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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A098474 Triangle read by rows, T(n,k) = C(n,k)*C(2*k,k)/(k+1), n >= 0, 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 1, 3, 6, 5, 1, 4, 12, 20, 14, 1, 5, 20, 50, 70, 42, 1, 6, 30, 100, 210, 252, 132, 1, 7, 42, 175, 490, 882, 924, 429, 1, 8, 56, 280, 980, 2352, 3696, 3432, 1430, 1, 9, 72, 420, 1764, 5292, 11088, 15444, 12870, 4862, 1, 10, 90, 600, 2940, 10584, 27720
Offset: 0

Views

Author

Paul Barry, Sep 09 2004

Keywords

Comments

A Catalan scaled binomial matrix.
From Philippe Deléham, Sep 01 2005: (Start)
Table U(n,k), k >= 0, n >= 0, read by antidiagonals, begins:
row k = 0: 1, 1, 2, 5, 14, ... is A000108
row k = 1: 1, 2, 6, 20, 70, ... is A000984
row k = 2: 1, 3, 12, 50, 280, ... is A007854
row k = 3: 1, 4, 20, 104, 548, ... is A076035
row k = 4: 1, 5, 30, 185, 1150, ... is A076036
G.f. for row k: 1/(1-(k+1)*x*C(x)) where C(x) is the g.f. = for Catalan numbers A000108.
U(n,k) = Sum_{j=0..n} A106566(n,j)*(k+1)^j. (End)
This sequence gives the coefficients (increasing powers of x) of the Jensen polynomials for the Catalan sequence A000108 of degree n and shift 0. For the definition of Jensen polynomials for a sequence see a comment in A094436. - Wolfdieter Lang, Jun 25 2019

Examples

			Rows begin:
  1;
  1, 1;
  1, 2,  2;
  1, 3,  6,   5;
  1, 4, 12,  20,  14;
  1, 5, 20,  50,  70,  42;
  1, 6, 30, 100, 210, 252, 132;
  ...
Row 3: t*(1 - 3*t + 6*t^2 - 5*t^3)/(1 - 4*t)^(9/2) = 1/2*Sum_{k >= 1} k*(k+1)*(k+2)*(k+3)/4!*binomial(2*k,k)*t^k. - _Peter Bala_, Jun 13 2016

Links

Crossrefs

Row sums are A007317.
Antidiagonal sums are A090344.
Principal diagonal is A000108.
Mirror image of A124644.
Cf. A000984, A007854, A011973, A076035, A076036, A098443, A098473, A106566, A267633.

Programs

  • Maple
    p := proc(n) option remember; if n = 0 then 1 else normal((x*(1 + 4*x)*diff(p(n-1, x), x) + (2*x + n + 1)*p(n-1, x))/(n + 1)) fi end:
row := n -> local k; seq(coeff(p(n), x, k), k = 0..n):
for n from 0 to 6 do row(n) od;  # Peter Luschny, Jun 21 2023
  • Mathematica
    Table[Binomial[n, k] Binomial[2 k, k]/(k + 1), {n, 0, 10}, {k, 0, n}] // Flatten (* or *)
Table[(-1)^k*CatalanNumber[k] Pochhammer[-n, k]/k!, {n, 0, 10}, {k, 0, n}] // Flatten (* Michael De Vlieger, Feb 17 2017 *)
  • Python
    from functools import cache
@cache
def A098474row(n: int) -> list[int]:
    if n == 0: return [1]
    a = A098474row(n - 1) + [0]
    row = [0] * (n + 1)
    row[0] = 1; row[1] = n
    for k in range(2, n + 1):
        row[k] = (a[k] * (n + k + 1) + a[k - 1] * (4 * k - 2)) // (n + 1)
    return row  # Peter Luschny, Jun 22 2023
  • Sage
    def A098474(n,k):
    return (-1)^k*catalan_number(k)*rising_factorial(-n,k)/factorial(k)
for n in range(7): [A098474(n,k) for k in (0..n)] # Peter Luschny, Feb 05 2015

Formula

G.f.: 2/(1-x+(1-x-4*x*y)^(1/2)). - Vladeta Jovovic, Sep 11 2004
E.g.f.: exp(x*(1+2*y))*(BesselI(0, 2*x*y)-BesselI(1, 2*x*y)). - Vladeta Jovovic, Sep 11 2004
G.f.: 1/(1-x-xy/(1-xy/(1-x-xy/(1-xy/(1-x-xy/(1-xy/(1-x-xy/(1-xy/(1-... (continued fraction). - Paul Barry, Feb 11 2009
Sum_{k=0..n} T(n,k)*x^(n-k) = A126930(n), A005043(n), A000108(n), A007317(n+1), A064613(n), A104455(n) for x = -2, -1, 0, 1, 2, 3 respectively. - Philippe Deléham, Dec 12 2009
T(n,k) = (-1)^k*Catalan(k)*Pochhammer(-n,k)/k!. - Peter Luschny, Feb 05 2015
O.g.f.: [1 - sqrt(1-4tx/(1-x))]/(2tx) = 1 + (1+t) x + (1+2t+2t^2) x^2 + (1+3t+6t^2+5t^3) x^3 + ... , generating the polynomials of this entry, reverse of A124644. See A011973 for a derivation and the inverse o.g.f., connected to the Fibonacci, Chebyshev, and Motzkin polynomials. See also A267633. - Tom Copeland, Jan 25 2016
From Peter Bala, Jun 13 2016: (Start)
The o.g.f. F(x,t) = ( 1 - sqrt(1 - 4*t*x/(1 - x)) )/(2*t*x) satisfies the partial differential equation d/dx(x*(1 - x)*F) - x*t*(1 + 4*t)*dF/dt - 2*x*t*F = 1. This gives a recurrence for the row polynomials: (n + 2)*R(n+1,t) = t*(1 + 4*t)*R'(n,t) + (2*t + n + 2)*R(n,t), where the prime ' indicates differentiation with respect to t.
Equivalently, setting Q(n,t) = t^(n+2)*R(n,-t)/(1 - 4*t)^(n + 3/2) we have t^2*d/dt(Q(n,t)) = (n + 2)*Q(n+1,t).
This leads to the following expansions:
Q(0,t) = (1/2)*Sum_{k >= 1} k*binomial(2*k,k)*t^(k+1)
Q(1,t) = (1/2)*Sum_{k >= 1} k*(k+1)/2!*binomial(2*k,k)*t^(k+2)
Q(2,t) = (1/2)*Sum_{k >= 1} k*(k+1)*(k+2)/3!*binomial(2*k,k) *t^(k+3) and so on. (End)
Sum_{k=0..n} T(n,k)*x^k = A007317(n+1), A162326(n+1), A337167(n) for x = 1, 2, 3 respectively. - Sergii Voloshyn, Mar 31 2022

Extensions

New name using a formula of Paul Barry by Peter Luschny, Feb 05 2015

A124644 Triangle read by rows. T(n, k) = binomial(n, k) * CatalanNumber(n - k).

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 5, 6, 3, 1, 14, 20, 12, 4, 1, 42, 70, 50, 20, 5, 1, 132, 252, 210, 100, 30, 6, 1, 429, 924, 882, 490, 175, 42, 7, 1, 1430, 3432, 3696, 2352, 980, 280, 56, 8, 1, 4862, 12870, 15444, 11088, 5292, 1764, 420, 72, 9, 1, 16796, 48620, 64350, 51480, 27720
Offset: 0

Views

Author

Farkas Janos Smile (smile_farkasjanos(AT)yahoo.com.au), Dec 21 2006

Keywords

Comments

Equal to A091867*A007318. - Philippe Deléham, Dec 12 2009
Exponential Riordan array [exp(2x)*(Bessel_I(0,2x)-Bessel_I(1,2x)),x]. - Paul Barry, Mar 03 2011
From Tom Copeland, Nov 04 2014: (Start)
O.g.f: G(x,t) = C[Pinv(x,t)] = {1 - sqrt[1 - 4 *x /(1-x*t)]}/2 where C(x) = [1 - sqrt(1-4x)]/2, an o.g.f. for the shifted Catalan numbers A000108 with inverse Cinv(x) = x*(1-x), and Pinv(x,t)= -P(-x,t) = x/(1-t*x) with inverse P(x,t) = 1/(1+t*x). This puts this array in a family of arrays formed from the composition of C and P and their inverses. -G(-x,t) is the comp. inverse of the o.g.f. of A030528.
This is an Appell sequence with lowering operator d/dt p(n,t) = n*p(n-1,t) and (p(.,t)+a)^n = p(n,t+a). The e.g.f. has the form e^(x*t)/w(t) where 1/w(t) is the e.g.f. of the first column, which is the Catalan sequence A000108. (End)

Examples

			From _Paul Barry_, Jan 28 2009: (Start)
Triangle begins
   1,
   1,  1,
   2,  2,  1,
   5,  6,  3,  1,
  14, 20, 12,  4,  1,
  42, 70, 50, 20,  5,  1 (End)

Links

Crossrefs

Cf. A098474 (mirror image), A000108, A091867, A030528, A104597.
Row sums give A007317(n+1).

Programs

  • Maple
    m:=n->binomial(2*n, n)/(n+1): T:=proc(n, k) if k<=n then binomial(n, k)*m(n-k) else 0 fi end: for n from 0 to 10 do seq(T(n, k), k=0..n) od;
  • Mathematica
    Table[Binomial[n, #] Binomial[2 #, #]/(# + 1) &[n - k], {n, 0, 10}, {k, 0, n}] // Flatten (* or *)
Table[Abs[(-1)^k*CatalanNumber[#] Pochhammer[-n, #]/#!] &[n - k], {n, 0, 10}, {k, 0, n}] // Flatten (* Michael De Vlieger, Feb 17 2017 *)
  • Sage
    def A124644(n,k):
    return (-1)^(n-k)*catalan_number(n-k)*rising_factorial(-n,n-k)/factorial(n-k)
for n in range(7): [A124644(n,k) for k in (0..n)] # Peter Luschny, Feb 05 2015

Formula

T(n,k) = [x^(n-k)]F(-n,n-k+1;1;-1-x). - Paul Barry, Sep 05 2008
G.f.: 1/(1-xy-x/(1-x/(1-xy-x/(1-x/(1-xy-x/(1-x.... (continued fraction). - Paul Barry, Jan 06 2009
G.f.: 1/(1-x-xy-x^2/(1-2x-xy-x^2/(1-2x-xy-x^2/(1-.... (continued fraction). - Paul Barry, Jan 28 2009
T(n,k) = Sum_{i = 0..n} C(n,i)*(-1)^(n-i)*Sum{j = 0..i} C(j,k)*C(i,j)*A000108(i-j). - Paul Barry, Aug 03 2009
Sum_{k = 0..n} T(n,k)*x^k = A126930(n), A005043(n), A000108(n), A007317(n+1), A064613(n), A104455(n) for x = -2, -1, 0, 1, 2, 3 respectively. T(n,k)= A007318(n,k)*A000108(n-k). - Philippe Deléham, Dec 12 2009
E.g.f.: exp(2*x + x*y)*(Bessel_I(0,2*x) - Bessel_I(1,2*x)). - Paul Barry, Mar 10 2010
From Tom Copeland, Nov 08 2014: (Start)
O.g.f.: G(x,t) = C[P(x,t)] = [1 - sqrt(1-4*x / (1-t*x))] / 2 = Sum_{n >= 1} (C. + t)^(n-1) * x^n] = x + (1 + t) x^2 + (2 + 2t + t^2) x^3 + ... umbrally, where (C.)^n = C_n = (1,1,2,5,8,...) = A000108(x), C(x)= x*A000108(x)= G(x,0), and P(x,t) = x/(1 + t*x), a special linear fractional (Mobius) transformation. P(x,-t)= -P(-x,t) is the inverse of P(x,t).
Inverse o.g.f.: Ginv(x,t) = P[Cinv(x),-t] = x*(1-x) / [1 - t*x(1-x)] = -A030528(-x,t), where Cinv(x) = x*(1-x) is the inverse of C(x).
G(x,t) = x*A091867(x,t+1), and Ginv(x,t) = x*A104597(x,-(t+1)). (End)
T(n, k) = (-1)^(n-k)*Catalan(n-k)*Pochhammer(-n,n-k)/(n-k)!. - Peter Luschny, Feb 05 2015
Recurrence: T(n, 0) = Catalan(n) = 1/(n+1)*binomial(2*n, n) and, for 1 <= k <= n, T(n, k) = (n/k) * T(n-1, k-1). - Peter Bala, Feb 04 2024

Extensions

Name brought in line with the Maple program by Peter Luschny, Jun 21 2023

A346762 G.f. A(x) satisfies: A(x) = 1 / (1 - 2*x) + x * (1 - 2*x) * A(x)^3.

Original entry on oeis.org

1, 3, 11, 50, 271, 1655, 10900, 75388, 539295, 3954593, 29557251, 224308078, 1723659436, 13384272660, 104855628776, 827760536528, 6578127170319, 52581460222645, 422478996770305, 3410174204693310, 27640220748529799, 224866485110361767, 1835589569664256976
Offset: 0

Views

Author

Ilya Gutkovskiy, Aug 02 2021

Keywords

Comments

Second binomial transform of A001764.

Crossrefs

Cf. A001764, A064613, A188687, A346763.

Programs

  • Mathematica
    nmax = 22; A[] = 0; Do[A[x] = 1/(1 - 2 x) + x (1 - 2 x) A[x]^3 + O[x]^(nmax + 1) // Normal,nmax + 1]; CoefficientList[A[x], x]
Table[Sum[Binomial[n, k] Binomial[3 k, k] 2^(n - k)/(2 k + 1), {k, 0, n}], {n, 0, 22}]
Table[2^n HypergeometricPFQ[{1/3, 2/3, -n}, {1, 3/2}, -27/8], {n, 0, 22}]

Formula

a(n) = Sum_{k=0..n} binomial(n,k) * binomial(3*k,k) * 2^(n-k) / (2*k + 1).
a(n) ~ 35^(n + 3/2) / (81 * sqrt(Pi) * n^(3/2) * 4^(n+1)). - Vaclav Kotesovec, Nov 26 2021

A349533 G.f. A(x) satisfies A(x) = 1 / ((1 - 2 * x) * (1 - x * A(x)^2)).

Original entry on oeis.org

1, 3, 13, 74, 499, 3719, 29494, 243888, 2078431, 18122369, 160885449, 1449268478, 13213370392, 121696581804, 1130565483472, 10581614352704, 99685591788687, 944490400760597, 8994266558594671, 86040075341770806, 826423263373253923, 7967095415955791687
Offset: 0

Views

Author

Ilya Gutkovskiy, Nov 21 2021

Keywords

Links

Crossrefs

Cf. A001764, A064613, A199475, A349253, A349534, A349535.

Programs

  • Mathematica
    nmax = 21; A[] = 0; Do[A[x] = 1/((1 - 2 x) (1 - x A[x]^2)) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
a[n_] := a[n] = 2^n + Sum[Sum[a[i] a[j] a[n - i - j - 1], {j, 0, n - i - 1}], {i, 0, n - 1}]; Table[a[n], {n, 0, 21}]
Table[Sum[Binomial[n + k, 2 k] Binomial[3 k, k] 2^(n - k)/(2 k + 1), {k, 0, n}], {n, 0, 21}]

Formula

a(n) = 2^n + Sum_{i=0..n-1} Sum_{j=0..n-i-1} a(i) * a(j) * a(n-i-j-1).
a(n) = Sum_{k=0..n} binomial(n+k,2*k) * binomial(3*k,k) * 2^(n-k) / (2*k+1).
a(n) = 2^n*F([1/3, 2/3, -n, 1+n], [1/2, 1, 3/2], -3^3/2^5), where F is the generalized hypergeometric function. - Stefano Spezia, Nov 21 2021
a(n) ~ 177^(1/4) * (43 + 3*sqrt(177))^(n + 1/2) / (9 * sqrt(Pi) * n^(3/2) * 2^(3*n + 5/2)). - Vaclav Kotesovec, Nov 22 2021

A349581 G.f. A(x) satisfies: A(x) = 1 / (1 - 2*x) + x * (1 - 2*x)^2 * A(x)^4.

Original entry on oeis.org

1, 3, 12, 66, 460, 3681, 31848, 289176, 2714044, 26103468, 255876048, 2546717454, 25666830724, 261407935366, 2686191839232, 27815564456544, 289960011573212, 3040424427011492, 32046741183678288, 339345854532800136, 3608307717155678256, 38511520730570169033
Offset: 0

Views

Author

Ilya Gutkovskiy, Nov 22 2021

Keywords

Comments

Second binomial transform of A002293.

Crossrefs

Cf. A002293, A064613, A346646, A346762, A349582, A349584, A349590, A349591.

Programs

  • Mathematica
    nmax = 21; A[] = 0; Do[A[x] = 1/(1 - 2 x) + x (1 - 2 x)^2 A[x]^4 + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
Table[Sum[Binomial[n, k] Binomial[4 k, k] 2^(n - k)/(3 k + 1), {k, 0, n}], {n, 0, 21}]
  • PARI
    a(n) = sum(k=0, n, binomial(n,k)*binomial(4*k,k)*2^(n-k)/(3*k+1)); \\ Michel Marcus, Nov 23 2021

Formula

a(n) = Sum_{k=0..n} binomial(n,k) * binomial(4*k,k) * 2^(n-k) / (3*k+1).
a(n) = 2^n*F([1/4, 1/2, 3/4, -n], [2/3, 1, 4/3], -2^7/3^3), where F is the generalized hypergeometric function. - Stefano Spezia, Nov 22 2021
a(n) ~ 2^(n - 10) * 155^(n + 3/2) / (sqrt(Pi) * n^(3/2) * 3^(3*n + 3/2)). - Vaclav Kotesovec, Nov 26 2021

A349582 G.f. A(x) satisfies: A(x) = 1 / (1 - 2*x) + x * (1 - 2*x)^3 * A(x)^5.

Original entry on oeis.org

1, 3, 13, 85, 733, 7292, 78267, 880250, 10226237, 121713373, 1476272394, 18180126906, 226704989103, 2856790765238, 36321840773980, 465362291912648, 6002272018481901, 77873186277771107, 1015583616140910999, 13306207249869273003, 175064043975233981626
Offset: 0

Views

Author

Ilya Gutkovskiy, Nov 22 2021

Keywords

Comments

Second binomial transform of A002294.

Crossrefs

Cf. A002294, A064613, A346647, A346762, A349581, A349584, A349590, A349591.

Programs

  • Mathematica
    nmax = 20; A[] = 0; Do[A[x] = 1/(1 - 2 x) + x (1 - 2 x)^3 A[x]^5 + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
Table[Sum[Binomial[n, k] Binomial[5 k, k] 2^(n - k)/(4 k + 1), {k, 0, n}], {n, 0, 20}]
  • PARI
    a(n) = sum(k=0, n, binomial(n,k)*binomial(5*k,k)*2^(n-k)/(4*k+1)); \\ Michel Marcus, Nov 23 2021

Formula

a(n) = Sum_{k=0..n} binomial(n,k) * binomial(5*k,k) * 2^(n-k) / (4*k+1).
a(n) = 2^n*F([1/5, 2/5, 3/5, 4/5, -n], [1/2, 3/4, 1, 5/4], -5^5/2^9), where F is the generalized hypergeometric function. - Stefano Spezia, Nov 22 2021
a(n) ~ 3637^(n + 3/2) / (78125 * sqrt(Pi) * n^(3/2) * 2^(8*n + 7/2)). - Vaclav Kotesovec, Nov 26 2021

A349584 G.f. A(x) satisfies: A(x) = 1 / (1 - 2*x) + x * (1 - 2*x)^4 * A(x)^6.

Original entry on oeis.org

1, 3, 14, 107, 1106, 13173, 168820, 2264298, 31356818, 444803666, 6429510234, 94356870748, 1402149248128, 21055387206719, 319007902203196, 4870481885025752, 74858763620576738, 1157339247553310574, 17985974981514604660, 280813589679135551721
Offset: 0

Views

Author

Ilya Gutkovskiy, Nov 22 2021

Keywords

Comments

Second binomial transform of A002295.

Crossrefs

Cf. A002295, A064613, A346648, A346762, A349581, A349582, A349590, A349591.

Programs

  • Mathematica
    nmax = 19; A[] = 0; Do[A[x] = 1/(1 - 2 x) + x (1 - 2 x)^4 A[x]^6 + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
Table[Sum[Binomial[n, k] Binomial[6 k, k] 2^(n - k)/(5 k + 1), {k, 0, n}], {n, 0, 19}]
  • PARI
    a(n) = sum(k=0, n, binomial(n,k)*binomial(6*k,k)*2^(n-k)/(5*k+1)); \\ Michel Marcus, Nov 23 2021

Formula

a(n) = Sum_{k=0..n} binomial(n,k) * binomial(6*k,k) * 2^(n-k) / (5*k+1).
a(n) = 2^n*F([1/6, 1/3, 1/2, 2/3, 5/6, -n], [2/5, 3/5, 4/5, 1, 6/5], -3^6*(2/5)^5), where F is the generalized hypergeometric function. - Stefano Spezia, Nov 22 2021
a(n) ~ 2^(n - 15/2) * 26453^(n + 3/2) / (6561 * sqrt(3*Pi) * n^(3/2) * 5^(5*n + 3/2)). - Vaclav Kotesovec, Nov 26 2021

A349590 G.f. A(x) satisfies: A(x) = 1 / (1 - 2*x) + x * (1 - 2*x)^5 * A(x)^7.

Original entry on oeis.org

1, 3, 15, 132, 1595, 22134, 329718, 5136028, 82579819, 1359902823, 22818697128, 388728802702, 6705324823466, 116878939752376, 2055505806198352, 36427660285955808, 649894104351874395, 11662729497015257677, 210383830525447606431, 3812719304673511150854
Offset: 0

Views

Author

Ilya Gutkovskiy, Nov 22 2021

Keywords

Comments

Second binomial transform of A002296.

Crossrefs

Cf. A002296, A064613, A346649, A346762, A349581, A349582, A349584, A349591.

Programs

  • Mathematica
    nmax = 19; A[] = 0; Do[A[x] = 1/(1 - 2 x) + x (1 - 2 x)^5 A[x]^7 + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
Table[Sum[Binomial[n, k] Binomial[7 k, k] 2^(n - k)/(6 k + 1), {k, 0, n}], {n, 0, 19}]
  • PARI
    a(n) = sum(k=0, n, binomial(n,k)*binomial(7*k,k)*2^(n-k)/(6*k+1)); \\ Michel Marcus, Nov 23 2021

Formula

a(n) = Sum_{k=0..n} binomial(n,k) * binomial(7*k,k) * 2^(n-k) / (6*k+1).
a(n) = 2^n*F([1/7, 2/7, 3/7, 4/7, 5/7, 6/7, -n], [1/3, 1/2, 2/3, 5/6, 1, 7/6], -7^7/(2^7*3^6)), where F is the generalized hypergeometric function. - Stefano Spezia, Nov 22 2021
a(n) ~ 916855^(n + 3/2) / (282475249 * sqrt(Pi) * n^(3/2) * 3^(6*n + 3/2) * 4^(3*n + 1)). - Vaclav Kotesovec, Nov 26 2021

A349591 G.f. A(x) satisfies: A(x) = 1 / (1 - 2*x) + x * (1 - 2*x)^6 * A(x)^8.

Original entry on oeis.org

1, 3, 16, 160, 2216, 35110, 596016, 10573748, 193586424, 3629709697, 69342483276, 1344897261828, 26411276859800, 524117511080056, 10493756451964088, 211719733855698808, 4300202981875132408, 87854045612854431128, 1804215079309443709632
Offset: 0

Views

Author

Ilya Gutkovskiy, Nov 22 2021

Keywords

Comments

Second binomial transform of A007556.

Crossrefs

Cf. A007556, A064613, A346650, A346762, A349581, A349582, A349584, A349590.

Programs

  • Mathematica
    nmax = 18; A[] = 0; Do[A[x] = 1/(1 - 2 x) + x (1 - 2 x)^6 A[x]^8 + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
Table[Sum[Binomial[n, k] Binomial[8 k, k] 2^(n - k)/(7 k + 1), {k, 0, n}], {n, 0, 18}]
  • PARI
    a(n) = sum(k=0, n, binomial(n,k)*binomial(8*k,k)*2^(n-k)/(7*k+1)); \\ Michel Marcus, Nov 23 2021

Formula

a(n) = Sum_{k=0..n} binomial(n,k) * binomial(8*k,k) * 2^(n-k) / (7*k+1).
a(n) = 2^n*F([1/8, 1/4, 3/8, 1/2, 5/8, 3/4, 7/8, -n], [2/7, 3/7, 4/7, 5/7, 6/7, 1, 8/7], -2^23/7^7), where F is the generalized hypergeometric function. - Stefano Spezia, Nov 22 2021
a(n) ~ 2^(n - 67/2) * 9212151^(n + 3/2) / (sqrt(Pi) * n^(3/2) * 7^(7*n + 3/2)). - Vaclav Kotesovec, Nov 26 2021

A349603 a(n) = Sum_{k=0..n} binomial(n,k) * A000108(k) * k^(n-k).

Original entry on oeis.org

1, 1, 4, 20, 126, 937, 7938, 74909, 775022, 8688827, 104608026, 1342844846, 18273663268, 262347913479, 3957524475778, 62511713866200, 1030842278673510, 17700339693712731, 315740112103311666, 5839137279831300536, 111749137533005481700, 2209538389126578658875
Offset: 0

Views

Author

Vaclav Kotesovec, Nov 23 2021

Keywords

Crossrefs

Cf. A000248, A007317, A064613, A104455, A104498, A154623, A292632.

Programs

  • Mathematica
    Join[{1}, Table[Sum[Binomial[n, j]*CatalanNumber[j]*j^(n-j), {j, 0, n}], {n, 1, 25}]]
  • PARI
    a(n) = sum(k=0, n, binomial(n,k)*(binomial(2*k,k)/(k+1))*k^(n-k)); \\ Michel Marcus, Nov 23 2021
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