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A260513 a(n) = (8*n+13*n^3+3*n^5)/24; also the sum of triangular numbers taken in successive groups of increasing size (see Example). 1
1, 9, 46, 164, 460, 1091, 2289, 4376, 7779, 13045, 20856, 32044, 47606, 68719, 96755, 133296, 180149, 239361, 313234, 404340, 515536, 649979, 811141, 1002824, 1229175, 1494701, 1804284, 2163196, 2577114, 3052135, 3594791, 4212064, 4911401, 5700729, 6588470 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
LINKS
FORMULA
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>6. - Colin Barker, Aug 07 2015
G.f.: x*(x^4+3*x^3+7*x^2+3*x+1) / (x-1)^6. - Colin Barker, Aug 07 2015
EXAMPLE
The first ten triangular numbers are 1,3,6,10,15,21,28,36,45,and 55. Take them in groups, respectively, of 1, 2, 3, and 4 = (1), (3, 6), (10, 15, 21), and (28, 36, 45, 55). Summing each group separately = 1, 9, 46, 164.
MATHEMATICA
Table[1/24*(8*x+13*x^3+3*x^5), {x, 50}]
Module[{nn=40}, Total/@TakeList[Accumulate[Range[(nn(nn+1))/2]], Range[nn]]] (* or *) LinearRecurrence[{6, -15, 20, -15, 6, -1}, {1, 9, 46, 164, 460, 1091}, 40] (* Harvey P. Dale, Aug 09 2023 *)
PROG
(PARI) Vec(x*(x^4+3*x^3+7*x^2+3*x+1)/(x-1)^6 + O(x^100)) \\ Colin Barker, Aug 07 2015
CROSSREFS
Sequence in context: A280852 A009294 A061620 * A001781 A258477 A320755
KEYWORD
nonn,easy
AUTHOR
Harvey P. Dale, Jul 27 2015
STATUS
approved

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Last modified March 28 08:22 EDT 2024. Contains 371236 sequences. (Running on oeis4.)