A000108 -id:A000108 - cp's OEIS frontend

A251568 Expansion of e.g.f. exp(xC(x)^2) where C(x) = 1 + xC(x)^2 is the g.f. of the Catalan numbers, A000108.

1, 1, 5, 43, 529, 8501, 169021, 4010455, 110676833, 3484717129, 123320412181, 4847038223171, 209536628422705, 9882471447634813, 505033804901100749, 27802319803528367791, 1640388588050579832001, 103275015543414629215505, 6910877628962983581031333

Offset: 0

Paul D. Hanna, Dec 05 2014

			E.g.f.: A(x) = 1 + x + 5*x^2/2! + 43*x^3/3! + 529*x^4/4! + 8501*x^5/5! + ...
where
log(A(x)) = x + 2*x^2 + 5*x^3 + 14*x^4 + 42*x^5 + 132*x^6 + ... + A000108(n)*x^n + ...

G. C. Greubel, Table of n, a(n) for n = 0..360

Cf. A080893, A250917.
Cf. A082579, A380511, A380514.
Cf. A001517, A251569, A000108, A382032.

CatalanNumber := n -> binomial(2*n,n)/(n+1):
a := n -> `if`(n=0, 1, n!*CatalanNumber(n)*hypergeom([1-n], [2+n], -1)):
seq(simplify(a(n)), n=0..9); # Peter Luschny, May 04 2017

Flatten[{1,Table[Sum[n!/k!*Binomial[2*n-1,n-k]*2*k/(n+k),{k,1,n}],{n,1,20}]}] (* Vaclav Kotesovec, Feb 14 2015 *)
a[0] = 1; a[n_] := (2n)!/(n+1)! Hypergeometric1F1[1-n, n+2, -1];
Table[a[n], {n, 0, 20}] (* Jean-François Alcover, May 03 2017, after Vladimir Kruchinin *)

{a(n)=my(C=1);for(i=1,n,C=1+x*C^2 +x*O(x^n));n!*polcoef(exp(x*C^2),n)}
for(n=0,20,print1(a(n),", "))

{a(n) = if(n==0, 1, sum(k=1, n, n!/k! * binomial(2*n-1, n-k) * 2*k/(n+k) ))}
for(n=0, 20, print1(a(n), ", "))

my(N=20, x='x+O('x^N)); Vec(serlaplace(exp(serreverse(x*(1-x))^2/x))) \\ Seiichi Manyama, Mar 15 2025

a(n) = Sum_{k=0..n} (n!/k!) * binomial(2*n-1, n-k) * 2*k/(n+k) for n > 0 with a(0)=1.
E.g.f. A(x) satisfies: A'(x)/A(x) = C'(x) = C(x)^2 / sqrt(1-4*x) where C(x) = (1-sqrt(1-4*x))/(2*x) is the Catalan function.
Recurrence equation: a(n) = -(n^2 - 5*n +1)*a(n-1) + n*(2*n - 3)*(2*n - 4)*a(n-2) with a(0) = 1, a(1) = 1. It appears that a(n) - 1 is divisible by n*(n - 1) for n >= 2. - Peter Bala, Feb 14 2015
a(n) ~ 2^(2*n+1/2) * n^(n-1) / exp(n-1). - Vaclav Kotesovec, Feb 14 2015
a(n) are special values of the hypergeometric function 1F1: a(n) = 4^n*Gamma(n+1/2)*exp(-1)*hypergeom([2*n+1], [n+2], 1)/(sqrt(Pi)*(n+1)), for n>=1. - Karol A. Penson, Jun 01 2015
a(n) = ((2*n)!/(n+1)!)*hypergeometric([1-n],[n+2],-1), a(0)=1. - Vladimir Kruchinin, May 03 2017
From Seiichi Manyama, Mar 15 2025: (Start)
E.g.f.: exp( (1/x) * Series_Reversion( x*(1-x) )^2 ).
E.g.f.: exp( Series_Reversion( x/(1+x)^2 ) ). (End)

A333093 a(n) is equal to the n-th order Taylor polynomial (centered at 0) of c(x)^n evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4x))/(2x) is the o.g.f. of the Catalan numbers A000108.

Original entry on oeis.org

1, 2, 8, 41, 232, 1377, 8399, 52138, 327656, 2077934, 13270633, 85226594, 549837391, 3560702069, 23132584742, 150695482041, 984021596136, 6438849555963, 42208999230224, 277144740254566, 1822379123910857, 11998811140766701, 79095365076843134

Offset: 0

Peter Bala, Mar 07 2020

The sequence satisfies the Gauss congruences: a(n*p^k) == a(n*p^(k-1)) ( mod p^k ) for all prime p and positive integers n and k.
We conjecture that the sequence satisfies the stronger supercongruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k. Examples of these congruences are given below.
More generally, for each integer m, we conjecture that the sequence
  {a_m(n) : n >= 0}, defined by setting a_m(n) = the n-th order Taylor polynomial of c(x)^(m*n) evaluated at x = 1, satisfies the same supercongruences. For cases, see A099837 (m = -2), A100219 (m = -1), A000012 (m = 0), A333094 (m = 2), A333095 (m = 3), A333096 (m = 4), A333097 (m = 5).

			n-th order Taylor polynomial of c(x)^n:
  n = 0: c(x)^0 = 1 + O(x)
  n = 1: c(x)^1 = 1 + x + O(x^2)
  n = 2: c(x)^2 = 1 + 2*x + 5*x^2 + O(x^3)
  n = 3: c(x)^3 = 1 + 3*x + 9*x^2 + 28*x^3 + O(x^4)
  n = 4: c(x)^4 = 1 + 4*x + 14*x^2 + 48*x^3 + 165*x^4 + O(x^5)
Setting x = 1 gives a(0) = 1, a(1) = 1 + 1 = 2, a(2) = 1 + 2 + 5 = 8, a(3) = 1 + 3 + 9 + 28 = 41 and a(4) = 1 + 4 + 14 + 48 + 165 = 232.
The triangle of coefficients of the n-th order Taylor polynomial of c(x)^n, n >= 0, in descending powers of x begins
                                        row sums
  n = 0 |   1                               1
  n = 1 |   1   1                           2
  n = 2 |   5   2    1                      8
  n = 3 |  28   9    3   1                 41
  n = 4 | 165  48   14   4   1            232
   ...
This is a Riordan array belonging to the Hitting time subgroup of the Riordan group. The first column sequence [1, 1, 5, 28, 165, ...] = [x^n] c(x)^n = A025174(n).
Examples of supercongruences:
a(13) - a(1) = 3560702069 - 2 = (3^2)*(13^3)*31*37*157 == 0 ( mod 13^3 ).
a(3*7) - a(3) = 11998811140766701 - 41 = (2^2)*5*(7^4)*32213*7756841 == 0 ( mod 7^3 ).
a(5^2) - a(5) = 22794614296746579502 - 1377 = (5^6)*7*53*6491*605796421 == 0 ( mod 5^6 ).

Peter Bala, Notes on A333093

Cf. A000108, A001006, A001764, A005554, A025174, A333090 through A333097, A372214, A372215.

seq(add(n/(n+k)*binomial(n+2*k-1,k), k = 0..n), n = 1..25);
#alternative program
c:= x -> (1/2)*(1-sqrt(1-4*x))/x:
G := (x,n) -> series(c(x)^n, x, 51):
seq(add(coeff(G(x, n), x, k), k = 0..n), n = 0..25);

Table[SeriesCoefficient[((1 + x)^2 * (1 - Sqrt[(1 - 3*x)/(1 + x)]) / (2*x))^n, {x, 0, n}], {n, 0, 25}] (* Vaclav Kotesovec, Mar 28 2020 *)

a(n) = Sum_{k = 0..n} n/(n+k)*binomial(n+2*k-1,k) for n >= 1.
a(n) = [x^n] ( (1 + x)*c(x/(1 + x)) )^n = [x^n] ( (1 + x)*(1 + x*M(x)) )^n, where M(x) = ( 1 - x - sqrt(1 - 2*x - 3*x^2) ) / (2*x^2) is the o.g.f. of the Motzkin numbers A001006.
O.g.f.: ( 1 + x*f'(x)/f(x) )/( 1 - x*f(x) ), where f(x) = 1 + x + 3*x^2 + 12*x^3 + 55*x^4 + ... = (1/x)*Revert( x/c(x) ) is the o.g.f. of A001764.
Row sums of the Riordan array ( 1 + x*f'(x)/f(x), x*f(x) ) belonging to the Hitting time subgroup of the Riordan group.
a(n) ~ 3^(3*n + 3/2) / (7 * sqrt(Pi*n) * 2^(2*n+1)). - Vaclav Kotesovec, Mar 28 2020
a(n) = Sum_{k = 0..n} n/(n+2*k)*binomial(n+2*k, k) for n >= 1. - Peter Bala, Apr 20 2024
D-finite with recurrence 2*n*(2*n-1)*(3991*n -21664)*a(n) +(-1329757*n^3 +9119565*n^2 -18270518*n +10657440)*a(n-1) +10*(947050*n^3 -6943257*n^2 +15944396*n -11260008)*a(n-2) +12*(-787878*n^3 +5778161*n^2 -13283386*n +9383340)*a(n-3) +9*(3*n-10)*(3*n-8)*(100503*n -141587)*a(n-4)=0, n>=5. - R. J. Mathar, Nov 22 2024

A026012 Second differences of Catalan numbers A000108.

Original entry on oeis.org

1, 2, 6, 19, 62, 207, 704, 2431, 8502, 30056, 107236, 385662, 1396652, 5088865, 18642420, 68624295, 253706790, 941630580, 3507232740, 13105289370, 49114150020, 184560753390, 695267483664, 2625197720454, 9933364416572, 37660791173152, 143048202990504

Offset: 0

N. J. A. Sloane, Clark Kimberling

nonn

Number of (s(0), s(1), ..., s(n)) such that s(i) is a nonnegative integer and |s(i) - s(i-1)| = 1 for i = 1,2,...,n, s(0) = s(2n) = 2.
Number of Dyck paths of semilength n+2 with no initial and no final UD's. Example: a(2)=6 because the only Dyck paths of semilength 4 with no initial and no final UD's are UUDUDUDD, UUDUUDDD, UUUDDUDD, UUUDUDDD, UUDDUUDD, UUUUDDDD. - Emeric Deutsch, Oct 26 2003
Number of branches of length 1 starting from the root in all ordered trees with n+1 edges. Example: a(1)=2 because the tree /\ has two branches of length 1 starting from the root and the path-tree of length 2 has none. a(n) = Sum_{k=0..n+1} (k*A127158(n+1,k)). - Emeric Deutsch, Mar 01 2007
Number of staircase walks from (0,0) to (n,n) that never cross y=x+2. Example: a(3) = 19 because up,up,up,right,right,right is not allowed but the other binomial(6,3)-1 = 19 paths are. - Mark Spindler, Nov 11 2012
Number of standard Young tableaux of skew shape (n+2,n)/(2), for n>=2. - Ran Pan, Apr 07 2015

S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (see pp. 188, 196).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Paul Barry, On a Central Transform of Integer Sequences, arXiv:2004.04577 [math.CO], 2020.
Reinis Cirpons, James East, and James D. Mitchell, Transformation representations of diagram monoids, arXiv:2411.14693 [math.RA], 2024. See pp. 3, 33.
Ryota Inagaki, Tanya Khovanova, and Austin Luo, On Chip-Firing on Undirected Binary Trees, Ann. Comb. (2025). See p. 15.
Jocelyn Quaintance and Harris Kwong, A combinatorial interpretation of the Catalan and Bell number difference tables, Integers, 13 (2013), #A29.
Murray Tannock, Equivalence classes of mesh patterns with a dominating pattern, MSc Thesis, Reykjavik Univ., May 2016. See Appendix B2.

T(2n, n), where T is the array defined in A026009.

Cf. A000108, A000245, A000344, A039599, A059346, A127158, A342912.

Differences[Table[CatalanNumber[n], {n, 0, 28}], 2] (* Jean-François Alcover, Sep 28 2012 *)
Table[Binomial[2n,n]-Binomial[2n,n-3],{n,0,26}] (* Mark Spindler, Nov 11 2012 *)

a(n) = 3*(3*n^2+3*n+2)*binomial(2*n, n)/((n+1)*(n+2)*(n+3)); /* Joerg Arndt, Aug 19 2012 */

Expansion of (1+x^1*C^3)*C^1, where C = (1-(1-4*x)^(1/2))/(2*x) is g.f. for Catalan numbers, A000108.
a(n) = 3*(3*n^2+3*n+2)*binomial(2*n, n)/((n+1)*(n+2)*(n+3)). - Emeric Deutsch, Oct 26 2003
a(n) = Sum_{k=0..2} A039599(n,k) = A000108(n) + A000245(n) + A000344(n). - Philippe Deléham, Nov 12 2008
a(n) = binomial(2*n,n)/(n+1)*hypergeom([-2,n+1/2],[n+2],4). - Peter Luschny, Aug 15 2012
a(n) = binomial(2*n,n) - binomial(2n,n-3). - Mark Spindler, Nov 11 2012
D-finite with recurrence (n+3)*a(n) + (-5*n-6)*a(n-1) + 2*(2*n-3)*a(n-2) = 0. - R. J. Mathar, Jun 20 2013
E.g.f.: exp(2*x)*(BesselI(0,2*x) - BesselI(3,2*x)). - Ilya Gutkovskiy, Feb 28 2017
Sum_{n>=0} a(n)/4^n = 6. - Amiram Eldar, Jul 10 2023
a(n) = C(n+2)+C(n)-2*C(n+1), C = A000108. - Alois P. Heinz, Apr 02 2025
Binomial transform of A342912. - Mélika Tebni, Apr 05 2025

A057513 Number of separate orbits to which permutations given in A057511/A057512 (induced by deep rotation of general parenthesizations/plane trees) partition each A000108(n) objects encoded by A014486 between (A014138(n-1)+1)-th and (A014138(n))-th terms.

Original entry on oeis.org

1, 1, 2, 4, 9, 21, 56, 153, 451, 1357, 4212, 13308, 42898, 140276, 465324, 1561955, 5300285, 18156813, 62732842, 218405402, 765657940

Offset: 0

Antti Karttunen Sep 03 2000

It is much faster to compute this sequence empirically with the given C-program than to calculate the terms with the formula in its present form.

A. Karttunen, Gatomorphisms (with the complete Scheme source)
Index entries for sequences related to rooted trees
A. Karttunen, C-program for computing empirically the initial terms of this sequence

CountCycles given in A057502, for other procedures, follow A057511 and A057501.
Similarly generated sequences: A001683, A002995, A003239, A038775, A057507. Cf. also A000081.
Occurs for first time in A073201 as row 12. Cf. A057546 and also A000081.

A057513 := proc(n) local i; `if`((0=n),1,(1/A003418(n-1))*add(A079216bi(n,i),i=1..A003418(n-1))); end;
# Or empirically:
DeepRotatePermutationCycleCounts := proc(upto_n) local u,n,a,r,b; a := []; for n from 0 to upto_n do b := []; u := (binomial(2*n,n)/(n+1)); for r from 0 to u-1 do b := [op(b),1+CatalanRank(n,DeepRotateL(CatalanUnrank(n,r)))]; od; a := [op(a),CountCycles(b)]; od; RETURN(a); end;

a(0)=1, a(n) = (1/A003418(n-1))*Sum_{i=1..A003418(n-1)} A079216(n, i) [Needs improvement.] - Antti Karttunen, Jan 03 2003

A080893 Expansion of e.g.f. exp(xC(x)) = exp((1-sqrt(1-4x))/2), where C(x) is the g.f. of the Catalan numbers A000108.

Original entry on oeis.org

1, 1, 3, 19, 193, 2721, 49171, 1084483, 28245729, 848456353, 28875761731, 1098127402131, 46150226651233, 2124008553358849, 106246577894593683, 5739439214861417731, 332993721039856822081, 20651350143685984386753

Offset: 0

Emanuele Munarini, Mar 31 2003

Essentially the same as A001517: a(n+1) = A001517(n).

For k >= 2, the difference a(n+k) - a(n) is divisible by k. It follows that for each k, the sequence formed by taking a(n) modulo k is periodic with period dividing k. For example, modulo 10 the sequence becomes [1, 1, 3, 9, 3, 1, 1, 3, 9, 3, ...], a purely periodic sequence of period 5. Cf. A047974. - Peter Bala, Feb 11 2025

W. Mlotkowski and A. Romanowicz, A family of sequences of binomial type, Probability and Mathematical Statistics, Vol. 33, Fasc. 2 (2013), pp. 401-408.
Eric Weisstein's World of Mathematics, Bell Polynomial.

Cf. A000108, A001517, A009235, A375173.

y[x_] := y[x] = 2(2x - 3)y[x - 1] + y[x - 2]; y[0] = 1; y[1] = 1; Table[y[n],{n,0,17}]
With[{nn=20},CoefficientList[Series[Exp[(1-Sqrt[1-4x])/2],{x,0,nn}], x] Range[0,nn]!] (* Harvey P. Dale, Oct 30 2011 *)

{a(n) = if( n<1, n = 1 - n); n! * polcoeff( exp( (1 - sqrt(1 - 4*x + x * O(x^n))) / 2), n)} /* Michael Somos, Apr 07 2012 */

A080893 = lambda n: hypergeometric([-n+1, n], [], -1)
[simplify(A080893(n)) for n in (0..19)] # Peter Luschny, Oct 17 2014

E.g.f.: exp((1-sqrt(1-4*x))/2).
D-finite with recurrence: a(n+2) = 2*(2*n + 1)*a(n+1) + a(n).
Recurrence: y(n+1) = Sum_{k = 0..n} binomial(n, k)*binomial(2k, k)*k!*y(n-k).
a(1 - n) = a(n).  a(n + 1) = A001517(n). - Michael Somos, Apr 07 2012
G.f.: 1 + x/Q(0), where Q(k)= 1 - x - 2*x*(k+1)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, May 17 2013
a(n) ~ 2^(2*n-3/2)*n^(n-1)/exp(n-1/2). - Vaclav Kotesovec, Jun 26 2013
a(n) = hypergeom([-n+1, n], [], -1). - Peter Luschny, Oct 17 2014
a(n) = Sum_{k=0..n} (-4)^(n-k) * Stirling1(n,k) * A009235(k) = (-4)^n * Sum_{k=0..n} (1/2)^k * Stirling1(n,k) * Bell_k(-1/2), where Bell_n(x) is n-th Bell polynomial. - Seiichi Manyama, Aug 02 2024

A085880 Triangle T(n,k) read by rows: multiply row n of Pascal's triangle (A007318) by the n-th Catalan number (A000108).

Original entry on oeis.org

1, 1, 1, 2, 4, 2, 5, 15, 15, 5, 14, 56, 84, 56, 14, 42, 210, 420, 420, 210, 42, 132, 792, 1980, 2640, 1980, 792, 132, 429, 3003, 9009, 15015, 15015, 9009, 3003, 429, 1430, 11440, 40040, 80080, 100100, 80080, 40040, 11440, 1430, 4862, 43758, 175032, 408408, 612612, 612612, 408408, 175032, 43758, 4862

Offset: 0

N. J. A. Sloane, Aug 17 2003

Coefficients of terms in the series reversion of (1-k*x-(k+1)*x^2)/(1+x). - Paul Barry, May 21 2005
Equals A131427 * A007318 as infinite lower triangular matrices. [Philippe Deléham, Sep 15 2008]
Sum_{k=0..n} T(n,k)*x^k = A168491(n), A000007(n), A000108(n), A151374(n), A005159(n), A151403(n), A156058(n), A156128(n), A156266(n), A156270(n), A156273(n), A156275(n) for x = -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 respectively. - Philippe Deléham, Nov 15 2013
Diagonal sums are A052709(n+1). - Philippe Deléham, Nov 15 2013

			Triangle starts:
[ 1]     1;
[ 2]     1,     1;
[ 3]     2,     4,      2;
[ 4]     5,    15,     15,      5;
[ 5]    14,    56,     84,     56,     14;
[ 6]    42,   210,    420,    420,    210,     42;
[ 7]   132,   792,   1980,   2640,   1980,    792,    132;
[ 8]   429,  3003,   9009,  15015,  15015,   9009,   3003,    429;
[ 9]  1430, 11440,  40040,  80080, 100100,  80080,  40040,  11440,  1430;
[10]  4862, 43758, 175032, 408408, 612612, 612612, 408408, 175032, 43758, 4862;
...

G. C. Greubel, Rows n = 0..100 of triangle, flattened
Paul Barry, Characterizations of the Borel triangle and Borel polynomials, arXiv:2001.08799 [math.CO], 2020.

Flat(List([0..10], n-> List([0..n], k-> Binomial(n,k)*Binomial(2*n,n)/( n+1) ))); # G. C. Greubel, Feb 07 2020

[Binomial(n,k)*Catalan(n): k in [0..n], n in [0..10]]; // G. C. Greubel, Feb 07 2020

seq(seq(binomial(n, k)*binomial(2*n, n)/(n+1), k = 0..n), n = 0..10); # G. C. Greubel, Feb 07 2020

Table[Binomial[n, k]*CatalanNumber[n], {n,0,10}, {k,0,n}]//Flatten (* G. C. Greubel, Feb 07 2020 *)

tabl(nn) = {for (n=0, nn, c =  binomial(2*n,n)/(n+1); for (k=0, n, print1(c*binomial(n, k), ", ");); print(););} \\ Michel Marcus, Apr 09 2015

[[binomial(n,k)*catalan_number(n) for k in (0..n)] for n in (0..10)] # G. C. Greubel, Feb 07 2020

Triangle given by [1, 1, 1, 1, 1, 1, ...] DELTA [1, 1, 1, 1, 1, 1, ...] where DELTA is Deléham's operator defined in A084938.
Sum_{k>=0} T(n, k) = A151374(n) (row sums). - Philippe Deléham, Aug 11 2005
G.f.: (1-sqrt(1-4*(x+y)))/(2*(x+y)). - Vladimir Kruchinin, Apr 09 2015

A086618 a(n) = Sum{k=0..n} binomial(n,k)^2*C(k), where C() = A000108() are the Catalan numbers.

Original entry on oeis.org

1, 2, 7, 33, 183, 1118, 7281, 49626, 349999, 2535078, 18758265, 141254655, 1079364105, 8350678170, 65298467487, 515349097713, 4100346740511, 32858696386766, 265001681344569

Offset: 0

Paul D. Hanna, Jul 24 2003

nonn

Main diagonal of square table A086617 of coefficients, T(n,k), of x^n*y^k in f(x,y) that satisfies f(x,y) = 1/[(1-x)(1-y)] + xy*f(x,y)^2.
a(n) is the number of permutations of length 2n which are invariant under the reverse-complement map and have no decreasing subsequences of length 4. - Eric S. Egge, Oct 21 2008
In 2012, Zhi-Wei Sun proved that for any odd prime p we have the congruence a(1) + ... + a(p-1) == 0 (mod p^2). - Zhi-Wei Sun, Aug 22 2013

			a(5) = 1118 = 1*1^2 + 1*5^2 + 2*10^2 + 5*10^2 + 14*5^2 + 42*1^2.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
D. Daly and L. Pudwell, Pattern avoidance in rook monoids, 2013.
T. Denton, Algebraic and Affine Pattern Avoidance, arXiv preprint arXiv:1303.3767 [math.CO], 2013.
Z.-W. Sun, Congruences for Franel numbers, arXiv preprint arXiv:1112.1034 [math.NT], 2011. See (1.22).
Z.-W. Sun, On sums of Apery polynomials and related congruences, J. Number Theory 132(2012), 2673-2699.

Cf. A086617 (table), A086615 (antidiagonal sums), A003046 (determinants).
Cf. A000108.
Cf. A228456.

Flatten[{1,RecurrenceTable[{(n+3)^2*(4*n+7)*a[n+2]==2*(20*n^3+117*n^2+220*n+135)*a[n+1]-9*(n+1)^2*(4*n+11)*a[n],a[1]==2,a[2]==7},a,{n,1,20}]}] (* Vaclav Kotesovec, Sep 11 2012 *)
Table[HypergeometricPFQ[{1/2, -n, -n}, {1, 2}, 4], {n, 0, 20}] (* Vladimir Reshetnikov, Oct 03 2016 *)

a(n)=sum(k=0,n-1,binomial(n-1,k)^2*binomial(2*k,k)/(k+1)) \\ Charles R Greathouse IV, Sep 12 2012

a(n)=sum(k=0,n-1,(4*k+3)*sum(i=0,k,binomial(k,i)^2*binomial(2*i,i)))/3/n^2 \\ Charles R Greathouse IV, Sep 12 2012

Recurrence: (n+3)^2*(4*n+7)*a(n+2) = 2*(20*n^3+117*n^2+220*n+135)*a(n+1) - 9*(n+1)^2*(4*n+11)*a(n). - Vaclav Kotesovec, Sep 11 2012
a(n) ~ 3^(5/2)/(8*Pi) * 9^n/n^2. - Vaclav Kotesovec, Oct 06 2012
G.f.: (1-(1-9*x)^(1/3)*hypergeom([1/3,1/3],[1],-27*x*(1-x)^2/(1-9*x)^2))/(6*x). - Mark van Hoeij, May 02 2013
a(n) = hypergeom([1/2,-n,-n], [1,2], 4). - Vladimir Reshetnikov, Oct 03 2016
D-finite with recurrence (n+1)^2*a(n) +(-19*n^2+8*n+6)*a(n-1) +9*(11*n^2-30*n+21)*a(n-2) -81*(n-2)^2*a(n-3)=0. - R. J. Mathar, Aug 01 2022

Edited by N. J. A. Sloane, Sep 14 2012. The formula in the new definition was first sent in by Michael Somos, Feb 19 2004

Minor edits Vaclav Kotesovec, Mar 31 2014

A162481 Expansion of (1/(1-x)^3)*c(x/(1-x)^3), c(x) the g.f. of A000108.

Original entry on oeis.org

1, 4, 14, 54, 235, 1119, 5658, 29800, 161621, 896198, 5056824, 28938519, 167548937, 979653821, 5776252440, 34305807512, 205039491091, 1232333298174, 7443336041318, 45157243590384, 275051410542141, 1681362181696823, 10311616254855422, 63428758470722109

Offset: 0

Paul Barry, Jul 04 2009

Seiichi Manyama, Table of n, a(n) for n = 0..1000

Cf. A000108, A086616, A162482, A360045.

a[n_] := Sum[Binomial[n + 2*k + 2, n - k] * CatalanNumber[k], {k, 0, n}]; Array[a, 22, 0] (* Amiram Eldar, Jun 30 2020 *)

G.f.: 1/((1-x)^3-x-x^2/((1-x)^3-2*x-x^2/((1-x)^3-2*x-x^2/((1-x)^3-2*x-x^2/(1-... (continued fraction);
a(n) = Sum_{k=0..n} C(n+2k+2,n-k)*A000108(k).
Conjecture: (n+1)*a(n) +2*(1-4*n)*a(n-1) +2*(5*n-3)*a(n-2) +4*(2-n)*a(n-3) +(n-3)*a(n-4) = 0. - R. J. Mathar, Dec 11 2011
G.f. A(x) satisfies: A(x) = 1/(1 - x)^3 + x * A(x)^2. - Ilya Gutkovskiy, Jun 30 2020
a(n) = binomial(n+2,2) + Sum_{k=0..n-1} a(k) * a(n-k-1). - Seiichi Manyama, Jan 23 2023
G.f.: (1 - sqrt(1 - 4*x/(1-x)^3))/(2*x). - Vaclav Kotesovec, Jan 24 2023

A197433 Sum of distinct Catalan numbers: a(n) = Sum_{k>=0} A030308(n,k)*C(k+1) where C(n) is the n-th Catalan number (A000108). (C(0) and C(1) not treated as distinct.)

Original entry on oeis.org

0, 1, 2, 3, 5, 6, 7, 8, 14, 15, 16, 17, 19, 20, 21, 22, 42, 43, 44, 45, 47, 48, 49, 50, 56, 57, 58, 59, 61, 62, 63, 64, 132, 133, 134, 135, 137, 138, 139, 140, 146, 147, 148, 149, 151, 152, 153, 154, 174, 175, 176, 177, 179, 180, 181, 182, 188, 189, 190, 191, 193, 194, 195, 196

Offset: 0

Philippe Deléham, Oct 15 2011

Replace 2^k with A000108(k+1) in binary expansion of n.
From Antti Karttunen, Jun 22 2014: (Start)
On the other hand, A244158 is similar, but replaces 10^k with A000108(k+1) in decimal expansion of n.
This sequence gives all k such that A014418(k) = A239903(k), which are precisely all nonnegative integers k whose representations in those two number systems contain no digits larger than 1. From this also follows that this is a subsequence of A244155.
(End)

Antti Karttunen, Table of n, a(n) for n = 0..8191

Characteristic function: A176137.
Subsequence of A244155.
Cf. A000108, A030308, A197432, A014418, A239903, A244158, A244159, A244230, A244231, A244232, A244315, A244316.
Cf. also A060112.
Other sequences that are built by replacing 2^k in binary representation with other numbers: A022290 (Fibonacci), A029931 (natural numbers), A059590 (factorials), A089625 (primes), A197354 (odd numbers).

nmax = 63;
a[n_] := If[n == 0, 0, SeriesCoefficient[(1/(1-x))*Sum[CatalanNumber[k+1]* x^(2^k)/(1 + x^(2^k)), {k, 0, Log[2, n] // Ceiling}], {x, 0, n}]];
Table[a[n], {n, 0, nmax}] (* Jean-François Alcover, Nov 18 2021, after Ilya Gutkovskiy *)

For all n, A244230(a(n)) = n. - Antti Karttunen, Jul 18 2014

G.f.: (1/(1 - x))*Sum_{k>=0} Catalan number(k+1)*x^(2^k)/(1 + x^(2^k)). - Ilya Gutkovskiy, Jul 23 2017

Name clarified by Antti Karttunen, Jul 18 2014

A368025 Array read by ascending antidiagonals: A(n,k) is the determinant of the n-th order Hankel matrix of Catalan numbers M(n) whose generic element is given by M(i,j) = A000108(i+j+k) with i,j = 0, ..., n-1.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 5, 1, 1, 1, 4, 14, 14, 1, 1, 1, 5, 30, 84, 42, 1, 1, 1, 6, 55, 330, 594, 132, 1, 1, 1, 7, 91, 1001, 4719, 4719, 429, 1, 1, 1, 8, 140, 2548, 26026, 81796, 40898, 1430, 1, 1, 1, 9, 204, 5712, 111384, 884884, 1643356, 379236, 4862, 1

Offset: 0

Stefano Spezia, Dec 08 2023

This array is a variant of the triangles A078920 and A123352 extended to the trivial cases (here for k=0).

			The array begins:
  1, 1, 1,   1,    1,      1,        1, ...
  1, 1, 2,   5,   14,     42,      132, ...
  1, 1, 3,  14,   84,    594,     4719, ...
  1, 1, 4,  30,  330,   4719,    81796, ...
  1, 1, 5,  55, 1001,  26026,   884884, ...
  1, 1, 6,  91, 2548, 111384,  6852768, ...
  1, 1, 7, 140, 5712, 395352, 41314284, ...
  ...

Arthur T. Benjamin, Naiomi T. Cameron, Jennifer J. Quinn, and Carl R. Yerger, Catalan determinants-a combinatorial approach, Congressus Numerantium 200, 27-34 (2010). On ResearchGate.
Jishe Feng, The explicit formula of Hankel determinant with Catalan elements, arXiv:2010.06586 [math.GM], 2020.
M. E. Mays and Jerzy Wojciechowski, A determinant property of Catalan numbers. Discrete Math. 211, No. 1-3, 125-133 (2000).
Wikipedia, Hankel matrix.

Cf. A000108 (n=1), A005700 (n=2), A006149 (n=3), A006150 (n=4), A006151 (n=5).
Cf. A000012 (k=0 or k=1 or n=0), A000330, A078920, A091962, A123352, A335857 (k=6).
Cf. A355400, A368026 (permanent), A378112.
Antidiagonal sums give A355503.

A:= proc(n, k) option remember; `if`(k=0, 1, 2^n*mul(
      (2*(k-i)+2*n-3)/(k+2*n-1-i), i=0..n-1)*A(n, k-1))
    end:
seq(seq(A(d-k, k), k=0..d), d=0..10);  # Alois P. Heinz, Dec 20 2023

A[n_,k_]:=If[n==0,1,Det[Table[CatalanNumber[i+j+k],{i,0,n-1},{j,0,n-1}]]]; Table[A[n-k,k],{n,0,11},{k,0,n}]//Flatten

For an explicit formula of A(n,k), see equation (5) in Feng, 2020.
A(n,2) = n + 1.
A(n,3) = A000330(n+1).
A(n,4) = A006858(n+1).
A(n,5) = A091962(n+1).
Diagonal: A(n,n) = A123352(2*n-1,n-1) = A355400(n).

cp's OEIS Frontend

A251568 Expansion of e.g.f. exp(x*C(x)^2) where C(x) = 1 + x*C(x)^2 is the g.f. of the Catalan numbers, A000108.

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A333093 a(n) is equal to the n-th order Taylor polynomial (centered at 0) of c(x)^n evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. of the Catalan numbers A000108.

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A026012 Second differences of Catalan numbers A000108.

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A057513 Number of separate orbits to which permutations given in A057511/A057512 (induced by deep rotation of general parenthesizations/plane trees) partition each A000108(n) objects encoded by A014486 between (A014138(n-1)+1)-th and (A014138(n))-th terms.

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A080893 Expansion of e.g.f. exp(x*C(x)) = exp((1-sqrt(1-4*x))/2), where C(x) is the g.f. of the Catalan numbers A000108.

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A085880 Triangle T(n,k) read by rows: multiply row n of Pascal's triangle (A007318) by the n-th Catalan number (A000108).

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A086618 a(n) = Sum{k=0..n} binomial(n,k)^2*C(k), where C() = A000108() are the Catalan numbers.

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A251568 Expansion of e.g.f. exp(xC(x)^2) where C(x) = 1 + xC(x)^2 is the g.f. of the Catalan numbers, A000108.

A333093 a(n) is equal to the n-th order Taylor polynomial (centered at 0) of c(x)^n evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4x))/(2x) is the o.g.f. of the Catalan numbers A000108.

A080893 Expansion of e.g.f. exp(xC(x)) = exp((1-sqrt(1-4x))/2), where C(x) is the g.f. of the Catalan numbers A000108.