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%I #45 Sep 08 2022 08:46:16
%S 2,7,20,47,94,167,272,415,602,839,1132,1487,1910,2407,2984,3647,4402,
%T 5255,6212,7279,8462,9767,11200,12767,14474,16327,18332,20495,22822,
%U 25319,27992,30847,33890,37127,40564,44207,48062,52135,56432,60959,65722,70727,75980,81487,87254
%N a(n) = n^3 + (n+1)*(n+2).
%C For n>1, many consecutive terms of the sequence are generated by floor(sqrt(n^2 + 2)^3) + n^2 + 2.
%C It appears that this is a subsequence of A000037 (the nonsquares).
%C The primes in the sequence belong to A045326.
%C Inverse binomial transform is 2, 5, 8, 6, 0, 0, 0, ... (0 continued).
%H Bruno Berselli, <a href="/A270109/b270109.txt">Table of n, a(n) for n = 0..1000</a>
%H MathsSmart, <a href="https://www.youtube.com/watch?v=iMc3z8TvV8g">Number pattern and Puzzle - 7, 20, 47, 94, 167</a>.
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).
%F O.g.f.: (2 - x + 4*x^2 + x^3)/(1 - x)^4.
%F E.g.f.: (2 + x)*(1 + x)^2*exp(x).
%F a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4), n>3.
%F 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.
%F 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).
%F a(n) + a(-n) = 2*A059100(n) = A255843(n).
%F a(n) - a(-n) = 4*A229183(n).
%t Table[n^3 + (n + 1) (n + 2), {n, 0, 50}]
%o (PARI) vector(50, n, n--; n^3+(n+1)*(n+2))
%o (Sage) [n^3+(n+1)*(n+2) for n in (0..50)]
%o (Maxima) makelist(n^3+(n+1)*(n+2), n, 0, 50);
%o (Magma) [n^3+(n+1)*(n+2): n in [0..50]];
%Y Subsequence of A001651, A047212.
%Y Cf. A000037, A045326.
%Y Cf. A027444: numbers of the form n^3+n*(n+1); A085490: numbers of the form n^3+(n-1)*n.
%Y Cf. A008865: numbers of the form n+(n+1)*(n+2); A130883: numbers of the form n^2+(n+1)*(n+2).
%K nonn,easy
%O 0,1
%A _Bruno Berselli_, Mar 11 2016, at the suggestion of Giuseppe Amoruso in BASE Cinque forum.
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