A189234 -id:A189234 - cp's OEIS frontend

A216605 Expansion of g.f. (6 + 5x - 20x^2 - 12x^3 + 12x^4 + 3x^5)/(1 + x - 5x^2 - 4x^3 + 6x^4 + 3*x^5 - x^6).

6, -1, 11, -4, 31, -16, 98, -64, 327, -256, 1126, -1024, 3958, -4083, 14116, -16189, 50887, -63768, 184958, -249547, 676626, -970771, 2488156, -3756867, 9188406, -14474916, 34049481, -55564474, 126540536, -212637571, 471398623, -811660849, 1759603367

Offset: 0

Roman Witula, Sep 10 2012

Previous name was: a(n) is equal to the rational part (considering of the ring Z(sqrt(13))) of the numbers A(n) = ((sqrt(13) - 1)/2)*A(n-1) + A(n-2) + ((3-sqrt(13))/2)*A(n-3), with A(0) = 6, A(1) = sqrt(13) - 1, and A(2) = 11 - sqrt(13).
The Berndt-type sequence number 1 for the argument 2*Pi/13 defined by the following relation: a(n) + b(n)*sqrt(13) = A(n) = 2*(c(1)^n + c(3)^n + c(4)^n), where c(j) := 2*cos(2*Pi*j/13), j=1..6. The b(n) = A216486(n), n=0,1,..., are all positive integers. We note that we also have a(n) - b(n)*sqrt(13) = B(n) = 2*(c(2)^n + c(5)^n + c(6)^n) and the following recurrence relation holds: B(n) = -((sqrt(13)+ 1)/2)*B(n-1) + B(n-2) + ((3+sqrt(13))/2)*B(n-3), with B(0) = 6, B(1) = -sqrt(13) - 1, and B(2) = 11 + sqrt(13).
We note that 4*a(n) - a(n+2) is divisible by 13 for every n = 0,1,... .

			We have 4*a(2*n-1)=a(2*n+1) for every n = 1,2,...,5 and a(13) - 4*a(11) = 13. Further we have c(1)^3 + c(3)^3 + c(4)^3 = 4*(c(1) + c(3) + c(4)) since A(3) = 4*sqrt(13) - 4, c(2)^3 + c(5)^3 + c(6)^3 = 4*(c(2) + c(5) + c(6)) since B(3) = - 4*sqrt(13) - 4, 2 + c(1)^4 + c(3)^4 + c(4)^4 = 3*(c(1)^2 + c(3)^2 + c(4)^2) and 2 + c(2)^4 + c(5)^4 + c(6)^4 = 3*(c(2)^2 + c(5)^2 + c(6)^2).

R. Witula and D. Slota, Quasi-Fibonacci numbers of order 13, Thirteenth International Conference on Fibonacci Numbers and Their Applications, Congressus Numerantium, 201 (2010), 89-107.
R. Witula, On some applications of formulas for sums of the unimodular complex numbers, Wyd. Pracowni Komputerowej Jacka Skalmierskiego, Gliwice 2011 (in Polish).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
A. Akbary and Q. Wang, A generalized Lucas sequence and permutation binomials, Proc. Am. Math. Soc. 134 (2006) 15-22. Sequence a(n) with l=13 is the unsigned version.
Russell A. Gordon, Lucas Type Sequences and Sums of Binomial Coefficients, Integers (2023) Vol 23, Art. No. A84. See p. 21.
R. Witula and D. Slota, Quasi-Fibonacci numbers of order 13, (abstract) see p. 15.
Index entries for linear recurrences with constant coefficients, signature (-1,5,4,-6,-3,1).

Cf. A094649, A189234, A216486.

I:=[6,-1,11,-4,31,-16]; [n le 6 select I[n] else -Self(n-1)+5*Self(n-2)+4*Self(n-3)-6*Self(n-4)-3*Self(n-5)+Self(n-6): n in [1..35]]; // Vincenzo Librandi, Aug 30 2017

LinearRecurrence[{-1, 5, 4, -6, -3, 1}, {6, -1, 11, -4, 31, -16}, 30]
CoefficientList[Series[(6 + 5 x - 20 x^2 - 12 x^3 + 12 x^4 + 3 x^5)/(1 + x - 5 x^2 - 4 x^3 + 6 x^4 + 3 x^5 - x^6), {x, 0, 33}], x] (* Vincenzo Librandi, Aug 30 2017 *)

G.f.: (6 + 5*x - 20*x^2 - 12*x^3 + 12*x^4 + 3*x^5)/(1 + x - 5*x^2 - 4*x^3 + 6*x^4 + 3*x^5 - x^6). - Bruno Berselli, Sep 11 2012

a(n) = -a(n-1) + 5*a(n-2) + 4*a(n-3) - 6*a(n-4) - 3*a(n-5) + a(n-6), which from the following decomposition can be generated (see Witula-Slota's and Witula's references for details): X^6 + X^5 - 5*X^4 - 4*X^3 + 6*X^2 + 3*X - 1 = ((X - c(1))*(X - c(3))*(X - c(4)))*((X - c(2))*(X - c(5))*(X - c(6))) = (X^3 + ((1 - sqrt(13))/2)*X^2 - X + (sqrt(13) - 3)/2)*(X^3 + ((1 + sqrt(13))/2)*X^2 - X - (sqrt(13) + 3)/2). - Roman Witula, Sep 11 2012

New name using existing g.f. from Joerg Arndt, Feb 15 2024

A189235 Expansion of (5-16x+6x^2+10x^3-2x^4)/(1-4x+2x^2+5x^3-2x^4-x^5).

Original entry on oeis.org

5, 4, 12, 25, 64, 159, 411, 1068, 2808, 7423, 19717, 52529, 140251, 375015, 1003770, 2688570, 7204696, 19313075, 51782613, 138861732, 372414289, 998851473, 2679146955, 7186319506, 19276417059, 51707411684, 138702360471, 372064319188

Offset: 0

L. Edson Jeffery, Apr 18 2011

Same as A062883 preceded by 5.
Let U be the unit-primitive matrix (see [Jeffery])
U=U_(11,2)=
(0 0 1 0 0)
(0 1 0 1 0)
(1 0 1 0 1)
(0 1 0 1 1)
(0 0 1 1 1).
Then a(n)=Trace(U^n).
Evidently one of a class of accelerator sequences for Catalan's constant based on traces of successive powers of a unit-primitive matrix U_(N,r) (0Formulae given below are special cases of general one's defined and discussed in Witula-Slota's paper. For example a(n) = A(n;1), where A(n;d) := Sum_{k=1..5} (1 + 2d*cos(2Pi*k/11))^n, n=0,1,..., d in C. - Roman Witula, Jul 26 2012

R. Witula and D. Slota, Quasi-Fibonacci Numbers of Order 11, 10 (2007), J. Integer Seq., Article 07.8.5.

L. E. Jeffery, Unit-primitive matrices
Index entries for linear recurrences with constant coefficients, signature (4, -2, -5, 2, 1).

Cf. A189234, A189236, A189237.

u = {{0, 0, 1, 0, 0}, {0, 1, 0, 1, 0}, {1, 0, 1, 0, 1}, {0, 1, 0, 1, 1}, {0, 0, 1, 1, 1}}; a[n_] := Tr[ MatrixPower[u, n] ]; Table[a[n], {n, 0, 27}] (* Jean-François Alcover, Oct 14 2013 *)

Vec((5-16*x+6*x^2+10*x^3-2*x^4)/(1-4*x+2*x^2+5*x^3-2*x^4-x^5)+O(x^99)) \\ Charles R Greathouse IV, Sep 25 2012

G.f.: (5-16*x+6*x^2+10*x^3-2*x^4)/(1-4*x+2*x^2+5*x^3-2*x^4-x^5).
a(n)=4*a(n-1)-2*a(n-2)-5*a(n-3)+2*a(n-4)+a(n-5), {a(m)}=5,4,12,25,64, m=0..4.
a(n)=Sum_{k=1..5} ((x_k)^2-1)^n; x_k=2*(-1)^(k-1)*cos(k*Pi/11).

A189236 Expansion of (5-8x-15x^2+4x^3+4x^4)/(1-2x-5x^2+2x^3+4x^4+x^5).

Original entry on oeis.org

5, 2, 14, 32, 114, 347, 1142, 3649, 11826, 38111, 123139, 397443, 1283406, 4143479, 13378435, 43194542, 139463234, 450284986, 1453839839, 4694021537, 15155624819, 48933074467, 157990585613, 510105367936, 1646980994190, 5317619734147

Offset: 0

L. Edson Jeffery, Apr 18 2011

(Start) Let U be the unit-primitive matrix (see [Jeffery])
U=U_(11,3)=
(0 0 0 1 0)
(0 0 1 0 1)
(0 1 0 1 1)
(1 0 1 1 1)
(0 1 1 1 1).
Then a(n)=Trace(U^n). (End)
Evidently one of a class of accelerator sequences for Catalan's constant based on traces of successive powers of a unit-primitive matrix U_(N,r) (0

L. E. Jeffery, Unit-primitive matrices
Index entries for linear recurrences with constant coefficients, signature (2, 5, -2, -4, -1).

Cf. A062883, A189234, A189235, A189237.

CoefficientList[Series[ (5-8x-15x^2+4x^3+4x^4)/ (1-2x-5x^2+2x^3+4x^4+x^5), {x,0,29}],x]  (* Harvey P. Dale, Apr 19 2011 *)
LinearRecurrence[{2, 5, -2, -4, -1}, {5, 2, 14, 32, 114}, 30] (* T. D. Noe, Apr 19 2011 *)

Vec((5-8*x-15*x^2+4*x^3+4*x^4)/(1-2*x-5*x^2+2*x^3+4*x^4+x^5)+O(x^99)) \\ Charles R Greathouse IV, Sep 25 2012

G.f.: (5-8*x-15*x^2+4*x^3+4*x^4)/(1-2*x-5*x^2+2*x^3+4*x^4+x^5).
a(n)=2*a(n-1)+5*a(n-2)-2*a(n-3)-4*a(n-4)-a(n-5), {a(m)}={5,2,14,32,114}, m=0..4.
a(n)=Sum_{k=1..5} ((x_k)^3-2*(x_k))^n; x_k=2*(-1)^(k-1)*cos(k*Pi/11).
Series expansion of g.f. at x=infinity gives -A062883 and all but the first term of -A189235.

A189237 Expansion of (5-12x-9x^2+8x^3+x^4)/(1-3x-3x^2+4x^3+x^4-x^5).

Original entry on oeis.org

5, 3, 15, 42, 155, 533, 1884, 6604, 23219, 81555, 286555, 1006734, 3537032, 12426742, 43659386, 153390077, 538911123, 1893376346, 6652069455, 23370962220, 82110068595, 288480349402, 1013528712002, 3560868017067, 12510529683224

Offset: 0

L. Edson Jeffery, Apr 18 2011

(Start) Let U be the unit-primitive matrix (see [Jeffery])
U=U_(11,4)=
(0 0 0 0 1)
(0 0 0 1 1)
(0 0 1 1 1)
(0 1 1 1 1)
(1 1 1 1 1).
Then a(n)=Trace(U^n). (End)
Evidently one of a class of accelerator sequences for Catalan's constant based on traces of successive powers of a unit-primitive matrix U_(N,r) (0

L. E. Jeffery, Unit-primitive matrices
Index entries for linear recurrences with constant coefficients, signature (3, 3, -4, -1, 1).

A189234, A189235, A189236.

CoefficientList[Series[(5-12x-9x^2+8x^3+x^4)/(1-3x-3x^2+4x^3+x^4-x^5), {x,0,30}],x] (* or *) LinearRecurrence[{3,3,-4,-1,1},{5,3,15,42,155},30] (* Harvey P. Dale, Oct 01 2011 *)

Vec((5-12*x-9*x^2+8*x^3+x^4)/(1-3*x-3*x^2+4*x^3+x^4-x^5)+O(x^99)) \\ Charles R Greathouse IV, Sep 26 2012

G.f.: (5-12*x-9*x^2+8*x^3+x^4)/(1-3*x-3*x^2+4*x^3+x^4-x^5).
a(n)=3*a(n-1)+3*a(n-2)-4*a(n-3)-a(n-4)+a(n-5), {a(m)}={5,3,15,42,155}, m=0..4.
a(n)=Sum_{k=1..5} ((x_k)^4-3*(x_k)^2+1)^n; x_k=2*(-1)^(k-1)*cos(k*Pi/11).
Series expansion of g.f. at x=infinity gives -A189234(n+1).

A209235 Rectangular array read by antidiagonals, with entry k in row n given by T(n,k) = 2^{k-1}Sum_{j=1..n} (cos((2j-1)Pi/(2n+1)))^{k-1}.

Original entry on oeis.org

1, 2, 1, 3, 1, 1, 4, 1, 3, 1, 5, 1, 5, 4, 1, 6, 1, 7, 4, 7, 1, 8, 1, 11, 4, 19, 16, 18, 1, 9, 1, 13, 4, 25, 16, 38, 29, 1, 10, 1, 15, 4, 31, 16, 58, 57, 47, 1, 11, 1, 17, 4, 37, 16, 78, 64, 117, 76, 1, 12, 1, 19, 4, 43, 16, 98, 64, 187, 193, 123, 1

Offset: 1

L. Edson Jeffery, Jan 12 2013

Antidiagonal sums: {1,3,5,9,16,26,46,78,136,...}.

			Array begins as
.1..1...1..1...1...1
.2..1...3..4...7..11
.3..1...5..4..13..16
.4..1...7..4..19..16
.5..1...9..4..25..16
.6..1..11..4..31..16

Rows: cf. A000012, A000032, A094649, A189234, A216605, etc.

Cf. A185095, A186740.

T(n,k) = 2^{k-1}*Sum_{j=1..n} (cos((2*j-1)*Pi/(2*n+1)))^{k-1}.
Empirical g.f. for row n: F(x) = (Sum_{u=0..n-1} A122765(n,n-1-u)*x^u)/(Sum_{v=0..n} A108299(n,v)*x^v).
Empirical: odd column first differences tend to A000984 = {1, 2, 6, 20, 70, 252, ...} (central binomial coefficients).

A376498 Array read by ascending antidiagonals: A(n, k) = 2^kSum_{j=1..n} cos((2j - 1)Pi/(2n + 1))^k.

Original entry on oeis.org

0, 1, 0, 2, 1, 0, 3, 1, 1, 0, 4, 1, 3, 1, 0, 5, 1, 5, 4, 1, 0, 6, 1, 7, 4, 7, 1, 0, 7, 1, 9, 4, 13, 11, 1, 0, 8, 1, 11, 4, 19, 16, 18, 1, 0, 9, 1, 13, 4, 25, 16, 38, 29, 1, 0, 10, 1, 15, 4, 31, 16, 58, 57, 47, 1, 0, 11, 1, 17, 4, 37, 16, 78, 64, 117, 76, 1, 0

Offset: 0

Cheng-Jun Li, Sep 25 2024

It is only a conjecture that the A(n, k) are always integers.

			Array starts:
[0] 0, 0,  0, 0,  0,  0,   0,  0,   0,   0,    0,    0, ...  [A000004]
[1] 1, 1,  1, 1,  1,  1,   1,  1,   1,   1,    1,    1, ...  [A000012]
[2] 2, 1,  3, 4,  7, 11,  18, 29,  47,  76,  123,  199, ...  [A000032]
[3] 3, 1,  5, 4, 13, 16,  38, 57, 117, 193,  370,  639, ...  [A096975]
[4] 4, 1,  7, 4, 19, 16,  58, 64, 187, 247,  622,  925, ...  [A094649]
[5] 5, 1,  9, 4, 25, 16,  78, 64, 257, 256,  874, 1013, ...  [A189234]
[6] 6, 1, 11, 4, 31, 16,  98, 64, 327, 256, 1126, 1024, ...  [A216605]
[7] 7, 1, 13, 4, 37, 16, 118, 64, 397, 256, 1378, 1024, ...
[8] 8, 1, 15, 4, 43, 16, 138, 64, 467, 256, 1630, 1024, ...
[9] 9, 1, 17, 4, 49, 16, 158, 64, 537, 256, 1882, 1024, ...

Rows: A000004 (n=0), A000012 (n=1), A000032 (n=2), A096975 (n=3), A094649 (n=4), A189234 (n=5), A216605 (n=6, with alternate signs).
Columns: A001477 (k=0), A057427 (k=1).
Cf. A180870.

A(n, k) = 2^k*sum(j=1, n, (cos((2*j-1)*Pi/(2*n+1)))^k, x=0)

A(n + k, 2*k - 1) = A(k, 2*k-1) = 4^(k-1).

Let P_n(x) be the polynomial: Sum_{k=0..n} x^k*A180870(n, k). Let R_n(x) be the polynomial Product_{k=0..n} x-Roots(P_n, k)^m. A(n, k) = abs([x^1] R_n(x))/2^(m*(n-1)), for n > 0. - Thomas Scheuerle, Oct 07 2024

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A216605 Expansion of g.f. (6 + 5*x - 20*x^2 - 12*x^3 + 12*x^4 + 3*x^5)/(1 + x - 5*x^2 - 4*x^3 + 6*x^4 + 3*x^5 - x^6).

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A189235 Expansion of (5-16*x+6*x^2+10*x^3-2*x^4)/(1-4*x+2*x^2+5*x^3-2*x^4-x^5).

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A189236 Expansion of (5-8*x-15*x^2+4*x^3+4*x^4)/(1-2*x-5*x^2+2*x^3+4*x^4+x^5).

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A189237 Expansion of (5-12*x-9*x^2+8*x^3+x^4)/(1-3*x-3*x^2+4*x^3+x^4-x^5).

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A209235 Rectangular array read by antidiagonals, with entry k in row n given by T(n,k) = 2^{k-1}*Sum_{j=1..n} (cos((2*j-1)*Pi/(2*n+1)))^{k-1}.

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A376498 Array read by ascending antidiagonals: A(n, k) = 2^k*Sum_{j=1..n} cos((2*j - 1)*Pi/(2*n + 1))^k.

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A216605 Expansion of g.f. (6 + 5x - 20x^2 - 12x^3 + 12x^4 + 3x^5)/(1 + x - 5x^2 - 4x^3 + 6x^4 + 3*x^5 - x^6).

A189235 Expansion of (5-16x+6x^2+10x^3-2x^4)/(1-4x+2x^2+5x^3-2x^4-x^5).

A189236 Expansion of (5-8x-15x^2+4x^3+4x^4)/(1-2x-5x^2+2x^3+4x^4+x^5).

A189237 Expansion of (5-12x-9x^2+8x^3+x^4)/(1-3x-3x^2+4x^3+x^4-x^5).

A209235 Rectangular array read by antidiagonals, with entry k in row n given by T(n,k) = 2^{k-1}Sum_{j=1..n} (cos((2j-1)Pi/(2n+1)))^{k-1}.

A376498 Array read by ascending antidiagonals: A(n, k) = 2^kSum_{j=1..n} cos((2j - 1)Pi/(2n + 1))^k.