A119259 -id:A119259 - cp's OEIS frontend

A098665 a(n) = Sum_{k = 0..n} binomial(n,k) * binomial(n+1,k+1) * 4^k.

1, 6, 43, 332, 2661, 21810, 181455, 1526040, 12939145, 110413406, 947052723, 8157680228, 70518067309, 611426078346, 5315138311383, 46308989294640, 404274406256145, 3535479068797110, 30966952059306555, 271616893912241532, 2385412594943633781, 20973327081776664546

Offset: 0

Paul Barry, Sep 20 2004

Fifth binomial transform of A098664.

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Cf. A098664, A119259.

Table[SeriesCoefficient[((1+3*x)-Sqrt[1-10*x+9*x^2])/(8*x*Sqrt[1-10*x+9*x^2]),{x,0,n}],{n,0,20}] (* Vaclav Kotesovec, Oct 15 2012 *)
a[n_] := 4^n*HypergeometricPFQ[{-n, -n - 1}, {1}, 1/4]; Flatten[Table[a[n], {n,0,21}]] (* Detlef Meya, May 21 2024 *)

my(x='x+O('x^66)); Vec(((1+3*x)-sqrt(1-10*x+9*x^2))/(8*x*sqrt(1-10*x+9*x^2))) \\ Joerg Arndt, May 12 2013

G.f.: ((1+3*x)-sqrt(1-10*x+9*x^2))/(8*x*sqrt(1-10*x+9*x^2)).
E.g.f.: exp(5x)*(BesselI(0, 4x)+BesselI(1, 4x)/2).
Recurrence: (n+1)*(2*n-1)*a(n) = 4*(5*n^2-2)*a(n-1) - 9*(n-1)*(2*n+1)*a(n-2). - Vaclav Kotesovec, Oct 15 2012
a(n) ~ 9^(n+1)/(4*sqrt(2*Pi*n)). - Vaclav Kotesovec, Oct 15 2012
From Peter Bala, Jan 07 2022: (Start)
The following formulas assume an offset of 1:
a(n) = (1/4) * Sum_{k = 0..n} binomial(n,k)*A119259(k).
a(n) = (1/4) * Sum_{k = 0..n} binomial(n,k)*binomial(2*n-k-1,n-k)*3^k.
a(n) = (1/4) * [x^n] ((1 + 3*x)/(1 - x))^n.
The Gauss congruences a(n*p^k) == a(n^p^(k-1)) (mod p^k) hold for prime p >= 3 and positive integers n and k. (End)
a(n) = 4^n*hypergeom([-n, -n - 1], [1], 1/4). - Detlef Meya, May 21 2024

A333565 O.g.f.: (1 + 4x)/((1 + x)sqrt(1 - 8*x)).

Original entry on oeis.org

1, 7, 33, 223, 1537, 11007, 80385, 595455, 4456449, 33615871, 255148033, 1946337279, 14908784641, 114597822463, 883479412737, 6828492980223, 52895475040257, 410544577183743, 3191929428770817, 24855137310736383, 193811815161921537, 1513167009951514623, 11827298001565515777

Offset: 0

Peter Bala, Apr 11 2020

This 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, since the power series E(x) := exp( Sum_{n >= 1} a(n)*x^n/n ) has integer coefficients. See Stanley, Ex. 5.2 (a), p. 72, and its solution on p. 104.

We conjecture that this sequence satisfies the stronger congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 3 and positive integers n and k. The particular case when n = k = 1 follows from the corresponding result for A333564. Some examples of these congruences are given below.

			Examples of congruences:
a(11) - a(1) = 1946337279 - 7 = (2^3)*(11^3)*182789 == 0 ( mod 11^3 ).
a(2*11) - a(2) = 11827298001565515777 - 33 = (2^5)*(3^2)*(11^3)*107* 288357478039 == 0 ( mod 11^3 ).
a(5^2) - a(5) = 5680983691406772011007 - 11007 = (2^8)*(3^3)*(5^6)*7* 19*1123*352183001 == 0 ( mod 5^6 ).

R. P. Stanley. Enumerative combinatorics. Vol. 2, (volume 62 of Cambridge Studies in Advanced Mathematics). Cambridge University Press, Cambridge, 1999.

Cf. A000984, A113647, A115137, A119259, A333564.

a := proc (n) option remember; `if`(n = 0, 1, `if`(n = 1, 7, `if`(n = 2, 33, ((3*n+4)*a(n-1)+(36*n-76)*a(n-2)+(32*n-80)*a(n-3))/n)))
end proc:
seq(a(n), n = 0..25);

a[n_] := (-1)^n - 2^(n+2) Binomial[2n, n-1] Hypergeometric2F1[1, 2n +1, n + 2, 2];
Table[Simplify[a[n]], {n, 0, 22}] (* Peter Luschny, Apr 13 2020 *)
CoefficientList[Series[(1+4x)/((1+x)Sqrt[1-8x]),{x,0,30}],x] (* Harvey P. Dale, Jan 24 2021 *)

a(n) = (2^n)*binomial(2*n,n) + 3*sum_{k = 0..n-1} (-1)^(n+k+1)*2^k* binomial(2*k,k).
a(n) = 4*A333564(n) + (-1)^n for n >= 1.
a(n) = 2*A119259(n) - (-1)^n.
a(n) = (-1)^n + 4*Sum_{k = 1..n} (3*k-1)*2^(k-1)*A000108(k-1).
a(n) ~ 8^n * 4/(3*sqrt(Pi*n)).
Congruences: a(p) == 7 ( mod p^3 ) for all prime p >= 3.
O.g.f. A(x) = 1 + 7*x + 33*x^2 + ... satisfies the differential equation (x + 1)*(4*x + 1)*(8*x - 1)*A'(x) + (16*x^2 - 4*x + 7)*A(x) = 0. Cf. A333564.
P-recursive: n*(3*n - 4)*a(n) = (21*n^2 - 40*n + 12)*a(n-1) + 4*(3*n - 1)*(2*n - 3)*a(n-2) with a(0) = 1 and a(1) = 7.
Alternative form: (a(n) + a(n-1))/(a(n) - a(n-2)) = P(n)/Q(n), where P(n) = 4*(3*n - 1)*(2*n - 3) and Q(n) = (21*n^2 - 40*n + 12).
Also, n*a(n) = (3*n + 4)*a(n-1) + 4*(9*n - 19)*a(n-2) + 16*(2*n - 5)*a(n-3) with a(0) = 1, a(1) = 7 and a(2) = 33.
exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 7*x + 41*x^2 + 247*x^3 + ... is the o.g.f. of the second diagonal of triangle A113647. See also A115137.

A333560 Square array read by antidiagonals: T(n,k) = Sum_{j = 0..nk} binomial(n+j-1,j)2^j; n,k >= 0.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 17, 7, 1, 1, 111, 129, 15, 1, 1, 769, 2815, 769, 31, 1, 1, 5503, 65537, 47103, 4097, 63, 1, 1, 40193, 1579007, 3080193, 647167, 20481, 127, 1, 1, 297727, 38862849, 208470015, 109051905, 7929855, 98305, 255, 1, 1, 2228225, 970522623, 14413725697, 19012780031, 3271557121, 90177535, 458753, 511, 1

Offset: 0

Peter Bala, Mar 26 2020

We conjecture that each column sequence satisfies the following supercongruences:

Column k: T(n*p^j, k) == T(n*p^(j-1),k) ( mod p^(3*j) ) for prime p >= 5 and positive integers n and j. Some examples are given below.

			Square array begins
      |k=0    k=1       k=2           k=3             k=4
  - - - - - - - - - - - - - - - - - - - - - - - - - - - -
  n=0 | 1      1         1             1               1
  n=1 | 1      3         7            15              31
  n=2 | 1     17       129           769            4097
  n=3 | 1    111      2815         47103          647167
  n=4 | 1    769     65537       3080193       109051905
  n=5 | 1   5503   1579007     208470015     19012780031
  n=6 | 1  40193  38862849   14413725697   3385776406529
  n=7 | 1 297727 970522623 1011196362751 611732191969279
  ...
Examples of congruences for column k = 1:
T(5,1) - T(1,1) = 5503 - 3 = (2^2)*(5^3)*11 == 0 ( mod 5^3 ).
T(7,1) - T(1,1) = 297727 - 3 = (2^2)*(7^4)*31 == 0 ( mod 7^3 ).
T(2*11,1) - T(2,1) = 5913649000782757889 - 17 = (2^4)*(3^2)*(11^3)*107*288357478039 == 0 ( mod 11^3 ).
T(5^2,1) - T(5,1) = 2840491845703386005503 - 5503 = (2^7)*(3^3)*(5^6)*7*19*1123*352183001 == 0 ( mod 5^6 ).

A119259 (column 1), A333561 (column 2), A333562 (column 3). Cf. A333563.

T := (n, k) -> add(binomial(n+j-1, j)*2^j, j = 0..n*k):
T_col := k -> seq(T(n, k), n = 0..7):
seq(print(T_col(k)), k = 0..10);

T(n,k) = Sum_{j = 0..n*k} binomial(n+j-1,j)*2^j.

Conjectural o.g.f. for column k: 2^(k+1)*x*f'(k,(2^k)*x)/(2*f(k,(2^k)*x) - 1) + 1/(1 + x), where f(k,x) = Sum_{n >= 0} 1/((k+1)*n+1)*C((k+1)*n+1,n)* x^n.

A333561 a(n) = Sum_{j = 0..2n} binomial(n+j-1,j)2^j.

Original entry on oeis.org

1, 7, 129, 2815, 65537, 1579007, 38862849, 970522623, 24494735361, 623210135551, 15956734640129, 410649406472191, 10612705274626049, 275241225206890495, 7159857331658817537, 186731505521384226815, 4880983719142471237633, 127836403093194475044863

Offset: 0

Peter Bala, Mar 27 2020

Column 2 of the square array A333560. Compare with A119259(n) = Sum_{j = 0..n} binomial(n+j-1,j)*2^j.

We conjecture that this sequence satisfies the supercongruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k. Some examples are given below.

			Examples of supercongruences:
a(11) - a(1) = 410649406472191 - 7 = (2^3)*3*(11^3)*12855290711 == 0 ( mod 11^3 ).
a(3*7) - a(3) = 61103847305642669128888090623 - 2815 = (2^8)*(7^5)* 87326419*162627033103121 == 0 ( mod 7^3 ).
a(5^2) - a(5) = 29754989698128108780761000609579007 - 1579007 = (2^11)*(5^6)*179*751*10267*673710468794491483 == 0 ( mod 5^6 ).

Cf. A001764, A062747, A119259, A333560, A333562.

seq(add( binomial(n+j-1,j)*2^j, j = 0..2*n), n = 0..25);

Table[(-1)^n - 2^(2*n+1) * Binomial[3*n, 2*n+1] * Hypergeometric2F1[1, 3*n+1, 2*n+2, 2], {n, 0, 20}] (* Vaclav Kotesovec, Mar 28 2020 *)

a(n) = sum(j = 0, 2*n, binomial(n+j-1,j)*2^j); \\ Michel Marcus, Mar 28 2020

Conjectural o.g.f.: 1/(1 + x) + 8*x*f'(4*x)/(2*f(4*x) - 1), where f(x) = 1 + x + 3*x^2 + 12*x^3 + 55*x^4 + ... is the o.g.f. of A001764.
exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 7*x + 89*x^2 + 1447*x^3 + ... appears to be the o.g.f. of A062747.
Conjectural recurrence: n*(n - 1)*(2*n - 1)*(3098*n - 6455)*a(n) = (n - 1)*(172988*n^3 - 585840*n^2 + 550321*n - 169824)*a(n-1) - 12*(11825*n^4 - 168518*n^3 + 627675*n^2 - 853766*n + 350744)*a(n-2) - 36*(n - 3)*(3*n - 7)*(3*n - 8)*(991*n - 724)*a(n-3) with a(1) = 7, a(2) = 129, a(3) = 2815.
From Vaclav Kotesovec, Mar 28 2020: (Start)
a(n) ~ 3^(3*n + 1/2) / (4*sqrt(Pi*n)).
Recurrence: n*(2*n - 1)*(7*n^2 - 20*n + 14)*a(n) = (364*n^4 - 1411*n^3 + 1818*n^2 - 868*n + 120)*a(n-1) + 6*(3*n - 5)*(3*n - 4)*(7*n^2 - 6*n + 1)*a(n-2). (End)
From Peter Bala, Mar 05 2022: (Start)
a(n) = Sum_{k = 0..2*n} binomial(3*n, 2*n-k)*binomial(n+k-1,k).
a(n) = [x^(2*n)] ( (1 + x^3)/(1 - x) )^n.
The o.g.f. satisfies the algebraic equation (108*x^3 + 212*x^2 + 100*x - 4)*A(x)^3 - (216*x^2 + 208*x - 8)*A(x)^2 + (48*x^2 + 155*x - 5)*A(x) + 8*x^2 - 40*x + 1 = 0. (End)
a(n) = binomial(3*n, 2*n)*hypergeom([-2*n, n], [n + 1], -1). - Peter Luschny, Mar 07 2022

A333562 a(n) = Sum_{j = 0..3n} binomial(n+j-1,j)2^j.

Original entry on oeis.org

1, 15, 769, 47103, 3080193, 208470015, 14413725697, 1011196362751, 71695889072129, 5124481173422079, 368599603785760769, 26648859989512290303, 1934777421539431153665, 140966705275001764839423, 10301634747725237826093057, 754776795329691207916847103

Offset: 0

Peter Bala, Mar 27 2020

Column 3 of the square array A333560. Compare with A119259(n) = Sum_{j = 0..n} binomial(n+j-1,j)*2^j.

We conjecture that this sequence satisfies the congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k. Some examples are given below.

			Examples of congruences:
a(11) - a(1) = 26648859989512290303 - 15 = (2^4)*3*(11^3)*417118394526551 == 0 ( mod 11^3 ).
a(3*7) - a(3) = 121414496850169263529624169428526563327 - 47103 = (2^11)*(7^4)*24691554473186884926207539141513 == 0 ( mod 7^3 ).
a(5^2) - a(5) = 3682696038139661781421472944275523824848470015 - 208470015 = (2^16)*(5^7)*71*1315737187*37481160881*205425986821331 == 0 ( mod 5^6 ).

Cf. A002293, A062752, A119259, A333560, A333561.

seq(add( binomial(n+j-1,j)*2^j, j = 0..3*n), n = 0..25);

Table[(-1)^n - 2^(3*n+1) * Binomial[4*n, 3*n+1] * Hypergeometric2F1[1, 4*n+1, 3*n+2, 2], {n, 0, 15}] (* Vaclav Kotesovec, Mar 28 2020 *)

a(n) = sum(j = 0, 3*n, binomial(n+j-1,j)*2^j); \\ Michel Marcus, Mar 28 2020

Conjectural o.g.f.: 1/(1 + x) + 16*x*f'(8*x)/(2*f(8*x) - 1), where f(x) = 1 + x + 4*x^2 + 22*x^3 + 140*x^4 + ... is the o.g.f. of A002293.
exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 15*x + 497*x^2 + 22031*x^3 + ... appears to be the o.g.f. of A062752.
a(n) ~ 2^(11*n + 3/2) / (5*sqrt(Pi*n) * 3^(3*n + 1/2)). - Vaclav Kotesovec, Mar 28 2020

A259557 a(n) = binomial(4n-1, 2n).

Original entry on oeis.org

1, 3, 35, 462, 6435, 92378, 1352078, 20058300, 300540195, 4537567650, 68923264410, 1052049481860, 16123801841550, 247959266474052, 3824345300380220, 59132290782430712, 916312070471295267, 14226520737620288370

Offset: 0

Vladimir Kruchinin, Jun 30 2015

Essentially the same as A100033.

V. V. Kruchinin and D. V. Kruchinin, A Generating Function for the Diagonal T_{2n,n} in Triangles, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.6.

Cf. A001700, A079489, A100033.

[Binomial(4*n-1, 2*n): n in [0..20]]; // Vincenzo Librandi, Jul 01 2015

Table[Binomial[4 n - 1, 2 n], {n, 0, 30}] (* Vincenzo Librandi, Jul 01 2015 *)

vector(20, n, n--; binomial(4*n-1, 2*n)) \\ Michel Marcus, Jul 01 2015

G.f. A(x)=1+x*B(x)'/B(x), where B(x) is g.f. of A079489.
a(n) = A100033(n-1) for n>0.
D-finite with recurrence n*(2*n-1)*a(n) -2*(4*n-1)*(4*n-3)*a(n-1)=0. - R. J. Mathar, Jul 06 2015
a(n) = [x^(2*n)] 1/(1 - x)^(2*n). - Ilya Gutkovskiy, Oct 10 2017
From Peter Bala, Jun 11 2023: (Start)
a(n) = (1/2) * [x^n] ( (1 + x)^2/( 1 - x) )^(2*n) for n >= 1.
Right-hand side of the identity (1/2)*Sum_{k = 0..n} binomial(4*n,k)*binomial(3*n-k-1,n-k) = binomial(4*n-1,2*n) for n >= 1.
a(n) = [x^n] E(x)^n, where E(x) = exp( Sum_{k >= 1} A119259(k)*x^k/k ). (End)
a(n) = Sum_{k = 0..2*n} (-1)^k*binomial(-n, k)*binomial(-3*n-k, 2*n-k) = Sum_{k = 0..2*n} (-1)^k*binomial(n+k-1, k)*binomial(5*n-1, 2*n-k). - Peter Bala, Jun 08 2024

A207327 Riordan array (1, x*(1+x)^2/(1-x)).

Original entry on oeis.org

1, 0, 1, 0, 3, 1, 0, 4, 6, 1, 0, 4, 17, 9, 1, 0, 4, 32, 39, 12, 1, 0, 4, 48, 111, 70, 15, 1, 0, 4, 64, 240, 268, 110, 18, 1, 0, 4, 80, 432, 769, 530, 159, 21, 1, 0, 4, 96, 688, 1792, 1905, 924, 217, 24, 1, 0, 4

Offset: 0

Philippe Deléham, Feb 17 2012

Triangle T(n,k), read by rows, given by (0, 3, -5/3, 4/15, -3/5, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938.

Row sums are A077995(n).

			Triangle begins :
1
0, 1
0, 3, 1
0, 4, 6, 1
0, 4, 17, 9, 1
0, 4, 32, 39, 12, 1
0, 4, 48, 111, 70, 15, 1
0, 4, 64, 240, 268, 110, 18, 1
0, 4, 80, 432, 769, 530, 159, 21, 1
0, 4, 96, 688, 1792, 1905, 924, 217, 24, 1
0, 4, 112, 1008, 3584, 5503, 3999, 1477, 284, 27, 1
0, 4, 128, 1392, 6400, 13440, 13842, 7483, 2216, 360, 30, 1

Cf. Diagonals : A000012, A008585, A022266, A000007, A113311

T(2*n,n) = A119259(n).
G.f.: (1-x)/(1-(1+y)*x-2*y*x^2-y*x^3).
T(n,k) = T(n-1,k) + T(n-1,k-1) + 2*T(n-2,k-1) + T(n-3,k-1), T(0,0) = 1, T(1,0) = 0.

cp's OEIS Frontend

A098665 a(n) = Sum_{k = 0..n} binomial(n,k) * binomial(n+1,k+1) * 4^k.

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A333565 O.g.f.: (1 + 4*x)/((1 + x)*sqrt(1 - 8*x)).

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A333560 Square array read by antidiagonals: T(n,k) = Sum_{j = 0..n*k} binomial(n+j-1,j)*2^j; n,k >= 0.

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A333561 a(n) = Sum_{j = 0..2*n} binomial(n+j-1,j)*2^j.

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A333562 a(n) = Sum_{j = 0..3*n} binomial(n+j-1,j)*2^j.

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A259557 a(n) = binomial(4*n-1, 2*n).

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A207327 Riordan array (1, x*(1+x)^2/(1-x)).

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A333565 O.g.f.: (1 + 4x)/((1 + x)sqrt(1 - 8*x)).

A333560 Square array read by antidiagonals: T(n,k) = Sum_{j = 0..nk} binomial(n+j-1,j)2^j; n,k >= 0.

A333561 a(n) = Sum_{j = 0..2n} binomial(n+j-1,j)2^j.

A333562 a(n) = Sum_{j = 0..3n} binomial(n+j-1,j)2^j.

A259557 a(n) = binomial(4n-1, 2n).