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5th binomial transform of (1,4,0,0,0,0,...).
7

%I #25 Sep 08 2022 08:45:09

%S 1,9,65,425,2625,15625,90625,515625,2890625,16015625,87890625,

%T 478515625,2587890625,13916015625,74462890625,396728515625,

%U 2105712890625,11138916015625,58746337890625,308990478515625,1621246337890625

%N 5th binomial transform of (1,4,0,0,0,0,...).

%H Vincenzo Librandi, <a href="/A081040/b081040.txt">Table of n, a(n) for n = 0..300</a>

%H Silvana Ramaj, <a href="https://digitalcommons.georgiasouthern.edu/cgi/viewcontent.cgi?article=3464&amp;context=etd">New Results on Cyclic Compositions and Multicompositions</a>, Master's Thesis, Georgia Southern Univ., 2021. See p. 67.

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (10,-25).

%F a(n) = 10*a(n-1) - 25*a(n-2), a(0)=1, a(1)=9.

%F a(n) = (4n+5)*5^(n-1).

%F a(n) = Sum_{k=0..n} (k+1)*4^k*binomial(n, k).

%F G.f.: (1-x)/(1-5x)^2.

%t CoefficientList[Series[(1 - x) / (1 - 5 x)^2, {x, 0, 30}], x] (* _Vincenzo Librandi_, Aug 06 2013 *)

%t LinearRecurrence[{10,-25},{1,9},30] (* _Harvey P. Dale_, Jan 10 2021 *)

%o (Magma) [(4*n+5)*5^(n-1): n in [0..25]]; // _Vincenzo Librandi_, Aug 06 2013

%o (PARI) a(n)=(4*n+5)*5^(n-1) \\ _Charles R Greathouse IV_, Oct 07 2015

%Y Cf. A081039, A081041.

%K nonn,easy

%O 0,2

%A _Paul Barry_, Mar 03 2003