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A126935 Define an array by d(m, 0) = 1, d(m, 1) = m; d(m, k) = (m - k + 1) d(m+1, k-1) - (k-1) (m+1) d(m+2, k-2). Sequence gives d(n,3). 2

%I #11 Oct 18 2022 15:30:54

%S 0,-12,-24,-30,-24,0,48,126,240,396,600,858,1176,1560,2016,2550,3168,

%T 3876,4680,5586,6600,7728,8976,10350,11856,13500,15288,17226,19320,

%U 21576,24000,26598,29376,32340,35496,38850,42408,46176,50160,54366,58800,63468,68376,73530

%N Define an array by d(m, 0) = 1, d(m, 1) = m; d(m, k) = (m - k + 1) d(m+1, k-1) - (k-1) (m+1) d(m+2, k-2). Sequence gives d(n,3).

%D V. van der Noort and N. J. A. Sloane, Paper in preparation, 2007.

%H G. C. Greubel, <a href="/A126935/b126935.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).

%F a(n) = n*(n+2)*(n-5).

%F From _G. C. Greubel_, Jan 29 2020: (Start)

%F G.f.: (-6)*x*(2 - 4*x + x^2)/(1-x)^4.

%F E.g.f.: x*(-12 + x^2)*exp(x). (End)

%p seq( n*(n+2)*(n-5), n=0..50); # _G. C. Greubel_, Jan 29 2020

%t T[n_, k_]:= T[n, k]= If[k==0, 1, If[k==1, n, (n-k+1)*T[n+1, k-1] - (k-1)*(n+1)* T[n+2, k-2]]]; Table[T[n, 3], {n,0,50}] (* _G. C. Greubel_, Jan 29 2020 *)

%o (PARI) vector(50, n, my(m=n-1); m*(m+2)*(m-5) ) \\ _G. C. Greubel_, Jan 29 2020

%o (Magma) [n*(n+2)*(n-5): n in [0..50]]; // _G. C. Greubel_, Jan 29 2020

%o (Sage) [n*(n+2)*(n-5) for n in (0..50)] # _G. C. Greubel_, Jan 29 2020

%Y A row of A105937.

%K sign,easy

%O 0,2

%A Vincent v.d. Noort, Mar 21 2007

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Last modified September 18 08:04 EDT 2024. Contains 375997 sequences. (Running on oeis4.)