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 1-4 of 4 results.

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)

A052918 a(0) = 1, a(1) = 5, a(n+1) = 5*a(n) + a(n-1).

Original entry on oeis.org

1, 5, 26, 135, 701, 3640, 18901, 98145, 509626, 2646275, 13741001, 71351280, 370497401, 1923838285, 9989688826, 51872282415, 269351100901, 1398627786920, 7262490035501, 37711077964425, 195817879857626
Offset: 0

Views

Author

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

Keywords

Comments

A087130(n)^2 - 29*a(n-1)^2 = 4*(-1)^n, n >= 1. - Gary W. Adamson, Jul 01 2003, corrected Oct 07 2008, corrected by Jianing Song, Feb 01 2019
a(p-1) == 29^((p-1)/2) (mod p), for odd primes p. - Gary W. Adamson, Feb 22 2009 [See A087475 for more info about this congruence. - Jason Yuen, Apr 05 2025]
For more information about this type of recurrence, follow the Khovanova link and see A054413, A086902 and A178765. - Johannes W. Meijer, Jun 12 2010
Binomial transform of A015523. - Johannes W. Meijer, Aug 01 2010
For positive n, a(n) equals the permanent of the n X n tridiagonal matrix with 5's along the main diagonal and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
a(n) equals the number of words of length n on alphabet {0,1,...,5} avoiding runs of zeros of odd lengths. - Milan Janjic, Jan 28 2015
From Michael A. Allen, Feb 15 2023: (Start)
Also called the 5-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 5 kinds of squares available. (End)

Crossrefs

Row 5 of A073133, A172236, and A352361.
Cf. A087130, A099365 (squares), A100237, A175184 (Pisano periods), A201005 (prime subsequence).

Programs

  • GAP
    a:=[1,5];; for n in [3..30] do a[n]:=5*a[n-1]+a[n-2]; od; a; # G. C. Greubel, Oct 16 2019
  • Magma
    I:=[1, 5]; [n le 2 select I[n] else 5*Self(n-1)+Self(n-2): n in [1..30]]; // Vincenzo Librandi, Feb 23 2013
    
  • Magma
    R:=PowerSeriesRing(Integers(), 22); Coefficients(R!( 1/(1 - 5*x - x^2) )); // Marius A. Burtea, Oct 16 2019
    
  • Maple
    spec := [S,{S=Sequence(Union(Z,Z,Z,Z,Z,Prod(Z,Z)))},unlabeled]: seq(combstruct[count](spec,size=n), n=0..30);
    a[0]:=1: a[1]:=5: for n from 2 to 26 do a[n]:=5*a[n-1]+a[n-2] od: seq(a[n], n=0..30); # Zerinvary Lajos, Jul 26 2006
    with(combinat):a:=n->fibonacci(n,5):seq(a(n),n=1..30); # Zerinvary Lajos, Dec 07 2008
  • Mathematica
    LinearRecurrence[{5, 1}, {1, 5}, 30] (* Vincenzo Librandi, Feb 23 2013 *)
    Table[Fibonacci[n+1, 5], {n,0,30}] (* Vladimir Reshetnikov, May 08 2016 *)
  • PARI
    Vec(1/(1-5*x-x^2)+O(x^30)) \\ Charles R Greathouse IV, Nov 20 2011
    
  • Sage
    [lucas_number1(n,5,-1) for n in range(1, 22)] # Zerinvary Lajos, Apr 24 2009
    

Formula

G.f.: 1/(1 - 5*x - x^2).
a(3n) = A041047(5n), a(3n+1) = A041047(5n+3), a(3n+2) = 2*A041047(5n+4). - Henry Bottomley, May 10 2000
a(n) = Sum_{alpha=RootOf(-1+5*z+z^2)} (1/29)*(5+2*alpha)*alpha^(-1-n).
a(n-1) = (((5 + sqrt(29))/2)^n - ((5 - sqrt(29))/2)^n)/sqrt(29). - Gary W. Adamson, Jul 01 2003
a(n) = U(n, 5*i/2)*(-i)^n with i^2 = -1 and Chebyshev's U(n, x/2) = S(n, x) polynomials. See triangle A049310.
Let M = {{0, 1}, {1, 5}}, then a(n) is the lower-right term of M^n. - Roger L. Bagula, May 29 2005
a(n) = F(n, 5), the n-th Fibonacci polynomial evaluated at x = 5. - T. D. Noe, Jan 19 2006
a(n) = denominator of n-th convergent to [1, 4, 5, 5, 5, ...], for n > 0. Continued fraction [1, 4, 5, 5, 5, ...] = 0.807417596..., the inradius of a right triangle with legs 2 and 5. n-th convergent = A100237(n)/A052918(n), the first few being: 1/1, 4/5, 21/26, 109/135, 566/701, ... - Gary W. Adamson, Dec 21 2007
From Johannes W. Meijer, Jun 12 2010: (Start)
a(2n+1) = 5*A097781(n), a(2n) = A097835(n).
Limit_{k->oo} a(n+k)/a(k) = (A087130(n) + a(n-1)*sqrt(29))/2.
Limit_{n->oo} A087130(n)/a(n-1) = sqrt(29). (End)
From L. Edson Jeffery, Jan 07 2012: (Start)
Define the 2 X 2 matrix A = {{1, 1}, {5, 4}}. Then:
a(n) is the upper-left term of (1/5)*(A^(n+2) - A^(n+1));
a(n) is the upper-right term of A^(n+1);
a(n) is the lower-left term of (1/5)*A^(n+1);
a(n) is the lower-right term of (Sum_{k=0..n} A^k). (End)
Sum_{n>=0} (-1)^n/(a(n)*a(n+1)) = (sqrt(29) - 5)/2. - Vladimir Shevelev, Feb 23 2013
G.f.: x/(1 - 5*x - x^2) = Sum_{n >= 0} x^(n+1) *( Product_{k = 1..n} (m*k + 5 - m + x)/(1 + m*k*x) ) for arbitrary m (a telescoping series). - Peter Bala, May 08 2024

A238379 Expansion of (1 - x)/(1 - 36*x + x^2).

Original entry on oeis.org

1, 35, 1259, 45289, 1629145, 58603931, 2108112371, 75833441425, 2727895778929, 98128414600019, 3529895029821755, 126978092658983161, 4567681440693572041, 164309553772309610315, 5910576254362452399299, 212616435603275976764449
Offset: 0

Views

Author

Bruno Berselli, Feb 25 2014

Keywords

Comments

First bisection of A041611.

Crossrefs

Cf. similar sequences with g.f. (1-x)/(1-k*x+x^2): A122367 (k=3), A079935 (k=4), A004253 (k=5), A001653 (k=6), A049685 (k=7), A070997 (k=8), A070998 (k=9), A138288 (k=10), A078922 (k=11), A077417 (k=12), A085260 (k=13), A001570 (k=14), A160682 (k=15), A157456 (k=16), A161595 (k=17). From 18 to 38, even k only, except k=27 and k=31: A007805 (k=18), A075839 (k=20), A157014 (k=22), A159664 (k=24), A153111 (k=26), A097835 (k=27), A159668 (k=28), A157877 (k=30), A111216 (k=31), A159674 (k=32), A077420 (k=34), this sequence (k=36), A097315 (k=38).

Programs

  • Magma
    [n le 2 select 35^(n-1) else 36*Self(n-1)-Self(n-2): n in [1..20]];
    
  • Magma
    R:=PowerSeriesRing(Integers(), 20); Coefficients(R!( (1 - x)/(1 - 36*x + x^2))); // Marius A. Burtea, Jan 14 2020
    
  • Mathematica
    CoefficientList[Series[(1 - x)/(1 - 36 x + x^2), {x, 0, 20}], x] (* or *) LinearRecurrence[{36, -1}, {1, 35}, 20]
  • PARI
    a(n)=([0,1; -1,36]^n*[1;35])[1,1] \\ Charles R Greathouse IV, May 10 2016
  • Sage
    m = 20; L. = PowerSeriesRing(ZZ, m); f = (1-x)/(1-36*x+x^2)
    print(f.coefficients())
    

Formula

G.f.: (1 - x)/(1 - 36*x + x^2).
a(n) = a(-n-1) = 36*a(n-1) - a(n-2).
a(n) = ((19-sqrt(323))/38)*(1+(18+sqrt(323))^(2*n+1))/(18+sqrt(323))^n.
a(n+1) - a(n) = 34*A144128(n+1).
323*a(n+1)^2 - ((a(n+2)-a(n))/2)^2 = 34.
Sum_{n>0} 1/(a(n) - 1/a(n)) = 1/34.
See also Tanya Khovanova in Links field:
a(n) = 35*a(n-1) + 34*Sum_{i=0..n-2} a(i).
a(n+2)*a(n) - a(n+1)^2 = 36-2 = 34 = 34*1,
a(n+3)*a(n) - a(n+1)*a(n+2) = 36*(36-2) = 1224 = 34*36.
Generalizing:
a(n+4)*a(n) - a(n+1)*a(n+3) = 44030 = 34*1295,
a(n+5)*a(n) - a(n+1)*a(n+4) = 1583856 = 34*46584,
a(n+6)*a(n) - a(n+1)*a(n+5) = 56974786 = 34*1675729, etc.,
where 1, 36, 1295, 46584, 1675729, ... is the sequence A144128, which is the second bisection of A041611.
a(n)^2 - 36*a(n)*a(n+1) + a(n+1)^2 + 34 = 0 (see comments by Colin Barker in similar sequences).

A097834 Chebyshev polynomials S(n,27) + S(n-1,27) with Diophantine property.

Original entry on oeis.org

1, 28, 755, 20357, 548884, 14799511, 399037913, 10759224140, 290100013867, 7821941150269, 210902311043396, 5686540457021423, 153325690028535025, 4134107090313424252, 111467565748433919779, 3005490168117402409781
Offset: 0

Views

Author

Wolfdieter Lang, Sep 10 2004

Keywords

Comments

(5*a(n))^2 - 29*b(n)^2 = -4 with b(n) = A097835(n) give all positive solutions of this Pell equation.
From Klaus Purath, Sep 24 2024: (Start)
a(n) = (t(i+2n+1) - t(i))/(t(i+n+1) - t(i+n)), where (t) is any sequence satisfying t(i) = 28t(i-1) - 28t(i-2) + t(i-3) or t(i) = 27t(i-1) - t(i-2) without regard to initial values and including this sequence itself, as long as t(i+n+1) - t(i+n) != 0 for integer i and n >= 0.
a(n) = (Sum_{i .. i+2n} t(i))/t(i+n), where (t) is any recurrence of the form (27,-1) without regard to initial values and including this sequence itself, as long as t(i+n) != 0 for integer i and n >= 0.
a(n) = t(n) - t(n-1) = (t(n+1) - t(n-2))/28, where (t) is any third order recurrence with constant coefficients (28,-28,1) and initial values t(0) = x, t(1) = x + 1, t(2) = x + 29 for any integer x.
a(n) = t(n-1) + t(n) = (t(n-2) + t(n+1))/26, where (t) is any third order recurrence with constant coefficients (26,26,-1) and initial values t(0) = x, t(1) = 1 - x, t(2) = x + 27 for any integer x.
a(n) = (t(i+4n+2) - t(i))/(t(i+2n+2) - t(i+2n)), where (t) is any recurrence of the form (5,1) without regard to initial values, as long as t(i+2n+2) - t(i+2n) != 0 for nonnegative integer i and n. (End)

Examples

			All positive solutions of Pell equation x^2 - 29*y^2 = -4 are
(5=5*1,1), (140=5*28,26), (3775=5*755,701), (101785=5*20357,18901), ...
		

Crossrefs

A087130(2*n + 1) = 5 * a(n). - Michael Somos, Nov 01 2008

Programs

  • Mathematica
    a[n_] := -2/5*I*(-1)^n*ChebyshevT[2*n + 1, 5*I/2]; Table[a[n], {n, 0, 15}] (* Jean-François Alcover, Jun 21 2013, from 2nd formula *)
  • PARI
    {a(n) = (-1)^n * subst(2 * I / 5 * poltchebi(2*n + 1), 'x, -5/2 * I)}; /* Michael Somos, Nov 04 2008 */

Formula

a(n) = S(n, 27) + S(n-1, 27) = S(2*n, sqrt(29)), with S(n, x)=U(n, x/2) Chebyshev's polynomials of the second kind, A049310. S(-1, x) = 0 = U(-1, x). S(n, 27)=A097781(n).
a(n) = (-2/5)*i*((-1)^n)*T(2*n+1, 5*i/2) with the imaginary unit i and Chebyshev's polynomials of the first kind. See the T-triangle A053120.
G.f.: (1+x)/(1-27*x+x^2).
a(n) = - a(-1-n) for all n in Z. - Michael Somos, Nov 01 2008
From Peter Bala, Aug 26 2022: (Start)
a(n) = (2/5)*(5/2 o 5/2 o ... o 5/2) (2*n+1 terms), where the binary operation o is defined on real numbers by x o y = x*sqrt(1 + y^2) + y*sqrt(1 + x^2). The operation o is commutative and associative with identity 0.
The aerated sequence (b(n))n>=1 = [1, 0, 28, 0, 755, 0, 20357, 0, ...], with o.g.f. x*(1 + x^2)/(1 - 27*x^2 + x^4), is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -25, 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,5) + (1/5)*( 1 + (-1)^(n+1) )*F(n+1,5), where F(n,x) is the n-th Fibonacci polynomial - see A168561 (but with row indexing starting at n = 1).
Exp( Sum_{n >= 1} 10*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 10*A052918(n)*x^n.
Exp( Sum_{n >= 1} (-10)*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 10*A052918(n)*(-x)^n.
(End)
a(n) = (a(n-1)*a(n-2) - 783)/a(n-3) for n >= 3. - Klaus Purath, Sep 24 2024
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