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A027484
a(n) = n*(n^4-1)/2.
1
15, 120, 510, 1560, 3885, 8400, 16380, 29520, 49995, 80520, 124410, 185640, 268905, 379680, 524280, 709920, 944775, 1238040, 1599990, 2042040, 2576805, 3218160, 3981300, 4882800, 5940675, 7174440, 8605170, 10255560, 12149985
OFFSET
2,1
COMMENTS
Row sums in a pandiagonal magic 4D-cube with entries (0..n^4-1).
Can be computed from the fourth power of a matrix constructed with the Stirling numbers of the first kind (see A027479).
LINKS
FORMULA
From Chai Wah Wu, Apr 08 2021: (Start)
a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6) for n > 7.
G.f.: 15*x^2*(x + 1)^2/(x - 1)^6. (End)
MATHEMATICA
Table[(m^5 - m)/2, {m, 34}] (* Zerinvary Lajos, Mar 21 2007 *)
LinearRecurrence[{6, -15, 20, -15, 6, -1}, {15, 120, 510, 1560, 3885, 8400}, 30] (* Harvey P. Dale, Aug 02 2024 *)
PROG
(Magma) [n*(n^4 - 1)/2: n in [2..50]]; // Vincenzo Librandi, Dec 29 2012
(PARI) a(n)=n*(n^4-1)/2 \\ Charles R Greathouse IV, Oct 21 2022
CROSSREFS
First subdiagonal of A027479.
Sequence in context: A259746 A139615 A196506 * A185542 A226989 A126898
KEYWORD
nonn,easy
STATUS
approved