%I #10 Jul 27 2020 21:08:47

%S 4,4,39,13,152,28,391,49,804,76,1439,109,2344,148,3567,193,5156,244,

%T 7159,301,9624,364,12599,433,16132,508,20271,589,25064,676,30559,769,

%U 36804,868,43847,973,51736,1084,60519,1201,70244,1324,80959,1453,92712

%N First minimum value > 0 of the form x^3-k^2 when k > n^3.

%F Let k be the smallest integer>n^3 such that A070923(k-1)> A070923(k) and such that A070923(k) < A070923(k+1), then a(n)= A070923(k); for n>=1 a(2n-1) = 8n^3-9n^2+6n-1, a(2n)=3n^2+1.

%F From _Chai Wah Wu_, Jul 27 2020: (Start)

%F a(n) = 4*a(n-2) - 6*a(n-4) + 4*a(n-6) - a(n-8) for n > 8.

%F G.f.: x*(-x^7 + x^6 + 20*x^4 - 3*x^3 + 23*x^2 + 4*x + 4)/((x - 1)^4*(x + 1)^4). (End)

%e Let n=2 then n^3=8 and A070923(9)= 44, A070923(10)=25, A070923(11)=4, A070923(12)=72 so the first minimum is 4, hence a(2)=4

%o (PARI) for(n=1,100,s=n^3+1; while(ceil(s^(2/3))^3-s^2>ceil((s+1)^(2/3))^3-(s+1)^2,s++); print1(ceil(s^(2/3))^3-s^2,","))

%Y Cf. A070923.

%K easy,nonn

%O 1,1

%A _Benoit Cloitre_, May 25 2002