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A092286
Fourth diagonal (m=3) of triangle A084938; a(n) = A084938(n+3,n) = (n^3 + 9*n^2 + 26*n)/6.
2
0, 6, 16, 31, 52, 80, 116, 161, 216, 282, 360, 451, 556, 676, 812, 965, 1136, 1326, 1536, 1767, 2020, 2296, 2596, 2921, 3272, 3650, 4056, 4491, 4956, 5452, 5980, 6541, 7136, 7766, 8432, 9135, 9876, 10656, 11476, 12337, 13240, 14186, 15176
OFFSET
0,2
COMMENTS
If X is an n-set and Y a fixed (n-4)-subset of X then a(n-4) is equal to the number of 3-subsets of X intersecting Y. - Milan Janjic, Aug 15 2007
For n>=0, A092286(n) is the number of 2 X 2 matrices with all terms in {0,1,...,n} and (sum of terms) = n + 1. A092286(n) is also the number of 2 X 2 matrices with all terms in {0,1,...,n} and (sum of terms) = 3n - 1. - Clark Kimberling, Mar 20 2012
LINKS
Guillaume Aupy, Julien Herrmann. Periodicity in optimal hierarchical checkpointing schemes for adjoint computations. Optimization Methods and Software, Volume 32, 2017 - Issue 3. Preprint
FORMULA
a(n) = A084938(n+3, n) = sum_{k=0..3} A090238(3, k)*binomial(n, k).
From Gary Detlefs, Aug 02 2010: (Start)
a(n) = 1/2 * sum_{k=1..n} (k+3)(k+2).
a(n) = 1/6 * n *(n^2 + 9n + 26). (End)
G.f.: x*(6 - 8*x + 3*x^2)/(1-x)^4. - Colin Barker, Mar 18 2012
MAPLE
a:=n->(n^3 + 9*n^2 + 26*n)/6: seq(a(n), n=3..45);
MATHEMATICA
q=60; (Transpose[NestList[Accumulate, Range[q], q]]-Range[q])[[4]] (* Vladimir Joseph Stephan Orlovsky, Apr 08 2011 *)
Table[(n^3 + 9*n^2 + 26*n)/6, {n, 0, 100}] (* T. D. Noe, Apr 12 2011 *)
CROSSREFS
Sequence in context: A005891 A108182 A244242 * A301723 A288113 A276917
KEYWORD
easy,nonn
AUTHOR
Philippe Deléham, Jan 30 2004
STATUS
approved