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A069778
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q-factorial numbers 3!_q.
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19
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1, 6, 21, 52, 105, 186, 301, 456, 657, 910, 1221, 1596, 2041, 2562, 3165, 3856, 4641, 5526, 6517, 7620, 8841, 10186, 11661, 13272, 15025, 16926, 18981, 21196, 23577, 26130, 28861, 31776, 34881, 38182, 41685, 45396, 49321, 53466, 57837, 62440, 67281, 72366
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OFFSET
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0,2
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COMMENTS
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Number of proper n-colorings of the 4-cycle with one vertex color fixed (offset 2). - Michael Somos, Jul 19 2002
If Y is a 4-subset of an n-set X then, for n>=5, a(n-5) is the number of 5-subsets of X having at least two elements in common with Y. - Milan Janjic, Dec 08 2007
Binomial transform of 1, 5, 10, 6, 0, 0, 0 (0 continued). - Philippe Deléham, Mar 17 2014
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REFERENCES
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T. A. Gulliver, Sequences from Cubes of Integers, Int. Math. Journal, 4 (2003), 439-445.
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LINKS
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FORMULA
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a(n) = (n + 1)*(n^2 + n + 1).
a(n) = (n+1)^3-2*T(n) where T(n) =n*(n+1)/2= A000217(n) is the n-th triangular number. - Herman Jamke (hermanjamke(AT)fastmail.fm), Sep 14 2006
a(n) = n^8 mod (n^3+n), with offset 1..a(1)=1. - Gary Detlefs, May 02 2010
a(n) = 4*a(n-1)-6*a(n-2)+ 4*a(n-3)- a(n-4), n>3. - Harvey P. Dale, Jul 11 2011
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EXAMPLE
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For 2-colorings only 1212 is proper so a(2-2)=1. The proper 3-colorings are: 1212,1313,1213,1312,1232,1323 so a(3-2)=6.
a(0) = 1*1 = 1;
a(1) = 1*1 + 5*1 = 6;
a(2) = 1*1 + 5*2 + 10*1 = 21;
a(3) = 1*1 + 5*3 + 10*3 + 6*1 = 52;
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MAPLE
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(n+1)*(n^2+n+1) ;
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MATHEMATICA
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LinearRecurrence[{4, -6, 4, -1}, {1, 6, 21, 52}, 41] (* or *) Table[(n + 1) (n^2 + n + 1), {n, 0, 41}] (* Harvey P. Dale, Jul 11 2011 *)
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PROG
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(PARI) a(n)=(n+1)*(n^2+n+1)
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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