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A100638
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Successive powers of the matrix A=[1,2;3,4] written by rows in groups of 4.
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3
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1, 2, 3, 4, 7, 10, 15, 22, 37, 54, 81, 118, 199, 290, 435, 634, 1069, 1558, 2337, 3406, 5743, 8370, 12555, 18298, 30853, 44966, 67449, 98302, 165751, 241570, 362355, 528106, 890461, 1297782, 1946673, 2837134, 4783807, 6972050, 10458075
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OFFSET
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1,2
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COMMENTS
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Consider the matrix A = [1, 2; 3, 4]. Then the sequence gives a(1) = A_{1,1} = A_11, a(2) = A_12, a(3) = A_21, a(4) = A_22, a(5)=(A^2)_11, a(6)=(A^2)_12, a(7)=(A^2)_21, a(8)=(A^2)_22, a(9)=(A^3)_11, a(10)=(A^3)_12, ...
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LINKS
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Alois P. Heinz, Table of n, a(n) for n = 1..1000
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FORMULA
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a(4n-3) = A124610(n), a(4n-2) = 2 A015535(n), a(4n-1) = 3 A015535(n), a(4n) = a(4n-3) + a(4n-1). - M. F. Hasler, Dec 01 2008
a(n) = 5*a(n-4)+2*a(n-8). a(4n+1)=A124610(n+1), n>=0. G.f.: x*(1+2x+3x^2+4x^3+2x^4+2x^7) / (1-5x^4-2x^8). - R. J. Mathar, Dec 04 2008
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MAPLE
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a:= proc(n) local r, m; (<<1|2>, <3|4>>^iquo(n+3, 4, 'r'))[iquo(r+2, 2, 'm'), m+1] end: seq(a(n), n=1..50); # Alois P. Heinz, Dec 01 2008
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MATHEMATICA
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LinearRecurrence[{0, 0, 0, 5, 0, 0, 0, 2}, {1, 2, 3, 4, 7, 10, 15, 22}, 50] (* Jean-François Alcover, May 18 2018, after R. J. Mathar *)
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PROG
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A100638(n)=([1, 2; 3, 4]^((n-1)\4+1))[(n-1)%4\2+1, 2-n%2] /* M. F. Hasler, Dec 01 2008 */
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CROSSREFS
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Sequence in context: A078159 A129490 A018132 * A319437 A270659 A159288
Adjacent sequences: A100635 A100636 A100637 * A100639 A100640 A100641
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KEYWORD
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easy,nonn
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AUTHOR
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Simone Severini, Dec 04 2004
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EXTENSIONS
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Edited by Benoit Jubin, M. F. Hasler and N. J. A. Sloane, Dec 01 2008
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STATUS
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approved
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