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%I #13 Jul 08 2017 12:48:05
%S 1,2,3,2,4,3,4,4,4,5,5,5,5,6,6,5,7,6,6,7,7,7,7,7,7,8,8,8,8,8,8,9,8,9,
%T 9,8,10,9,9,9,10,10,9,10,10,10,11,10,10,11,11,10,11,11,11,12,11,11,12,
%U 11,12,12,12,11,13,12,12,12,13,12,13,12,13,13,13,13,13,13,14,13,13,14,14
%N Number of squares between consecutive cubes.
%C a(n) = sum {A010052(k): n^3 <= k < (n+1)^3}. - _Reinhard Zumkeller_, Jul 24 2015, corrected.
%H Reinhard Zumkeller, <a href="/A144968/b144968.txt">Table of n, a(n) for n = 0..10000</a>
%F a(n) = ceiling((n+1)^(3/2)) - ceiling(n^(3/2)).
%t Last[#]-First[#]&/@Partition[Table[Ceiling[n^(3/2)],{n,0,90}],2,1] (* _Harvey P. Dale_, Jul 10 2013 *)
%o (Haskell)
%o a144968 n = a144968_list !! n
%o a144968_list = zipWith (-) (tail a185549_list) a185549_list
%o -- _Reinhard Zumkeller_, Jul 24 2015
%o (PARI) for(n=0,50, print1(ceil((n+1)^(3/2)) - ceil(n^(3/2)), ", ")) \\ _G. C. Greubel_, Jul 08 2017
%Y Cf. A000290, A000578, A077115.
%Y Cf. A010052, A185549.
%K nonn
%O 0,2
%A _Reinhard Zumkeller_, Sep 27 2008, Sep 29 2008
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