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)

A054413 a(n) = 7*a(n-1) + a(n-2), with a(0)=1 and a(1)=7.

Original entry on oeis.org

1, 7, 50, 357, 2549, 18200, 129949, 927843, 6624850, 47301793, 337737401, 2411463600, 17217982601, 122937341807, 877779375250, 6267392968557, 44749530155149, 319514104054600, 2281348258537349, 16288951913816043, 116304011655249650, 830417033500563593
Offset: 0

Views

Author

Henry Bottomley, May 10 2000

Keywords

Comments

In general, sequences with recurrence a(n) = k*a(n-1) + a(n-2) and a(0)=1 (and a(-1)=0) have the generating function 1/(1-k*x-x^2). If k is odd (k>=3) they satisfy a(3n) = b(5n), a(3n+1) = b(5n+3), a(3n+2) = 2*b(5n+4) where b(n) is the sequence of denominators of continued fraction convergents to sqrt(k^2+4). [If k is even then a(n) is the sequence of denominators of continued fraction convergents to sqrt(k^2/4+1).]
a(p-1) == 53^((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]
From Johannes W. Meijer, Jun 12 2010: (Start)
For the sequence given above k=7 which implies that it is associated with A041091.
For a similar statement about sequences with recurrence a(n) = k*a(n-1) + a(n-2) but with a(0) = 2, and a(-1) = 0, see A086902; a sequence that is associated with A041090.
For more information follow the Khovanova link and see A087130, A140455 and A178765.
(End)
For positive n, a(n) equals the permanent of the n X n tridiagonal matrix with 7'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,...,7} avoiding runs of zeros of odd lengths. - Milan Janjic, Jan 28 2015
From Michael A. Allen, Feb 21 2023: (Start)
Also called the 7-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 7 kinds of squares available. (End)

Crossrefs

Row n=7 of A073133, A172236 and A352361.
Cf. A099367 (squares).

Programs

Formula

a(3n) = A041091(5n), a(3n+1) = A041091(5n+3), a(3n+2) = 2*A041091(5n+4).
G.f.: 1/(1 - 7x - x^2).
a(n) = U(n, 7*i/2)*(-i)^n with i^2=-1 and Chebyshev's U(n, x/2) = S(n, x) polynomials. See A049310.
a(n) = F(n, 7), the n-th Fibonacci polynomial evaluated at x=7. - T. D. Noe, Jan 19 2006
From Sergio Falcon, Sep 24 2007: (Start)
a(n) = (sigma^n - (-sigma)^(-n))/(sqrt(53)) with sigma = (7+sqrt(53))/2;
a(n) = Sum_{i=0..floor((n-1)/2)} binomial(n-1-i,i)*7^(n-1-2i). (End)
a(n) = ((7 + sqrt(53))^n - (7 - sqrt(53))^n)/(2^n*sqrt(53)). Offset 1. a(3)=50. - Al Hakanson (hawkuu(AT)gmail.com), Jan 17 2009
From Johannes W. Meijer, Jun 12 2010: (Start)
a(2n+1) = 7*A097836(n), a(2n) = A097838(n).
Lim_{k->oo} a(n+k)/a(k) = (A086902(n) + A054413(n-1)*sqrt(53))/2.
Lim_{n->oo} A086902(n)/A054413(n-1) = sqrt(53).
(End)
Sum_{n>=0} (-1)^n/(a(n)*a(n+1)) = (sqrt(53)-7)/2. - Vladimir Shevelev, Feb 23 2013
From Kai Wang, Feb 24 2020: (Start)
Sum_{m>=0} 1/(a(m)*a(m+2)) = 1/49.
Sum_{m>=0} 1/(a(2*m)*a(2*m+2)) = (sqrt(53)-7)/14.
In general, for sequences with recurrence f(n)= k*f(n-1)+f(n-2) and f(0)=1,
Sum_{m>=0} 1/(f(m)*f(m+2)) = 1/(k^2).
Sum_{m>=0} 1/(f(2*m)*f(2*m+2)) = (sqrt(k^2+4) - k)/(2*k). (End)
E.g.f.: (1/53)*exp(7*x/2)*(53*cosh(sqrt(53)*x/2) + 7*sqrt(53)*sinh(sqrt(53)*x/2)). - Stefano Spezia, Feb 26 2020
G.f.: x/(1 - 7*x - x^2) = Sum_{n >= 0} x^(n+1) *( Product_{k = 1..n} (m*k + 7 - m + x)/(1 + m*k*x) ) for arbitrary m (a telescoping series). - Peter Bala, May 08 2024

Extensions

Formula corrected by Johannes W. Meijer, May 30 2010, Jun 02 2010
Extended by T. D. Noe, May 23 2011

A097837 Chebyshev polynomials S(n,51) + S(n-1,51) with Diophantine property.

Original entry on oeis.org

1, 52, 2651, 135149, 6889948, 351252199, 17906972201, 912904330052, 46540213860451, 2372638002552949, 120957997916339948, 6166485255730784399, 314369790044353664401, 16026692807006306100052, 817046963367277257438251
Offset: 0

Views

Author

Wolfdieter Lang, Sep 10 2004

Keywords

Comments

(7*a(n))^2 - 53*b(n)^2 = -4 with b(n)=A097838(n) gives all positive solutions of this Pell equation.

Examples

			All positive solutions of Pell equation x^2 - 53*y^2 = -4 are (7=7*1,1), (364=7*52,50), (18557=7*2651,2549), (946043=7*135149,129949), ...
		

Crossrefs

Programs

  • GAP
    a:=[1,52];; for n in [3..30] do a[n]:=51*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jan 12 2019
  • Magma
    m:=30; R:=PowerSeriesRing(Integers(), m); Coefficients(R!( (1+x)/(1-51*x+x^2) )); // G. C. Greubel, Jan 12 2019
    
  • Mathematica
    LinearRecurrence[{51,-1}, {1,52}, 30] (* G. C. Greubel, Jan 12 2019 *)
  • PARI
    my(x='x+O('x^30)); Vec((1+x)/(1-51*x+x^2)) \\ G. C. Greubel, Jan 12 2019
    
  • Sage
    ((1+x)/(1-51*x+x^2)).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Jan 12 2019
    

Formula

a(n) = S(n, 51) + S(n-1, 51) = S(2*n, sqrt(53)), 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, 51)=A097836(n).
a(n) = (-2/7)*i*((-1)^n)*T(2*n+1, 7*i/2) with the imaginary unit i and Chebyshev's polynomials of the first kind. See the T-triangle A053120.
G.f.: (1+x)/(1-51*x+x^2).
a(n) = 51*a(n-1) - a(n-2); a(0)=1, a(1)=52. - Philippe Deléham, Nov 18 2008
From Peter Bala, Aug 26 2022: (Start)
a(n) = (2/7)*(7/2 o 7/2 o ... o 7/2) (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, 52, 0, 2651, 0, 135149, 0, ...], with o.g.f. x*(1 + x^2)/(1 - 51*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 = -49, 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,7) + (1/7)*( 1 + (-1)^(n+1) )*F(n+1,7), where F(n,x) is the n-th Fibonacci polynomial - see A168561 (but with row indexing starting at n = 1).
Exp( Sum_{n >= 1} 14*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 14*A054413(n)*x^n.
Exp( Sum_{n >= 1} (-14)*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 14*A054413(n)*(-x)^n. (End)

A097836 Chebyshev polynomials S(n,51).

Original entry on oeis.org

1, 51, 2600, 132549, 6757399, 344494800, 17562477401, 895341852651, 45644872007800, 2326993130545149, 118631004785794799, 6047854250944989600, 308321935793408674801, 15718370871212897425251, 801328592496064360013000
Offset: 0

Views

Author

Wolfdieter Lang, Sep 10 2004

Keywords

Comments

Used for all positive integer solutions of Pell equation x^2 - 53*y^2 = -4. See A097837 with A097838.
a(n-1), with a(-1) := 0, and b(n) := A099368(n) give the proper and improper nonnegative solutions of the Pell equation b(n)^2 - 53*(7*a(n-1))^2 = +4, n >= 0. - Wolfdieter Lang, Jun 27 2013

Programs

  • GAP
    a:=[1,51];; for n in [2..30] do a[n]:=51*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jan 12 2019
  • Magma
    m:=30; R:=PowerSeriesRing(Integers(), m); Coefficients(R!( 1/(1-51*x+x^2) )); // G. C. Greubel, Jan 12 2019
    
  • Mathematica
    LinearRecurrence[{51,-1}, {1,51}, 30] (* G. C. Greubel, Jan 12 2019 *)
  • PARI
    my(x='x+O('x^30)); Vec(1/(1-51*x+x^2)) \\ G. C. Greubel, Jan 12 2019
    
  • Sage
    (1/(1-51*x+x^2)).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Jan 12 2019
    

Formula

a(n) = S(n, 51)=U(n, 51/2)= S(2*n+1, sqrt(53))/sqrt(53) with S(n, x) = U(n, x/2) Chebyshev's polynomials of the second kind, A049310. S(-1, x)= 0 = U(-1, x).
a(n) = 51*a(n-1) - a(n-2), n >= 1, a(-1)=0, a(0)=1, a(1)=51.
a(n) = (ap^(n+1) - am^(n+1))/(ap-am) with ap := (51+7*sqrt(53))/2 and am := (51-7*sqrt(53))/2 = 1/ap.
G.f.: 1/(1-51*x+x^2).
Showing 1-4 of 4 results.