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a(n+1) = 11*a(n) - a(n-1) - 3, a(0)=a(1)=1.
1

%I #21 Oct 01 2016 21:18:32

%S 1,1,7,73,793,8647,94321,1028881,11223367,122428153,1335486313,

%T 14567921287,158911647841,1733460204961,18909150606727,

%U 206267196469033,2250030010552633,24544062919609927,267734662105156561

%N a(n+1) = 11*a(n) - a(n-1) - 3, a(0)=a(1)=1.

%C The old entry with this sequence number was a duplicate of A039963.

%C The simultaneous equations (p+1)(p+2) == -1 (mod q), (q+1)(q+2) == -1 (mod p), where p and q are odd, have solutions {3, 3}, {3, 21}, {7, 73}, {21, 507}, {73, 793}, {793, 8647} and suggested this recurrence.

%H Vincenzo Librandi, <a href="/A092444/b092444.txt">Table of n, a(n) for n = 0..980</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (12,-12,1).

%H <a href="/index/Fi#FIXEDPOINTS">Index entries for sequences that are fixed points of mappings</a>

%F a(n) = 2*(A004190(n)-10*A004190(n-1))/3+1/3. G.f.: (1-11x+7x^2)/((1-x)(1-11x+x^2)). [From _R. J. Mathar_, Sep 20 2008]

%K nonn

%O 0,3

%A _N. J. A. Sloane_, Sep 19 2008, based on emails from _M. F. Hasler_ and _Jim Nastos_ on Apr 25 2008

%E More terms from _R. J. Mathar_, Sep 20 2008