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a(n) = ceiling(n*(n+1)*(n+2)/8).
5

%I #24 Aug 28 2024 04:14:33

%S 0,1,3,8,15,27,42,63,90,124,165,215,273,342,420,510,612,727,855,998,

%T 1155,1329,1518,1725,1950,2194,2457,2741,3045,3372,3720,4092,4488,

%U 4909,5355,5828,6327,6855,7410,7995,8610,9256,9933,10643,11385

%N a(n) = ceiling(n*(n+1)*(n+2)/8).

%C A lower bound to Honaker's triangle problem A047837.

%H Vincenzo Librandi, <a href="/A047866/b047866.txt">Table of n, a(n) for n = 0..10000</a>

%H <a href="/index/Rec#order_11">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1,0,0,0,0,1,-3,3,-1).

%F G.f.: x*(1 + 2*x^2 - x^3 + 3*x^4 - 2*x^5 + 3*x^6)/((1-x)^4*(1+x)*(1+x^2)*(1+x^4)). - _R. J. Mathar_, Mar 11 2012

%t Table[Ceiling[n*(n+1)*(n+2)/8], {n, 0, 100}] (* _Paolo Xausa_, Aug 28 2024 *)

%o (Magma) [Ceiling( n*(n+1)*(n+2)/8 ) : n in [0..50]]; // _Vincenzo Librandi_, Sep 28 2011

%o (PARI) a(n)=(n*(n+1)*(n+2)+6)\8 \\ _Charles R Greathouse IV_, Oct 07 2015

%Y Cf. A047837.

%K nonn,easy

%O 0,3

%A _Mike Keith_