A049124 -id:A049124 - cp's OEIS frontend

A101786 G.f. satisfies: A(x) = 1 + xA(x)/(1 - 2x^2*A(x)^2).

1, 1, 1, 3, 9, 25, 77, 247, 801, 2657, 8969, 30635, 105785, 368745, 1295493, 4582767, 16309953, 58357313, 209798289, 757461011, 2745281705, 9984464761, 36428252541, 133293594343, 489028250465, 1798543861537, 6629635284505

Offset: 0

Paul D. Hanna, Dec 16 2004

nonn

Formula may be derived using the Lagrange Inversion theorem (cf. A049124).

			Generated from Fibonacci polynomials (A011973) and coefficients of odd powers of 1/(1-x):
a(1) = 1*1/1
a(2) = 1*1/1 + 0*1*2/3
a(3) = 1*1/1 + 1*3*2/3
a(4) = 1*1/1 + 2*6*2/3 + 0*1*2^2/5
a(5) = 1*1/1 + 3*10*2/3 + 1*5*2^2/5
a(6) = 1*1/1 + 4*15*2/3 + 3*15*2^2/5 + 0*1*2^3/7
a(7) = 1*1/1 + 5*21*2/3 + 6*35*2^2/5 + 1*7*2^3/7
a(8) = 1*1/1 + 6*28*2/3 + 10*70*2^2/5 + 4*28*2^3/7 + 0*1*2^4/9
This process is equivalent to the formula:
a(n) = Sum_{k=0..[(n-1)/2]} C(n-k-1,k)*C(n,2*k)*2^k/(2*k+1).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

Cf. A011973, A001045, A049124, A101785.

Flatten[{1,Table[Sum[Binomial[n-k-1,k]*Binomial[n,2*k]*2^k/(2*k+1),{k,0,Floor[(n-1)/2]}],{n,1,20}]}] (* Vaclav Kotesovec, Sep 17 2013 *)
ShiftedReversion[ser_, n_, sgn_] := CoefficientList[(sgn/x)InverseSeries[Series[x sgn ser, {x, 0, n}]],x];
Jacobsthal := (2x^2 - 1)/((x + 1)(2x - 1)); (* with A001045(0) = 1 *)
ShiftedReversion[Jacobsthal, 27, -1] (* Peter Luschny, Jan 10 2019 *)

{a(n)=if(n==0,1,sum(k=0,(n-1)\2, binomial(n-k-1,k)*binomial(n,2*k)*2^k/ (2*k+1)))}

a(n) = Sum_{k=0..[(n-1)/2]} C(n-k-1, k)*C(n, 2*k)*2^k/(2*k+1) for n>0, with a(0)=1.
G.f.: A(x) = (1/x)*Series_Reversion( x*(1 - 2*x^2)/(1+x - 2*x^2) ).
Recurrence: 2*n*(n+1)*(31*n^2 - 127*n + 120)*a(n) = 3*n*(62*n^3 - 285*n^2 + 359*n - 88)*a(n-1) + (62*n^4 - 378*n^3 + 1009*n^2 - 1425*n + 792)*a(n-2) + (n-3)*(682*n^3 - 3135*n^2 + 4133*n - 1272)*a(n-3) - 9*(n-4)*(n-3)*(31*n^2 - 65*n + 24)*a(n-4). - Vaclav Kotesovec, Sep 17 2013
a(n) ~ c*d^n/(sqrt(Pi)*n^(3/2)), where d = 3/4 + 1/(4*sqrt(3/(35 - (176*2^(2/3))/(9959 + 465*sqrt(465))^(1/3) + 2*(19918 + 930*sqrt(465))^(1/3)))) + 1/2*sqrt(35/6 + (44*2^(2/3))/(3*(9959 + 465*sqrt(465))^(1/3)) - (9959 + 465*sqrt(465))^(1/3)/(3*2^(2/3)) + 127/2*sqrt(3/(35 - (176*2^(2/3))/ (9959 + 465*sqrt(465))^(1/3) + 2*(19918 + 930*sqrt(465))^(1/3)))) = 3.9027270552404829297969 = ... is the root of the equation 9 - 22*d - 2*d^2 - 6*d^3 + 2*d^4 = 0 and c = 0.68546565145612597016100560323891887595749... - Vaclav Kotesovec, Sep 17 2013

A307528 G.f. A(x) satisfies: A(x) = 1 + x^2A(x)^2/(1 - xA(x) - x^2A(x)^2 - x^3A(x)^3).

Original entry on oeis.org

1, 0, 1, 1, 4, 9, 27, 76, 226, 680, 2078, 6441, 20153, 63684, 202732, 649930, 2095854, 6794684, 22131765, 72393439, 237703654, 783198068, 2588645047, 8580674778, 28517805357, 95009277576, 317242351135, 1061500510809, 3558683892258, 11952025977378, 40209157279701

Offset: 0

Ilya Gutkovskiy, Apr 12 2019

nonn

			G.f.: A(x) = 1 + x^2 + x^3 + 4*x^4 + 9*x^5 + 27*x^6 + 76*x^7 + 226*x^8 + 680*x^9 + 2078*x^10 + ...

Cf. A000073, A049124, A049140, A307411, A307412, A307413, A307529.

terms = 31; CoefficientList[1/x InverseSeries[Series[x (1 - x - x^2 - x^3)/(1 - x - x^3), {x, 0, terms}], x], x]
terms = 30; A[] = 0; Do[A[x] = 1 + x^2 A[x]^2/(1 - x A[x] - x^2 A[x]^2 - x^3 A[x]^3) + O[x]^(terms + 1) // Normal, {terms + 1}]; CoefficientList[A[x], x]
terms = 31; t[n_] := t[n] = SeriesCoefficient[x^2/(1 - x - x^2 - x^3), {x, 0, n}]; A[] = 0; Do[A[x] = 1 + Sum[t[k] x^k A[x]^k, {k, 2, j}] + O[x]^j, {j, 1, terms}]; CoefficientList[A[x], x]

G.f. A(x) satisfies: A(x) = 1 + Sum_{k>=2} A000073(k)*x^k*A(x)^k.

G.f.: A(x) = (1/x)*Series_Reversion(x*(1 - x - x^2 - x^3)/(1 - x - x^3)).

A307529 G.f. A(x) satisfies: A(x) = (1 - x^2A(x)^2)/(1 - x^2A(x)^2 - x^3*A(x)^3).

Original entry on oeis.org

1, 0, 0, 1, 0, 1, 4, 1, 10, 23, 18, 92, 168, 241, 856, 1480, 2904, 8266, 14854, 33496, 83578, 161047, 380488, 884326, 1819714, 4321045, 9730466, 21019404, 49456092, 110408981, 246005440, 572574553, 1281705752, 2906696339, 6711882928, 15128432758, 34625418170

Offset: 0

Ilya Gutkovskiy, Apr 12 2019

nonn

			G.f.: A(x) = 1 + x^3 + x^5 + 4*x^6 + x^7 + 10*x^8 + 23*x^9 + 18*x^10 + 92*x^11 + 168*x^12 + ...

Cf. A000931, A049124, A049140, A067955, A307411, A307412, A307413, A307528.

terms = 37; CoefficientList[1/x InverseSeries[Series[x (1 - x^2 - x^3)/(1 - x^2), {x, 0, terms}], x], x]
terms = 36; A[] = 0; Do[A[x] = (1 - x^2 A[x]^2)/(1 - x^2 A[x]^2 - x^3 A[x]^3) + O[x]^(terms + 1) // Normal, {terms + 1}]; CoefficientList[A[x], x]
terms = 37; p[n_] := p[n] = SeriesCoefficient[(1 - x^2)/(1 - x^2 - x^3), {x, 0, n}]; A[] = 1; Do[A[x] = Sum[p[k] x^k A[x]^k, {k, 0, j}] + O[x]^j, {j, 1, terms}]; CoefficientList[A[x], x]

G.f. A(x) satisfies: A(x) = Sum_{k>=0} A000931(k)*x^k*A(x)^k.

G.f.: A(x) = (1/x)*Series_Reversion(x*(1 - x^2 - x^3)/(1 - x^2)).

A319120 T(n, k) = binomial(n - k - 1, k)binomial(2n - 2*k, n)/(n + 1), for n >= 1 and 0 <= k <= floor((n - 1)/2), triangle read by rows.

Original entry on oeis.org

1, 2, 5, 1, 14, 6, 42, 28, 1, 132, 120, 12, 429, 495, 90, 1, 1430, 2002, 550, 20, 4862, 8008, 3003, 220, 1, 16796, 31824, 15288, 1820, 30, 58786, 125970, 74256, 12740, 455, 1, 208012, 497420, 348840, 79968, 4900, 42

Offset: 1

Colin Defant, Sep 17 2018

T(n,0) = A000108(n).
Let L(u,v) be the set of integer partitions whose Young diagrams fit inside a u by v rectangle. Given lambda in L(u,v), let E(lambda) be the number of partitions whose Young diagrams fit inside the Young diagram of lambda. Also, for 1 <= i <= v, let x_i(lambda)-1 be the number of parts of lambda of length v+1-i. Let x_{v+1}(lambda) = u+v+1-Sum_{i=1..v} x_i(lambda) so that (x_1(lambda), ..., x_{v+1}(lambda)) is a composition of u+v+1 into v+1 parts. Let F(lambda) = Product_{i=1..v+1} Catalan(x_i(lambda)). Conjecturally, T(n,k) = Sum_{lambda in L(n-2k-1)} E(lambda) * F(lambda).
Conjecturally, T(n,k) is the number of permutations pi of [n] such that s(pi) has k descents and avoids the patterns 231, 312, and 321, where s is West's stack-sorting map.
Conjecturally, T(n,k) is the number of permutations pi of [n] that avoid the 4 patterns 4312, 4321, 4231, 3241 (more succinctly, that avoid 32x1 for all x) and contain k entries which are neither left-right maxima nor right-left minima (equivalently, contain k entries that serve as the "2" of a 321). - David Callan, Mar 05 2019

			Triangle begins:
       1;
       2;
       5,      1;
      14,      6;
      42,     28,      1;
     132,    120,     12;
     429,    495,     90,     1;
    1430,   2002,    550,    20;
    4862,   8008,   3003,   220,    1;
   16796,  31824,  15288,  1820,   30;
   58786, 125970,  74256, 12740,  455,  1;
  208012, 497420, 348840, 79968, 4900, 42;
  ...

Colin Defant, Stack-sorting preimages of permutation classes, arXiv:1809.03123 [math.CO], 2018.

Cf. A000108. Row sums give A049124.

Flatten[Table[Table[(1/(n + 1)) Binomial[n - k - 1, k] Binomial[2 n - 2 k, n], {k, 0, Floor[(n - 1)/2]}], {n, 1, 12}]]

T(n,k) = binomial(n-k-1,k) * binomial(2*n-2*k,n)/(n+1);
tabf(nn) = for (n=1, nn, for (k=0, (n-1)\2, print1(T(n,k), ", ")); print); \\ Michel Marcus, Sep 20 2018

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A101786 G.f. satisfies: A(x) = 1 + x*A(x)/(1 - 2*x^2*A(x)^2).

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A307528 G.f. A(x) satisfies: A(x) = 1 + x^2*A(x)^2/(1 - x*A(x) - x^2*A(x)^2 - x^3*A(x)^3).

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A307529 G.f. A(x) satisfies: A(x) = (1 - x^2*A(x)^2)/(1 - x^2*A(x)^2 - x^3*A(x)^3).

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A319120 T(n, k) = binomial(n - k - 1, k)*binomial(2*n - 2*k, n)/(n + 1), for n >= 1 and 0 <= k <= floor((n - 1)/2), triangle read by rows.

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A101786 G.f. satisfies: A(x) = 1 + xA(x)/(1 - 2x^2*A(x)^2).

A307528 G.f. A(x) satisfies: A(x) = 1 + x^2A(x)^2/(1 - xA(x) - x^2A(x)^2 - x^3A(x)^3).

A307529 G.f. A(x) satisfies: A(x) = (1 - x^2A(x)^2)/(1 - x^2A(x)^2 - x^3*A(x)^3).

A319120 T(n, k) = binomial(n - k - 1, k)binomial(2n - 2*k, n)/(n + 1), for n >= 1 and 0 <= k <= floor((n - 1)/2), triangle read by rows.