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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A098614 Product of Fibonacci and Catalan numbers: a(n) = A000045(n+1)*A000108(n).

Original entry on oeis.org

1, 1, 4, 15, 70, 336, 1716, 9009, 48620, 267410, 1494844, 8465184, 48466796, 280073300, 1631408400, 9568812015, 56466198990, 335002137360, 1997007404700, 11955535480350, 71850862117320, 433322055191220, 2621615826231480, 15906988165723200, 96775058652983100
Offset: 0

Views

Author

Paul D. Hanna, Oct 09 2004

Keywords

Comments

Radius of convergence: r = (sqrt(5)-1)/8; A(r) = sqrt(2+2/sqrt(5)). More generally, given {S} such that: S(n) = b*S(n-1) + c*S(n-2), |b|>0, |c|>0, then Sum_{n>=0} S(n)*Catalan(n)*x^n = sqrt( (1-2*b*x - sqrt(1-4*b*x-16*c*x^2))/(2*b^2+8*c) )/x.
a(n) is also the number of nonnesting permutations of {1,1,2,2,...,n,n} that avoid the patterns 1223, 1332, 2113, or the patterns 1123, 1132, 2133. - Amya Luo, Dec 11 2024

Examples

			Sequence has the factored form: {1*1, 1*1, 2*2, 3*5, 5*14, 8*42, 13*132, 21*429, ...}.
		

Crossrefs

Programs

  • Magma
    [Fibonacci(n+1)*Catalan(n): n in [0..40]]; // G. C. Greubel, Jul 31 2024
    
  • Mathematica
    With[{nn=30},Times@@@Thread[{Fibonacci[Range[nn]],CatalanNumber[ Range[ 0,nn-1]]}]] (* Harvey P. Dale, Nov 14 2011 *)
  • PARI
    {a(n)=local(X=x+O(x^(n+3)), A); A = sqrt( (1-2*x - sqrt(1-4*X-16*x^2)) / (10*x^2)); polcoeff(A, n)}
    for(n=0, 30, print1(a(n), ", "))
    
  • PARI
    {a(n)=binomial(2*n,n)/(n+1)*round(((1+sqrt(5))^(n+1)-(1-sqrt(5))^(n+1))/(2^(n+1)*sqrt(5)))}
    
  • SageMath
    [fibonacci(n+1)*catalan_number(n) for n in range(41)] # G. C. Greubel, Jul 31 2024

Formula

G.f.: A(x) = sqrt( (1-2*x - sqrt(1-4*x-16*x^2))/10 )/x.
G.f. satisfies: A(x) = sqrt( 1 + 2*x*A(x)^2 + 5*x^2*A(x)^4 ).
a(n) == 1 (mod 2) iff n = 2^k - 1 for k>=0.
n*(n+1)*a(n) -2*n*(2*n-1)*a(n-1) -4*(2*n-1)*(2*n-3)*a(n-2) = 0. - R. J. Mathar, Nov 17 2018
Sum_{n>=0} a(n)/8^n = 2*sqrt(2/5). - Amiram Eldar, May 06 2023

A098616 Product of Pell and Catalan numbers: a(n) = A000129(n+1)*A000108(n).

Original entry on oeis.org

1, 2, 10, 60, 406, 2940, 22308, 175032, 1408550, 11561836, 96425836, 814773960, 6960289532, 60012947800, 521582661000, 4564643261040, 40190674554630, 355772529165900, 3164408450118300, 28266363849505320, 253466716153665300, 2280803103062033160, 20588945107316958840
Offset: 0

Views

Author

Paul D. Hanna, Oct 09 2004

Keywords

Comments

Radius of convergence: r = (sqrt(2)-1)/4, where A(r) = sqrt(2+sqrt(2)).
More generally, given {S} such that: S(n) = b*S(n-1) + c*S(n-2), |b|>0, |c|>0, then Sum_{n>=0} S(n)*Catalan(n)*x^n = sqrt( (1-2*b*x - sqrt(1-4*b*x-16*c*x^2))/(2*b^2+8*c) )/x.

Examples

			Sequence begins: [1*1, 2*1, 5*2, 12*5, 29*14, 70*42, 169*132, 408*429,...].
		

Crossrefs

Programs

  • Mathematica
    With[{nn=30},Times@@@Thread[{LinearRecurrence[{2,1},{1,2},nn], CatalanNumber[ Range[0,nn-1]]}]] (* Harvey P. Dale, Jan 04 2012 *)
    a[n_] := Fibonacci[n + 1, 2] * CatalanNumber[n]; Array[a, 25, 0] (* Amiram Eldar, May 05 2023 *)
  • PARI
    a(n) = binomial(2*n,n)/(n+1)*round(((1+sqrt(2))^(n+1)-(1-sqrt(2))^(n+1))/(2*sqrt(2)))

Formula

G.f.: A(x) = sqrt( (1-4*x - sqrt(1-8*x-16*x^2))/16 )/x.
Run lengths of zeros (mod 10) equal (5^k - (-1)^k)/2 - 1 starting at index (5^k + (-1)^k)/2:
a(n) == 0 (mod 10) for n = (5^k + (-1)^k)/2 through n = 5^k - 1 when k>=1.
a(n) ~ 2^(2*n-3/2) * (1+sqrt(2))^(n+1) / (sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, May 09 2014
A(-x) = 1/x * series reversion( x*(2*x + sqrt(1 - 4*x^2)) ). Compare with the o.g.f. B(x) of the central binomial numbers A000984, which satisfies B(-x) = 1/x * series reversion( x*(2*x + sqrt(1 + 4*x^2)) ). See also A214377. - Peter Bala, Oct 19 2015
n*(n+1)*a(n) -4*n*(2*n-1)*a(n-1) -4*(2*n-1)*(2*n-3)*a(n-2)=0. - R. J. Mathar, Nov 17 2018
Sum_{n>=0} a(n)/16^n = 2*sqrt(3-sqrt(7)). - Amiram Eldar, May 05 2023
G.f. A(x) satisfies A(x) = sqrt( 1 + 4*x*A(x)^2 + 8*x^2*A(x)^4 ). - Paul D. Hanna, Dec 14 2024

A112705 Triangle built from partial sums of Catalan numbers A000108 multiplied by powers.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 4, 3, 1, 1, 9, 11, 4, 1, 1, 23, 51, 22, 5, 1, 1, 65, 275, 157, 37, 6, 1, 1, 197, 1619, 1291, 357, 56, 7, 1, 1, 626, 10067, 11497, 3941, 681, 79, 8, 1, 1, 2056, 64979, 107725, 46949, 9431, 1159, 106, 9, 1, 1, 6918, 431059, 1045948, 587621, 140681, 19303, 1821, 137, 10, 1
Offset: 0

Views

Author

Wolfdieter Lang, Oct 31 2005

Keywords

Comments

The column sequences (without leading zeros) begin with A000012 (powers of 1), A112705 (partial sums Catalan), A112696-A112704, for m=0..10.

Examples

			Triangle starts:
  1;
  1, 1;
  1, 2,  1;
  1, 4,  3,   1;
  1, 9,  11,  4,   1;
  1, 23, 51,  22,  5,  1;
  1, 65, 275, 157, 37, 6, 1;
  ...
		

Crossrefs

Row sums give A112706.

Programs

  • Mathematica
    col[m_] := col[m] = CatalanNumber[#]*m^#& /@ Range[0, 20] // Accumulate;
    T[n_, m_] := If[m == 0, 1, col[m][[n - m + 1]]];
    Table[T[n, m], {n, 0, 10}, {m, 0, n}] // Flatten (* Jean-François Alcover, Aug 29 2022 *)
  • PARI
    t(n, m) = if (m==0, 1, if (n==m, 1, sum(kk=0, n-m, m^kk*binomial(2*kk, kk)/(kk+1))));
    tabl(nn) = {for (n=0, nn, for (m=0, n, print1(t(n, m), ", ");); print(););} \\ Michel Marcus, Nov 25 2015

Formula

a(n, m) = sum(C(k)*m^k, k=0..n-m), n>m>0, with C(n):=A000108(n); a(n, n)=1; a(n, 0)=1; a(n, m)=0 if n
G.f. for column m>=0 (without leading zeros): c(m*x)/(1-x), where c(x):=(1-sqrt(1-4*x))/(2*x) is the o.g.f. of Catalan numbers A000108.

A158835 Triangle, read by rows, that transforms diagonals in the array A158825 of coefficients of successive iterations of x*C(x) where C(x) is the Catalan function (A000108).

Original entry on oeis.org

1, 1, 1, 4, 2, 1, 27, 11, 3, 1, 254, 94, 21, 4, 1, 3062, 1072, 217, 34, 5, 1, 45052, 15212, 2904, 412, 50, 6, 1, 783151, 257777, 47337, 6325, 695, 69, 7, 1, 15712342, 5074738, 906557, 116372, 12035, 1082, 91, 8, 1, 357459042, 113775490, 19910808, 2483706
Offset: 1

Author

Paul D. Hanna, Mar 28 2009, Mar 29 2009

Keywords

Comments

Conjecture: n-th reversed row polynomial is t_n where we start with vector v of fixed length m with elements v_i = 1, then set t := v and for i=1..m-1, for j=1..i, for k=j+1..i+1 apply v_k := v_k + z*v_{k-1} and t_{i+1} := v_{i+1} (after ending each cycle for j). - Mikhail Kurkov, Sep 03 2024

Examples

			Triangle T begins:
  1;
  1,1;
  4,2,1;
  27,11,3,1;
  254,94,21,4,1;
  3062,1072,217,34,5,1;
  45052,15212,2904,412,50,6,1;
  783151,257777,47337,6325,695,69,7,1;
  15712342,5074738,906557,116372,12035,1082,91,8,1;
  357459042,113775490,19910808,2483706,246596,20859,1589,116,9,1;
  9094926988,2861365660,492818850,60168736,5801510,470928,33747,2232,144,10,1;
  ...
Array A158825 of coefficients in iterations of x*C(x) begins:
  1,1,2,5,14,42,132,429,1430,4862,16796,58786,208012,742900,...;
  1,2,6,21,80,322,1348,5814,25674,115566,528528,2449746,...;
  1,3,12,54,260,1310,6824,36478,199094,1105478,6227712,...;
  1,4,20,110,640,3870,24084,153306,993978,6544242,43652340,...;
  1,5,30,195,1330,9380,67844,500619,3755156,28558484,...;
  1,6,42,315,2464,19852,163576,1372196,11682348,100707972,...;
  1,7,56,476,4200,38052,351792,3305484,31478628,303208212,...;
  1,8,72,684,6720,67620,693048,7209036,75915708,807845676,...;
  1,9,90,945,10230,113190,1273668,14528217,167607066,...;
  1,10,110,1265,14960,180510,2212188,27454218,344320262,...;
  ...
This triangle transforms diagonals of A158825 into each other:
T*A158831 = A158832; T*A158832 = A158833; T*A158833 = A158834;
where:
A158831 = [1,1,6,54,640,9380,163576,3305484,...];
A158832 = [1,2,12,110,1330,19852,351792,7209036,...];
A158833 = [1,3,20,195,2464,38052,693048,14528217,...];
A158834 = [1,4,30,315,4200,67620,1273668,27454218,...].
		

Crossrefs

Cf. columns: A158836, A158837, A158838, A158839, row sums: A158840.

Programs

  • PARI
    {T(n, k)=local(F=x, CAT=serreverse(x-x^2+x*O(x^(n+2))), M, N, P, m=max(n, k)); M=matrix(m+2, m+2, r, c, F=x; for(i=1, r+c-2, F=subst(F, x, CAT)); polcoeff(F, c)); N=matrix(m+1, m+1, r, c, M[r, c]); P=matrix(m+1, m+1, r, c, M[r+1, c]); (P~*N~^-1)[n+1, k+1]}
    for(n=0,12,for(k=0,n,print1(T(n,k),", "));print(""))

Extensions

Edited by N. J. A. Sloane, Oct 04 2010, to make entries, offset, b-file and link to b-file all consistent.

A236830 Riordan array (1/(1-x*C(x)^3), x*C(x)), C(x) the g.f. of A000108.

Original entry on oeis.org

1, 1, 1, 4, 2, 1, 16, 7, 3, 1, 65, 27, 11, 4, 1, 267, 108, 43, 16, 5, 1, 1105, 440, 173, 65, 22, 6, 1, 4597, 1812, 707, 267, 94, 29, 7, 1, 19196, 7514, 2917, 1105, 398, 131, 37, 8, 1, 80380, 31307, 12111, 4597, 1680, 575, 177, 46, 9, 1, 337284, 130883, 50503, 19196, 7085, 2488, 808, 233, 56, 10, 1
Offset: 0

Author

Philippe Deléham, Feb 01 2014

Keywords

Comments

T(n+3,n) = A011826(n+5).

Examples

			Triangle begins:
      1;
      1,    1;
      4,    2,    1;
     16,    7,    3,    1;
     65,   27,   11,    4,   1;
    267,  108,   43,   16,   5,   1;
   1105,  440,  173,   65,  22,   6,  1;
   4597, 1812,  707,  267,  94,  29,  7, 1;
  19196, 7514, 2917, 1105, 398, 131, 37, 8, 1;
Production matrix is:
   1  1
   3  1   1
   6  1   1   1
  10  1   1   1   1
  15  1   1   1   1   1
  21  1   1   1   1   1   1
  28  1   1   1   1   1   1   1
  36  1   1   1   1   1   1   1   1
  45  1   1   1   1   1   1   1   1   1
  55  1   1   1   1   1   1   1   1   1   1
  66  1   1   1   1   1   1   1   1   1   1   1
  78  1   1   1   1   1   1   1   1   1   1   1   1
  91  1   1   1   1   1   1   1   1   1   1   1   1   1
  ...
		

Programs

  • GAP
    Flat(List([0..12], n-> List([0..n], k-> Sum([0..n-k], j-> Binomial(2*n-k-j-2, n-k-j)*Fibonacci(2*j-1) )))); # G. C. Greubel, Jul 18 2019
  • Magma
    [(&+[Binomial(2*n-k-j-2, n-k-j)*Fibonacci(2*j-1): j in [0..n-k]]): k in [0..n], n in [0..12]]; // G. C. Greubel, Jul 18 2019
    
  • Maple
    A236830 := (n,k) -> add(combinat:-fibonacci(2*i-1)*binomial(2*n-2-k-i,n-k-i), i = 0..n-k): seq(seq(A236830(n, k), k = 0..n), n = 0..10); # Peter Bala, Feb 18 2018
  • Mathematica
    (* The function RiordanArray is defined in A256893. *)
    c[x_] := (1 - Sqrt[1 - 4 x])/(2 x);
    RiordanArray[1/(1 - # c[#]^3)&, # c[#]&, 11] // Flatten (* Jean-François Alcover, Jul 16 2019 *)
    Table[Sum[Binomial[2*n-k-j-2, n-k-j]*Fibonacci[2*j-1], {j,0,n-k}], {n, 0, 12}, {k, 0, n}]//Flatten (* G. C. Greubel, Jul 18 2019 *)
  • PARI
    T(n,k) = sum(j=0,n-k, binomial(2*n-k-j-2, n-k-j)*fibonacci(2*j -1));
    for(n=0,12, for(k=0,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Jul 18 2019
    
  • Sage
    [[sum( binomial(2*n-k-j-2, n-k-j)*fibonacci(2*j -1) for j in (0..n-k) ) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Jul 18 2019
    

Formula

Sum_{k=0..n} T(n,k) = A026726(n).
G.f.: 1/((x^2*C(x)^4-x*C(x))*y-x*C(x)^3+1), where C(x) the g.f. of A000108. - Vladimir Kruchinin, Apr 22 2015
From Peter Bala, Feb 18 2018: (Start)
T(n,k) = Sum_{i = 0..n-k} Fibonacci(2*i-1)*binomial(2*n-2-k-i,n-k-i).
The n-th row polynomial of row reverse triangle is the n-th degree Taylor polynomial of the rational function (1 - 3*x + 2*x^2)/(1 - 3*x + x^2) * 1/(1 - x)^n about 0. For example, for n = 4, (1 - 3*x + 2*x^2)/(1 - 3*x + x^2) * 1/(1 - x)^4 = 1 + 4*x + 11*x^2 + 27*x^3 + 65*x^4 + O(x^5), giving row 4 as (65, 27, 11, 4, 1). (End)

A051960 a(n) = C(n)*(3n+2) where C(n) = Catalan numbers = A000108.

Original entry on oeis.org

2, 5, 16, 55, 196, 714, 2640, 9867, 37180, 140998, 537472, 2057510, 7904456, 30458900, 117675360, 455657715, 1767883500, 6871173870, 26747767200, 104268528210, 406975466040, 1590307356300, 6220814327520, 24357232569150, 95452906901976, 374369872911804
Offset: 0

Author

Barry E. Williams, Jan 05 2000

Keywords

Comments

If Y is a fixed 2-subset of a 2n-set X then a(n-1) is the number of n-subsets of X intersecting Y. - Milan Janjic, Oct 21 2007
a(n-1) is the number of vertices in the n-dimensional halohedron (or equivalently, n-cycle cubeahedron). - Vincent Pilaud, May 12 2020

References

  • Albert H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.

Crossrefs

Cf. A000108 and A051924.
Cf. A024482 and A097613.
Half A028283.

Programs

  • Magma
    [Catalan(n)*(3*n+2): n in [0..30]]; // Vincenzo Librandi, Oct 01 2015
  • Maple
    a := n -> 4^n*(2+3*n)*GAMMA(1/2+n)/(sqrt(Pi)*GAMMA(2+n)):
    seq(a(n), n=0..25); # Peter Luschny, Dec 14 2015
  • Mathematica
    Table[CatalanNumber[n] (3n+2), {n,0,30}] (* Michael De Vlieger, Sep 30 2015 *)
  • Maxima
    a(n):=sum(binomial(n-k+1,k)*2^(n-2*k+1)*binomial(n,k),k,0,(n+1)/2); /* Vladimir Kruchinin, Sep 30 2015 */
    
  • PARI
    a(n) = (3*n+2)*binomial(2*n, n)/(n+1);
    vector(30, n, a(n-1)) \\ Altug Alkan, Sep 30 2015
    

Formula

(n+1)*a(n) - 2*(n+2)*a(n-1) - 4*(2*n-3)*a(n-2) = 0. - conjectured by R. J. Mathar, Oct 02 2014, verified by Robert Israel, Sep 30 2015
G.f.: (1 + 2*x)/(2*x*sqrt(1-4*x)) - 1/(2*x). - Vladimir Kruchinin, Sep 30 2015.
a(n) = Sum_{k=0..(n+1)/2} (binomial(n-k+1,k)*2^(n-2*k+1)*binomial(n,k)). - Vladimir Kruchinin, Sep 30 2015.
a(n) = 4^n*(2+3*n)*Gamma(n + 1/2)/(sqrt(Pi)*Gamma(n+2)). - Peter Luschny, Dec 14 2015
a(n - 1) = A051924(n) + A000108(n - 1). - F. Chapoton, Mar 05 2022
Sum_{n>=0} a(n)/8^n = 5*sqrt(2) - 4. - Amiram Eldar, May 06 2023
E.g.f.: exp(2*x)*(2*BesselI(0,2*x) + BesselI(1,2*x)). - Stefano Spezia, May 14 2025
a(n) = 2*binomial(2*n, n) + binomial(2*n, n-1) = 2*A000984(n) + A001791(n). - Peter Bala, Aug 23 2025

A076036 G.f.: 1/(1 - 5*x*C(x)), where C(x) = (1 - sqrt(1 - 4*x))/(2*x) = g.f. for the Catalan numbers A000108.

Original entry on oeis.org

1, 5, 30, 185, 1150, 7170, 44760, 279585, 1746870, 10916150, 68219860, 426353130, 2664633580, 16653699860, 104084695500, 650526003825, 4065775405350, 25411052086350, 158818913483700, 992617612224750, 6203857867325700, 38774103465635100, 242338116077385600
Offset: 0

Author

N. J. A. Sloane, Oct 29 2002

Keywords

Comments

Numbers have the same parity as the Catalan numbers, that is, a(n) is even except for n of the form 2^m - 1. Follows from C(x) = 1/(1 - x*C(x)) = 1/(1 - 5*x*C(x)) (mod 2). - Peter Bala, Jul 24 2016

Crossrefs

Programs

  • PARI
    C(x) = (1 - sqrt(1 - 4*x))/(2*x);
    my(x = 'x + O('x^25)); Vec(1/(1 - 5*x*C(x))) \\ Michel Marcus, Jan 21 2020

Formula

a(n) = Sum_{k = 0..n} A106566(n, k)*5^k. - Philippe Deléham, Sep 01 2005
a(n) = Sum{k = 0..n} A039599(n,k)*4^k. - Philippe Deléham, Sep 08 2007
a(0) = 1, a(n) = (25*a(n-1) - 5*A000108(n-1))/4 for n >= 1. - Philippe Deléham, Nov 27 2007
a(n) = Sum_{k = 0..n} A116395(n,k)*3^k. - Philippe Deléham, Sep 27 2009
D-finite with recurrence: +4*n*a(n) +(-41*n+24)*a(n-1) +50*(2*n-3)*a(n-2)=0. - R. J. Mathar, Jan 20 2020
a(n) = 5*A076025(n), n>0. - R. J. Mathar, Jan 20 2020

A132460 Irregular triangle read by rows of the initial floor(n/2) + 1 coefficients of 1/C(x)^n, where C(x) is the g.f. of the Catalan sequence (A000108).

Original entry on oeis.org

1, 1, 1, -2, 1, -3, 1, -4, 2, 1, -5, 5, 1, -6, 9, -2, 1, -7, 14, -7, 1, -8, 20, -16, 2, 1, -9, 27, -30, 9, 1, -10, 35, -50, 25, -2, 1, -11, 44, -77, 55, -11, 1, -12, 54, -112, 105, -36, 2, 1, -13, 65, -156, 182, -91, 13, 1, -14, 77, -210, 294, -196, 49, -2
Offset: 0

Author

Paul D. Hanna, Aug 21 2007

Keywords

Comments

The length of row n is A008619(n).
Essentially equals a signed version of A034807, the triangle of Lucas polynomials. The initial n coefficients of 1/C(x)^n consist of row n followed by floor((n-1)/2) zeros for n > 0.
For the following formula for 1/C(x)^n see the W. Lang reference, proposition 1 on p. 411:
1/C(x)^n = (sqrt(x))^n*(S(n,1/sqrt(x)) - sqrt(x)*S(n-1,1/sqrt(x))*C(x)), n >= 0, with the Chebyshev polynomials S(n,x) with coefficients given in A049310. See also the coefficient array A115139 for P(n,x) = (sqrt(x)^(n-1))*S(n-1, 1/sqrt(x)). - Wolfdieter Lang, Sep 14 2013
This triangular array is composed of interleaved rows of reversed, A127677 (cf. A156308, A217476, A263916) and reversed, signed A111125. - Tom Copeland, Nov 07 2015
It seems that the n-th row lists the coefficients of the HOMFLYPT (HOMFLY) polynomial reduced to one variable for link family n, see Jablan's slide 38. - Andrey Zabolotskiy, Jan 16 2018
For n >= 1 row n gives the coefficients of the Girard-Waring formula for the sum of x1^n + x2^n in terms of the elementary symmetric functions e_1(x1,x2) = x1 + x2 and e_2(x1,x2) = x1*x2. This is an array using the partitions of n, in the reverse Abramowitz-Stegun order, with all partitions with parts larger than 2 eliminated. E.g., n = 4: x1^4 + x2^4 = 1*e1^4 - 4*e1^3*e2 + 2*e1*e2^2. See also A115131, row n = 4, with the mentioned partitions omitted. - Wolfdieter Lang, May 03 2019
Row n lists the coefficients of the n-th Faber polynomial for the replicable function given in A154272 with offset -1. - Ben Toomey, May 12 2020

Examples

			The irregular triangle T(n,k) begins:
n\k 0    1    2    3    4    5    6   7 ...
-------------------------------------------------
0:  1
1:  1
2:  1   -2
3:  1   -3
4:  1   -4    2
5:  1   -5    5
6:  1   -6    9   -2
7:  1   -7   14   -7
8:  1   -8   20  -16    2
9:  1   -9   27  -30    9
10: 1  -10   35  -50   25   -2
11: 1  -11   44  -77   55  -11
12: 1  -12   54 -112  105  -36    2
13: 1  -13   65 -156  182  -91   13
14: 1  -14   77 -210  294 -196   49  -2
... (reformatted - _Wolfdieter Lang_, May 03 2019)
		

Crossrefs

Cf. A000108, A008619, A034807 (Lucas polynomials), A111125, A115131 (Waring numbers), A127677, A132461 (row squared sums), A156308, A217476, A263916.

Programs

  • Mathematica
    T[0, 0] = 1; T[n_, k_] := (-1)^k (Binomial[n-k, k] + Binomial[n-k-1, k-1]);
    Table[T[n, k], {n, 0, 14}, {k, 0, n/2}] // Flatten (* Jean-François Alcover, Jun 04 2018 *)
  • PARI
    {T(n,k)=if(k>n\2,0,(-1)^k*(binomial(n-k, k)+binomial(n-k-1, k-1)))}

Formula

T(n,k) = (-1)^k*( C(n-k,k) + C(n-k-1,k-1) ) for n >= 0, 0 <= k <= floor(n/2).
T(0,0) = 1; T(n,k) = (-1)^k*n*binomial(n-k,k)/(n-k), k = 0..floor(n/2). - Wolfdieter Lang, May 03 2019

A176137 Number of partitions of n into distinct Catalan numbers, cf. A000108.

Original entry on oeis.org

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

Author

Reinhard Zumkeller, Apr 09 2010

Keywords

Comments

a(n) <= 1;
a(A000108(n)) = 1; a(A141351(n)) = 1; a(A014138(n)) = 1.
A197433 gives all such numbers k that a(k) = 1, in other words, this is the characteristic function of A197433, and all three sequences mentioned above are its subsequences. - Antti Karttunen, Jun 25 2014

Examples

			56 = 42+14 = A000108(5)+A000108(4), all other sums of distinct Catalan numbers are not equal 56, therefore a(56)=1.
		

Crossrefs

When right-shifted (prepended with 1) this sequence is the first differences of A244230.

Programs

Formula

a(n) = f(n,1,1) with f(m,k,c) = if c>m then 0^m else f(m-c,k+1,c') + f(m,k+1,c') where c'=2*c*(2*k+1)/(k+2).

A246596 Run Length Transform of Catalan numbers A000108.

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 2, 5, 1, 1, 1, 2, 2, 2, 5, 14, 1, 1, 1, 2, 1, 1, 2, 5, 2, 2, 2, 4, 5, 5, 14, 42, 1, 1, 1, 2, 1, 1, 2, 5, 1, 1, 1, 2, 2, 2, 5, 14, 2, 2, 2, 4, 2, 2, 4, 10, 5, 5, 5, 10, 14, 14, 42, 132, 1, 1, 1, 2, 1, 1, 2, 5, 1, 1, 1, 2, 2, 2, 5, 14, 1, 1, 1, 2, 1, 1, 2, 5
Offset: 0

Author

N. J. A. Sloane, Sep 06 2014

Keywords

Comments

The Run Length Transform of a sequence {S(n), n>=0} is defined to be the sequence {T(n), n>=0} given by T(n) = Product_i S(i), where i runs through the lengths of runs of 1's in the binary expansion of n. E.g. 19 is 10011 in binary, which has two runs of 1's, of lengths 1 and 2. So T(19) = S(1)*S(2). T(0)=1 (the empty product).

Examples

			From _Omar E. Pol_, Feb 15 2015: (Start)
Written as an irregular triangle in which row lengths are the terms of A011782:
1;
1;
1,2;
1,1,2,5;
1,1,1,2,2,2,5,14;
1,1,1,2,1,1,2,5,2,2,2,4,5,5,14,42;
1,1,1,2,1,1,2,5,1,1,1,2,2,2,5,14,2,2,2,4,2,2,4,10,5,5,5,10,14,14,42,132;
...
Right border gives the Catalan numbers. This is simply a restatement of the theorem that this sequence is the Run Length Transform of A000108.
(End)
		

Crossrefs

Cf. A000108.
Cf. A003714 (gives the positions of ones).
Run Length Transforms of other sequences: A005940, A069739, A071053, A227349, A246588, A246595, A246660, A246661, A246674.

Programs

  • Maple
    Cat:=n->binomial(2*n,n)/(n+1);
    ans:=[];
    for n from 0 to 100 do lis:=[]; t1:=convert(n, base, 2); L1:=nops(t1); out1:=1; c:=0;
    for i from 1 to L1 do
    if out1 = 1 and t1[i] = 1 then out1:=0; c:=c+1;
    elif out1 = 0 and t1[i] = 1 then c:=c+1;
    elif out1 = 1 and t1[i] = 0 then c:=c;
    elif out1 = 0 and t1[i] = 0 then lis:=[c, op(lis)]; out1:=1; c:=0;
    fi;
    if i = L1 and c>0 then lis:=[c, op(lis)]; fi;
    od:
    a:=mul(Cat(i), i in lis);
    ans:=[op(ans), a];
    od:
    ans;
  • Mathematica
    f = CatalanNumber; Table[Times @@ (f[Length[#]]&) /@ Select[ Split[ IntegerDigits[n, 2]], #[[1]] == 1&], {n, 0, 87}] (* Jean-François Alcover, Jul 11 2017 *)
  • Python
    from operator import mul
    from functools import reduce
    from gmpy2 import divexact
    from re import split
    def A246596(n):
        s, c = bin(n)[2:], [1, 1]
        for m in range(1, len(s)):
            c.append(divexact(c[-1]*(4*m+2),(m+2)))
        return reduce(mul,(c[len(d)] for d in split('0+',s))) if n > 0 else 1
    # Chai Wah Wu, Sep 07 2014
    
  • Sage
    # uses[RLT from A246660]
    A246596_list = lambda len: RLT(lambda n: binomial(2*n, n)/(n+1), len)
    A246596_list(88) # Peter Luschny, Sep 07 2014
    
  • Scheme
    ; using MIT/GNU Scheme
    (define (A246596 n) (fold-left (lambda (a r) (* a (A000108 r))) 1 (bisect (reverse (binexp->runcount1list n)) (- 1 (modulo n 2)))))
    (define A000108 (EIGEN-CONVOLUTION 1 *))
    ;; Note: EIGEN-CONVOLUTION can be found from my IntSeq-library and other functions are as in A227349. - Antti Karttunen, Sep 08 2014

Formula

a(n) = A069739(A005940(1+n)). - Antti Karttunen, May 29 2017
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