OFFSET
1,1
FORMULA
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.
From Chai Wah Wu, Jul 27 2020: (Start)
a(n) = 4*a(n-2) - 6*a(n-4) + 4*a(n-6) - a(n-8) for n > 8.
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)
EXAMPLE
PROG
(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, ", "))
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Benoit Cloitre, May 25 2002
STATUS
approved