A364923
G.f. satisfies A(x) = 1 + x*A(x)^4 / (1 - 2*x*A(x)^3).
Original entry on oeis.org
1, 1, 6, 48, 442, 4419, 46626, 511032, 5761650, 66394596, 778518552, 9258850440, 111417705702, 1354135251538, 16598001854700, 204945037918800, 2546849778687138, 31828936270676172, 399777371427582024, 5043824569861127808, 63892650400004356776
Offset: 0
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A364923 := proc(n)
add( 3^k*(-2)^(n-k)*binomial(n,k)*binomial(3*n+k+1,n)/(3*n+k+1),k=0..n) ;
end proc:
seq(A364923(n),n=0..80); # R. J. Mathar, Aug 16 2023
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a(n) = sum(k=0, n, 3^k*(-2)^(n-k)*binomial(n, k)*binomial(3*n+k+1, n)/(3*n+k+1));
A224071
Number of Schroeder paths of semilength n in which there are no (2,0)-steps at level 1.
Original entry on oeis.org
1, 2, 5, 15, 52, 201, 841, 3726, 17213, 82047, 400600, 1993377, 10071777, 51532938, 266462229, 1390174911, 7308741084, 38682855225, 205940368441, 1102091393574, 5925177392573, 31987877317887, 173337754977904
Offset: 0
a(2) = 5 because we have HH, UDH, HUD, UDUD and UUDD.
G.f. = 1 + 2*x + 5*x^2 + 15*x^3 + 52*x^4 + 201*x^5 + 841*x^6 + ...
- Vincenzo Librandi, Table of n, a(n) for n = 0..200
- J. Bloom and S. Elizalde, Pattern avoidance in matchings and partitions, arXiv:1211.3442 [math.CO], 2012; Theorem 6.1.
- Paul Barry, A study of Integer Sequences, Riordan Arrays, Pascal-like Arrays and Hankel Transforms, Ph.D Thesis, University College, Cork, Republic of Ireland, 2009.
- Arnauld Mesinga Mwafise, Computational and Combinatorial Enumeration of Poset Matrices, 2024. See p. 8.
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CoefficientList[Series[4/(3-5*x+Sqrt[x^2-6*x+1]), {x, 0, 20}], x] (* Vaclav Kotesovec, May 23 2013 *)
a[ n_] := SeriesCoefficient[ (3 - 5 x - Sqrt[ 1 - 6 x + x^2]) / (2 - 6 x + 6 x^2), {x, 0, n}]; (* Michael Somos, Mar 28 2014 *)
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a(n):=sum((k+1)*((-1)^floor((k+2)/3)+(-1)^floor((k+1)/3))*sum(binomial(n+1,n-k-i)*binomial(n+i,n),i,0,n-k),k,0,n)/(2*(n+1)); /* Vladimir Kruchinin, Mar 08 2016*/
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z='z+O('z^66); Vec(4/(3-5*z+sqrt(1-6*z+z^2))) /* Joerg Arndt, Mar 30 2013 */
A336729
G.f. A(x) satisfies: A(x) = 1 + x * A(x) / (1 + 3 * x * A(x)).
Original entry on oeis.org
1, 1, -2, 1, 10, -38, 28, 289, -1262, 1054, 11044, -51302, 45604, 482068, -2319176, 2140129, 22753378, -111964106, 105927508, 1130780062, -5652760340, 5444054956, 58291068808, -294808277414, 287740874260, 3088109246572, -15758505143192, 15541351662484, 167103084713608
Offset: 0
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a[0] = 1; a[n_] := Sum[(-3)^(n - k) * Binomial[n, k] * Binomial[n , k - 1], {k, 1, n}] / n; Array[a, 29, 0] (* Amiram Eldar, Aug 02 2020 *)
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{a(n) = local(A=1+x*O(x^n)); for(i=0, n, A=1+x*A/(1+3*x*A)); polcoef(A, n)}
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{a(n) = if(n==0, 1, sum(k=1, n, (-3)^(n-k)*binomial(n, k)*binomial(n, k-1))/n)}
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N=40; x='x+O('x^N); Vec(2/(1-4*x+sqrt(1+4*x+16*x^2)))
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{a(n) = sum(k=0, n, (-3)^k*4^(n-k)*binomial(n, k)*binomial(n+k, n)/(k+1))}
A371363
Expansion of (1/x) * Series_Reversion( x * (1-3*x)^3 / (1-2*x) ).
Original entry on oeis.org
1, 7, 85, 1261, 20788, 365845, 6731758, 127938625, 2491921516, 49480794460, 997897366717, 20384025765619, 420869454302620, 8769197604091246, 184151509243984300, 3893585866824069577, 82817275938125471548, 1770880435886367151060
Offset: 0
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my(N=20, x='x+O('x^N)); Vec(serreverse(x*(1-3*x)^3/(1-2*x))/x)
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a(n) = sum(k=0, n, 2^(n-k)*binomial(3*n+k+2, k)*binomial(3*n+1, n-k))/(n+1);
A234939
Coefficients of Hilbert series for suboperad of bicolored noncrossing configurations generated by a triangle with colored base and at least one more colored edge and a triangle with one colored non-base edge.
Original entry on oeis.org
1, 2, 8, 38, 200, 1124, 6608, 40142, 249992, 1587548, 10241264, 66926204, 442120016, 2947660616, 19808372384, 134030802782, 912385334792, 6244056445868, 42935538999728, 296493196682036, 2055313327353200, 14297177397185912, 99769106353379168, 698228176760193068
Offset: 1
A364924
G.f. satisfies A(x) = 1 + x*A(x)^5 / (1 - 2*x*A(x)^4).
Original entry on oeis.org
1, 1, 7, 67, 743, 8970, 114445, 1517976, 20722023, 289224355, 4108588558, 59207805442, 863439906413, 12718638581368, 188960182480440, 2828238875318256, 42605850936335463, 645497106959662857, 9829072480785776101, 150345303724987825021
Offset: 0
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a(n) = sum(k=0, n, 3^k*(-2)^(n-k)*binomial(n, k)*binomial(4*n+k+1, n)/(4*n+k+1));
A371362
Expansion of (1/x) * Series_Reversion( x * (1-3*x)^2 / (1-2*x) ).
Original entry on oeis.org
1, 4, 31, 298, 3199, 36742, 441748, 5489554, 69945295, 908836768, 11996580199, 160418984656, 2168512922692, 29584600414168, 406823494817560, 5632906243123090, 78465351036084655, 1098851032467132484, 15461857967408794333, 218490450548650811914
Offset: 0
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my(N=30, x='x+O('x^N)); Vec(serreverse(x*(1-3*x)^2/(1-2*x))/x)
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a(n) = sum(k=0, n, 2^(n-k)*binomial(2*n+k+1, k)*binomial(2*n, n-k))/(n+1);
A243693
Number of Hyposylvester classes of 3-multiparking functions of length n.
Original entry on oeis.org
1, 1, 5, 32, 233, 1833, 15180, 130392, 1151057, 10378883, 95182445, 885053524, 8324942620, 79071217228, 757310811912, 7305728683824, 70923966744609, 692370887676567, 6792525607165935, 66933512163735000, 662190712902022017, 6574831459429388169, 65494637699437417584
Offset: 0
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a := proc(n) option remember; if n <= 1 then return 1 fi;
(a(n - 2)*(-800*n^3 + 3024*n^2 - 3184*n + 672) + a(n - 1)*(3275*n^3 - 7467*n^2 +
5038*n - 1008))/(300*n^3 - 234*n^2 - 192*n) end:
seq(a(n), n = 0..22); # Peter Luschny, Apr 13 2024
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a[n_] := 3^(n - Boole[n>0]) Hypergeometric2F1[1 - n, -2 n, 2, 1/3];
Table[a[n], {n, 0, 22}] (* Peter Luschny, Apr 12 2024 *)
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a(n) = sum(k=0, n, 3^k*(-2)^(n-k)*binomial(n, k)*binomial(2*n+k+1, n)/(2*n+k+1)); \\ Seiichi Manyama, Aug 12 2023
Name clarified by
Jun Yan, Apr 12 2024
A250307
Number of 2-colored Schroeder paths of semilength n in which there are no (2,0)-steps at level 1.
Original entry on oeis.org
1, 3, 10, 37, 152, 690, 3422, 18257, 103144, 608730, 3713524, 23235490, 148281656, 961255200, 6311395814, 41878914665, 280365966232, 1891270498050, 12842102343820, 87703053156406, 601999871121200, 4150859861430252, 28736613316786220, 199671324115916570
Offset: 0
a(2) = 10 because we have H1H1, H1H2, H2H1, H2H2, UDH1, UDH2, H1UD, H2UD, UDUD and UUDD.
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CoefficientList[Series[1/(1 - 2 x - x (1 + 2 x - Sqrt[1 - 8 x + 4 x^2])/(6 x)), {x, 0, 30}], x] (* Vincenzo Librandi, Apr 21 2015 *)
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a(n):=sum((k+1)*(sum(2^j*(-1)^(-k+j)*binomial(k-j,j),j,0,k))*sum(binomial(j,-n-k+2*j-2)*4^(-n-k+2*j-2)*3^(n-j+1)*binomial(n+1,j),j,0,n+1),k,0,n)/(n+1); /* Vladimir Kruchinin, Mar 08 2016 */
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