%I #28 Feb 22 2022 03:50:57
%S 1,9,18,50,75,147,196,324,405,605,726,1014,1183,1575,1800,2312,2601,
%T 3249,3610,4410,4851,5819,6348,7500,8125,9477,10206,11774,12615,14415,
%U 15376,17424,18513,20825,22050,24642,26011,28899,30420,33620,35301
%N Group the natural numbers such that the n-th group sum is divisible by the n-th triangular number: (1), (2, 3, 4), (5, 6, 7), (8, 9, 10, 11, 12), (13, 14, 15, 16, 17), (18, 19, 20, 21, 22, 23, 24), ... Sequence contains the group sum.
%C The number of terms in the groups is given by A063196. i.e., 1, 3, 3, 5, 5, 7, 7, 9, 9, 11, 11, ...
%C Also the arithmetic mean of the n-th group is T(n), the n-th triangular number.
%H <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (1,3,-3,-3,3,1,-1).
%F a(n) = n*(n+1)*(2*n+1+(-1)^n)/4. - _Wesley Ivan Hurt_, Sep 19 2014
%F a(n) = a(n-1)+3*a(n-2)-3*a(n-3)-3*a(n-4)+3*a(n-5)+a(n-6)-a(n-7) for n>7. - _Colin Barker_, Sep 19 2014
%F G.f.: x*(x^4+8*x^3+6*x^2+8*x+1) / ((x-1)^4*(x+1)^3). - _Colin Barker_, Sep 19 2014
%F From _Amiram Eldar_, Feb 22 2022: (Start)
%F Sum_{n>=1} 1/a(n) = 4*(1-log(2)).
%F Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/2 - 4. (End)
%t Table[n*(n + 1)*(2*n + 1 + (-1)^n)/4, {n, 1, 40}] (* _Amiram Eldar_, Feb 22 2022 *)
%o (Haskell)
%o a086500 n = a086500_list !! (n-1)
%o a086500_list = scanl1 (+) $ tail a181900_list
%o -- _Reinhard Zumkeller_, Mar 31 2012
%o (PARI) Vec(x*(x^4+8*x^3+6*x^2+8*x+1)/((x-1)^4*(x+1)^3) + O(x^100)) \\ _Colin Barker_, Sep 20 2014
%Y Cf. A001082, A022998, A063196, A181900.
%K nonn,easy
%O 1,2
%A _Amarnath Murthy_, Jul 28 2003
%E More terms from _Ray Chandler_, Sep 17 2003