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a(n) = floor(n*(n+1)/6) - 1.
3

%I #24 Jan 14 2021 07:06:19

%S 0,1,2,4,6,8,11,14,17,21,25,29,34,39,44,50,56,62,69,76,83,91,99,107,

%T 116,125,134,144,154,164,175,186,197,209,221,233,246,259,272,286,300,

%U 314,329,344,359,375,391,407,424,441,458,476,494,512,531,550,569,589,609,629,650,671,692

%N a(n) = floor(n*(n+1)/6) - 1.

%C a(n) = A001840(n-1)-1.

%H Bernhard Ganter, <a href="http://ceur-ws.org/Vol-2668/paper1.pdf">Notes on Integer Partitions</a>, Technische Universität Dresden (2020); See Prop. 3.

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (2,-1,1,-2,1).

%F G.f.: x^3*(-1-x^2+x^3) / ( (1+x+x^2)*(x-1)^3 ). - _R. J. Mathar_, Jan 14 2021

%t Table[Floor[n (n + 1)/6] - 1, {n, 2, 80}] (* _Robert P. P. McKone_, Dec 19 2020 *)

%o (Python)

%o def A339573(n):

%o return n*(n+1)//6-1 # _Chai Wah Wu_, Dec 10 2020

%Y Cf. A001840.

%K nonn,easy

%O 2,3

%A _N. J. A. Sloane_, Dec 10 2020