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

0,1

Comments

For n>1, many consecutive terms of the sequence are generated by floor(sqrt(n^2 + 2)^3) + n^2 + 2.

It appears that this is a subsequence of A000037 (the nonsquares).

The primes in the sequence belong to A045326.

Inverse binomial transform is 2, 5, 8, 6, 0, 0, 0, ... (0 continued).

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).

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) = 2*A059100(n) = A255843(n).

a(n) - a(-n) = 4*A229183(n).

Mathematica

Table[n^3 + (n + 1) (n + 2), {n, 0, 50}]

Prog

(PARI) vector(50, n, n--; n^3+(n+1)*(n+2))
(Sage) [n^3+(n+1)*(n+2) for n in (0..50)]
(Maxima) makelist(n^3+(n+1)*(n+2), n, 0, 50);
(Magma) [n^3+(n+1)*(n+2): n in [0..50]];

Crossrefs

Subsequence of A001651, A047212.

Cf. A000037, A045326.

Cf. A027444: numbers of the form n^3+n*(n+1); A085490: numbers of the form n^3+(n-1)*n.

Cf. A008865: numbers of the form n+(n+1)*(n+2); A130883: numbers of the form n^2+(n+1)*(n+2).

Sequence in context: A090145 A244307 A368881 * A360421 A123203 A309298

Adjacent sequences: A270106 A270107 A270108 * A270110 A270111 A270112

Keyword

nonn,easy

Author

Bruno Berselli, Mar 11 2016, at the suggestion of Giuseppe Amoruso in BASE Cinque forum.

Status

approved