) PROG (Perl) use bignum; my @seq = (1, 1, 1); print "1 1\n2 1\n3 1\n"; for ( my \$i = 3; \$i < 400; \$i++ ) { my \$next = ( \$seq[\$i-1] + \$seq[\$i-3] ) / gcd( \$seq[\$i-1], \$seq[\$i-3] ); my \$ind = \$i+1; print "\$ind \$next\n"; push( @seq, \$next ); } sub gcd { my (\$x, \$y) = @_; (\$x, \$y) = (\$y, \$x % \$y) while \$y; return \$x; } (Haskell) a214551 n = a214551_list !! n a214551_list = 1 : 1 : 1 : zipWith f a214551_list (drop 2 a214551_list) where f u v = (u + v) `div` gcd u v -- Reinhard Zumkeller, Jul 23 2012 (Sage) def A214551Rec(): x, y, z = 1, 1, 1 yield x while True: x, y, z = y, z, (z + x)//gcd(z, x) yield x A214551 = A214551Rec(); print([next(A214551) for _ in range(65)]) # Peter Luschny, Oct 18 2012 (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 (Python) from math import gcd def aupton(nn): alst = [1, 1, 1] for n in range(3, nn+1): alst.append((alst[n-1] + alst[n-3])//gcd(alst[n-1], alst[n-3])) return alst print(aupton(64)) # Michael S. Branicky, Mar 28 2022 CROSSREFS Similar to A000930. Cf. A341312, A341313, which are also similar. Cf. also A214320, A214321, A214322, A214323 (gcd's), A219898 (records), A214324, A214325, A214330, A214331, A214809, A227836, A227837. Starting with a(2) = 3 gives A214626. - Reinhard Zumkeller, Jul 23 2012 Sequence in context: A122453 A017849 A134536 * A343435 A211010 A131731 Adjacent sequences: A214548 A214549 A214550 * A214552 A214553 A214554 KEYWORD nonn,nice AUTHOR Reed Kelly, Jul 20 2012 STATUS approved

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Last modified December 6 21:00 EST 2022. Contains 358648 sequences. (Running on oeis4.)