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 1-10 of 248 results. Next

A383991 Series expansion of the exponential generating function exp(-tridend(-x)) - 1 where tridend(x) = (1 - 3*x - sqrt(1-6*x+x^2)) / (4*x) (A001003).

Original entry on oeis.org

0, 1, -5, 49, -743, 15421, -407909, 13135165, -498874991, 21838772377, -1082819193029, 59983280191561, -3671752681190615, 246130081055714389, -17932045676505509093, 1410893903131294766101, -119227840965746009631839, 10769985399394862863318705
Offset: 0

Views

Author

Michael De Vlieger, May 16 2025

Keywords

Comments

The series -tridend(-x) is the inverse for the substitution of the series trias(x), given by the suspension of the Koszul dual of trias. - Bérénice Delcroix-Oger, May 28 2025

Crossrefs

Programs

  • Mathematica
    nn = 19; f[x_] := Exp[x] - 1;
    Range[0, nn]! * CoefficientList[Series[f[(1 + 3*x - Sqrt[1 + 6*x + x^2])/(4*x)], {x, 0, nn}], x]

A186826 Riordan array (s(x),x*S(x)) where s(x) is the g.f. of the little Schroeder numbers A001003, and S(x) is the g.f. of the large Schroeder numbers A006318.

Original entry on oeis.org

1, 1, 1, 3, 3, 1, 11, 11, 5, 1, 45, 45, 23, 7, 1, 197, 197, 107, 39, 9, 1, 903, 903, 509, 205, 59, 11, 1, 4279, 4279, 2473, 1061, 347, 83, 13, 1, 20793, 20793, 12235, 5483, 1949, 541, 111, 15, 1, 103049, 103049, 61463, 28435, 10717, 3285, 795, 143, 17, 1, 518859, 518859, 312761, 148249, 58351, 19199, 5197, 1117, 179, 19, 1
Offset: 0

Views

Author

Paul Barry, Feb 27 2011

Keywords

Comments

Reverse of A144944. Inverse of A186827.

Examples

			Triangle begins
       1;
       1,      1;
       3,      3,      1;
      11,     11,      5,      1;
      45,     45,     23,      7,     1;
     197,    197,    107,     39,     9,     1;
     903,    903,    509,    205,    59,    11,    1;
    4279,   4279,   2473,   1061,   347,    83,   13,    1;
   20793,  20793,  12235,   5483,  1949,   541,  111,   15,   1;
  103049, 103049,  61463,  28435, 10717,  3285,  795,  143,  17,  1;
  518859, 518859, 312761, 148249, 58351, 19199, 5197, 1117, 179, 19, 1;
Production matrix of this triangle begins
  1, 1;
  2, 2, 1;
  2, 2, 2, 1;
  2, 2, 2, 2, 1;
  2, 2, 2, 2, 2, 1;
  2, 2, 2, 2, 2, 2, 1;
  2, 2, 2, 2, 2, 2, 2, 1;
  2, 2, 2, 2, 2, 2, 2, 2, 1;
  2, 2, 2, 2, 2, 2, 2, 2, 2, 1;
For instance, 107=1*45+2*23+2*7+2*1.
		

Crossrefs

Cf. A001003, A006318, A010683 (row sums), A144944 (row reverse), A186827 (inverse), A186828 (diagonal sums), A239204.

Programs

  • Haskell
    a186826 n k = a186826_tabl !! n !! k
    a186826_row n = a186826_tabl !! n
    a186826_tabl = map reverse a144944_tabl
    -- Reinhard Zumkeller, May 11 2013
    
  • Mathematica
    t[, 0]=1; t[p, p_]:= t[p, p]= t[p, p-1]; t[p_, q_]:= t[p, q]= t[p, q -1] + t[p-1, q] + t[p-1, q-1];
    Table[t[p, q], {p,0,10}, {q,p,0,-1}]//Flatten (* Jean-François Alcover, Jul 16 2019 *)
  • SageMath
    @CachedFunction
    def t(n,k):
        if (k<0 or k>n): return 0
        elif (k==0): return 1
        elif (kA186826(n,k): return t(n+2,n-k)
    flatten([[A186826(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Mar 11 2023

Formula

Riordan array ((1+x+sqrt(1-6*x+x^2))/(4*x), (1-x-sqrt(1-6*x+x^2))/2).
Sum_{k=0..n} T(n,k) = A010683(n).
Sum_{k=0..floor(n/2)} T(n-k, k) = A186828(n).
R(n,k) = k*Sum_{i=0..n-k} (A001003(i)/(n-i))*Sum_{m=0..n-k-i} binomial(n-i,m)*binomial(2*(n-i)-m-k-1, n-i-1), k>0, R(n,0) = A001003(n). - Vladimir Kruchinin, Mar 09 2011
Sum_{k=0..n} (-1)^k*T(n, k) = A239204(n-2). - G. C. Greubel, Mar 11 2023

A383987 Series expansion of the exponential generating function -tridend(-(1-exp(x))) where tridend(x) = (1 - 3*x - sqrt(1+6*x+x^2)) / (4*x) (A001003).

Original entry on oeis.org

0, 1, -5, 49, -725, 14401, -360005, 10863889, -384415925, 15612336481, -715930020005, 36592369889329, -2062911091119125, 127170577711282561, -8510569547826528005, 614491222512504748369, -47615614242877583230325, 3941408640018910366196641
Offset: 0

Views

Author

Michael De Vlieger, May 16 2025

Keywords

Crossrefs

Composition of A001003 with exp(x)-1.

Programs

  • Mathematica
    nn = 17; f[x_] := (1 + 3*x - Sqrt[1 + 6*x + x^2])/(4*x); Range[0, nn]! * CoefficientList[Series[f[-(1 - Exp[x])], {x, 0, nn}], x]

A092840 Primes in A001003.

Original entry on oeis.org

3, 11, 197, 103049
Offset: 1

Views

Author

Eric W. Weisstein, Mar 07 2004

Keywords

Comments

The next term is too large to include.

Crossrefs

Programs

  • PARI
    s(m)= { if (m==1, return(a1)); if (m==2, return(a2)); r = (3*(2*m - 3)*a2 - (m - 3)*a1)/m; a1=a2; a2=r; return(r); }
    { a1=1; a2=1; n=0; for (m=1, 300, a=s(m); if (isprime(a), n++; print1(a, ", "))); } \\ Harry J. Smith, Jun 21 2009

A104858 Partial sums of the little Schroeder numbers (A001003).

Original entry on oeis.org

1, 2, 5, 16, 61, 258, 1161, 5440, 26233, 129282, 648141, 3294864, 16943733, 87983106, 460676625, 2429478144, 12893056497, 68802069506, 368961496469, 1987323655056, 10746633315501, 58321460916482, 317537398625945
Offset: 0

Views

Author

Emeric Deutsch, Apr 24 2005

Keywords

Comments

The subsequence of primes begins: 2, 5, 61, no more through a(30). [Jonathan Vos Post, Feb 12 2010]

Crossrefs

Programs

  • Maple
    G:=(1+z-sqrt(1-6*z+z^2))/4/z/(1-z): Gser:=series(G,z=0,29): 1,seq(coeff(Gser,z^n),n=1..27);
  • Mathematica
    CoefficientList[Series[(1+x-Sqrt[1-6*x+x^2])/4/x/(1-x), {x, 0, 20}], x] (* Vaclav Kotesovec, Oct 17 2012 *)

Formula

G.f.: (1 + z- sqrt(1 - 6*z + z^2))/(4*z*(1 - z)).
Recurrence: (n+1)*a(n) = (7*n-2)*a(n-1) - (7*n-5)*a(n-2) + (n-2)*a(n-3). - Vaclav Kotesovec, Oct 17 2012
a(n) ~ sqrt(24 + 17*sqrt(2))*(3 + 2*sqrt(2))^n/(8*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 17 2012
Define a triangle T(n,1) = T(n,n) = 1 for n >= 1 and all other elements by T(r,c) = T(r,c-1) + T(r-1,c-1) + T(r-1,c). Its second column is A005408, its third column is A059993, and the sum of all terms in its row n is a(n-1). - J. M. Bergot, Dec 01 2012

A176479 a(n) = (n+1)*A001003(n).

Original entry on oeis.org

1, 2, 9, 44, 225, 1182, 6321, 34232, 187137, 1030490, 5707449, 31760676, 177435297, 994551222, 5590402785, 31500824304, 177880832001, 1006362234162, 5703029112297, 32367243171740, 183945502869345, 1046646207221582, 5961966567317649, 33995080211156904
Offset: 0

Views

Author

Paul Barry, Apr 18 2010

Keywords

Comments

Central coefficients T(2n,n) of the Riordan array ((1-x)/(1-2x), x(1-x)/(1-2x)), A105306.
a(n) counts the bi-degree sequences of directed trees (i.e., digraphs whose underlying graph is a tree) with n edges. - Nikos Apostolakis, Dec 31 2016
a(n) is also the number of Dyck paths having exactly n peaks in level 1 and n peaks in level 2 and no other peaks. a(2) = 9: /\/\//\/\\, /\//\/\\/\, //\/\\/\/\, /\/\//\\//\\, /\//\\/\//\\, /\//\\//\\/\, //\\/\/\//\\, //\\/\//\\/\, //\\//\\/\/\. - Alois P. Heinz, Jun 20 2017
For n>0, a(n) is the number of ordered trees with n+1 leaves, no node of outdegree 1, and having one of its leaves marked. - Juan B. Gil, Jan 03 2024

Crossrefs

Row n=2 of A288972.

Programs

  • Maple
    a:= proc(n) option remember; `if`(n<2, n+1,
         (6*n-3)/n*a(n-1) -(n-2)/(n-1)*a(n-2))
        end:
    seq(a(n), n=0..25);  # Alois P. Heinz, Jun 22 2017
  • Mathematica
    a[n_] := Sum[Binomial[n - 1, k - 1]*Binomial[n + k, n], {k, 0, n}]; Array[a, 25, 0] (* or *)
    CoefficientList[ Series[1/4 - (x - 3)/(4 Sqrt[x^2 - 6x +1]), {x, 0, 25}], x] (* Robert G. Wilson v, Dec 31 2016 *)
    Table[(n+1)Hypergeometric2F1[1-n, -n, 2, 2], {n,0,21}] (* Peter Luschny, Jan 02 2017 *)

Formula

E.g.f.: 1+exp(3*x)*Bessel_I(1,2*sqrt(2)*x)/sqrt(2) +int(exp(3*x) *Bessel_I(1,2*sqrt(2)*x) /(sqrt(2)*x),x).
G.f.: 1/4 - (x-3)/(4*sqrt(x^2-6*x+1)). - Dmitry Kruchinin, Aug 31 2012
Conjecture: n*(n-1)*a(n) -3*(2*n-1)*(n-1)*a(n-1) +n*(n-2)*a(n-2) = 0. - R. J. Mathar, Dec 03 2014
a(n) = Sum_{k=0..n} binomial(n-1,n-k) * binomial(n+k,n). - Nikos Apostolakis, Dec 31 2016
a(n) = (n+1)*hypergeom([1-n, -n], [2], 2). - Peter Luschny, Jan 02 2017

A092839 Indices of primes in A001003.

Original entry on oeis.org

3, 4, 6, 10, 216
Offset: 1

Views

Author

Eric W. Weisstein, Mar 07 2004

Keywords

Comments

a(6), if it exists, is > 1.5*10^6. - Robert Price, Apr 16 2014

Crossrefs

Programs

  • PARI
    s(m)= { if (m==1, return(a1)); if (m==2, return(a2)); r = (3*(2*m - 3)*a2 - (m - 3)*a1)/m; a1=a2; a2=r; return(r); }
    { a1=1; a2=1; n=0; for (m=1, 300, a=s(m); if (isprime(a), n++; print1(m, ", "))); } \\ Harry J. Smith, Jun 21 2009

A162548 A Chebyshev transform of the little Schroeder numbers A001003.

Original entry on oeis.org

1, 1, 2, 9, 37, 156, 695, 3203, 15118, 72739, 355475, 1759624, 8804341, 44457125, 226256114, 1159387253, 5976713665, 30974296468, 161285018771, 843388543471, 4427120165182, 23319497761799, 123221525405447, 652989260163472
Offset: 0

Views

Author

Paul Barry, Jul 05 2009

Keywords

Comments

Hankel transform is Somos-4 variant A162546.

Crossrefs

Programs

  • Magma
    R:=PowerSeriesRing(Rationals(), 30); Coefficients(R!( (1+x+x^2 -Sqrt(1-6*x+3*x^2-6*x^3+x^4))/(4*x*(1+x^2)) )); // G. C. Greubel, Feb 26 2019
    
  • Mathematica
    CoefficientList[Series[(1+x+x^2 -Sqrt[1-6*x+3*x^2-6*x^3+x^4])/(4*x*(1+x^2)), {x,0,30}], x] (* G. C. Greubel, Feb 26 2019 *)
  • PARI
    my(x='x+O('x^30)); Vec((1+x+x^2 -sqrt(1-6*x+3*x^2-6*x^3+x^4))/( 4*x*(1+x^2))) \\ G. C. Greubel, Feb 26 2019
    
  • Sage
    ((1+x+x^2 -sqrt(1-6*x+3*x^2-6*x^3+x^4))/(4*x*(1+x^2))).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Feb 26 2019

Formula

G.f.: (1/(1+x^2))*s(x/(1+x^2)), s(x) the g.f. of A001003.
G.f.: (1+x+x^2 - sqrt(1-6*x+3*x^2-6*x^3+x^4))/(4*x*(1+x^2)).
G.f.: 1/(1+x^2-x/(1-2*x/(1+x^2-x/(1-2*x/(1+x^2-x/(1-2*x/(1+x+x^2/(1-... (continued fraction);
a(n) = Sum_{k=0..floor(n/2)} (-1)^k*C(n-k,k)*A001003(n-2k).
Conjecture: (n+1)*a(n) +3*(-2*n+1)*a(n-1) +(4*n-5)*a(n-2) +12*(-n+2)*a(n-3) +(4*n-11)*a(n-4) +3*(-2*n+7)*a(n-5) +(n-5)*a(n-6)=0. - R. J. Mathar, Nov 15 2012. (Formula verified and used for computations. - Fung Lam, Feb 19 2014)

A172094 The Riordan square of the little Schröder numbers A001003.

Original entry on oeis.org

1, 1, 1, 3, 4, 1, 11, 17, 7, 1, 45, 76, 40, 10, 1, 197, 353, 216, 72, 13, 1, 903, 1688, 1145, 458, 113, 16, 1, 4279, 8257, 6039, 2745, 829, 163, 19, 1, 20793, 41128, 31864, 15932, 5558, 1356, 222, 22, 1, 103049, 207905, 168584, 90776, 35318, 10070, 2066, 290, 25, 1
Offset: 0

Views

Author

Philippe Deléham, Jan 25 2010

Keywords

Comments

The Riordan square is defined in A321620.
Previous name was: Triangle, read by rows, given by [1,2,1,2,1,2,1,2,1,2,1,2,...] DELTA [1,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938.
Riordan array (f(x), f(x)-1) where f(x) is the g.f. of A001003. Equals A122538*A007318.

Examples

			Triangle begins:
     1
     1,      1
     3,      4,      1
    11,     17,      7,     1
    45,     76,     40,    10,     1
   197,    353,    216,    72,    13,     1
   903,   1688,   1345,   458,   113,    16,    1
  4279,   8257,   6039,  2745,   829,   163,   19,   1
20793,  41128,  31864, 15932,  5558,  1356,  222,  22,  1
103049, 207905, 168584, 90776, 35318, 10070, 2066, 290, 25, 1
.
Production matrix begins:
1, 1
2, 3, 1
0, 2, 3, 1
0, 0, 2, 3, 1
0, 0, 0, 2, 3, 1
0, 0, 0, 0, 2, 3, 1
0, 0, 0, 0, 0, 2, 3, 1
0, 0, 0, 0, 0, 0, 2, 3, 1
... - _Philippe Deléham_, Sep 24 2014
		

Crossrefs

T(n, 0) = A001003(n) (little Schröder), A109980 (row sums).
Diagonals: A239204, A000012, A016777.

Programs

  • Maple
    T := (n, k) -> local j; add((binomial(n-1, j)*binomial(n+1, k+j+1) - binomial(n, j)*binomial(n, k+j+1))*2^j, j = 0..n-k):
    for n from 0 to 9 do seq(T(n, k), k = 0..n) od;  # Peter Luschny, Jan 24 2025
  • Mathematica
    DELTA[r_, s_, m_] := Module[{p, q, t, x, y}, q[k_] := x r[[k+1]] + y s[[k+1]]; p[0, ] = 1; p[, -1] = 0; p[n_ /; n >= 1, k_ /; k >= 0] := p[n, k] = p[n, k-1] + q[k] p[n-1, k+1] // Expand; t[n_, k_] := Coefficient[p[n, 0], x^(n-k)*y^k]; t[0, 0] = p[0, 0]; Table[t[n, k], {n, 0, m}, {k, 0, n}]];
    nmax = 9;
    DELTA[Table[{1, 2}, (nmax+1)/2] // Flatten, Prepend[Table[0, {nmax}], 1], nmax] // Flatten (* Jean-François Alcover, Aug 07 2018 *)
    (* Function RiordanSquare defined in A321620. *)
    RiordanSquare[(1 + x - Sqrt[1 - 6x + x^2])/(4x), 11] // Flatten  (* Peter Luschny, Nov 27 2018 *)

Formula

T(0, 0) = 1, T(n, k) = 0 if k>n, T(n, 0) = T(n-1, 0) + 2*T(n-1, 1), T(n, k) = T(n-1, k-1) + 3*T(n-1, k) + 2*T(n-1, k+1) for k>0.
Sum_{0<=k<=n} T(n, k) = A109980(n).
Sum_{k>=0} T(m, k)*T(n, k)*2^k = T(m+n, 0) = A001003(m+n).
T(n, k) = Sum_{j=0..n-k} (binomial(n-1, j)*binomial(n+1, k+j+1) - binomial(n, j)*binomial(n, k+j+1))*2^j. (Cigler) - Peter Luschny, Jan 24 2025

Extensions

New name by Peter Luschny, Nov 27 2018

A268138 a(n) = (Sum_{k=0..n-1} A001850(k)*A001003(k+1))/n.

Original entry on oeis.org

1, 5, 51, 747, 13245, 264329, 5721415, 131425079, 3159389817, 78729848397, 2019910325499, 53087981674275, 1423867359013749, 38855956977763857, 1076297858301372687, 30203970496501504239, 857377825323716359665, 24586286492003180067989, 711463902659879056604995, 20756358426519694831851227
Offset: 1

Views

Author

Zhi-Wei Sun, Jan 26 2016

Keywords

Comments

Conjecture: (i) All the terms are odd integers. Also, p | a(p) for any odd prime p.
(ii) Let D_n(x) = Sum_{k=0..n} binomial(n,k)*binomial(n+k,k)*x^k = Sum_{k=0..n} binomial(n,k)^2*x^k*(x+1)^(n-k) for n >= 0, and s_n(x) = Sum_{k=1..n} (binomial(n,k)*binomial(n,k-1)/n)*x^(k-1)*(x+1)^(n-k) = (Sum_{k=0..n} binomial(n,k)*binomial(n+k,k)*x^k/(k+1))/(x+1) for n > 0. Then, for any positive integer n, all the coefficients of the polynomial (1/n)*Sum_{k=0..n-1} D_k(x)*s_{k+1}(x) are integral and the polynomial is irreducible over the field of rational numbers.
The conjecture was essentially proved by the author in arXiv:1602.00574, except for the irreducibility of (Sum_{k=0..n-1} D_k(x)*s_{k+1}(x))/n. - Zhi-Wei Sun, Feb 01 2016

Examples

			a(3) = 51 since (A001850(0)*A001003(1) + A001850(1)*A001003(2) + A001850(2)*A001003(3))/3 = (1*1 + 3*3 + 13*11)/3 = 153/3 = 51.
		

Crossrefs

Programs

  • Maple
    A001850 := n -> LegendreP(n, 3); seq(((3*(2*n+1)*A001850(n)*A001850(n-1)-n*A001850(n-1)^2)/(n+1) - A001850(n)^2)/4, n=1..20); # Mark van Hoeij, Nov 12 2022
    # Alternative (which also gives an integer for n = 0):
    f := n -> hypergeom([-n, -n], [1], 2):          # A001850
    h := n -> hypergeom([-n,  n], [1], 2):          # A182626
    g := n -> hypergeom([-n,  n, 1/2], [1, 1], -8): # A358388
    a := n -> (f(n)*((3*n + 1)*f(n) - (-1)^n*(6*n + 3)*h(n)) - n*g(n))/(2*n + 2):
    seq(simplify(a(n)), n = 1..20); # Peter Luschny, Nov 13 2022
  • Mathematica
    d[n_]:=Sum[Binomial[n,k]Binomial[n+k,k],{k,0,n}]
    s[n_]:=Sum[Binomial[n,k]Binomial[n,k-1]/n*2^(k-1),{k,1,n}]
    a[n_]:=Sum[d[k]s[k+1],{k,0,n-1}]/n
    Table[a[n],{n,1,20}]

Formula

a(n) = ((3*(2*n+1)*A001850(n)*A001850(n-1) - n*A001850(n-1)^2)/(n+1) - A001850(n)^2)/4. - Mark van Hoeij, Nov 12 2022
G.f.: (1-(1+1/x)*Int((1-34*x+x^2)^(1/2) * hypergeom([-1/2,1/2],[1], -32*x/(1-34*x+x^2))/((1-x)*(1+x)^2),x))/4. - Mark van Hoeij, Nov 28 2024
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