%I #28 Mar 24 2024 02:27:22
%S 0,5,36,217,1272,7421,43260,252145,1469616,8565557,49923732,290976841,
%T 1695937320,9884647085,57611945196,335787024097,1957110199392,
%U 11406874172261,66484134834180,387497934832825,2258503474162776,13163522910143837,76722633986700252
%N Expansion of x*(5+x)/(1-7*x+7*x^2-x^3).
%C Table of differences re Table A182441.
%C This is a sequence of differences between rows k and k+1 of table A182441. That is if A182441(k+1,0)-A182441(k,0) = 1, a(n) = A182441(k+1,n+1) - A182441(k,n+1) for n = 0 to 3. The remainder of the sequence is a continuation using the recursive formula D(n) = 6D(n-1)- D(n-2) + 6.
%C It appears that for n > 0, a(n) is divisible by A213005(n).
%C It appears that if p is a prime of the form 8*r +/- 1 then a(p-1) == 0 (mod p), and that if p is a prime of the form 8*r +/- 3 then a(p+1) == 0 (mod p).
%H Colin Barker, <a href="/A212329/b212329.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (7,-7,1).
%F a(n) = 7*a(n-1) - 7*a(n-2) + a(n-3).
%F From _Colin Barker_, Mar 05 2016: (Start)
%F a(n) = (-6+(5-3*sqrt(2))*(3+2*sqrt(2))^n + (3-2*sqrt(2))^n*(5+3*sqrt(2)))/4.
%F G.f.: x*(5+x) / ((1-x)*(1-6*x+x^2)).
%F (End)
%t m = 12; n = 1; c = 0;
%t list3 = Reap[While[c < 22, t = 6 n - m + 6; Sow[t];m = n; n = t;c++]][[2,1]]
%t CoefficientList[ Series[x (5 + x)/(1 - 7x + 7x^2 - x^3), {x, 0, 20}], x] (* or *)
%t LinearRecurrence[{7, -7, 1}, {0, 5, 36}, 21] (* _Robert G. Wilson v_, Jun 24 2014 *)
%o (PARI) concat(0, Vec(x^2*(5+x)/((1-x)*(1-6*x+x^2)) + O(x^40))) \\ _Colin Barker_, Mar 05 2016
%Y Cf. A182441, A213005.
%K nonn,easy
%O 1,2
%A _Kenneth J Ramsey_, May 14 2012