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A303219
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a(n) = n for n <= 11. Thereafter a(n) = a(n-a(n-2)) + a(n-a(n-4)) + a(n-a(n-6)).
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1
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1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 12, 12, 13, 14, 15, 16, 17, 18, 18, 18, 19, 20, 21, 22, 23, 24, 24, 24, 25, 26, 27, 28, 29, 30, 30, 30, 31, 32, 33, 34, 34, 34, 35, 36, 36, 36, 37, 38, 39, 40, 41, 42, 42, 42, 43, 44, 45, 46, 47, 48, 48, 48, 49, 50, 51, 52, 52, 52, 53, 54, 54, 54, 55, 56
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OFFSET
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1,2
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COMMENTS
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Let a_i(n) = n for n <= 6*i - 1. Thereafter a_i(n) = a_i(n-a_i(n-i)) + a_i(n-a_i(n-2*i)) + a_i(n-a_i(n-3*i)). This sequence is a_2(n).
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LINKS
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A. Isgur, R. Lech, S. Moore, S. Tanny, Y. Verberne, and Y. Zhang, Constructing New Families of Nested Recursions with Slow Solutions, SIAM J. Discrete Math., 30(2), 2016, 1128-1147. (20 pages); DOI:10.1137/15M1040505
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MATHEMATICA
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nmax = 100;
q[_] = 0; For[n = 1, n <= 11, n++, q[n] = n]; For[n = 12, n <= nmax, n++, q[n] = q[n - q[n-2]] + q[n - q[n-4]] + q[n - q[n-6]]]; Array[q, nmax] (* Jean-François Alcover, Feb 18 2019, from PARI *)
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PROG
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(PARI) q=vector(100); for(n=1, 11, q[n]=n); for(n=12, #q, q[n] = q[n-q[n-2]] + q[n-q[n-4]] + q[n-q[n-6]]); q
(GAP) a:=List([1..11], i->i);; for n in [12..100] do a[n]:=a[n-a[n-2]]+a[n-a[n-4]]+a[n-a[n-6]]; od; a; # Muniru A Asiru, May 19 2018
(Magma) [n le 11 select n else Self(n-Self(n-2))+Self(n-Self(n-4))+Self(n-Self(n-6)): n in [1..70]]; // Vincenzo Librandi, May 20 2018
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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