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 A254875 a(n) = floor((10*n^3 + 57*n^2 + 102*n + 72) / 72). 5

%I

%S 1,3,8,16,28,45,68,97,134,179,233,297,372,458,557,669,795,936,1093,

%T 1266,1457,1666,1894,2142,2411,2701,3014,3350,3710,4095,4506,4943,

%U 5408,5901,6423,6975,7558,8172,8819,9499,10213,10962,11747,12568,13427,14324,15260

%N a(n) = floor((10*n^3 + 57*n^2 + 102*n + 72) / 72).

%H G. C. Greubel, <a href="/A254875/b254875.txt">Table of n, a(n) for n = 0..2500</a>

%H Kyu-Hwan Lee, Se-jin Oh, <a href="http://arxiv.org/abs/1601.06685">Catalan triangle numbers and binomial coefficients</a>, arXiv:1601.06685 [math.CO], 2016.

%H <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (2,0,-1,-1,0,2,-1).

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

%F a(n) - 2*a(n+1) + 2*a(n+3) - a(n+4) = -1 if n == 0 (mod 3) else -2 for all n in Z.

%F a(n) = -A254874(-4-n) for all n in Z.

%e G.f. = 1 + 3*x + 8*x^2 + 16*x^3 + 28*x^4 + 45*x^5 + 68*x^6 + 97*x^7 + ...

%t a[ n_] := Quotient[ 10 n^3 + 57 n^2 + 102 n + 72, 72];

%t Table[Floor[(10n^3+57n^2+102n+72)/72],{n,0,60}] (* or *) LinearRecurrence[ {2,0,-1,-1,0,2,-1},{1,3,8,16,28,45,68},60] (* _Harvey P. Dale_, Jan 07 2017 *)

%o (PARI) {a(n) = (10*n^3 + 57*n^2 + 102*n + 72) \ 72};

%o (PARI) {a(n) = polcoeff( (-1)^(n<0) * (if( n<0, n = -4 - n; x, x^2) + 1 + x + x^2 + x^3) / ((1 - x)^2 * (1 - x^2) * (1 - x^ 3)) + x * O(x^n), n)};

%o (MAGMA) [Floor((10*n^3 +57*n^2 +102*n +72)/72): n in [0..30]]; // _G. C. Greubel_, Aug 03 2018

%Y Cf. A254874.

%K nonn,easy

%O 0,2

%A _Michael Somos_, Feb 09 2015

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Last modified February 28 23:19 EST 2020. Contains 332353 sequences. (Running on oeis4.)