A270109 a(n) = n^3 + (n+1)*(n+2).
2, 7, 20, 47, 94, 167, 272, 415, 602, 839, 1132, 1487, 1910, 2407, 2984, 3647, 4402, 5255, 6212, 7279, 8462, 9767, 11200, 12767, 14474, 16327, 18332, 20495, 22822, 25319, 27992, 30847, 33890, 37127, 40564, 44207, 48062, 52135, 56432, 60959, 65722, 70727, 75980, 81487, 87254
Offset: 0
Links
- Bruno Berselli, Table of n, a(n) for n = 0..1000
- MathsSmart, Number pattern and Puzzle - 7, 20, 47, 94, 167.
- Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).
Crossrefs
Programs
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Magma
[n^3+(n+1)*(n+2): n in [0..50]];
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Mathematica
Table[n^3 + (n + 1) (n + 2), {n, 0, 50}]
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Maxima
makelist(n^3+(n+1)*(n+2), n, 0, 50);
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PARI
vector(50, n, n--; n^3+(n+1)*(n+2))
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Sage
[n^3+(n+1)*(n+2) for n in (0..50)]
Formula
O.g.f.: (2 - x + 4*x^2 + x^3)/(1 - x)^4.
E.g.f.: (2 + x)*(1 + x)^2*exp(x).
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4), n>3.
a(n+h) - a(n) + a(n-h) = n^3 + n^2 + (6*h^2+3)*n + (2*h^2+2) for any h. This identity becomes a(n) = n^3 + n^2 + 3*n + 2 if h=0.
a(h*a(n) + n) = (h*a(n))^3 + (3*n+1)*(h*a(n))^2 + (3*n^2+2*n+3)*(h*a(n)) + a(n) for any h, therefore a(h*a(n) + n) is always a multiple of a(n).
a(n) - a(-n) = 4*A229183(n).
Comments