A190987 -id:A190987 - cp's OEIS frontend

A190958 a(n) = 2a(n-1) - 10a(n-2), with a(0) = 0, a(1) = 1.

0, 1, 2, -6, -32, -4, 312, 664, -1792, -10224, -2528, 97184, 219648, -532544, -3261568, -1197696, 30220288, 72417536, -157367808, -1038910976, -504143872, 9380822016, 23803082752, -46202054656, -330434936832, -198849327104, 2906650714112, 7801794699264

Offset: 0

Vladimir Joseph Stephan Orlovsky, May 24 2011

For the difference equation a(n) = c*a(n-1) - d*a(n-2), with a(0) = 0, a(1) = 1, the solution is a(n) = d^((n-1)/2) * ChebyshevU(n-1, c/(2*sqrt(d))) and has the alternate form a(n) = ( ((c + sqrt(c^2 - 4*d))/2)^n - ((c - sqrt(c^2 - 4*d))/2)^n )/sqrt(c^2 - 4*d). In the case c^2 = 4*d then the solution is a(n) = n*d^((n-1)/2). The generating function is x/(1 - c*x + d^2) and the exponential generating function takes the form (2/sqrt(c^2 - 4*d))*exp(c*x/2)*sinh(sqrt(c^2 - 4*d)*x/2) for c^2 > 4*d, (2/sqrt(4*d - c^2))*exp(c*x/2)*sin(sqrt(4*d - c^2)*x/2) for 4*d > c^2, and x*exp(sqrt(d)*x) if c^2 = 4*d. - G. C. Greubel, Jun 10 2022

Vincenzo Librandi, Table of n, a(n) for n = 0..190
Index entries for linear recurrences with constant coefficients, signature (2,-10).

Sequences of the form a(n) = c*a(n-1) - d*a(n-2), with a(0)=0, a(1)=1:
c/d...1.......2.......3.......4.......5.......6.......7.......8.......9......10
.A128834,A107920,A106852,A106853,A106854,A145934,A145976,A145978,A146078,A146080
.A000027,A009545,A088137,A088138,A045873,A088139,A089181,A090591,A127357,A190958
.A001906,A000225,A057083,A049072,A190959,A190960,A190961,A190962,A190963,A190964
.A001353,A007070,A003462,A001787,A099456,A190965,A168175,A190966,A190967,A190968
.A004254,A107839,A116415,A002450,A030191,A001047,A099450,A190969,A190970,A190971
.A001109,A154244,A138395,A084326,A003463,A030192,A081179,A006516,A027471,A106392
.A004187,A186446,A190972,A190973,A190974,A003464,A030240,A152268,A099459,A016127
.A001090,A190975,A190976,A099156,A190977,A190978,A023000,A057084,A154245,A154235
.A018913,A190979,A190980,A190981,A190982,A190983,A190984,A023001,A057085,A178869
A004189,A190869,A190985,A190986,A190987,A190988,A190989,A190990,A002452,A057086

I:=[0,1]; [n le 2 select I[n] else 2*Self(n-1)-10*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Sep 17 2011

Mathematica
```
LinearRecurrence[{2,-10}, {0,1}, 50]
```

a(n)=([0,1; -10,2]^n*[0;1])[1,1] \\ Charles R Greathouse IV, Apr 08 2016

[lucas_number1(n,2,10) for n in (0..50)] # G. C. Greubel, Jun 10 2022

G.f.: x / ( 1 - 2*x + 10*x^2 ). - R. J. Mathar, Jun 01 2011
E.g.f.: (1/3)*exp(x)*sin(3*x). - Franck Maminirina Ramaharo, Nov 13 2018
a(n) = 10^((n-1)/2) * ChebyshevU(n-1, 1/sqrt(10)). - G. C. Greubel, Jun 10 2022
a(n) = (1/3)*10^(n/2)*sin(n*arctan(3)) = Sum_{k=0..floor(n/2)} (-1)^k*3^(2*k)*binomial(n,2*k+1). - Gerry Martens, Oct 15 2022

A342133 Square array T(n,k), n>=0, k>=0, read by antidiagonals, where column k is the expansion of g.f. 1/(1 - 2kx + k*x^2).

Original entry on oeis.org

1, 1, 0, 1, 2, 0, 1, 4, 3, 0, 1, 6, 14, 4, 0, 1, 8, 33, 48, 5, 0, 1, 10, 60, 180, 164, 6, 0, 1, 12, 95, 448, 981, 560, 7, 0, 1, 14, 138, 900, 3344, 5346, 1912, 8, 0, 1, 16, 189, 1584, 8525, 24960, 29133, 6528, 9, 0, 1, 18, 248, 2548, 18180, 80750, 186304, 158760, 22288, 10, 0

Offset: 0

Seiichi Manyama, Mar 01 2021

			Square array begins:
  1, 1,   1,    1,     1,     1, ...
  0, 2,   4,    6,     8,    10, ...
  0, 3,  14,   33,    60,    95, ...
  0, 4,  48,  180,   448,   900, ...
  0, 5, 164,  981,  3344,  8525, ...
  0, 6, 560, 5346, 24960, 80750, ...

Seiichi Manyama, Antidiagonals n = 0..139, flattened
Index entries for sequences related to Chebyshev polynomials.

Columns 0..5 give A000007, A000027(n+1), A007070, A138395, A099156(n+1), A190987(n+1).
Rows 0..2 give A000012, A005843, A033991.
Main diagonal gives (-1) * A109520(n+1).
Cf. A342120, A342129, A342134.

T:= (n, k)-> (<<0|1>, <-k|2*k>>^(n+1))[1, 2]:
seq(seq(T(n, d-n), n=0..d), d=0..12);  # Alois P. Heinz, Mar 01 2021

T[n_, k_] := Sum[If[k == j == 0, 1, (2*k)^j] * (-2)^(j - n) * Binomial[j, n - j], {j, 0, n}]; Table[T[k, n - k], {n, 0, 10}, {k, 0, n}] // Flatten (* Amiram Eldar, Apr 27 2021 *)

T(n, k) = sum(j=0, n\2, (2*k)^(n-j)*(-2)^(-j)*binomial(n-j, j));

T(n, k) = sum(j=0, n, (2*k)^j*(-2)^(j-n)*binomial(j, n-j));

T(n, k) = round(sqrt(k)^n*polchebyshev(n, 2, sqrt(k)));

T(0,k) = 1, T(1,k) = 2*k and T(n,k) = k*(2*T(n-1,k) - T(n-2,k)) for n > 1.
T(n,k) = Sum_{j=0..floor(n/2)} (2*k)^(n-j) * (-1/2)^j * binomial(n-j,j) = Sum_{j=0..n} (2*k)^j * (-1/2)^(n-j) * binomial(j,n-j).
T(n,k) = sqrt(k)^n * U(n, sqrt(k)) where U(n, x) is a Chebyshev polynomial of the second kind.

A220673 Coefficients of formal series in powers of (tan(x))^2 for tan(5*x)/tan(x).

Original entry on oeis.org

5, 40, 376, 3560, 33720, 319400, 3025400, 28657000, 271443000, 2571145000, 24354235000, 230686625000, 2185095075000, 20697517625000, 196049700875000, 1857009420625000, 17589845701875000, 166613409915625000, 1578184870646875000

Offset: 0

Wolfdieter Lang, Jan 16 2013

Formally Sum_{n>=0} a(n)*(tan(x))^(2*n) = tan(5*x)/ tan(x).
Convergence holds for x from the two open intervals (1-sqrt(6/5), 1-2/sqrt(5)) and (1+2/sqrt(5), 1+sqrt(6/5)), namely (-0.095445115, 0.105572809) and (1.894427191, 2.095445115) (10 digits).
These intervals follow from the denominator of the o.g.f. G(5,x) = (5 - 10*x + x^2)/(1 - 10*x + 5*x^2).
If one replaces x by (tan(x))^2 in this o.g.f. one obtains the formula for tan(5*x)/tan(x) in terms of (tan(x))^2. This formula is the special (n=5) solution of a general recurrence derivable from the addition theorem for tan(n*x) = tan(x + (n-1)*x), namely, with Q(n,x) := tan(n,x)/tan(x), Q(n,x) = (1 + Q(n-1,x))/(1 - v*Q(n-1,x)), where v = v(x) =(tan(x))^2, and the input is Q(1,x) = 1. Read as function of v the solution for Q(5,x) is just G(5,v) with replaced v=v(x).
See the irregular triangles A034867 and A034839 whose row polynomials N(n,x) and D(n,x), respectively, give for n >= 1 the solution to the recurrences N(n,x) = D(n-1,x) + N(n-1,x), D(n,x) = D(n-1,x) + x*N(n-1,x), with inputs N(1,x) = 1 and D(1,x) = 1. The proof by the Pascal triangle A007318 recurrence is trivial. Therefore, Q(n,x) from the preceding comment is given by Q(n,x) = N(n,-v)/D(n,-v) with v=v(x) = (tan(x))^2.
One also has, with Chebyshev's S polynomials (see A049310) Q(n,x) = tan(n*x)/tan(x) = (S(n,y) + S(n-2,y))/(S(n,y) - S(n-2,y)) = y*S(n-1,y)/(S(n,y) - S(n-2,y)) = 1/(1 - (2/y)*S(n-2,y)/S(n-1,y)), where y = y(x) = 2/sqrt(1 + (tan(x))^2). This derives from sin(n*x)/cos(n*x) in terms of Chebyshev S polynomials with argument 2*cos(x) = y(x). Note that S(n-2,y)/S(n-1,y) has the continued fraction representation 1/(y-1/(y- ... -1/(y )..(n-1)brackets..), i.e. (n-1) y's.
These calculations have been motivated by e-mails from Thomas Olsen.

			Q(5,x=0.1) = tan(0.5)/tan(0.1) = 5.444802663 (Maple 10 digits);
G(5,tan(0.1)^2) = 5.444802664;
Sum_{n>=0} a(n)*(tan(0.1))^(2*n) = 5.444802664.

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. 2*A000012 (case n=2), A080923(n+1), n>=0 (case n=3), A077445(n+1), n>=0 (case n=4), A034867, A034839.

m:=30; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((5-10*x+x^2)/(1-10*x+5*x^2))); // G. C. Greubel, Mar 06 2018

CoefficientList[Series[(5-10*x+x^2)/(1-10*x+5*x^2), {x,0,50}], x] (* G. C. Greubel, Mar 06 2018 *)

my(x='x+O('x^30)); Vec((5-10*x+x^2)/(1-10*x+5*x^2)) \\ G. C. Greubel, Mar 06 2018

O.g.f.: G(5,x) = (5 - 10*x + x^2)/(1 - 10*x + 5*x^2).
a(n) = delta(n,0)/5 - 8*b(n) + 24*b(n+1)/5, n>=0, with Kronecker's delta and b(n):= A190987(n).
E.g.f.: (1 + 8*exp(5*x)*(3*cosh(2*sqrt(5)*x) + sqrt(5)*sinh(2*sqrt(5)*x)))/5. - Stefano Spezia, May 23 2025

cp's OEIS Frontend

A190958 a(n) = 2*a(n-1) - 10*a(n-2), with a(0) = 0, a(1) = 1.

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A342133 Square array T(n,k), n>=0, k>=0, read by antidiagonals, where column k is the expansion of g.f. 1/(1 - 2*k*x + k*x^2).

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Maple

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A220673 Coefficients of formal series in powers of (tan(x))^2 for tan(5*x)/tan(x).

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Comments

Examples

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Crossrefs

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Magma

Mathematica

PARI

Formula

A190958 a(n) = 2a(n-1) - 10a(n-2), with a(0) = 0, a(1) = 1.

A342133 Square array T(n,k), n>=0, k>=0, read by antidiagonals, where column k is the expansion of g.f. 1/(1 - 2kx + k*x^2).