A215097 a(n) = n^3 - a(n-2) for n >= 2 and a(0)=0, a(1)=1.
0, 1, 8, 26, 56, 99, 160, 244, 352, 485, 648, 846, 1080, 1351, 1664, 2024, 2432, 2889, 3400, 3970, 4600, 5291, 6048, 6876, 7776, 8749, 9800, 10934, 12152, 13455, 14848, 16336, 17920, 19601, 21384, 23274, 25272, 27379, 29600, 31940, 34400, 36981, 39688, 42526
Offset: 0
Links
- Index entries for linear recurrences with constant coefficients, signature (4,-7,8,-7,4,-1).
Crossrefs
Cf. A000217 (n^2 - a(n-1)).
Cf. A125577 (n^2 - a(n-1) with a(0)=1).
Cf. A011934 (n^3 - a(n-1)).
Cf. A153026 (n^3 - a(n-1) with a(1)=0).
Cf. A194274 (n^2 - a(n-2)).
Cf. A187093 (n^2 - a(n-2) with a(0)=a(1)=1, a(-1)=0).
Cf. A107386 ((n-2)^2 - a(n-1) with a(0)=0, a(1)=a(2)=1, a(3)=2).
Cf. A206481 ((n-1)^3 - a(n-2)).
Programs
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Mathematica
RecurrenceTable[{a[0] == 0, a[1] == 1, a[n] == n^3 - a[n - 2]}, a[n], {n, 0, 43}] (* Bruno Berselli, Aug 07 2012 *) -
Python
prpr = 0 prev = 1 for n in range(2,77): print(prpr, end=',') curr = n*n*n - prpr prpr = prev prev = curr
Formula
G.f.: (x+4*x^2+x^3)/((-1+x)^4*(1+x^2)). - David Scambler, Aug 06 2012
a(n) = (n*(n^2-3)-(1-(-1)^n)*i^(n+1))/2, where i=sqrt(-1). - Bruno Berselli, Aug 07 2012