A083043 -id:A083043 - cp's OEIS frontend

A244419 Coefficient triangle of polynomials related to the Dirichlet kernel. Rising powers. Riordan triangle ((1+z)/(1+z^2), 2*z/(1+z^2)).

1, 1, 2, -1, 2, 4, -1, -4, 4, 8, 1, -4, -12, 8, 16, 1, 6, -12, -32, 16, 32, -1, 6, 24, -32, -80, 32, 64, -1, -8, 24, 80, -80, -192, 64, 128, 1, -8, -40, 80, 240, -192, -448, 128, 256, 1, 10, -40, -160, 240, 672, -448, -1024, 256, 512, -1, 10, 60, -160, -560, 672, 1792, -1024, -2304, 512, 1024

Offset: 0

Wolfdieter Lang, Jul 29 2014

This is the row reversed version of A180870. See also A157751 and A228565.
The Dirichlet kernel is D(n,x) = Sum_{k=-n..n} exp(i*k*x) = 1 + 2*Sum_{k=1..n} T(n,x) = S(n, 2*y) + S(n-1, 2*y) = S(2*n, sqrt(2*(1+y))) with y = cos(x), n >= 0, with the Chebyshev polynomials T (A053120) and S (A049310). This triangle T(n, k) gives in row n the coefficients of the polynomial Dir(n,y) = D(n,x=arccos(y)) = Sum_{m=0..n} T(n,m)*y^m. See A180870, especially the Peter Bala comments and formulas.
This is the Riordan triangle ((1+z)/(1+z^2), 2*z/(1+z^2)) due to the o.g.f. for Dir(n,y) given by (1+z)/(1 - 2*y*z + z^2) = G(z)/(1 - y*F(z)) with G(z) = (1+z)/(1+z^2) and F(z) = 2*z/(1+z^2) (see the Peter Bala formula under A180870). For Riordan triangles and references see the W. Lang link 'Sheffer a- and z- sequences' under A006232.
The A- and Z- sequences of this Riordan triangle are (see the mentioned W. Lang link in the preceding comment also for the references): The A-sequence has o.g.f. 1+sqrt(1-x^2) and is given by A(2*k+1) = 0 and A(2*k) [2, -1/2, -1/8, -1/16, -5/128, -7/256, -21/1024, -33/2048, -429/32768, -715/65536, ...], k >= 0. (See A098597 and A046161.)
  The Z-sequence has o.g.f. sqrt((1-x)/(1+x)) and is given by
  [1, -1, 1/2, -1/2, 3/8, -3/8, 5/16, -5/16, 35/128, -35/128, ...]. (See A001790 and A046161.)
The column sequences are A057077, 2*(A004526 with even numbers signed), 4*A008805 (signed), 8*A058187 (signed), 16*A189976 (signed), 32*A189980 (signed) for m = 0, 1, ..., 5.
The row sums give A005408 (from the o.g.f. due to the Riordan property), and the alternating row sums give A033999.
The row polynomials Dir(n, x), n >= 0, give solutions to the diophantine equation (a + 1)*X^2 - (a - 1)*Y^2 = 2 by virtue of the identity (a + 1)*Dir(n, -a)^2 - (a - 1)*Dir(n, a)^2 = 2, which is easily proved inductively using the recurrence Dir(n, a) = (1 + a)*(-1)^(n-1)*Dir(n-1, -a) + a*Dir(n-1, a) given below by Wolfdieter Lang. - Peter Bala, May 08 2025

			The triangle T(n,m) begins:
  n\m  0   1   2    3    4    5    6     7     8    9    10 ...
  0:   1
  1:   1   2
  2:  -1   2   4
  3:  -1  -4   4    8
  4:   1  -4 -12    8   16
  5:   1   6 -12  -32   16   32
  6:  -1   6  24  -32  -80   32   64
  7:  -1  -8  24   80  -80 -192   64   128
  8:   1  -8 -40   80  240 -192 -448   128   256
  9:   1  10 -40 -160  240  672 -448 -1024   256  512
  10: -1  10  60 -160 -560  672 1792 -1024 -2304  512  1024
  ...
Example for A-sequence recurrence: T(3,1) = Sum_{j=0..2} A(j)*T(2,j) = 2*(-1) + 0*2 + (-1/2)*4 = -4. Example for Z-sequence recurrence: T(4,0) = Sum_{j=0..3} Z(j)*T(3,j) = 1*(-1) + (-1)*(-4) + (1/2)*4 + (-1/2)*8 = +1. (For the A- and Z-sequences see a comment above.)
Example for the alternate recurrence: T(4,2) = 2*T(3,1) - T(3,2) = 2*(-4) - 4 = -12. T(4,3) = 0*T(3,2) + T(3,3) = T(3,3) = 8. - _Wolfdieter Lang_, Jul 30 2014

Wikipedia, Dirichlet kernel.

Dir(n, x) : A005408 (x = 1), A002878 (x = 3/2), A001834 (x = 2), A030221 (x = 5/2), A002315 (x = 3), A033890 (x = 7/2), A057080 (x = 4), A057081 (x = 9/2), A054320 (x = 5), A077416 (x = 6), A028230 (x = 7), A159678 (x = 8), A049629 (x = 9), A083043 (x = 10),
(-1)^n * Dir(n, x): A122367 (x = -3/2); A079935 (x = -2), A004253 (x = -5/2), A001653 (x = -3), A049685 (x = -7/2), A070997 (x = -4), A070998 (x = -9/2), A072256(n+1) (x = -5).
Cf. A180870 (row reversed), A157751, A228565, A049310, A053120, A057077, A004526, A008805, A058187, A189976, A189980, A005408, A033999.

T[n_, k_] := T[n, k] = Which[k == 0, (-1)^Quotient[n, 2], (0 <= n && n < k) || (n == -1 && k == 1), 0, True, 2 T[n-1, k-1] - T[n-2, k]];
Table[T[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jun 28 2019, from Sage *)

def T(n, k):
    if k == 0: return (-1)^(n//2)
    if (0 <= n and n < k) or (n == -1 and k == 1): return 0
    return 2*T(n-1, k-1) - T(n-2, k)
for n in range(11): [T(n,k) for k in (0..n)] # Peter Luschny, Jul 29 2014

T(n, m) = [y^m] Dir(n,y) for n >= m >= 0 and 0 otherwise, with the polynomials Dir(y) defined in a comment above.
T(n, m) = 2^m*(S(n,m) + S(n-1,m)) with the entries S(n,m) of A049310 given there explicitly.
O.g.f. for polynomials Dir(y) see a comment above (Riordan triangle ((1+z)/(1+z^2), 2*z/(1+z^2))).
O.g.f. for column m: ((1 + x)/(1 + x^2))*(2*x/(1 + x^2))^m, m >= 0, (Riordan property).
Recurrence for the polynomials: Dir(n, y) = 2*y*Dir(n-1, y) - Dir(n-2, y), n >= 1, with input D(-1, y) = -1 and D(0, y) = 1.
Triangle three-term recurrence: T(n,m) = 2*T(n-1,m-1) - T(n-2,m) for n >= m >= 1 with T(n,m) = 0 if 0 <= n < m, T(0,0) = 1, T(-1,1) = 0 and T(n,0) = A057077(n) = (-1)^(floor(n/2)).
From Wolfdieter Lang, Jul 30 2014: (Start)
In analogy to A157751 one can derive a recurrence for the row polynomials Dir(n, y) = Sum_{m=0..n} T(n,m)*y^m also using a negative argument but only one recursive step: Dir(n,y) = (1+y)*(-1)^(n-1)*Dir(n-1,-y) + y*Dir(n-1,y), n >= 1, Dir(0,y) = 1 (Dir(-1,y) = -1). See also A180870 from where this formula can be derived by row reversion.
This entails another triangle recurrence T(n,m) = (1 + (-1)^(n-m))*T(n-1,m-1) - (-1)^(n-m)*T(n-1,m), for n >= m >= 1 with T(n,m) = 0 if n < m and T(n,0) = (-1)^floor(n/2). (End)
From Peter Bala, Aug 14 2022: (Start)
The row polynomials Dir(n,x), n >= 0, are related to the Chebyshev polynomials of the first kind T(n,x) by the binomial transform as follows:
(2^n)*(x - 1)^(n+1)*Dir(n,x) = (-1) * Sum_{k = 0..2*n+1} binomial(2*n+1,k)*T(k,-x).
Note that Sum_{k = 0..2*n} binomial(2*n,k)*T(k,x) = (2^n)*(1 + x)^n*T(n,x). (End)
From Peter Bala, May 04 2025: (Start)
For n >= 1, the n-th row polynomial Dir(n, x) = (-1)^n * (U(n, -x) - U(n-1, -x)) = U(2*n, sqrt((1+x)/2)), where U(n, x) denotes the n-th Chebyshev polynomial of the second kind.
For n >= 1 and x < 1, Dir(n, x) = (-1)^n * sqrt(2/(1 - x )) * T(2*n+1, sqrt((1 - x)/2)), where T(n, x) denotes the n-th Chebyshev polynomial of the first kind.
Dir(n, x)^2 - 2*x*Dir(n, x)*Dir(n+1, x) + Dir(n+1, x)^2 = 2*(1 + x).
Dir(n, x) = (-1)^n * R(n, -2*(x+1)), where R(n, x) is the n-th row polynomial of the triangle A085478.
Dir(n, x) = Sum_{k = 0..n} (-1)^(n+k) * binomial(n+k, 2*k) * (2*x + 2)^k. (End)

A075839 Numbers k such that 11*k^2 - 2 is a square.

Original entry on oeis.org

1, 19, 379, 7561, 150841, 3009259, 60034339, 1197677521, 23893516081, 476672644099, 9509559365899, 189714514673881, 3784780734111721, 75505900167560539, 1506333222617099059, 30051158552174420641, 599516837820871313761, 11960285597865251854579

Offset: 1

Gregory V. Richardson, Oct 14 2002

Lim_{n -> infinity} a(n)/a(n-1) = 10 + 3*sqrt(11).

Positive values of x (or y) satisfying x^2 - 20xy + y^2 + 18 = 0. - Colin Barker, Feb 18 2014

A. H. Beiler, "The Pellian", ch. 22 in Recreations in the Theory of Numbers: The Queen of Mathematics Entertains. Dover, New York, New York, pp. 248-268, 1966.
L. E. Dickson, History of the Theory of Numbers, Vol. II, Diophantine Analysis. AMS Chelsea Publishing, Providence, Rhode Island, 1999, pp. 341-400.
Peter G. L. Dirichlet, Lectures on Number Theory (History of Mathematics Source Series, V. 16); American Mathematical Society, Providence, Rhode Island, 1999, pp. 139-147.

G. C. Greubel, Table of n, a(n) for n = 1..750 (terms 1..200 from Vincenzo Librandi)
Tanya Khovanova, Recursive Sequences
J. J. O'Connor and E. F. Robertson, Pell's Equation
Eric Weisstein's World of Mathematics, Pell Equation.
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (20,-1).

Row 20 of array A094954.
Cf. A075844, A221762, A041015.
Cf. similar sequences listed in A238379.

a:=[1,19];; for n in [3..20] do a[n]:=20*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Dec 06 2019

I:=[1,19]; [n le 2 select I[n] else 20*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Feb 20 2014

seq(coeff(series( x*(1-x)/(1-20*x+x^2), x, n+1), x, n), n = 1..20); # G. C. Greubel, Dec 06 2019

LinearRecurrence[{20,-1},{1,19},20] (* Harvey P. Dale, Apr 13 2012 *)
Rest@CoefficientList[Series[x*(1-x)/(1-20x+x^2), {x, 0, 20}], x] (* Vincenzo Librandi, Feb 20 2014 *)
a[c_, n_] := Module[{},
   p := Length[ContinuedFraction[ Sqrt[ c]][[2]]];
   d := Denominator[Convergents[Sqrt[c], n p]];
   t := Table[d[[1 + i]], {i, 0, Length[d] - 1, p}];
   Return[t];
] (* Complement of A041015 *)
a[11, 20] (* Gerry Martens, Jun 07 2015 *)

a(n)=subst(poltchebi(n+1)+poltchebi(n),x,10)/11

def A075839_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( x*(1-x)/(1-20*x+x^2) ).list()
a=A075839_list(20); a[1:] # G. C. Greubel, Dec 06 2019

11*a(n)^2 - 9*A083043(n)^2 = 2.
a(n) = ((3+sqrt(11))*(10+3*sqrt(11))^(n-1) - (3-sqrt(11))*(10-3*sqrt(11))^(n-1) )/(2*sqrt(11)). - Dean Hickerson, Dec 09 2002
From Michael Somos, Oct 29 2002: (Start)
G.f.: x*(1-x)/(1-20*x+x^2).
a(n) = 20*a(n-1) - a(n-2), n>1. (End)
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i) then a(n) = q(n, 18). - Benoit Cloitre, Dec 06 2002
a(-n+1) = a(n). - Michael Somos, Apr 18 2003
E.g.f.: (1/11)*exp(10*x)*(11*cosh(3*sqrt(11)*x) - 3*sqrt(11)*sinh(3*sqrt(11)*x)) - 1. - Stefano Spezia, Dec 06 2019

More terms from Colin Barker, Feb 18 2014

Offset changed to 1 by G. C. Greubel, Dec 06 2019

A131215 Numbers which are both 11-gonal and centered 11-gonal.

Original entry on oeis.org

1, 606, 241396, 96075211, 38237692791, 15218505655816, 6056927013322186, 2410641732796574421, 959429352726023297581, 381850471743224475863026, 151975528324450615370186976, 60485878422659601692858553631

Offset: 1

Richard Choulet, Sep 27 2007

nonn

A centered 11-gonal number is defined by (11*r^2 - 11*r + 2)/2 = A069125(r); a 11-gonal number by (9*p^2 - 7*p)/2 = A051682(p).
A number is both these numbers iff exist p and r such that (18*p - 7)^2 = 99*(2*r - 1) + 22.
The Diophantine equation X^2 = 99*Y^2 + 22 is such that : X is given by the sequence 11, 209, 4169, 83171,... in A131216; Y is given by the sequence 1, 21, 419, 8359,... in A083043.
The first equation is such that : p is given by 1, 12, 232, 4621,... which satisfies a(n+2) = 20*a(n+1) - a(n) - 7 and a(n+1) = 10*a(n) - 7/2 + sqrt(396*a(n)^2 - 308*a(n) + 33)/2 with g.f.  (1 -9*x +x^2)/( (1-x) * (1 -20*x + x^2) ); r is given by 1, 11, 210, 4180,... which satisfies a(n+2) = 20*a(n+1) - a(n) - 9 and a(n+1) = 10*a(n) - 9/2 + sqrt(396*a(n)^2 - 396*a(n) + 121)/2 with g.f. (1 - 10*x)/( (1-x)*(1 -20*x +x^2) ).

G. C. Greubel, Table of n, a(n) for n = 1..380
Index entries for linear recurrences with constant coefficients, signature (399,-399,1).

Cf. A128922.

a:=[1,606,241396];; for n in [4..20] do a[n]:=399*a[n-1]-399*a[n-2] +a[n-3]; od; a; # G. C. Greubel, Dec 06 2019

R:=PowerSeriesRing(Integers(), 20); Coefficients(R!( x*(1+207*x+x^2)/((1-x)*(1-398*x+x^2)) )); // G. C. Greubel, Dec 06 2019

seq(coeff(series(x*(1+207*x+x^2)/((1-x)*(1-398*x+x^2)), x, n+1), x, n), n = 1..20); # G. C. Greubel, Dec 06 2019

LinearRecurrence[{399,-399,1},{1,606,241396},20] (* Harvey P. Dale, Mar 04 2015 *)

my(x='x+O('x^20)); Vec(x*(1+207*x+x^2)/((1-x)*(1-398*x+x^2))) \\ G. C. Greubel, Dec 06 2019

def A131215_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( x*(1+207*x+x^2)/((1-x)*(1-398*x+x^2)) ).list()
a=A131215_list(20); a[1:] # G. C. Greubel, Dec 06 2019

a(n+2) = 398*a(n+1) - a(n) + 209.
a(n+1) = 199*a(n) + 209/2 + (5/2)*sqrt(6336*a(n)^2 + 6688*a(n) + 1617).
G.f.: z*(1 +207*z +z^2)/((1-z)*(1-398*z+z^2)).
a(1)=1, a(2)=606, a(3)=241396, a(n) = 399*a(n-1) - 399*a(n-2) + a(n-3). - Harvey P. Dale, Mar 04 2015

More terms from Paolo P. Lava, Sep 26 2008

A131216 Numbers X such that 99*X^2 - 2178 is a square.

Original entry on oeis.org

11, 209, 4169, 83171, 1659251, 33101849, 660377729, 13174452731, 262828676891, 5243399085089, 104605153024889, 2086859661412691, 41632588075228931, 830564901843165929, 16569665448788089649, 330562744073918627051

Offset: 1

Richard Choulet, Sep 27 2007

G. C. Greubel, Table of n, a(n) for n = 1..750
Index entries for linear recurrences with constant coefficients, signature (20,-1).

Cf. A083043.

a:=[11,209];; for n in [3..20] do a[n]:=20*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Dec 06 2019

R:=PowerSeriesRing(Integers(), 20); Coefficients(R!( 11*x*(1-x)/(1-20*x+x^2) )); // G. C. Greubel, Dec 06 2019

seq(coeff(series(11*x*(1-x)/(1-20*x+x^2), x, n+1), x, n), n = 0..20); # G. C. Greubel, Dec 06 2019

LinearRecurrence[{20, -1}, {11, 209}, 20] (* G. C. Greubel, Dec 06 2019 *)

my(x='x+O('x^20)); Vec(11*x*(1-x)/(1-20*x+x^2)) \\ G. C. Greubel, Dec 06 2019

def A131216_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( 11*x*(1-x)/(1-20*x+x^2) ).list()
A131216_list(20) # G. C. Greubel, Dec 06 2019

a(n+2) = 20*a(n+1) - a(n).
a(n+1) = 10*a(n+1)+ sqrt(99*a(n)^2 -2178).
G.f.: 11*z*(1-z)/(1-20*z+z^2) - Richard Choulet, Oct 09 2007
a(n) = 11*A075839(n). - R. J. Mathar, Aug 22 2012

A159680 The general form of the recurrences are the a(j), b(j) and n(j) solutions of the 2 equations problem: 9n(j) + 1 = a(j)a(j) and 11n(j) + 1 = b(j)b(j) with positive integer numbers.

Original entry on oeis.org

0, 40, 15960, 6352080, 2528111920, 1006182192120, 400457984351880, 159381271589856160, 63433345634778399840, 25246312181370213280200, 10047968814839710107119800, 3999066341994023252420400240, 1591618356144806414753212175760, 633460106679290959048526025552280

Offset: 1

Paul Weisenhorn, Apr 19 2009

Colin Barker, Table of n, a(n) for n = 1..350
Index entries for linear recurrences with constant coefficients, signature (399,-399,1).

Cf. A075839, A083043, A157456.

R:=PowerSeriesRing(Integers(), 30); [0] cat Coefficients(R!(40*x^2/((1-x)*(1-398*x+x^2)))); // G. C. Greubel, Jun 03 2018

for a from 1 by 2 to 100000 do b:=sqrt((9*a*a-2)/7): if (trunc(b)=b) then
n:=(a*a-1)/7: La:=[op(La),a]:Lb:=[op(Lb),b]:Ln:=[op(Ln),n]: end if: end do:

LinearRecurrence[{399,-399,1}, {0,40,15960}, 50] (* G. C. Greubel, Jun 03 2018 *)

a(n) = round((-20+(10+3*sqrt(11))*(199+60*sqrt(11))^(-n)+(10-3*sqrt(11))*(199+60*sqrt(11))^n)/198) \\ Colin Barker, Jul 26 2016

concat(0, Vec(-40*x^2/((x-1)*(x^2-398*x+1)) + O(x^20))) \\ Colin Barker, Jul 26 2016

[(10/99)*(chebyshev_U(n, 199) -397*chebyshev_U(n-1, 199) -1) for n in (1..30)] # G. C. Greubel, Jun 26 2022

The a(j) recurrence is a(1)=1; a(2)=19; a(t+2) = 20*a(t+1) - a(t) resulting in terms 1, 19, 379, 7561, ... (A075839).
The b(j) recurrence is b(1)=1; b(2)=21; b(t+2) = 20*b(t+1) - b(t) resulting in terms 1, 21, 419, 8359, ... (A083043).
The n(j) recurrence is n(0)=n(1)=0; n(2)=40; n(t+3) = 399*(n(t+2) - n(t+1)) + n(t) resulting in terms 0, 0, 40, 15960, 6352080 as listed above
G.f.: 40*x^2/((1-x)*(1-398*x+x^2)). - R. J. Mathar, Apr 20 2009
a(n) = (-20 + (10 + 3*sqrt(11))*(199 + 60*sqrt(11))^(-n) + (10 - 3*sqrt(11))*(199 + 60*sqrt(11))^n)/198. - Colin Barker, Jul 26 2016
From G. C. Greubel, Jun 26 2022: (Start)
a(n) = (10/99)*( ChebyshevU(n, 199) - 397*ChebyshevU(n-1, 199) - 1 ).
E.g.f.: (10/99)*(exp(199*x)*( (3*sqrt(11)/10)*sinh(60*sqrt(11)*x) + cosh(60*sqrt(11)*x) ) - exp(x)). (End)

More terms from R. J. Mathar, Apr 20 2009

cp's OEIS Frontend

A244419 Coefficient triangle of polynomials related to the Dirichlet kernel. Rising powers. Riordan triangle ((1+z)/(1+z^2), 2*z/(1+z^2)).

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A075839 Numbers k such that 11*k^2 - 2 is a square.

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A131215 Numbers which are both 11-gonal and centered 11-gonal.

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A131216 Numbers X such that 99*X^2 - 2178 is a square.

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A159680 The general form of the recurrences are the a(j), b(j) and n(j) solutions of the 2 equations problem: 9*n(j) + 1 = a(j)*a(j) and 11*n(j) + 1 = b(j)*b(j) with positive integer numbers.

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A159680 The general form of the recurrences are the a(j), b(j) and n(j) solutions of the 2 equations problem: 9n(j) + 1 = a(j)a(j) and 11n(j) + 1 = b(j)b(j) with positive integer numbers.