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.

Showing 1-8 of 8 results.

A195500 Denominators a(n) of Pythagorean approximations b(n)/a(n) to sqrt(2).

Original entry on oeis.org

3, 228, 308, 5289, 543900, 706180, 1244791, 51146940, 76205040, 114835995824, 106293119818725, 222582887719576, 3520995103197240, 17847666535865852, 18611596834765355, 106620725307595884, 269840171418387336, 357849299891217865
Offset: 1

Views

Author

Clark Kimberling, Sep 20 2011

Keywords

Comments

For each positive real number r, there is a sequence (a(n),b(n),c(n)) of primitive Pythagorean triples such that the limit of b(n)/a(n) is r and
|r-b(n+1)/a(n+1)| < |r-b(n)/a(n)|. Peter Shiu showed how to find (a(n),b(n)) from the continued fraction for r, and Peter J. C. Moses incorporated Shiu's method in the Mathematica program shown below.
Examples:
r...........a(n)..........b(n)..........c(n)
sqrt(2).....A195500.......A195501.......A195502
sqrt(3).....A195499.......A195503.......A195531
sqrt(5).....A195532.......A195533.......A195534
sqrt(6).....A195535.......A195536.......A195537
sqrt(8).....A195538.......A195539.......A195540
sqrt(12)....A195680.......A195681.......A195682
e...........A195541.......A195542.......A195543
pi..........A195544.......A195545.......A195546
tau.........A195687.......A195688.......A195689
1...........A046727.......A084159.......A001653
2...........A195614.......A195615.......A007805
3...........A195616.......A195617.......A097315
4...........A195619.......A195620.......A078988
5...........A195622.......A195623.......A097727
1/2.........A195547.......A195548.......A195549
3/2.........A195550.......A195551.......A195552
5/2.........A195553.......A195554.......A195555
1/3.........A195556.......A195557.......A195558
2/3.........A195559.......A195560.......A195561
1/4.........A195562.......A195563.......A195564
5/4.........A195565.......A195566.......A195567
7/4.........A195568.......A195569.......A195570
1/5.........A195571.......A195572.......A195573
2/5.........A195574.......A195575.......A195576
3/5.........A195577.......A195578.......A195579
4/5.........A195580.......A195611.......A195612
sqrt(1/2)...A195625.......A195626.......A195627
sqrt(1/3)...{1}+A195503...{0}+A195499...{1}+A195531
sqrt(2/3)...A195631.......A195632.......A195633
sqrt(3/4)...A195634.......A195635.......A195636

Examples

			For r=sqrt(2), the first five fractions b(n)/a(n) can be read from the following five primitive Pythagorean triples (a(n), b(n), c(n)) = (A195500, A195501, A195502):
(3,4,5); |r - b(1)/a(1)| = 0.08...
(228,325,397); |r - b(2)/a(2)| = 0.011...
(308,435,533); |r - b(3)/a(3)| = 0.0018...
(5289,7480,9161); |r - b(4)/a(4)| = 0.000042...
(543900,769189,942061); |r - b(5)/a(5)| = 0.0000003...
		

Crossrefs

Programs

  • Maple
    Shiu := proc(r,n)
            t := r+sqrt(1+r^2) ;
            cf := numtheory[cfrac](t,n+1) ;
            mn := numtheory[nthconver](cf,n) ;
            (mn-1/mn)/2 ;
    end proc:
    A195500 := proc(n)
            Shiu(sqrt(2),n) ;
            denom(%) ;
    end proc: # R. J. Mathar, Sep 21 2011
  • Mathematica
    r = Sqrt[2]; z = 18;
    p[{f_, n_}] := (#1[[2]]/#1[[
          1]] &)[({2 #1[[1]] #1[[2]], #1[[1]]^2 - #1[[
             2]]^2} &)[({Numerator[#1], Denominator[#1]} &)[
         Array[FromContinuedFraction[
            ContinuedFraction[(#1 + Sqrt[1 + #1^2] &)[f], #1]] &, {n}]]]];
    {a, b} = ({Denominator[#1], Numerator[#1]} &)[
      p[{r, z}]]  (* A195500, A195501 *)
    Sqrt[a^2 + b^2] (* A195502 *)

A085478 Triangle read by rows: T(n, k) = binomial(n + k, 2*k).

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 6, 5, 1, 1, 10, 15, 7, 1, 1, 15, 35, 28, 9, 1, 1, 21, 70, 84, 45, 11, 1, 1, 28, 126, 210, 165, 66, 13, 1, 1, 36, 210, 462, 495, 286, 91, 15, 1, 1, 45, 330, 924, 1287, 1001, 455, 120, 17, 1, 1, 55, 495, 1716, 3003, 3003, 1820, 680, 153, 19, 1
Offset: 0

Views

Author

Philippe Deléham, Aug 14 2003

Keywords

Comments

Coefficient array for Morgan-Voyce polynomial b(n,x). A053122 (unsigned) is the coefficient array for B(n,x). Reversal of A054142. - Paul Barry, Jan 19 2004
This triangle is formed from even-numbered rows of triangle A011973 read in reverse order. - Philippe Deléham, Feb 16 2004
T(n,k) is the number of nondecreasing Dyck paths of semilength n+1, having k+1 peaks. T(n,k) is the number of nondecreasing Dyck paths of semilength n+1, having k peaks at height >= 2. T(n,k) is the number of directed column-convex polyominoes of area n+1, having k+1 columns. - Emeric Deutsch, May 31 2004
Riordan array (1/(1-x), x/(1-x)^2). - Paul Barry, May 09 2005
The triangular matrix a(n,k) = (-1)^(n+k)*T(n,k) is the matrix inverse of A039599. - Philippe Deléham, May 26 2005
The n-th row gives absolute values of coefficients of reciprocal of g.f. of bottom-line of n-wave sequence. - Floor van Lamoen (fvlamoen(AT)planet.nl), Sep 24 2006
Unsigned version of A129818. - Philippe Deléham, Oct 25 2007
T(n, k) is also the number of idempotent order-preserving full transformations (of an n-chain) of height k >=1 (height(alpha) = |Im(alpha)|) and of waist n (waist(alpha) = max(Im(alpha))). - Abdullahi Umar, Oct 02 2008
A085478 is jointly generated with A078812 as a triangular array of coefficients of polynomials u(n,x): initially, u(1,x) = v(1,x) = 1; for n>1, u(n,x) = u(n-1,x)+x*v(n-1)x and v(n,x) = u(n-1,x)+(x+1)*v(n-1,x). See the Mathematica section. - Clark Kimberling, Feb 25 2012
Per Kimberling's recursion relations, see A102426. - Tom Copeland, Jan 19 2016
Subtriangle of the triangle given by (0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Mar 26 2012
T(n,k) is also the number of compositions (ordered partitions) of 2*n+1 into 2*k+1 parts which are all odd. Proof: The o.g.f. of column k, x^k/(1-x)^(2*k+1) for k >= 0, is the o.g.f. of the odd-indexed members of the sequence with o.g.f. (x/(1-x^2))^(2*k+1) (bisection, odd part). Thus T(n,k) is obtained from the sum of the multinomial numbers A048996 for the partitions of 2*n+1 into 2*k+1 parts, all of which are odd. E.g., T(3,1) = 3 + 3 from the numbers for the partitions [1,1,5] and [1,3,3], namely 3!/(2!*1!) and 3!/(1!*2!), respectively. The number triangle with the number of these partitions as entries is A152157. - Wolfdieter Lang, Jul 09 2012
The matrix elements of the inverse are T^(-1)(n,k) = (-1)^(n+k)*A039599(n,k). - R. J. Mathar, Mar 12 2013
T(n,k) = A258993(n+1,k) for k = 0..n-1. - Reinhard Zumkeller, Jun 22 2015
The n-th row polynomial in descending powers of x is the n-th Taylor polynomial of the algebraic function F(x)*G(x)^n about 0, where F(x) = (1 + sqrt(1 + 4*x))/(2*sqrt(1 + 4*x)) and G(x) = ((1 + sqrt(1 + 4*x))/2)^2. For example, for n = 4, (1 + sqrt(1 + 4*x))/(2*sqrt(1 + 4*x)) * ((1 + sqrt(1 + 4*x))/2)^8 = (x^4 + 10*x^3 + 15*x^2 + 7*x + 1) + O(x^5). - Peter Bala, Feb 23 2018
Row n also gives the coefficients of the characteristc polynomial of the tridiagonal n X n matrix M_n given in A332602: Phi(n, x) := Det(M_n - x*1_n) = Sum_{k=0..n} T(n, k)*(-x)^k, for n >= 0, with Phi(0, x) := 1. - Wolfdieter Lang, Mar 25 2020
It appears that the largest root of the n-th degree polynomial is equal to the sum of the distinct diagonals of a (2*n+1)-gon including the edge, 1. The largest root of x^3 - 6*x^2 + 5*x - 1 is 5.048917... = the sum of (1 + 1.80193... + 2.24697...). Alternatively, the largest root of the n-th degree polynomial is equal to the square of sigma(2*n+1). Check: 5.048917... is the square of sigma(7), 2.24697.... Given N = 2*n+1, sigma(N) (N odd) can be defined as 1/(2*sin(Pi/(2*N))). Relating to the 9-gon, the largest root of x^4 - 10*x^3 + 15*x^2 - 7*x + 1 is 8.290859..., = the sum of (1 + 1.879385... + 2.532088... + 2.879385...), and is the square of sigma(9), 2.879385... Refer to A231187 for a further clarification of sigma(7). - Gary W. Adamson, Jun 28 2022
For n >=1, the n-th row is given by the coefficients of the minimal polynomial of -4*sin(Pi/(4*n + 2))^2. - Eric W. Weisstein, Jul 12 2023
Denoting this lower triangular array by L, then L * diag(binomial(2*k,k)^2) * transpose(L) is the LDU factorization of A143007, the square array of crystal ball sequences for the A_n X A_n lattices. - Peter Bala, Feb 06 2024
T(n, k) is the number of occurrences of the periodic substring (01)^k in the periodic string (01)^n (see Proposition 4.7 at page 7 in Fang). - Stefano Spezia, Jun 09 2024

Examples

			Triangle begins as:
  1;
  1    1;
  1    3    1;
  1    6    5    1;
  1   10   15    7    1;
  1   15   35   28    9    1;
  1   21   70   84   45   11    1;
  1   28  126  210  165   66   13    1;
  1   36  210  462  495  286   91   15    1;
  1   45  330  924 1287 1001  455  120   17    1;
  1   55  495 1716 3003 3003 1820  680  153   19    1;
...
From _Philippe Deléham_, Mar 26 2012: (Start)
(0, 1, 0, 1, 0, 0, 0, ...) DELTA (1, 0, 1, -1, 0, 0, 0, ...) begins:
  1
  0, 1
  0, 1,  1
  0, 1,  3,   1
  0, 1,  6,   5,   1
  0, 1, 10,  15,   7,   1
  0, 1, 15,  35,  28,   9,  1
  0, 1, 21,  70,  84,  45, 11,  1
  0, 1, 28, 126, 210, 165, 66, 13, 1. (End)
		

Crossrefs

Programs

  • GAP
    Flat(List([0..12], n-> List([0..n], k-> Binomial(n+k, 2*k) ))); # G. C. Greubel, Aug 01 2019
  • Haskell
    a085478 n k = a085478_tabl !! n !! k
    a085478_row n = a085478_tabl !! n
    a085478_tabl = zipWith (zipWith a007318) a051162_tabl a025581_tabl
    -- Reinhard Zumkeller, Jun 22 2015
    
  • Magma
    [Binomial(n+k, 2*k): k in [0..n], n in [0..12]]; // G. C. Greubel, Aug 01 2019
    
  • Maple
    T := (n,k) -> binomial(n+k,2*k): seq(seq(T(n,k), k=0..n), n=0..11);
  • Mathematica
    (* First program *)
    u[1, x_]:= 1; v[1, x_]:= 1; z = 13;
    u[n_, x_]:= u[n-1, x] + x*v[n-1, x];
    v[n_, x_]:= u[n-1, x] + (x+1)*v[n-1, x];
    Table[Expand[u[n, x]], {n, 1, z/2}]
    Table[Expand[v[n, x]], {n, 1, z/2}]
    cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
    TableForm[cu]
    Flatten[%]   (* A085478 *)
    Table[Expand[v[n, x]], {n, 1, z}]
    cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
    TableForm[cv]
    Flatten[%]   (* A078812 *) (*Clark Kimberling, Feb 25 2012 *)
    (* Second program *)
    Table[Binomial[n + k, 2 k], {n, 0, 12}, {k, 0, n}] // Flatten (* G. C. Greubel, Aug 01 2019 *)
    CoefficientList[Table[Fibonacci[2 n + 1, Sqrt[x]], {n, 0, 10}], x] // Flatten (* Eric W. Weisstein, Jul 03 2023 *)
    Join[{{1}}, CoefficientList[Table[MinimalPolynomial[-4 Sin[Pi/(4 n + 2)]^2, x], {n, 20}], x]] (* Eric W. Weisstein, Jul 12 2023 *)
  • PARI
    T(n,k) = binomial(n+k,n-k)
    
  • Sage
    [[binomial(n+k,2*k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Aug 01 2019
    

Formula

T(n, k) = (n+k)!/((n-k)!*(2*k)!).
G.f.: (1-z)/((1-z)^2-tz). - Emeric Deutsch, May 31 2004
Row sums are A001519 (Fibonacci(2n+1)). Diagonal sums are A011782. Binomial transform of A026729 (product of lower triangular matrices). - Paul Barry, Jun 21 2004
T(n, 0) = 1, T(n, k) = 0 if n=0} T(n-1-j, k-1)*(j+1). T(0, 0) = 1, T(0, k) = 0 if k>0; T(n, k) = T(n-1, k-1) + T(n-1, k) + Sum_{j>=0} (-1)^j*T(n-1, k+j)*A000108(j). For the column k, g.f.: Sum_{n>=0} T(n, k)*x^n = (x^k) / (1-x)^(2*k+1). - Philippe Deléham, Feb 15 2004
Sum_{k=0..n} T(n,k)*x^(2*k) = A000012(n), A001519(n+1), A001653(n), A078922(n+1), A007805(n), A097835(n), A097315(n), A097838(n), A078988(n), A097841(n), A097727(n), A097843(n), A097730(n), A098244(n), A097733(n), A098247(n), A097736(n), A098250(n), A097739(n), A098253(n), A097742(n), A098256(n), A097767(n), A098259(n), A097770(n), A098262(n), A097773(n), A098292(n), A097776(n) for x=0,1,2,...,27,28 respectively. - Philippe Deléham, Dec 31 2007
T(2*n,n) = A005809(n). - Philippe Deléham, Sep 17 2009
A183160(n) = Sum_{k=0..n} T(n,k)*T(n,n-k). - Paul D. Hanna, Dec 27 2010
T(n,k) = 2*T(n-1,k) + T(n-1,k-1) - T(n-2,k). - Philippe Deléham, Feb 06 2012
O.g.f. for column k: x^k/(1-x)^(2*k+1), k >= 0. [See the o.g.f. of the triangle above, and a comment on compositions. - Wolfdieter Lang, Jul 09 2012]
E.g.f.: (2/sqrt(x + 4))*sinh((1/2)*t*sqrt(x + 4))*cosh((1/2)*t*sqrt(x)) = t + (1 + x)*t^3/3! + (1 + 3*x + x^2)*t^5/5! + (1 + 6*x + 5*x^2 + x^3)*t^7/7! + .... Cf. A091042. - Peter Bala, Jul 29 2013
T(n, k) = A065941(n+3*k, 4*k) = A108299(n+3*k, 4*k) = A194005(n+3*k, 4*k). - Johannes W. Meijer, Sep 05 2013
Sum_{k=0..n} (-1)^k*T(n,k)*A000108(k) = A000007(n) for n >= 0. - Werner Schulte, Jul 12 2017
Sum_{k=0..floor(n/2)} T(n-k,k)*A000108(k) = A001006(n) for n >= 0. - Werner Schulte, Jul 12 2017
From Peter Bala, Jun 26 2025: (Start)
The n-th row polynomial b(n, x) = (-1)^n * U(2*n, (i/2)*sqrt(x)), where U(n,x) is the n-th Chebyshev polynomial of the second kind.
b(n, x) = (-1)^n * Dir(n, -1 - x/2), where Dir(n, x) is the n-th row polynomial of the triangle A244419.
b(n, -1 - x) is the n-th row polynomial of A098493. (End)

A094954 Array T(k,n) read by antidiagonals. G.f.: x(1-x)/(1-kx+x^2), k>1.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 5, 1, 1, 4, 11, 13, 1, 1, 5, 19, 41, 34, 1, 1, 6, 29, 91, 153, 89, 1, 1, 7, 41, 169, 436, 571, 233, 1, 1, 8, 55, 281, 985, 2089, 2131, 610, 1, 1, 9, 71, 433, 1926, 5741, 10009, 7953, 1597, 1, 1, 10, 89, 631, 3409, 13201, 33461, 47956, 29681
Offset: 1

Views

Author

Ralf Stephan, May 31 2004

Keywords

Comments

Also, values of polynomials with coefficients in A098493 (see Fink et al.). See A098495 for negative k.
Number of dimer tilings of the graph S_{k-1} X P_{2n-2}.

Examples

			1,1,1,1,1,1,1,1,1,1,1,1,1,1, ...
1,2,5,13,34,89,233,610,1597, ...
1,3,11,41,153,571,2131,7953, ...
1,4,19,91,436,2089,10009,47956, ...
1,5,29,169,985,5741,33461,195025, ...
1,6,41,281,1926,13201,90481,620166, ...
		

Crossrefs

Rows are first differences of rows in array A073134.
Rows 2-14 are A000012, A001519, A079935/A001835, A004253, A001653, A049685, A070997, A070998, A072256, A078922, A077417, A085260, A001570. Other rows: A007805 (k=18), A075839 (k=20), A077420 (k=34), A078988 (k=66).
Columns include A028387. Diagonals include A094955, A094956. Antidiagonal sums are A094957.

Programs

  • Mathematica
    max = 14; row[k_] := Rest[ CoefficientList[ Series[ x*(1-x)/(1-k*x+x^2), {x, 0, max}], x]]; t = Table[ row[k], {k, 2, max+1}]; Flatten[ Table[ t[[k-n+1, n]], {k, 1, max}, {n, 1, k}]] (* Jean-François Alcover, Dec 27 2011 *)
  • PARI
    T(k,n)=polcoeff(x*(1-x)/(1-k*x+x*x),n)

Formula

Recurrence: T(k, 1) = 1, T(k, 2) = k-1, T(k, n) = kT(k, n-1) - T(k, n-2).
For n>3, T(k, n) = [k(k-2) + T(k, n-1)T(k, n-2)] / T(k, n-3).
T(k, n+1) = S(n, k) - S(n-1, k) = U(n, k/2) - U(n-1, k/2), with S, U = Chebyshev polynomials of second kind.
T(k+2, n+1) = Sum[i=0..n, k^(n-i) * C(2n-i, i)] (from comments by Benoit Cloitre).

A041025 Denominators of continued fraction convergents to sqrt(17).

Original entry on oeis.org

1, 8, 65, 528, 4289, 34840, 283009, 2298912, 18674305, 151693352, 1232221121, 10009462320, 81307919681, 660472819768, 5365090477825, 43581196642368, 354014663616769, 2875698505576520, 23359602708228929, 189752520171407952, 1541379764079492545
Offset: 0

Views

Author

Keywords

Comments

a(2*n+1) with b(2*n+1) := A041024(2*n+1), n >= 0, give all (positive integer) solutions to Pell equation b^2 - 17*a^2 = +1, a(2*n) with b(2*n) := A041024(2*n), n >= 0, give all (positive integer) solutions to Pell equation b^2 - 17*a^2 = -1 (cf. Emerson reference).
Bisection: a(2*n) = T(2*n+1,sqrt(17))/sqrt(17) = A078988(n), n >= 0 and a(2*n+1) = 8*S(n-1,66), n >= 0, with T(n,x), resp. S(n,x), Chebyshev's polynomials of the first, resp. second kind. S(-1,x)=0. See A053120, resp. A049310. - Wolfdieter Lang, Jan 10 2003
Sqrt(17) = 8/2 + 8/65 + 8/(65*4289) + 8/(4289*283009) + ... . - Gary W. Adamson, Dec 26 2007
For positive n, a(n) equals the permanent of the n X n tridiagonal matrix with 8's along the main diagonal and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
De Moivre's formula: a(n) = (r^n - s^n)/(r-s), for r > s, gives sequences with integers if r and s are conjugates. With r=4+sqrt(17) and s=4-sqrt(17), a(n+1)/a(n) converges to r=4+sqrt(17). - Sture Sjöstedt, Nov 11 2011
a(n) equals the number of words of length n on alphabet {0,1,...,8} avoiding runs of zeros of odd lengths. - Milan Janjic, Jan 28 2015
From Michael A. Allen, Feb 21 2023: (Start)
Also called the 8-metallonacci sequence; the g.f. 1/(1-k*x-x^2) gives the k-metallonacci sequence.
a(n) is the number of tilings of an n-board (a board with dimensions n X 1) using unit squares and dominoes (with dimensions 2 X 1) if there are 8 kinds of squares available. (End)

Crossrefs

Programs

Formula

G.f.: 1/(1 - 8*x - x^2).
a(n) = ((-i)^n)*S(n, 8*i), with S(n, x) := U(n, x/2) Chebyshev's polynomials of the second kind and i^2 = -1. See A049310.
a(n) = F(n, 8), the n-th Fibonacci polynomial evaluated at x=8. - T. D. Noe, Jan 19 2006
From Sergio Falcon, Sep 24 2007: (Start)
a(n) = ((4 + sqrt(17))^n - (4 - sqrt(17))^n)/(2*sqrt(17));
a(n) = Sum_{i=0..floor((n-1)/2)} binomial(n-1-i,i)*8^(n-1-2i). (End)
Let T be the 2 X 2 matrix [0, 1; 1, 8]. Then T^n * [1, 0] = [a(n-2), a(n-1)]. - Gary W. Adamson, Dec 26 2007
a(n) = 8*a(n-1) + a(n-2), n > 1; a(0)=1, a(1)=8. - Philippe Deléham, Nov 20 2008
a(p-1) == 68^((p-1)/2) (mod p) for odd primes p. - Gary W. Adamson, Feb 22 2009 [Corrected by Jason Yuen, Apr 05 2025. See A087475 for more info about this congruence.]
Sum_{n>=0} (-1)^n/(a(n)*a(n+1)) = sqrt(17) - 4. - Vladimir Shevelev, Feb 23 2013
G.f.: x/(1 - 8*x - x^2) = Sum_{n >= 0} x^n *( Product_{k = 1..n} (m*k + 8 - m + x)/(1 + m*k*x) ) for arbitrary m (a telescoping series). - Peter Bala, May 08 2024

A097316 Chebyshev U(n,x) polynomial evaluated at x=33.

Original entry on oeis.org

1, 66, 4355, 287364, 18961669, 1251182790, 82559102471, 5447649580296, 359462313197065, 23719065021425994, 1565098829100918539, 103272803655639197580, 6814439942443086121741, 449649763397588044837326
Offset: 0

Views

Author

Wolfdieter Lang, Aug 31 2004

Keywords

Comments

Used to form integer solutions of Pell equation a^2 - 17*b^2 =-1. See A078989 with A078988.

Crossrefs

Chebyshev sequence U(n, m): A000027 (m=1), A001353 (m=2), A001109 (m=3), A001090 (m=4), A004189 (m=5), A004191 (m=6), A007655 (m=7), A077412 (m=8), A049660 (m=9), A075843 (m=10), A077421 (m=11), A077423 (m=12), A097309 (m=13), A097311 (m=14), A097313 (m=15), A029548 (m=16), A029547 (m=17), A144128 (m=18), A078987 (m=19), this sequence (m=33).

Programs

  • GAP
    m:=33;; a:=[1,2*m];; for n in [3..20] do a[n]:=2*m*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Dec 22 2019
  • Magma
    m:=33; I:=[1, 2*m]; [n le 2 select I[n] else 2*m*Self(n-1) -Self(n-2): n in [1..20]]; // G. C. Greubel, Dec 22 2019
    
  • Maple
    seq( simplify(ChebyshevU(n, 33)), n=0..20); # G. C. Greubel, Dec 22 2019
  • Mathematica
    LinearRecurrence[{66, -1},{1, 66},14] (* Ray Chandler, Aug 11 2015 *)
    ChebyshevU[Range[21] -1, 33] (* G. C. Greubel, Dec 22 2019 *)
  • PARI
    vector( 21, n, polchebyshev(n-1, 2, 33) ) \\ G. C. Greubel, Dec 22 2019
    
  • Sage
    [chebyshev_U(n,33) for n in (0..20)] # G. C. Greubel, Dec 22 2019
    

Formula

a(n) = 66*a(n-1) - a(n-2), n>=1, a(0)=1, a(-1):=0.
a(n) = S(n, 66) with S(n, x) := U(n, x/2), Chebyshev's polynomials of the second kind. See A049310.
G.f.: 1/(1-66*x+x^2).
a(n) = Sum_{k=0..floor(n/2)} (-1)^k*binomial(n-k, k)*66^(n-2*k).
a(n) = ((33+8*sqrt(17))^(n+1) - (33-8*sqrt(17))^(n+1))/(16*sqrt(17)).

A188647 Array read by antidiagonals of a(n) = a(n-1)*k-((k-1)/(k^n)) where a(0)=1 and k=(sqrt(x^2+1)+x)^2 for integers x>=1.

Original entry on oeis.org

1, 5, 1, 29, 17, 1, 169, 305, 37, 1, 985, 5473, 1405, 65, 1, 5741, 98209, 53353, 4289, 101, 1, 33461, 1762289, 2026009, 283009, 10301, 145, 1, 195025, 31622993, 76934989, 18674305, 1050601, 21169, 197, 1, 1136689, 567451585, 2921503573, 1232221121, 107151001, 3090529, 39005, 257, 1
Offset: 0

Views

Author

Charles L. Hohn, Apr 06 2011

Keywords

Comments

Conjecture: Given function f(x, y)=(sqrt(x^2+y)+x)^2; constant k=f(x, y); and initial term a(0)=1; then for all integers x>=1 and y=[+-]1, k may be irrational, but sequence a(n)=a(n-1)*k-((k-1)/(k^n)) always produces integer sequences; y=1 results shown here; y=-1 results are A188646.
Also square array A(n,k), n >= 1, k >= 0, read by antidiagonals, where A(n,k) is (1/sqrt(n^2+1)) * T_{2*k+1}(sqrt(n^2+1)), with T the Chebyshev polynomial of the first kind. - Seiichi Manyama, Jan 02 2019

Examples

			Square array begins:
     | 0    1       2          3             4
-----+---------------------------------------------
   1 | 1,   5,     29,       169,          985, ...
   2 | 1,  17,    305,      5473,        98209, ...
   3 | 1,  37,   1405,     53353,      2026009, ...
   4 | 1,  65,   4289,    283009,     18674305, ...
   5 | 1, 101,  10301,   1050601,    107151001, ...
   6 | 1, 145,  21169,   3090529,    451196065, ...
   7 | 1, 197,  39005,   7722793,   1529074009, ...
   8 | 1, 257,  66305,  17106433,   4413393409, ...
   9 | 1, 325, 105949,  34539049,  11259624025, ...
  10 | 1, 401, 161201,  64802401,  26050404001, ...
  11 | 1, 485, 235709, 114554089,  55673051545, ...
  12 | 1, 577, 333505, 192765313, 111418017409, ...
  13 | 1, 677, 459005, 311204713, 210996336409, ...
  14 | 1, 785, 617009, 484968289, 381184458145, ...
  15 | 1, 901, 812701, 733055401, 661215159001, ...
  ...
		

Crossrefs

Row 1 is A001653, row 2 is A007805, row 3 is A097315, row 4 is A078988, row 5 is A097727, row 6 is A097730, row 7 is A097733, row 8 is A097736, row 9 is A097739, row 10 is A097742, row 11 is A097767, row 12 is A097770, row 13 is A097773.
Column 1 is A053755.
A(n,n) gives A323012.
Cf. A188645, A188646 (f(x, y) as above with y=-1).

Formula

A(n,k) = 2 * A188645(n,k) - A(n,k-1).
A(n,k) = Sum_{j=0..k} binomial(2*k+1,2*j)*(n^2+1)^(k-j)*n^(2*j). - Seiichi Manyama, Jan 02 2019

Extensions

Edited and extended by Seiichi Manyama, Jan 02 2019

A078989 Chebyshev sequence with Diophantine property.

Original entry on oeis.org

1, 67, 4421, 291719, 19249033, 1270144459, 83810285261, 5530208682767, 364909962777361, 24078527334623059, 1588817894122344533, 104837902484740116119, 6917712746098725319321, 456464203340031130959067, 30119719707695955917979101, 1987445036504593059455661599
Offset: 0

Views

Author

Wolfdieter Lang, Jan 10 2003

Keywords

Comments

One fourth of bisection (even part) of A041024.
(4*a(n))^2 - 17*A078988(n)^2= -1 (Pell -1 equation, see A077232-3).

Examples

			(x,y) = (4,1), (268,65), (17684,4289), ... give the positive integer solutions to x^2 - 17*y^2 =-1.
		

Crossrefs

Cf. A097316 for S(n, 66).
Cf. A041024.
Cf. similar sequences of the type (1/k)*sinh((2*n+1)*arcsinh(k)) listed in A097775.

Programs

  • GAP
    a:=[1,67];; for n in [3..20] do a[n]:=66*a[n-1]-a[n-2]; od; a; # Muniru A Asiru, Apr 05 2018
  • Mathematica
    LinearRecurrence[{66, -1}, {1, 67}, 20] (* Bruno Berselli, Apr 03 2018 *)
  • PARI
    x='x+O('x^99); Vec((1+x)/(1-66*x+x^2)) \\ Altug Alkan, Apr 05 2018
    

Formula

G.f.: (1 + x)/(1 - 66*x + x^2).
a(n) = 66*a(n-1) - a(n-2) for n>=1, a(-1)=-1, a(0)=1.
a(n) = S(2*n, 2*sqrt(17)) = -i*((-1)^n)*T(2*n+1, 4*i)/4 = S(n, 66) + S(n-1, 66) with i^2=-1 and S(n, x), resp. T(n, x), Chebyshev's polynomials of the second, resp. first, kind. See A049310 and A053120.
a(n) = A041024(2*n)/4.
a(n) = (1/4)*sinh((2*n + 1)*arcsinh(4)). - Bruno Berselli, Apr 03 2018

A195619 Denominators of Pythagorean approximations to 4.

Original entry on oeis.org

16, 1040, 68640, 4529184, 298857520, 19720067120, 1301225572416, 85861167712320, 5665535843440720, 373839504499375184, 24667741761115321440, 1627697116729111839840, 107403341962360266108016, 7086992872399048451289200
Offset: 1

Views

Author

Clark Kimberling, Sep 22 2011

Keywords

Comments

See A195500 for a discussion and references.

Crossrefs

Programs

  • Magma
    I:=[16, 1040, 68640]; [n le 3 select I[n] else 65*Self(n-1) +65*Self(n-2) -Self(n-3): n in [1..40]]; // G. C. Greubel, Feb 13 2023
    
  • Mathematica
    r = 4; z = 20;
    p[{f_, n_}] := (#1[[2]]/#1[[
          1]] &)[({2 #1[[1]] #1[[2]], #1[[1]]^2 - #1[[
             2]]^2} &)[({Numerator[#1], Denominator[#1]} &)[
         Array[FromContinuedFraction[
            ContinuedFraction[(#1 + Sqrt[1 + #1^2] &)[f], #1]] &, {n}]]]];
    {a, b} = ({Denominator[#1], Numerator[#1]} &)[
      p[{r, z}]]  (* A195619, A195620 *)
    Sqrt[a^2 + b^2] (* A078988 *)
    (* Peter J. C. Moses, Sep 02 2011 *)
    Table[(LucasL[2*n+1,8] - 8*(-1)^n)/34, {n,40}] (* G. C. Greubel, Feb 13 2023 *)
    LinearRecurrence[{65,65,-1},{16,1040,68640},20] (* Harvey P. Dale, May 01 2023 *)
  • PARI
    Vec(16*x/((x+1)*(x^2-66*x+1)) + O(x^20)) \\ Colin Barker, Jun 03 2015
    
  • SageMath
    A078989=BinaryRecurrenceSequence(66,-1,1,67)
    [4*(A078989(n) - (-1)^n)/17 for n in range(1,41)] # G. C. Greubel, Feb 13 2023

Formula

From Colin Barker, Jun 03 2015: (Start)
a(n) = 65*a(n-1) + 65*a(n-2) - a(n-3).
G.f.: 16*x/((1+x)*(1-66*x+x^2)). (End)
a(n) = ((4+sqrt(17))^(2*n+1) + (4-sqrt(17))^(2*n+1) - 8*(-1)^n)/34. - Colin Barker, Mar 03 2016
a(n) = 4*(A078989(n) - (-1)^n)/17. - G. C. Greubel, Feb 13 2023
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