Russell Walsmith - cp's OEIS frontend

A254308 a(n) = a(n-1) + (if a(n-1) is odd a(n-2) else a(n-3)) with a(0) = 0, a(1) = 1.

0, 1, 1, 2, 3, 5, 8, 11, 19, 30, 41, 71, 112, 153, 265, 418, 571, 989, 1560, 2131, 3691, 5822, 7953, 13775, 21728, 29681, 51409, 81090, 110771, 191861, 302632, 413403, 716035, 1129438, 1542841, 2672279, 4215120, 5757961, 9973081, 15731042, 21489003, 37220045

Offset: 0

Russell Walsmith, Feb 23 2015

nonn

Every third iteration is a tribonacci-type recursion: a(n) = a(n-1) + a(n-3) otherwise it is Fibonacci-type a(n) = a(n-1) + a(n-2).

			For n = 7, a(n-1) is even so 8 + 3 = 11.
G.f. = x + x^2 + 2*x^3 + 3*x^4 + 5*x^5 + 8*x^6 + 11*x^7 + 19*x^8 + 30*x^9 + ...

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Russell Walsmith, Fibonacci-Tribonacci Fusion
Index entries for linear recurrences with constant coefficients, signature (0,0,4,0,0,-1).

Cf. A001834, A001835, A052530.

a254308 n = a254308_list !! n
a254308_list = 0 : 1 : 1 : zipWith3 (\u v w -> u + if odd u then v else w)
               (drop 2 a254308_list) (tail a254308_list) a254308_list
-- Reinhard Zumkeller, Feb 24 2015

m:=60; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(x*(1+x+2*x^2-x^3+x^4)/(1-4*x^3+x^6))); // G. C. Greubel, Aug 03 2018

CoefficientList[Series[x*(1+x+2*x^2-x^3+x^4)/(1-4*x^3+x^6), {x, 0, 60}], x] (* G. C. Greubel, Aug 03 2018 *)
nxt[{a_,b_,c_}]:={b,c,If[OddQ[c],c+b,c+a]}; NestList[nxt,{0,1,1},50][[All,1]] (* or *) LinearRecurrence[{0,0,4,0,0,-1},{0,1,1,2,3,5},50] (* Harvey P. Dale, May 12 2022 *)

{a(n) = polcoeff( x * if( n<0, n=-n; -(1 - x + 2*x^2 + x^3 + x^4), (1 + x + 2*x^2 - x^3 + x^4)) / (1 - 4*x^3 + x^6) + x * O(x^n), n)}; /* Michael Somos, Feb 23 2015 */

Two identities: a(3n)/2 + a(3n-3)/2 = a(3n-1); a(3n)/2 - a(3n-3)/2 = a(3n-2).
G.f.: x * (1 + x + 2*x^2 - x^3 + x^4) / (1 - 4*x^3 + x^6). - Michael Somos, Feb 23 2015
0 = a(n) - 4*a(n+3) + a(n+6) for all n in Z. - Michael Somos, Feb 23 2015
a(3*n) = A052530(n). a(3*n-2) = A001835(n). a(3*n+2) = A001834(n). - Michael Somos, Feb 23 2015

A249580 List of quadruples (r,s,t,u): the matrix M = [[4,12,9][2,5,3][1,2,1]] is raised to successive negative powers, then (r,s,t,u) are the square roots of M[1,3], M[1,1], M[3,3], M[3,1] respectively.

Original entry on oeis.org

3, -1, -2, 1, -9, 4, 7, -3, 30, -13, -23, 10, -99, 43, 76, -33, 327, -142, -251, 109, -1080, 469, 829, -360, 3567, -1549, -2738, 1189, -11781, 5116, 9043, -3927, 38910, -16897, -29867, 12970, -128511, 55807, 98644, -42837

Offset: 1

Russell Walsmith, Nov 02 2014

The sequence, in reverse order, comprises numbers to the left of a(0) in A249579, where the terms would be labeled a(-1), a(-2), a(-3), ... .

This sequence 'factors' into four sequences with alternating signs. Ignoring signage, they are A052906, A003688, A052924 and A006190 (listed as crossrefs below).

			M^-1 = [[1,-6,9][-1,5,-6][1,-4,4]]. sqrt(M[1,3]) = 3, sqrt(M[1,1]) = -1, sqrt(M[3,3]) = -2, sqrt(M[3,1]) = 1. Then r = 3; s = -1; t = -2; ; u = 1.
M^-2 = [[16,-72,81][-12,55,-63][9,-42,49]]. sqrt(M[1,3]) = -9, sqrt(M[1,1]) = 4, sqrt(M[3,3]) = 7, sqrt(M[3,1]) = -3. Then r = -9; s = 4; t = 7; ; u = -3.

Colin Barker, Table of n, a(n) for n = 1..1000
Russell Walsmith, A sequence of matrices
Russell Walsmith, DCL-Chemy III: Hyper-Quadratics
Index entries for linear recurrences with constant coefficients, signature (0,0,0,-3,0,0,0,1).

Cf. A249579. Disregarding signage, a(4n) = A052906; a(4n+1) = A003688; a(4n+2) = A052924; a(4n+3) = A006190.

m[e_] := MatrixPower[{{4, 12, 9}, {2, 5, 3}, {1, 2, 1}}, -e]; g[e_, x_, y_] := (-1)^If[x == y, e, e + 1] Sqrt@ m[e][[x, y]]; f[e_] := {g[e, 1, 3], g[e, 1, 1], g[e, 3, 3], g[e, 3, 1]}; Array[f, 10] // Flatten (* Robert G. Wilson v, Dec 19 2014 *)

Vec(-x*(x^6+x^5+x^3-2*x^2-x+3)/(x^8-3*x^4-1) + O(x^100)) \\ Colin Barker, Nov 06 2014

a(n) = -3*a(n-4)+a(n-8). - Colin Barker, Nov 06 2014

G.f.: -x*(x^6+x^5+x^3-2*x^2-x+3) / (x^8-3*x^4-1). - Colin Barker, Nov 06 2014

A249579 List of quadruples (r,s,t,u): the matrix M = [[4,12,9][2,5,3][1,2,1]] is raised to successive powers, then (r,s,t,u) are the square roots of M[3,1], M[3,3], M[1,1], M[1,3] respectively.

Original entry on oeis.org

0, 1, 1, 0, 1, 1, 2, 3, 3, 4, 7, 9, 10, 13, 23, 30, 33, 43, 76, 99, 109, 142, 251, 327, 360, 469, 829, 1080, 1189, 1549, 2738, 3567, 3927, 5116, 9043, 11781, 12970, 16897, 29867, 38910, 42837, 55807, 98644, 128511, 141481, 184318, 325799, 424443, 467280

Offset: 0

Russell Walsmith, Nov 02 2014

			M^0 = [[1,0,0][0,1,0][0,0,1]]: r = sqrt(M[3,1]) = a(0) = 0, s = sqrt(M[3,3]) = a(1) = 1, t = sqrt(M[1,1]) = a(2) = 1, u = sqrt(M[1,3])u = a(3) = 0.
M^2 = [[49, 126, 81][21, 55, 36][9, 24, 16]]: r = sqrt(M[3, 1]) = a(8) = 3, s = sqrt(M[3, 3]) = a(9) = 4, t = sqrt(M[1, 1]) = a(10) = 7, u = sqrt(M[1, 3]) = a(11) = 9.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,3,0,0,0,1).

Cf. A249580
a(4n) = A006190
a(4n+2) = A052924.

I:=[0,1,1,0,1,1,2,3]; [n le 8 select I[n] else 3*Self(n-4)+Self(n-8): n in [1..50]]; // Vincenzo Librandi, Nov 14 2014

CoefficientList[Series[- x (3 x^6 - x^5 - 2 x^4 + x^3 + x + 1) / (x^8 + 3 x^4 - 1), {x, 0, 50}], x] (* Vincenzo Librandi, Nov 14 2014 *)

concat(0, Vec(-x*(3*x^6-x^5-2*x^4+x^3+x+1)/(x^8+3*x^4-1) + O(x^100))) \\ Colin Barker, Nov 13 2014

Some identities:
a4(n-1) + a(4n) = a(4n+1),
a(4n) + a(4n+1) = a(4n+2),
3a(4n) = a(4n+3).
a(n) = 3*a(n-4)+a(n-8). - Colin Barker, Nov 13 2014
G.f.: -x*(3*x^6-x^5-2*x^4+x^3+x+1) / (x^8+3*x^4-1). - Colin Barker, Nov 13 2014

More terms from Colin Barker, Nov 13 2014

A249581 List of quadruples (r,s,t,u): the matrix M = [[9,24,16][3,10,8][1,4,4]] is raised to successive powers, then (r,s,t,u) are the square roots of M[3,1], M[3,3], M[1,1], M[1,3] respectively.

Original entry on oeis.org

0, 1, 1, 0, 1, 2, 3, 4, 5, 8, 13, 20, 23, 36, 59, 92, 105, 164, 269, 420, 479, 748, 1227, 1916, 2185, 3412, 5597, 8740, 9967, 15564, 25531, 39868, 45465, 70996, 116461, 181860, 207391, 323852, 531243, 829564, 946025, 1477268, 2423293, 3784100

Offset: 0

Russell Walsmith, Nov 03 2014

The general form of these matrices is [[t^2,2tu,u^2][rt,st+ru,su][r^2,2rs,s^2]]. Different symmetries have different properties.

Iff |r * u - s * t| = 1 then terms to the left of a(0) are all integers.

			M^0 = [[1,0,0][0,1,0][0,0,1]]. r = sqrt(M[3,1]) = a(0) = 0; s = sqrt(M[3,3]) = a(1) = 1; t = sqrt(M[1,1]) = a(2) = 1; u = sqrt(M[1,3]) = a(3) = 0.
M^1 = [[9,24,16][3,10,8][1,4,4]]. r = sqrt(M[3,1]) = a(4) = 1; s = sqrt(M[3,3]) = a(5) = 2; t = sqrt(M[1,1]) = a(6) = 3; u = sqrt(M[1,3]) = a(7) = 4.

Colin Barker, Table of n, a(n) for n = 0..1000
Russell Walsmith, DCL-Chemy III: Hyper-Quadratics
Index entries for linear recurrences with constant coefficients, signature (0,0,0,5,0,0,0,-2).

Cf. A249579, A249580, A147722, A052984.

LinearRecurrence[{0,0,0,5,0,0,0,-2},{0,1,1,0,1,2,3,4},50] (* Harvey P. Dale, Aug 01 2016 *)

concat(0, Vec(x*(4*x^6-2*x^5-3*x^4+x^3+x+1)/(2*x^8-5*x^4+1) + O(x^100))) \\ Colin Barker, Nov 04 2014

a(4n) + a(4n + 1) = a(4n + 2).
a(4n + 1) + a(4n + 2) + a(4n + 3) - a(4n) = a(4n + 5)
4a(4n) = a(4n+3).
a(4n+1) = A147722(n), a(4n+2) = A052984(n).
a(n) = 5*a(n-4)-2*a(n-8). - Colin Barker, Nov 04 2014
G.f.: x*(4*x^6-2*x^5-3*x^4+x^3+x+1) / (2*x^8-5*x^4+1). - Colin Barker, Nov 04 2014

A249578 List of triples (r,s,t): the matrix M = [[4,12,9][2,7,6][1,4,4]] is raised to successive powers, then (r,s,t) are the square roots of M[3,1], M[1,1], M[1,3] respectively.

Original entry on oeis.org

0, 1, 0, 1, 2, 3, 4, 7, 12, 15, 26, 45, 56, 97, 168, 209, 362, 627, 780, 1351, 2340, 2911, 5042, 8733, 10864, 18817, 32592, 40545, 70226, 121635, 151316, 262087, 453948, 564719, 978122, 1694157, 2107560, 3650401, 6322680

Offset: 0

Russell Walsmith, Nov 03 2014

M is the 'Fibonacci matrix' F = [[1,2,1][1,1,0][1,0,0]]  taken to the third power and flipped on a vertical axis.
Sequence identities:
2a(3n-2) + a(3n-1) = a(3n+1)
2a(3n) + a(3n+1) = a(3n+3)
a(3n-2) + a(3n-1) + a(3n+1) = a(3n+2)
a(3n) + a(3n+1) + a(3n-3) = a(3n+2)
a(3n-1) * a(3n) + a(3n+1) * a(3n-2) = a(6n-2).

			M^0 = the 3 X 3 identity matrix = [[1,0,0][0,1,0][0,0,1]]. M[3,1] = 0; M[1,1] = 1; M[1,3] = 0. So the first triple is r = a(0) = 0; s = a(1) = 1; t = a(2) = 0.
M^1 = [[4,12,9][2,7,6][1,4,4]], so r = a(3) = 1; s = a(4) = 2; t = a(5) = 3.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,4,0,0,-1).

a(3n) = the n-th term of A001353.
a(3n+1) = n-th term of A001075.
a(3n+2) = n-th term of A005320.

I:=[0,1,0,1,2,3]; [n le 6 select I[n] else 4*Self(n-3)-Self(n-6): n in [1..40]]; // Vincenzo Librandi, Nov 04 2014

CoefficientList[Series[x (3 x^4 - 2 x^3 + x^2 + 1) / (x^6 - 4 x^3 + 1), {x, 0, 70}], x] (* Vincenzo Librandi, Nov 04 2014 *)
LinearRecurrence[{0,0,4,0,0,-1},{0,1,0,1,2,3},40] (* Harvey P. Dale, Jan 17 2017 *)

concat(0, Vec(x*(3*x^4-2*x^3+x^2+1)/(x^6-4*x^3+1) + O(x^100))) \\ Colin Barker, Nov 04 2014

a(n) = 4*a(n-3)-a(n-6).

G.f.: x*(3*x^4-2*x^3+x^2+1) / (x^6-4*x^3+1). - Colin Barker, Nov 04 2014

A249577 List of triples (r,s,t): the matrix M = [[1,4,4][1,3,2][1,2,1]] is raised to successive negative powers, then (r,s,t) are the square roots of M[3,1], M[1,1], M[1,3] respectively.

Original entry on oeis.org

2, -1, 1, -4, 3, -2, 10, -7, 5, -24, 17, -12, 58, -41, 29, -140, 99, -70, 338, -239, 169, -816, 577, -408, 1970, -1393, 985, -4756, 3363, -2378, 11482, -8119, 5741, -27720, 19601, -13860, 66922, -47321, 33461, -161564, 114243, -80782, 390050, -275807, 195025, -941664, 665857, -470832

Offset: 0

Russell Walsmith, Nov 01 2014

The sequence comprises, in reverse order, numbers to the right of a(0) in A249576.

			M^-1 = [[1,-4,4][-1,3,-2][1,-2,1]]. sqrt(M[1,3]) = 2; M[3,3] = M[1,1] = -1; M[3,1] = 1. Hence a(0) = 2; a(1) = -1; a(2) = 1.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,-2,0,0,1).

Cf. A249576, A000129, A001333, A163271, A077985.

LinearRecurrence[{0,0,-2,0,0,1},{2,-1,1,-4,3,-2},50] (* Harvey P. Dale, Aug 02 2024 *)

Vec(-(x^4+x^2-x+2)/(x^6-2*x^3-1) + O(x^100)) \\ Colin Barker, Nov 02 2014

a(n) = -2*a(n-3)+a(n-6); G.f.: -(x^4+x^2-x+2) / (x^6-2*x^3-1). - Colin Barker, Nov 02 2014

A249576 List of triples (r,s,t): the matrix M = [[1,4,4][1,3,2][1,2,1]] is raised to successive powers, then (r,s,t) are the square roots of M[3,1], M[1,1], M[1,3] respectively.

Original entry on oeis.org

0, 1, 0, 1, 1, 2, 2, 3, 4, 5, 7, 10, 12, 17, 24, 29, 41, 58, 70, 99, 140, 169, 239, 338, 408, 577, 816, 985, 1393, 1970, 2378, 3363, 4756, 5741, 8119, 11482, 13860, 19601, 27720, 33461, 47321, 66922, 80782, 114243, 161564, 195025, 275807, 390050, 470832, 665857, 941664

Offset: 0

Russell Walsmith, Nov 01 2014

Numbers to the left of a(0) are in A249577.
Some identities:
a(3n - 2) + a(3n - 1) = a(3n + 1).
a(3n) + a(3n + 1) = a(3(n + 1)).
a(3n - 2) + a(3n + 1) = a(3n + 2).
a(3n) + a(3n - 1) + a(3(n - 2)) = a(3n + 1).
a(3n - 1)a(3n) + a(3n + 2)a(3(n + 1)) = a(6n + 2).

			M^0 = the 3 X 3 identity matrix = [[1,0,0][0,1,0][0,0,1]]. M[3,1] = 0; M[1,1] = 1; M[1,3] = 0. So the first triple is r = a(0) = 0; s = a(1) = 1; t = a(2) = 0.
M^1 = [[1,4,4][1,3,2][1,2,1]], so r = a(3) = 1; s = a(4) = 1; t = a(5) = 2.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,2,0,0,1).

a(3n) = the n-th term of A000129, the Pell numbers.
a(3n+1) = n-th term of A001333.
a(3n+2) = n-th term of A163271.

LinearRecurrence[{0,0,2,0,0,1},{0,1,0,1,1,2},60] (* Harvey P. Dale, Dec 29 2021 *)

concat(0, Vec(-x*(2*x^4-x^3+x^2+1)/(x^6+2*x^3-1) + O(x^100))) \\ Colin Barker, Nov 02 2014

a(n) = -2*a(n-3)+a(n-6); G.f.: -x*(2*x^4-x^3+x^2+1) / (x^6+2*x^3-1). - Colin Barker, Nov 02 2014

A229526 The c coefficients of the transform ax^2 + (4a/k - b)x + 4a/k^2 + 2b/k + c= 0 for a,b,c = 1,-1,-1, k = 1,2,3...

Original entry on oeis.org

5, 1, 1, -1, -11, -5, -31, -11, -59, -19, -95, -29, -139, -41, -191, -55, -251, -71, -319, -89, -395, -109, -479, -131, -571, -155, -671, -181, -779, -209, -895, -239, -1019, -271, -1151, -305, -1291, -341, -1439, -379, -1595, -419, -1759, -461, -1931, -505

Offset: 1

Russell Walsmith, Sep 27 2013

The positive/negative roots of ax^2 + bx + c = 0 combine with the negative/positive roots of (ck^2 - bk + c)x^2 +(2ck - b)x + c = 0 to define a point on the hyperbola kxy + x + y = 0. To shift such points (roots) to the hyperbola’s other line, put the coefficients of these equations into the formula ax^2 + (4a/k - b)x + 4a/k^2 + 2b/k + c = 0. Let a,b,c = 1,-1,-1 and k = 1,2,3... Then the coefficients given by this last equation are the sequence 1,5,5; 1,3,1; 1,7/3,1/9... Clearing fractions, the c coefficients are the sequence above.

The n-th term = the (positive) n-4th term of A229525.

			For k = 5, the coefficients are 1, 9/5, -11/25. Clearing fractions gives 25, 45, -11 and -11 = a[5].

Colin Barker, Table of n, a(n) for n = 1..1000
Russell Walsmith, CL-Chemy III: Hyper-Quadratics

The a coefficients are A168077, b coefficients are A171621, the sum of a, b and c coefficients is A229525.

Vec(-x*(x^5+x^4-4*x^3-14*x^2+x+5)/((x-1)^3*(x+1)^3) + O(x^100)) \\ Colin Barker, Nov 02 2014

ax^2 + (4a/k - b)x + 4a/k^2 + 2b/k + c; a,b,c = 1,-1,-1, k = 1,2,3..n.
a(n) = 3*a(n-2)-3*a(n-4)+a(n-6). G.f.: -x*(x^5+x^4-4*x^3-14*x^2+x+5) / ((x-1)^3*(x+1)^3). - Colin Barker, Nov 02 2014
a(n) = (-5+3*(-1)^n)*(-4-2*n+n^2)/8. - Colin Barker, Nov 03 2014

More terms from Colin Barker, Nov 02 2014

A229525 Sum of coefficients of the transform ax^2 + (4a/k - b)x + 4a/k^2 + 2b/k + c = 0 for a,b,c = 1,-1,-1, k = 1,2,3...

Original entry on oeis.org

11, 5, 31, 11, 59, 19, 95, 29, 139, 41, 191, 55, 251, 71, 319, 89, 395, 109, 479, 131, 571, 155, 671, 181, 779, 209, 895, 239, 1019, 271, 1151, 305, 1291, 341, 1439, 379, 1595, 419, 1759, 461, 1931, 505, 2111, 551, 2299, 599, 2495, 649, 2699, 701, 2911, 755

Offset: 1

Russell Walsmith, Sep 26 2013

The positive/negative roots of ax^2 + bx + c = 0 combine with the negative/positive roots of (ck^2 - bk + c)x^2 +(2ck - b)x + c = 0 to define a point on the hyperbola kxy + x + y = 0. To shift such points (roots) to the hyperbola’s other line, put the coefficients of these equations into the formula Q = ax^2 + (4a/k - b)x + 4a/k^2 + 2b/k + c = 0. For a,b,c = 1,-1,-1 and k = 1,2,3..., the coefficients given by Q are the sequence 1,5,5; 1,3,1; 1,7/3,1/9... Clearing fractions and summing a+b+c gives the sequence.

The negative of the n-th term is the n+4th term of the c coefficient sequence A229526.

			For k = 5, the coefficients are 1, 9/5, -11/25. Clearing fractions, 25, 45, -11 and 25 + 45 -11 = 59 = a[5].

Colin Barker, Table of n, a(n) for n = 1..1000
Russell Walsmith, CL-Chemy III: Hyper-Quadratics
Index entries for linear recurrences with constant coefficients, signature (0,3,0,-3,0,1).

The a coefficients are A168077, b coefficients are A171621, c coefficients are A229526.

Vec(-x*(x^5-x^4-4*x^3-2*x^2+5*x+11)/((x-1)^3*(x+1)^3) + O(x^100)) \\ Colin Barker, Nov 02 2014

ax^2 + (4a/k - b)x + 4a/k^2 + 2b/k + c; a,b,c = 1,-1,-1, k = 1,2,3... n.
a(n) = 3*a(n-2)-3*a(n-4)+a(n-6). G.f.: -x*(x^5-x^4-4*x^3-2*x^2+5*x+11) / ((x-1)^3*(x+1)^3). - Colin Barker, Nov 02 2014
a(n) = -(-5+3*(-1)^n)*(4+6*n+n^2)/8. - Colin Barker, Nov 03 2014

More terms from Colin Barker, Nov 02 2014

A226593 Largest period of a recurrence sequence of pairs of permutations of length n.

Original entry on oeis.org

1, 3, 8, 18, 96, 216, 2112, 9720, 39024, 194256, 1116240

Offset: 1

Russell Walsmith, Jun 13 2013

The n! permutations (p) of the numbers 1,2,3..n may be paired (allowing duplication) in n!^2 ways. For a pair of permutations (p, p'), let p'' = p x p', p''' = p' x p'', and so on until the starting pair (p, p') is obtained. If p = p', this iterative process gives the Pisano periods. For most other pairs the periods have different lengths. The sequence gives the longest period that (p, p') generates for any p, p' of length n.

Period is invariant with respect to simultaneous conjugation of both p, p'. - Max Alekseyev, Feb 09 2024

			For n = 4: 3142 x 2341 = 1423; 2341 x 1423 = 2134... the sequence thus generated is of period = 18.

Russell Walsmith, An investigation of recursive sequences based on pairs of permutations of length n.

Cf. A001175 (Pisano periods: period of Fibonacci numbers (A000045) mod n).

Cf. A106291 (period of the Lucas sequence (A000032) mod n).

period(a,b)=my(n=matsize(a)[2], v=vector(n), aa=vector(n,i,a[i]), bb=vector(n,i,b[i]), id, nsteps); while(id!=n, for(i=1,n, v[i]=a[b[i]]); id=sum(i=1,n, b[i]==aa[i] && v[i]==bb[i]); for(i=1,n, a[i]=b[i]; b[i]=v[i]); nsteps++); nsteps
a(n)=my(a,b,m,p); for(k=1,n!, a=numtoperm(n,k); for(l=1,n!, b=numtoperm(n,l); p=period(a,b); if(p>m,m=p))); m \\ Ralf Stephan, Aug 13 2013

a(6) from Ralf Stephan, Aug 13 2013

Edited and a(7)-a(11) added by Max Alekseyev, Feb 13 2024

cp's OEIS Frontend

User: Russell Walsmith

A254308 a(n) = a(n-1) + (if a(n-1) is odd a(n-2) else a(n-3)) with a(0) = 0, a(1) = 1.

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A249580 List of quadruples (r,s,t,u): the matrix M = [[4,12,9][2,5,3][1,2,1]] is raised to successive negative powers, then (r,s,t,u) are the square roots of M[1,3], M[1,1], M[3,3], M[3,1] respectively.

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A249579 List of quadruples (r,s,t,u): the matrix M = [[4,12,9][2,5,3][1,2,1]] is raised to successive powers, then (r,s,t,u) are the square roots of M[3,1], M[3,3], M[1,1], M[1,3] respectively.

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A249581 List of quadruples (r,s,t,u): the matrix M = [[9,24,16][3,10,8][1,4,4]] is raised to successive powers, then (r,s,t,u) are the square roots of M[3,1], M[3,3], M[1,1], M[1,3] respectively.

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A249578 List of triples (r,s,t): the matrix M = [[4,12,9][2,7,6][1,4,4]] is raised to successive powers, then (r,s,t) are the square roots of M[3,1], M[1,1], M[1,3] respectively.

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A249577 List of triples (r,s,t): the matrix M = [[1,4,4][1,3,2][1,2,1]] is raised to successive negative powers, then (r,s,t) are the square roots of M[3,1], M[1,1], M[1,3] respectively.

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A249576 List of triples (r,s,t): the matrix M = [[1,4,4][1,3,2][1,2,1]] is raised to successive powers, then (r,s,t) are the square roots of M[3,1], M[1,1], M[1,3] respectively.

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