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%I #29 Feb 21 2022 01:20:11
%S 0,27,315,1638,5670,15345,35217,71820,134028,233415,384615,605682,
%T 918450,1348893,1927485,2689560,3675672,4931955,6510483,8469630,
%U 10874430,13796937,17316585,21520548,26504100,32370975,39233727,47214090,56443338,67062645,79223445
%N Sums of consecutive noncubes.
%C Relation with triangular numbers: a(n) = 3*((n^3+1) + ((n+1)^3-1)) * A000217(n). Example: a(3) = 3*(first term + last term)*A000217(3) = 3*(28+63)*6 = 1638.
%H Vincenzo Librandi, <a href="/A048396/b048396.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (6,-15,20,-15,6,-1).
%F a(n) = ( 6n^5 + 15n^4 + 18n^3 + 12n^2 + 3n ) / 2.
%F G.f.: 9*x*(1+x)*(3+14*x+3*x^2)/(1-x)^6. - _Colin Barker_, Mar 15 2012
%F a(n) = 6*a(n-1)-15*a(n-2)+20*a(n-3)-15*a(n-4)+6*a(n-5)-a(n-6). - _Wesley Ivan Hurt_, Apr 10 2015
%e Between 3^3 and 4^3 we have: 28 + 29 + ... + 62 + 63 = 1638 = a(3).
%p A048396:=n->(6*n^5+15*n^4+18*n^3+12*n^2+3*n)/2: seq(A048396(n), n=0..50); # _Wesley Ivan Hurt_, Apr 10 2015
%t Table[Total[Range[n^3+1,(n+1)^3-1]], {n,0,30}] (* _Harvey P. Dale_, Jan 08 2011 *)
%o (Magma) [(6*n^5+15*n^4+18*n^3+12*n^2+3*n)/2 : n in [0..50]]; // _Wesley Ivan Hurt_, Apr 10 2015
%o (PARI) a(n)=(6*n^5+15*n^4+18*n^3+12*n^2+3*n)/2 \\ _Charles R Greathouse IV_, Oct 07 2015
%Y Cf. A000217, A048395, A048397.
%K nonn,easy,nice
%O 0,2
%A _Patrick De Geest_, Mar 15 1999
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