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a(n) = (Sum_{i=0..n} 5^i) + 1 - (Sum_{i=0..n} 5^i) mod 2.
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%I #25 Mar 09 2024 15:52:56

%S 1,7,31,157,781,3907,19531,97657,488281,2441407,12207031,61035157,

%T 305175781,1525878907,7629394531,38146972657,190734863281,

%U 953674316407,4768371582031,23841857910157,119209289550781,596046447753907

%N a(n) = (Sum_{i=0..n} 5^i) + 1 - (Sum_{i=0..n} 5^i) mod 2.

%C Floretion Algebra Multiplication Program, FAMP Code: 1tesseq[ + 'ij' + 'ik' + 'ji' + 'jk' + 'ki' + 'kj' + e]

%C a(n) = term (1,1) in M^n, M = the 4 X 4 matrix [1, 1, 1, 2; 1, 1, 2, 1; 1, 2, 1, 1; 2, 1, 1, 1]. a(n)/a(n-1) tends to 5, a root to the charpoly x^4 - 4x^3 - 6x^2 + 4x + 5. - _Gary W. Adamson_, Mar 12 2009

%C This is 1+A003463(n+1) rounded down to the next odd integer. - _R. J. Mathar_, Sep 11 2019

%H Robert Munafo, <a href="http://www.mrob.com/pub/math/seq-floretion.html">Sequences Related to Floretions</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (5,1,-5).

%F a(n) = 4*a(n-1) + 5*a(n-2) - 2 (conjecture). - _Creighton Dement_, Apr 13 2005

%F (1/4) (5^(n+1) - 2(-1)^2 + 1). - _Ralf Stephan_, May 17 2007

%F From _R. J. Mathar_, Mar 19 2009: (Start)

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

%F a(n) = 5*a(n-1) + a(n-2) - 5*a(n-3). (End)

%t a = Table[Sum[5^i, {i, 0, n}] + 1 - Mod[Sum[5^i, {i, 0, n}], 2], {n, 0, 50}]

%Y Cf. A015531.

%K nonn,easy

%O 0,2

%A _Roger L. Bagula_, Mar 15 2005