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A101094
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a(n) = n*(n+1)*(n+2)*(n+3)*(1+3*n+n^2)/120.
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18
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1, 11, 57, 203, 574, 1386, 2982, 5874, 10791, 18733, 31031, 49413, 76076, 113764, 165852, 236436, 330429, 453663, 612997, 816431, 1073226, 1394030, 1791010, 2277990, 2870595, 3586401, 4445091, 5468617, 6681368, 8110344, 9785336
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OFFSET
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1,2
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COMMENTS
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LINKS
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FORMULA
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This sequence could be obtained from the general formula n*(n+1)*(n+2)*(n+3)*...*(n+k)*(n*(n+k)+(k-1)*k/6)/((k+3)!/6) at k=3. - Alexander R. Povolotsky, May 17 2008
a(k) = MagicNKZ(3,k,7) where MagicNKZ(n,k,z) = Sum_{j=0..k+1} (-1)^j*binomial(n+1-z,j)*(k-j+1)^n. (Cf. A101104.) - Danny Rorabaugh, Apr 23 2015
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MATHEMATICA
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Table[n*(n + 1)*(n + 2)*(n + 3)*(1 + 3*n + n^2)/120, {n, 31}] (* Michael De Vlieger, Apr 20 2015 *)
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PROG
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(PARI) a(n)=sum(l=1, n, sum(j=1, l, sum(m=1, j, sum(i=m*(m+1)/2-m+1, m*(m+1)/2, (2*i-1))))) \\ Alexander R. Povolotsky, May 17 2008
(Sage) [n*(n+1)*(n+2)*(n+3)*(1+3*n+n^2)/120 for n in range(1, 32)] # Danny Rorabaugh, Apr 20 2015
(Magma) [n*(n+1)*(n+2)*(n+3)*(1+3*n+n^2)/120 : n in [1..35]]; // Vincenzo Librandi, Apr 23 2015
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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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Cecilia Rossiter (cecilia(AT)noticingnumbers.net), Dec 15 2004
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EXTENSIONS
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STATUS
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approved
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