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a(n) = (5*n^2 - 3*n + 2)/2.
4

%I #27 Jun 30 2026 21:34:56

%S 1,2,8,19,35,56,82,113,149,190,236,287,343,404,470,541,617,698,784,

%T 875,971,1072,1178,1289,1405,1526,1652,1783,1919,2060,2206,2357,2513,

%U 2674,2840,3011,3187,3368,3554,3745,3941,4142,4348,4559,4775,4996,5222,5453,5689

%N a(n) = (5*n^2 - 3*n + 2)/2.

%C Binomial transform of [1, 1, 5, 0, 0, 0, 0, 0, ...]. - _Johannes W. Meijer_, Jul 07 2011

%H Vincenzo Librandi, <a href="/A192136/b192136.txt">Table of n, a(n) for n = 0..10000</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).

%F a(n) = (5*n^2 - 3*n + 2)/2.

%F a(n) = 2*a(n-1) - a(n-2) + 5.

%F a(n) = a(n-1) + 5*n - 4.

%F a(n) = 5*binomial(n+2,2) - 9*n - 4.

%F a(n) = A000217(n+1) - A000217(n) + 5*A000217(n-1); triangular numbers. - _Johannes W. Meijer_, Jul 07 2011

%F O.g.f.: (1-x+5*x^2)/(1-x)^3.

%F From _Elmo R. Oliveira_, Nov 16 2024: (Start)

%F E.g.f.: exp(x)*(2 + 2*x + 5*x^2)/2.

%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)

%t Table[(5n^2-3n+2)/2,{n,0,50}] (* _Harvey P. Dale_, Aug 08 2016 *)

%t (* Alternative: *)

%t LinearRecurrence[{3,-3,1},{1,2,8},50] (* _Harvey P. Dale_, Aug 08 2016 *)

%o (Magma) A192136:=func< n | (5*n^2-3*n+2)/2 >; [ A192136(n): n in [0..50] ]; // _Klaus Brockhaus_, Jun 27 2011

%o (PARI) a(n)=n*(5*n-3)/2+1 \\ _Charles R Greathouse IV_, Jun 17 2017

%Y Cf. A000124, A000217, A002522, A006137, A130883, A140063, A140064, A143689.

%K nonn,easy,changed

%O 0,2

%A _Eric Werley_, Jun 24 2011