cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A028230 Bisection of A001353. Indices of square numbers which are also octagonal.

Original entry on oeis.org

1, 15, 209, 2911, 40545, 564719, 7865521, 109552575, 1525870529, 21252634831, 296011017105, 4122901604639, 57424611447841, 799821658665135, 11140078609864049, 155161278879431551, 2161117825702177665, 30100488280951055759, 419245718107612602961, 5839339565225625385695
Offset: 1

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Author

Keywords

Comments

Chebyshev S-sequence with Diophantine property.
4*b(n)^2 - 3*a(n)^2 = 1 with b(n) = A001570(n), n>=0.
y satisfying the Pellian x^2 - 3*y^2 = 1, for even x given by A094347(n). - Lekraj Beedassy, Jun 03 2004
a(n) = L(n,-14)*(-1)^n, where L is defined as in A108299; see also A001570 for L(n,+14). - Reinhard Zumkeller, Jun 01 2005
Product x*y, where the pair (x, y) solves for x^2 - 3y^2 = -2, i.e., a(n) = A001834(n)*A001835(n). - Lekraj Beedassy, Jul 13 2006
Numbers n such that RootMeanSquare(1,3,...,2*A001570(k)-1) = n. - Ctibor O. Zizka, Sep 04 2008
As n increases, this sequence is approximately geometric with common ratio r = lim(n -> oo, a(n)/a(n-1)) = (2 + sqrt(3))^2 = 7 + 4 * sqrt(3). - Ant King, Nov 15 2011

References

  • R. L. Graham, D. E. Knuth, and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 329.
  • J. D. E. Konhauser et al., Which Way Did the Bicycle Go?, MAA 1996, p. 104.

Crossrefs

Programs

  • GAP
    a:=[1,15];; for n in [3..30] do a[n]:=14*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Dec 06 2019
  • Magma
    I:=[1,15]; [n le 2 select I[n] else 14*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Dec 06 2019
    
  • Maple
    seq(coeff(series((1+x)/(1-14*x+x^2), x, n+1), x, n), n = 0..30); # G. C. Greubel, Dec 06 2019
  • Mathematica
    LinearRecurrence[{14, - 1}, {1, 15}, 17] (* Ant King, Nov 15 2011 *)
    CoefficientList[Series[(1+x)/(1-14x+x^2), {x, 0, 30}], x] (* Vincenzo Librandi, Jun 17 2014 *)
  • PARI
    Vec((1+x)/(1-14*x+x^2)+O(x^99)) \\ Charles R Greathouse IV, Jun 16 2014
    
  • PARI
    isok(n) = ispolygonal(n^2, 8); \\ Michel Marcus, Jul 09 2017
    
  • Sage
    [(lucas_number2(n,14,1)-lucas_number2(n-1,14,1))/12 for n in range(1, 18)] # Zerinvary Lajos, Nov 10 2009
    

Formula

a(n) = 2*A001921(n)+1.
a(n) = 14*a(n-1) - a(n-2) for n>1.
a(n) = S(n, 14) + S(n-1, 14) = S(2*n, 4) with S(n, x) := U(n, x/2) Chebyshev's polynomials of the second kind. See A049310. S(-1, x) = 0, S(n, 14) = A007655(n+1) and S(n, 4) = A001353(n+1).
G.f.: x*(1+x)/(1-14*x+x^2).
a(n) = (ap^(2*n+1) - am^(2*n+1))/(ap - am) with ap := 2+sqrt(3) and am := 2-sqrt(3).
a(n+1) = Sum_{k=0..n} (-1)^k*binomial(2*n-k, k)*16^(n-k), n>=0.
a(n) = sqrt((4*A001570(n-1)^2 - 1)/3).
a(n) ~ 1/6*sqrt(3)*(2 + sqrt(3))^(2*n-1). - Joe Keane (jgk(AT)jgk.org), May 15 2002
4*a(n+1) = (A001834(n))^2 + 4*(A001835(n+1))^2 - (A001835(n))^2. E.g. 4*a(3) = 4*209 = 19^2 + 4*11^2 - 3^2 = (A001834(2))^2 + 4*(A001835(3))^2 - (A001835(2))^2. Generating floretion: 'i + 2'j + 3'k + i' + 2j' + 3k' + 4'ii' + 3'jj' + 4'kk' + 3'ij' + 3'ji' + 'jk' + 'kj' + 4e. - Creighton Dement, Dec 04 2004
a(n) = f(a(n-1),7) + f(a(n-2),7), where f(x,s) = s*x + sqrt((s^2-1)*x^2+1); f(0,s)=0. - Marcos Carreira, Dec 27 2006
From Ant King, Nov 15 2011: (Start)
a(n) = 1/6 * sqrt(3) * ( (tan(5*Pi/12)) ^ (2n-1) - (tan(Pi/12)) ^ (2n-1) ).
a(n) = floor (1/6 * sqrt(3) * (tan(5*Pi/12)) ^ (2n-1)). (End)
a(n) = A001353(n)^2-A001353(n-1)^2. - Antonio Alberto Olivares, Apr 06 2020
E.g.f.: 1 - exp(7*x)*(3*cosh(4*sqrt(3)*x) - 2*sqrt(3)*sinh(4*sqrt(3)*x))/3. - Stefano Spezia, Dec 12 2022
a(n) = sqrt(A036428(n)). - Bernard Schott, Dec 19 2022

Extensions

Additional comments from Wolfdieter Lang, Nov 29 2002
Incorrect recurrence relation deleted by Ant King, Nov 15 2011
Minor edits by Vaclav Kotesovec, Jan 28 2015

A057081 Even-indexed Chebyshev U-polynomials evaluated at sqrt(11)/2.

Original entry on oeis.org

1, 10, 89, 791, 7030, 62479, 555281, 4935050, 43860169, 389806471, 3464398070, 30789776159, 273643587361, 2432002510090, 21614379003449, 192097408520951, 1707262297685110, 15173263270645039, 134852107138120241, 1198495700972437130, 10651609201613813929
Offset: 0

Views

Author

Wolfdieter Lang, Aug 04 2000

Keywords

Comments

This is the m=11 member of the m-family of sequences S(n,m-2)+S(n-1,m-2) = S(2*n,sqrt(m)) (for S(n,x) see Formula). The m=4..10 instances are A005408, A002878, A001834, A030221, A002315, A033890 and A057080, resp. The m=1..3 (signed) sequences are: A057078, A057077 and A057079, resp.
General recurrence is a(n)=(a(1)-1)*a(n-1)-a(n-2), a(1)>=4, lim_{n->oo} a(n)= x*(k*x+1)^n, k =(a(1)-3), x=(1+sqrt((a(1)+1)/(a(1)-3)))/2. Examples in OEIS: a(1)=4 gives A002878. a(1)=5 gives A001834. a(1)=6 gives A030221. a(1)=7 gives A002315. a(1)=8 gives A033890. a(1)=9 gives A057080. a(1)=10 gives A057081. - Ctibor O. Zizka, Sep 02 2008
The primes in this sequence are 89, 389806471, 192097408520951, 7477414486269626733119, ... - Ctibor O. Zizka, Sep 02 2008
The aerated sequence (b(n))n>=1 = [1, 0, 10, 0, 89, 0, 791, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -7, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047. - Peter Bala, Mar 22 2015

Crossrefs

Programs

  • Maple
    A057081 := proc(n)
        option remember;
        if n <= 1 then
            op(n+1,[1,10]);
        else
            9*procname(n-1)-procname(n-2) ;
        end if;
    end proc: # R. J. Mathar, Apr 30 2017
  • Mathematica
    CoefficientList[Series[(1 + x)/(1 - 9*x + x^2), {x,0,50}], x] (* or *) LinearRecurrence[{9,-1}, {1,10}, 50] (* G. C. Greubel, Apr 12 2017 *)
  • PARI
    Vec((1+x)/(1-9*x+x^2) + O(x^30)) \\ Michel Marcus, Mar 22 2015
  • Sage
    [(lucas_number2(n,9,1)-lucas_number2(n-1,9,1))/7 for n in range(1, 20)] # Zerinvary Lajos, Nov 10 2009
    

Formula

a(n) = 9*a(n-1) - a(n-2), a(-1)=-1, a(0)=1.
a(n) = S(n, 9) + S(n-1, 9) = S(2*n, sqrt(11)) with S(n, x) := U(n, x/2), Chebyshev polynomials of 2nd kind, A049310. S(n, 9) = A018913(n).
G.f.: (1+x)/(1-9*x+x^2).
Let q(n, x) = Sum{i=0..n} x^(n-i)*binomial(2*n-i, i), a(n) = (-1)^n*q(n, -11). - Benoit Cloitre, Nov 10 2002
a(n) = L(n,-9)*(-1)^n, where L is defined as in A108299; see also A070998 for L(n,+9). - Reinhard Zumkeller, Jun 01 2005
From Peter Bala, Jun 08 2025: (Start)
a(n) = (1/sqrt(7)) * ( ((sqrt(11) + sqrt(7))/2)^(2*n+1) - ((sqrt(11) - sqrt(7))/2)^(2*n+1) ).
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 1/a(n)) = 1/11 (telescoping series: 11/(a(n) - 1/a(n)) = 1/A018913(n+1) + 1/A018913(n)).
Conjecture: for k >= 1, Sum_{n >= 1} (-1)^(n+1)/(a(k*n) - s(k)/a(k*n)) = 1/(1 + a(k)) where s(k) = a(0) + a(1) + ... + a(k-1).
Product_{n >= 1} (a(n) + 1)/(a(n) - 1) = sqrt(11/7) [telescoping product: ((a(n) + 1)/(a(n) - 1))^2 = (1 - 4/b(n+1))/(1 - 4/b(n)), where b(n) = 2 + A056918(n)]. (End)

A152198 Triangle read by rows, A007318 rows repeated.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 4, 6, 4, 1, 1, 4, 6, 4, 1, 1, 5, 10, 10, 5, 1, 1, 5, 10, 10, 5, 1, 1, 6, 15, 20, 15, 6, 1, 1, 6, 15, 20, 15, 6, 1, 1, 7, 21, 35, 35, 21, 7, 1, 1, 7, 21, 35, 35, 21, 7, 1, 1, 8, 28, 56, 70, 56, 28, 8, 1, 1, 8, 28, 56, 70, 56, 28, 8, 1
Offset: 0

Views

Author

Gary W. Adamson, Nov 28 2008

Keywords

Comments

Eigensequence of the triangle = A051163: (1, 2, 5, 12, 30, 76,...)
Another version of A152815. - Philippe Deléham, Dec 13 2008
Row sums : A016116(n); Diagonal sums: A000931(n+5). - Philippe Deléham, Dec 13 2008
Triangle, with zeros omitted, given by (1, 0, -1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Jan 16 2012
Sums along rising diagonals are A134816. - John Molokach, Jul 09 2013

Examples

			The triangle starts
1;
1;
1, 1;
1, 1;
1, 2, 1;
1, 2, 1;
1, 3, 3, 1;
1, 3, 3, 1;
1, 4, 6, 4, 1;
1, 4, 6, 4, 1;
1, 5, 10, 10, 5, 1;
1, 5, 10, 10, 5, 1;
...
Triangle (1,0,-1,0,0,...) DELTA (0,1,-1,0,0,...) begins:
1
1, 0
1, 1, 0
1, 1, 0, 0
1, 2, 1, 0, 0
1, 2, 1, 0, 0, 0
1, 3, 3, 1, 0, 0, 0
1, 3, 3, 1, 0, 0, 0, 0
1, 4, 6, 4, 1, 0, 0, 0, 0
1, 4, 6, 4, 1, 0, 0, 0, 0, 0
1, 5, 10, 10, 5, 1, 0, 0, 0, 0, 0...
		

Crossrefs

Programs

  • Mathematica
    t[n_, k_] := Binomial[ Floor[n/2], k]; Table[t[n, k], {n, 0, 17}, {k, 0, Floor[n/2]}] // Flatten (* Jean-François Alcover, Sep 13 2012 *)

Formula

Triangle read by rows, Pascal's triangle rows repeated.
Equals inverse binomial transform of A133156 unsigned.
G.f. : (1+x)/(1-(1+y)*x^2). - Philippe Deléham, Jan 16 2012
Sum_{k, 0<=k<=n} T(n,k)*x^k = A057077(n), A019590(n+1), A000012(n), A016116(n), A108411(n), A074872(n+1) for x = -2, -1, 0, 1, 2, 4 respectively. - Philippe Deléham, Jan 16 2012
T(n,k) = A065941(n-k, n-2*k) = abs(A108299(n-k, n-2*k)). - Johannes W. Meijer, Sep 05 2013

Extensions

More terms from Philippe Deléham, Dec 14 2008

A077417 Chebyshev T-sequence with Diophantine property.

Original entry on oeis.org

1, 11, 131, 1561, 18601, 221651, 2641211, 31472881, 375033361, 4468927451, 53252096051, 634556225161, 7561422605881, 90102515045411, 1073668757939051, 12793922580223201, 152453402204739361, 1816646903876649131, 21647309444315050211
Offset: 0

Views

Author

Wolfdieter Lang, Nov 29 2002

Keywords

Comments

7*a(n)^2 - 5*b(n)^2 = 2 with companion sequence b(n) = A077416(n), n>=0.
a(n) = L(n,12), where L is defined as in A108299; see also A077416 for L(n,-12). - Reinhard Zumkeller, Jun 01 2005
[a(n), A004191(n)] = the 2 X 2 matrix [1,10; 1,11]^(n+1) * [1,0]. - Gary W. Adamson, Mar 19 2008
Hankel transform of A174227. - Paul Barry, Mar 12 2010
Alternate denominators of the continued fraction convergents to sqrt(35), see A041059. - James R. Buddenhagen, May 20 2010
For positive n, a(n) equals the permanent of the (2n)X(2n) tridiagonal matrix with sqrt(10)'s along the main diagonal, and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
Positive values of x (or y) satisfying x^2 - 12xy + y^2 + 10 = 0. - Colin Barker, Feb 09 2014
a(n) = a(-1-n) for all n in Z. - Michael Somos, Jun 29 2019

Examples

			G.f. = 1 + 11*x + 131*x^2 + 1561*x^3 + 18601*x^4 221651*x^5 + 2641211*x^6 + ...
		

Crossrefs

Cf. A072256(n) with companion A054320(n-1), n>=1.
Row 12 of array A094954.
Cf. A004191.
Cf. A041059. [James R. Buddenhagen, May 20 2010]
Cf. similar sequences listed in A238379.

Programs

  • Magma
    I:=[1,11]; [n le 2 select I[n] else 12*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Feb 10 2014
    
  • Mathematica
    CoefficientList[Series[(1 - x)/(1 - 12 x + x^2), {x, 0, 30}], x] (* Vincenzo Librandi, Feb 10 2014 *)
    LinearRecurrence[{12,-1},{1,11},30] (* Harvey P. Dale, Apr 09 2015 *)
    a[ n_] := With[{x = Sqrt[7/2]}, ChebyshevT[2 n + 1, x]/x] // Expand; (* Michael Somos, Jun 29 2019 *)
  • PARI
    my(x='x+O('x^30)); Vec((1-x)/(1-12*x+x^2)) \\ G. C. Greubel, Jan 18 2018
    
  • PARI
    {a(n) = my(x = quadgen(56)/2); simplify(polchebyshev(2*n + 1, 1, x)/x)}; /* Michael Somos, Jun 29 2019 */

Formula

a(n) = 12*a(n-1) - a(n-2), a(-1)=1, a(0)=1.
a(n) = S(n, 12) - S(n-1, 12) = T(2*n+1, sqrt(14)/2)/(sqrt(14)/2) with S(n, x) := U(n, x/2), resp. T(n, x), Chebyshev's polynomials of the second, resp. first, kind. See A049310 and A053120. S(-1, x)=0, S(n, 12)=A004191(n).
G.f.: (1-x)/(1-12*x+x^2).
a(n) = (ap^(2*n+1) + am^(2*n+1))/sqrt(14) with ap := (sqrt(7)+sqrt(5))/sqrt(2) and am := (sqrt(7)-sqrt(5))/sqrt(2).
a(n) = sqrt((5*A077416(n)^2 + 2)/7).
a(n)*a(n+3) = 120 + a(n+1)*a(n+2). - Ralf Stephan, May 29 2004
E.g.f.: exp(6*x)*(7*cosh(sqrt(35)*x) + sqrt(35)*sinh(sqrt(35)*x))/7. - Stefano Spezia, Aug 29 2025

Extensions

More terms from Vincenzo Librandi, Feb 10 2014

A050935 a(n) = a(n-1) - a(n-3) with a(1)=0, a(2)=0, a(3)=1.

Original entry on oeis.org

0, 0, 1, 1, 1, 0, -1, -2, -2, -1, 1, 3, 4, 3, 0, -4, -7, -7, -3, 4, 11, 14, 10, -1, -15, -25, -24, -9, 16, 40, 49, 33, -7, -56, -89, -82, -26, 63, 145, 171, 108, -37, -208, -316, -279, -71, 245, 524, 595, 350, -174, -769, -1119, -945, -176, 943, 1888, 2064, 1121, -767, -2831, -3952
Offset: 1

Views

Author

Richard J. Palmaccio (palmacr(AT)pinecrest.edu), Dec 31 1999

Keywords

Comments

The Ze3 sums, see A180662, of triangle A108299 equal the terms of this sequence without the two leading zeros. - Johannes W. Meijer, Aug 14 2011

References

  • R. Palmaccio, "Average Temperatures Modeled with Complex Numbers", Mathematics and Informatics Quarterly, pp. 9-17 of Vol. 3, No. 1, March 1993.

Crossrefs

When run backwards this gives a signed version of A000931.
Cf. A099529.
Apart from signs, essentially the same as A078013.
Cf. A203400 (partial sums).

Programs

  • Haskell
    a050935 n = a050935_list !! (n-1)
    a050935_list = 0 : 0 : 1 : zipWith (-) (drop 2 a050935_list) a050935_list
    -- Reinhard Zumkeller, Jan 01 2012
    
  • Maple
    A050935 := proc(n) option remember; if n <= 1 then 0 elif n = 2 then 1 else A050935(n-1)-A050935(n-3); fi; end: seq(A050935(n), n=0..61);
  • Mathematica
    LinearRecurrence[{1,0,-1},{0,0,1},70] (* Harvey P. Dale, Jan 30 2014 *)
  • PARI
    a(n)=([0,1,0; 0,0,1; -1,0,1]^(n-1)*[0;0;1])[1,1] \\ Charles R Greathouse IV, Feb 06 2017

Formula

From Paul Barry, Oct 20 2004: (Start)
G.f.: x^2/(1-x+x^3).
a(n+2) = Sum_{k=0..floor(n/3)} binomial(n-2*k, k)*(-1)^k. (End)
G.f.: Q(0)*x^2/2, where Q(k) = 1 + 1/(1 - x*(12*k-1 + x^2)/( x*(12*k+5 + x^2 ) - 1/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Sep 12 2013

Extensions

Offset adjusted by Reinhard Zumkeller, Jan 01 2012

A070998 a(n) = 9*a(n-1) - a(n-2) for n > 0, a(0)=1, a(-1)=1.

Original entry on oeis.org

1, 8, 71, 631, 5608, 49841, 442961, 3936808, 34988311, 310957991, 2763633608, 24561744481, 218292066721, 1940066856008, 17242309637351, 153240719880151, 1361924169284008, 12104076803675921, 107574767063799281, 956068826770517608
Offset: 0

Views

Author

Joe Keane (jgk(AT)jgk.org), May 18 2002

Keywords

Comments

A Pellian sequence.
In general, Sum_{k=0..n} binomial(2n-k,k)j^(n-k) = (-1)^n*U(2n, i*sqrt(j)/2), i=sqrt(-1). - Paul Barry, Mar 13 2005
a(n) = L(n,9), where L is defined as in A108299; see also A057081 for L(n,-9). - Reinhard Zumkeller, Jun 01 2005
Number of 01-avoiding words of length n on alphabet {0,1,2,3,4,5,6,7,8} which do not end in 0. - Tanya Khovanova, Jan 10 2007
For positive n, a(n) equals the permanent of the (2n) X (2n) tridiagonal matrix with sqrt(7)'s along the main diagonal, and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
Positive values of x (or y) satisfying x^2 - 9xy + y^2 + 7 = 0. - Colin Barker, Feb 09 2014

Crossrefs

Row 9 of array A094954.
Cf. similar sequences listed in A238379.

Programs

  • Magma
    I:=[1,8]; [n le 2 select I[n] else 9*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Feb 10 2014
  • Mathematica
    CoefficientList[Series[(1 - x)/(1 - 9 x + x^2), {x, 0, 30}], x] (* Vincenzo Librandi, Feb 10 2014 *)
    LinearRecurrence[{9,-1},{1,8},30] (* Harvey P. Dale, Sep 24 2015 *)
  • Sage
    [lucas_number1(n, 9, 1) - lucas_number1(n-1, 9, 1) for n in range(1, 19)]  # Zerinvary Lajos, Nov 10 2009
    

Formula

a(n) ~ (1/11)*sqrt(11)*((1/2)*(sqrt(11) + sqrt(7)))^(2*n+1).
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i); then q(n, 7) = a(n). - Benoit Cloitre, Nov 10 2002
a(n)*a(n+3) = 63 + a(n+1)*a(n+2). - Ralf Stephan, May 29 2004
a(n) = (-1)^n*U(2n, i*sqrt(7)/2), U(n, x) Chebyshev polynomial of second kind, i=sqrt(-1). - Paul Barry, Mar 13 2005
G.f.: (1-x)/(1-9*x+x^2). - Philippe Deléham, Nov 03 2008
a(n) = A018913(n+1) - A018913(n). - R. J. Mathar, Jun 07 2013

Extensions

More terms from Vincenzo Librandi, Feb 10 2014

A130777 Coefficients of first difference of Chebyshev S polynomials.

Original entry on oeis.org

1, -1, 1, -1, -1, 1, 1, -2, -1, 1, 1, 2, -3, -1, 1, -1, 3, 3, -4, -1, 1, -1, -3, 6, 4, -5, -1, 1, 1, -4, -6, 10, 5, -6, -1, 1, 1, 4, -10, -10, 15, 6, -7, -1, 1, -1, 5, 10, -20, -15, 21, 7, -8, -1, 1, -1, -5, 15, 20, -35, -21, 28, 8, -9, -1, 1, 1, -6, -15, 35, 35, -56, -28, 36, 9, -10, -1, 1
Offset: 0

Views

Author

Philippe Deléham, Jul 14 2007

Keywords

Comments

Inverse of triangle in A061554.
Signed version of A046854.
From Paul Barry, May 21 2009: (Start)
Riordan array ((1-x)/(1+x^2),x/(1+x^2)).
This triangle is the coefficient triangle for the Hankel transforms of the family of generalized Catalan numbers that satisfy a(n;r)=r*a(n-1;r)+sum{k=1..n-2, a(k)*a(n-1-k;r)}, a(0;r)=a(1;r)=1. The Hankel transform of a(n;r) is h(n)=sum{k=0..n, T(n,k)*r^k} with g.f. (1-x)/(1-r*x+x^2). These sequences include A086246, A000108, A002212. (End)
From Wolfdieter Lang, Jun 11 2011: (Start)
The Riordan array ((1+x)/(1+x^2),x/(1+x^2)) with entries Phat(n,k)= ((-1)^(n-k))*T(n,k) and o.g.f. Phat(x,z)=(1+z)/(1-x*z+z^2) for the row polynomials Phat(n,x) is related to Chebyshev C and S polynomials as follows.
Phat(n,x) = (R(n+1,x)-R(n,x))/(x+2) = S(2*n,sqrt(2+x))
with R(n,x)=C_n(x) in the Abramowitz and Stegun notation, p. 778, 22.5.11. See A049310 for the S polynomials. Proof from the o.g.f.s.
Recurrence for the row polynomials Phat(n,x):
Phat(n,x) = x*Phat(n-1,x) - Phat(n-2,x) for n>=1; Phat(-1,x)=-1, Phat(0,x)=1.
The A-sequence for this Riordan array Phat (see the W. Lang link under A006232 for A- and Z-sequences for Riordan matrices) is given by 1, 0, -1, 0, -1, 0, -2, 0, -5,.., starting with 1 and interlacing the negated A000108 with zeros (o.g.f. 1/c(x^2) = 1-c(x^2)*x^2, with the o.g.f. c(x) of A000108).
The Z-sequence has o.g.f. sqrt((1-2*x)/(1+2*x)), and it is given by A063886(n)*(-1)^n.
The A-sequence of the Riordan array T(n,k) is identical with the one for the Riordan array Phat, and the Z-sequence is -A063886(n).
(End)
The row polynomials P(n,x) are the characteristic polynomials of the adjacency matrices of the graphs which look like P_n (n vertices (nodes), n-1 lines (edges)), but vertex no. 1 has a loop. - Wolfdieter Lang, Nov 17 2011
From Wolfdieter Lang, Dec 14 2013: (Start)
The zeros of P(n,x) are x(n,j) = -2*cos(2*Pi*j/(2*n+1)), j=1..n. From P(n,x) = (-1)^n*S(2*n,sqrt(2-x)) (see, e.g., the Lemma 6 of the W. Lang link).
The discriminants of the P-polynomials are given in A052750. (End)

Examples

			The triangle T(n,k) begins:
n\k  0   1   1   3    4    5    6    7    8    9  10  11  12  13 14 15 ...
0:   1
1:  -1   1
2:  -1  -1   1
3:   1  -2  -1   1
4:   1   2  -3  -1    1
5:  -1   3   3  -4   -1    1
6:  -1  -3   6   4   -5   -1    1
7:   1  -4  -6  10    5   -6   -1    1
8:   1   4 -10 -10   15    6   -7   -1    1
9:  -1   5  10 -20  -15   21    7   -8   -1    1
10: -1  -5  15  20  -35  -21   28    8   -9   -1   1
11:  1  -6 -15  35   35  -56  -28   36    9  -10  -1   1
12:  1   6 -21 -35   70   56  -84  -36   45   10 -11  -1   1
13: -1   7  21 -56  -70  126   84 -120  -45   55  11 -12  -1   1
14: -1  -7  28  56 -126 -126  210  120 -165  -55  66  12 -13  -1  1
15:  1  -8 -28  84  126 -252 -210  330  165 -220 -66  78  13 -14 -1  1
...  reformatted and extended - _Wolfdieter Lang_, Jul 31 2014.
---------------------------------------------------------------------------
From _Paul Barry_, May 21 2009: (Start)
Production matrix is
-1, 1,
-2, 0, 1,
-2, -1, 0, 1,
-4, 0, -1, 0, 1,
-6, -1, 0, -1, 0, 1,
-12, 0, -1, 0, -1, 0, 1,
-20, -2, 0, -1, 0, -1, 0, 1,
-40, 0, -2, 0, -1, 0, -1, 0, 1,
-70, -5, 0, -2, 0, -1, 0, -1, 0, 1 (End)
Row polynomials as first difference of S polynomials:
P(3,x) = S(3,x) - S(2,x) = (x^3 - 2*x) - (x^2 -1) = 1 - 2*x - x^2 +x^3.
Alternative triangle recurrence (see a comment above): T(6,2) = T(5,2) + T(5,1) = 3 + 3 = 6. T(6,3) = -T(5,3) + 0*T(5,1) = -(-4) = 4. - _Wolfdieter Lang_, Jul 31 2014
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964. Tenth printing, Wiley, 2002 (also electronically available).

Crossrefs

Cf. A066170, A046854, A057077 (first column).
Row sums: A010892(n+1); repeat(1,0,-1,-1,0,1). Alternating row sums: A061347(n+2); repeat(1,-2,1).

Programs

  • Maple
    A130777 := proc(n,k): (-1)^binomial(n-k+1,2)*binomial(floor((n+k)/2),k) end: seq(seq(A130777(n,k), k=0..n), n=0..11); # Johannes W. Meijer, Aug 08 2011
  • Mathematica
    T[n_, k_] := (-1)^Binomial[n - k + 1, 2]*Binomial[Floor[(n + k)/2], k];
    Table[T[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 14 2017, from Maple *)
  • Sage
    @CachedFunction
    def A130777(n,k):
        if n< 0: return 0
        if n==0: return 1 if k == 0 else 0
        h = A130777(n-1,k) if n==1 else 0
        return A130777(n-1,k-1) - A130777(n-2,k) - h
    for n in (0..9): [A130777(n,k) for k in (0..n)] # Peter Luschny, Nov 20 2012

Formula

Number triangle T(n,k) = (-1)^C(n-k+1,2)*C(floor((n+k)/2),k). - Paul Barry, May 21 2009
From Wolfdieter Lang, Jun 11 2011: (Start)
Row polynomials: P(n,x) = sum(k=0..n, T(n,k)*x^k) = R(2*n+1,sqrt(2+x)) / sqrt(2+x), with Chebyshev polynomials R with coefficients given in A127672 (scaled T-polynomials).
R(n,x) is called C_n(x) in Abramowitz and Stegun's handbook, p. 778, 22.5.11.
P(n,x) = S(n,x)-S(n-1,x), n>=0, S(-1,x)=0, with the Chebyshev S-polynomials (see the coefficient triangle A049310).
O.g.f. for row polynomials: P(x,z):= sum(n>=0, P(n,x)*z^n ) = (1-z)/(1-x*z+z^2).
(from the o.g.f. for R(2*n+1,x), n>=0, computed from the o.g.f. for the R-polynomials (2-x*z)/(1-x*z+z^2) (see A127672))
Proof of the Chebyshev connection from the o.g.f. for Riordan array property of this triangle (see the P. Barry comment above).
For the A- and Z-sequences of this Riordan array see a comment above. (End)
abs(T(n,k)) = A046854(n,k) = abs(A066170(n,k)) T(n,n-k) = A108299(n,k); abs(T(n,n-k)) = A065941(n,k). - Johannes W. Meijer, Aug 08 2011
From Wolfdieter Lang, Jul 31 2014: (Start)
Similar to the triangles A157751, A244419 and A180070 one can give for the row polynomials P(n,x) besides the usual three term recurrence another one needing only one recurrence step. This uses also a negative argument, namely P(n,x) = (-1)^(n-1)*(-1 + x/2)*P(n-1,-x) + (x/2)*P(n-1,x), n >= 1, P(0,x) = 1. Proof by computing the o.g.f. and comparing with the known one. This entails the alternative triangle recurrence T(n,k) = (-1)^(n-k)*T(n-1,k) + (1/2)*(1 + (-1)^(n-k))*T(n-1,k-1), n >= m >= 1, T(n,k) = 0 if n < k and T(n,0) = (-1)^floor((n+1)/2) = A057077(n+1). [P(n,x) recurrence corrected Aug 03 2014]
(End)

Extensions

New name and Chebyshev comments by Wolfdieter Lang, Jun 11 2010

A154955 a(1) = 1, a(2) = -1, followed by 0, 0, 0, ... .

Original entry on oeis.org

1, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
Offset: 1

Views

Author

Mats Granvik, Jan 18 2009

Keywords

Comments

Matrix inverse of A000012.
Moebius transform of the sequence A000035. Dirichlet inverse of A209229. Partial sums of a(n) is characteristic function of 1 (A063524). a(n)=(-1)^(n+1)*A019590(n). a(n) for n >= 1 is Dirichlet convolution of following functions b(n), c(n), a(n) = Sum_{d|n} b(d)*c(n/d): a(n) = A000012(n) * A092673(n). Examples of Dirichlet convolutions with function a(n), i.e. b(n) = Sum_{d|n} a(d)*c(n/d): a(n) * A000012(n) = A000035(n), a(n) * A000027(n) = A026741(n), a(n) * A008683(n) = A092673(n), a(n) * A036987(n-1) = A063524(n), a(n) * A000005(n) = A001227(n). - Jaroslav Krizek, Mar 21 2009
The Kn21 sums, see A180662, of triangle A108299 equal the terms of this sequence. - Johannes W. Meijer, Aug 14 2011
{a(n-1)}A132393.%20-%20_Wolfdieter%20Lang">{n>=1}, gives the alternating row sums of A132393. - _Wolfdieter Lang, May 09 2017
With offset 0 the alternating row sums of A097805. - Peter Luschny, Sep 07 2017

Crossrefs

Programs

Formula

G.f.: A(x) = x - x^2 = x / (1 + x / (1 - x)). - Michael Somos, Jan 03 2013
a(n) = (2/sqrt(3))*sin((2*Pi/3)*n!). - Lorenzo Pinlac, Jan 16 2022
a(n) = [n = 1] - [n = 2], where [] is the Iverson bracket. - Wesley Ivan Hurt, Jun 22 2024
Multiplicative with a(2) = -1, a(2^e) = 0 if e > 1, a(p^e) = 0 if p > 2. - Antti Karttunen, Dec 17 2024

Extensions

Keyword:mult added by Antti Karttunen, Dec 17 2024

A078922 a(n) = 11*a(n-1) - a(n-2) with a(1)=1, a(2) = 10.

Original entry on oeis.org

1, 10, 109, 1189, 12970, 141481, 1543321, 16835050, 183642229, 2003229469, 21851881930, 238367471761, 2600190307441, 28363725910090, 309400794703549, 3375045015828949, 36816094379414890, 401601993157734841
Offset: 1

Views

Author

Nick Renton (ner(AT)nickrenton.com), Jan 11 2003

Keywords

Comments

All positive integer solutions of Pell equation (3*b(n))^2 - 13*a(n)^2 = -4 together with b(n)=A097783(n-1), n >= 1.
a(n) = L(n-1,11), where L is defined as in A108299; see also A097783 for L(n,-11). - Reinhard Zumkeller, Jun 01 2005
Number of 01-avoiding words of length n on alphabet {0,1,2,3,4,5,6,7,8,9, A} which do not end in 0. - Tanya Khovanova, Jan 10 2007

Examples

			All positive solutions of the Pell equation x^2 - 13*y^2 = -4 are
(x,y)= (3=3*1,1), (36=3*12,10), (393=3*131,109), (4287=3*1429,1189 ), ...
		

Crossrefs

Row 11 of array A094954.
Cf. similar sequences listed in A238379.

Programs

  • GAP
    a:=[1,10];; for n in [3..30] do a[n]:=11*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jan 12 2019
  • Magma
    m:=30; R:=PowerSeriesRing(Integers(), m); Coefficients(R!( x*(1-x)/(1-11*x+x^2) )); // G. C. Greubel, Jan 12 2019
    
  • Mathematica
    LinearRecurrence[{11,-1},{1,10},20] (* Harvey P. Dale, Jan 26 2014 *)
    Table[Fibonacci[2n-1, 3], {n, 1, 20}] (* Vladimir Reshetnikov, Sep 16 2016 *)
  • PARI
    a(n)=([0,1;-1,11]^n*[1;1])[1,1] \\ Charles R Greathouse IV, Jun 11 2015
    
  • PARI
    my(x='x+O('x^30)); Vec(x*(1-x)/(1-11*x+x^2)) \\ G. C. Greubel, Jan 12 2019
    
  • Sage
    (x*(1-x)/(1-11*x+x^2)).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Jan 12 2019
    

Formula

a(1)=1, a(2)=10 and for n > 2, a(n) = ceiling(g*f^n) where f = (11+sqrt(117))/2 and g = (1-3/sqrt(13))/2. - Benoit Cloitre, Jan 12 2003
a(n)*a(n+3) = 99 + a(n+1)*a(n+2). - Ralf Stephan, May 29 2004
a(n) = S(n-1, 11) - S(n-2, 11) = T(2*n-1, sqrt(13)/2)/(sqrt(13)/2).
a(n+1) = ((-1)^n)*S(2*n, i*3), n >= 0, with the imaginary unit i and S(n, x) = U(n, x/2) Chebyshev's polynomials of the second kind, A049310.
G.f.: x*(1-x)/(1-11*x+x^2).
a(n) = A006190(2*n-1). - Vladimir Reshetnikov, Sep 16 2016

Extensions

More terms from Benoit Cloitre, Jan 12 2003
Definition adapted to offset by Georg Fischer, Jun 18 2021

A097783 Chebyshev polynomials S(n,11) + S(n-1,11) with Diophantine property.

Original entry on oeis.org

1, 12, 131, 1429, 15588, 170039, 1854841, 20233212, 220710491, 2407582189, 26262693588, 286482047279, 3125039826481, 34088956044012, 371853476657651, 4056299287190149, 44247438682433988, 482665526219583719, 5265073349732986921, 57433141320843272412
Offset: 0

Views

Author

Wolfdieter Lang, Aug 31 2004

Keywords

Comments

All positive integer solutions of Pell equation (3*a(n))^2 - 13*b(n)^2 = -4 together with b(n)=A078922(n+1), n>=0.

Examples

			All positive solutions to the Pell equation x^2 - 13*y^2 = -4 are (3=3*1,1), (36=3*12,10), (393=3*131,109), (4287=3*1429,1189 ), ...
		

Crossrefs

Programs

  • Magma
    I:=[1,12]; [n le 2 select I[n] else 11*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Mar 22 2015
  • Mathematica
    CoefficientList[Series[(1 + x) / (1 - 11 x + x^2), {x, 0, 33}], x] (* Vincenzo Librandi, Mar 22 2015 *)
  • PARI
    Vec((1+x)/(1-11*x+x^2) + O(x^30)) \\ Michel Marcus, Mar 22 2015
    
  • Sage
    [(lucas_number2(n,11,1)-lucas_number2(n-1,11,1))/9 for n in range(1, 19)] # Zerinvary Lajos, Nov 10 2009
    

Formula

a(n) = S(n, 11) + S(n-1, 11) = S(2*n, sqrt(13)), with S(n, x)=U(n, x/2) Chebyshev's polynomials of the 2nd kind, A049310. S(-1, x) = 0 = U(-1, x).
a(n) = (-2/3)*i*((-1)^n)*T(2*n+1, 3*i/2) with the imaginary unit i and Chebyshev's polynomials of the first kind. See the T-triangle A053120.
G.f.: (1+x)/(1-11*x+x^2).
a(n) = L(n,-11)*(-1)^n, where L is defined as in A108299; see also A078922 for L(n,+11). - Reinhard Zumkeller, Jun 01 2005
a(n) = 11*a(n-1) - a(n-2) with a(0)=1 and a(1)=12. - Philippe Deléham, Nov 17 2008
From Peter Bala, Mar 22 2015: (Start)
The aerated sequence (b(n))n>=1 = [1, 0, 12, 0, 131, 0, 1429, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -9, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047 for the connection with Chebyshev polynomials.
b(n) = 1/2*( (-1)^n - 1 )*F(n,3) + 1/3*( 1 + (-1)^(n+1) )*F(n+1,3), where F(n,x) is the n-th Fibonacci polynomial. The o.g.f. is x*(1 + x^2)/(1 - 11*x^2 + x^4).
Exp( Sum_{n >= 1} 6*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 6*A006190(n)*x^n.
Exp( Sum_{n >= 1} (-6)*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 6*A006190(n)*(-x)^n. Cf. A002315, A004146, A113224 and A192425. (End)
a(n) = A006497(2n+1)/3. - Adam Mohamed, Aug 22 2024
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