A234839 -id:A234839 - cp's OEIS frontend

A268303 Composite numbers n such that Sum_{k = 0..n} (-1)^k * C(n,k) * C(2*n,k) == -1 (mod n^3) (see A234839).

10, 25, 146, 586, 2186, 2386, 2594, 2642, 4162, 4226, 4258, 5186, 7745, 8258, 8354, 8458, 8714, 8746, 8842, 10306, 10378, 10786, 10826, 10834, 10898, 16418, 16546, 16706, 17026, 17674, 20546, 20642, 20738, 32834, 32906, 33322, 33505, 33802, 34058, 35338

Offset: 1

Michel Marcus, Jan 31 2016

nonn

A234839(p) == -1 (mod p^3) for all primes >= 5. But some composites also satisfy this property. They are the object of this sequence.

It appears that these composite are semiprimes with one factor always 2 or 5. See "3. Composite solutions of (1.5)" section in Chamberland et al. link.

Marc Chamberland and Karl Dilcher, A Binomial Sum Related to Wolstenholme's Theorem, J. Number Theory, Vol. 171, Issue 11 (Nov. 2009), pp. 2659-2672. See Table 1 p. 2666.

Cf. A234839.

isok(n) = Mod(sum(k=0, n, (-1)^k*binomial(n,k)*binomial(2*n,k)), n^3) == Mod(-1, n^3);
lista(nn) = forcomposite(n=2, nn, if (isok(n), print1(n, ", ")));

A348410 Number of nonnegative integer solutions to n = Sum_{i=1..n} (a_i + b_i), with b_i even.

Original entry on oeis.org

1, 1, 5, 19, 85, 376, 1715, 7890, 36693, 171820, 809380, 3830619, 18201235, 86770516, 414836210, 1988138644, 9548771157, 45948159420, 221470766204, 1069091485500, 5167705849460, 25009724705460, 121171296320475, 587662804774890, 2852708925078675, 13859743127937876

Offset: 0

César Eliud Lozada, Oct 17 2021

Suppose n objects are to be distributed into 2n baskets, half of these white and half black. White baskets may contain 0 or any number of objects, while black baskets may contain 0 or an even number of objects. a(n) is the number of distinct possible distributions.

			Some examples (semicolon separates white basket from black baskets):
For n=1: {{1 ; 0}} - Total possible ways: 1.
For n=2: {{0, 0 ; 0, 2}, {0, 0 ; 2, 0}, {0, 2 ; 0, 0}, {1, 1 ; 0, 0}, {2, 0 ; 0, 0}} - Total possible ways: 5.

Alois P. Heinz, Table of n, a(n) for n = 0..1441

Cf. A033715, A033716, A034297, A063020, A088218, A234839, A294860, A348474, A351856, A351857, A352373.

b:= proc(n, t) option remember; `if`(t=0, 1-signum(n),
      add(b(n-j, t-1)*(1+iquo(j, 2)), j=0..n))
    end:
a:= n-> b(n$2):
seq(a(n), n=0..25);  # Alois P. Heinz, Oct 17 2021

(* giveList=True produces the list of solutions *)
(* giveList=False gives the number of solutions *)
counter[objects_, giveList_: False] :=
  Module[{n = objects, nb, eq1, eqa, eqb, eqs, var, sol, var2, list},
   nb = n;
   eq1 = {Total[Map[a[#] + 2*b[#] &, Range[nb]]] - n == 0};
   eqa = {And @@ Map[0 <= a[#] <= n &, Range[nb]]};
   eqb = {And @@ Map[0 <= b[#] <= n &, Range[nb]]};
   eqs = {And @@ Join[eq1, eqa, eqb]};
   var = Flatten[Map[{a[#], b[#]} &, Range[nb]]];
   var = Join[Map[a[#] &, Range[nb]], Map[b[#] &, Range[nb]]];
   sol = Solve[eqs, var, Integers];
   var2 = Join[Map[a[#] &, Range[nb]], Map[2*b[#] &, Range[nb]]];
   list = Sort[Map[var2 /. # &, sol]];
   list = Map[StringReplace[ToString[#], {"," -> " ;"}, n] &, list];
   list = Map[StringReplace[#, {";" -> ","}, n - 1] &, list];
   Return[
    If[giveList, Print["Total: ", Length[list]]; list, Length[sol]]];
   ];
(* second program: *)
b[n_, t_] := b[n, t] = If[t == 0, 1 - Sign[n], Sum[b[n - j, t - 1]*(1 + Quotient[j, 2]), {j, 0, n}]];
a[n_] := b[n, n];
Table[a[n], {n, 0, 25}] (* Jean-François Alcover, Aug 16 2023, after Alois P. Heinz *)

Conjecture: D-finite with recurrence +7168*n*(2*n-1)*(n-1)*a(n) -64*(n-1)*(1759*n^2-5294*n+5112)*a(n-1) +12*(7561*n^3-75690*n^2+165271*n-101070)*a(n-2) +5*(110593*n^3-743946*n^2+1659971*n-1232778)*a(n-3) +2680*(4*n-15)*(2*n-7)*(4*n-13)*a(n-4)=0. - R. J. Mathar, Oct 19 2021
From Vaclav Kotesovec, Nov 01 2021: (Start)
Recurrence (of order 2): 16*(n-1)*n*(2*n - 1)*(51*n^2 - 162*n + 127)*a(n) = (n-1)*(5457*n^4 - 22791*n^3 + 32144*n^2 - 17536*n + 3072)*a(n-1) + 8*(2*n - 3)*(4*n - 7)*(4*n - 5)*(51*n^2 - 60*n + 16)*a(n-2).
a(n) ~ sqrt(3 + 5/sqrt(17)) * (107 + 51*sqrt(17))^n / (sqrt(Pi*n) * 2^(6*n+2)). (End)
From Peter Bala, Feb 21 2022: (Start)
a(n) = [x^n] ( (1 - x)*(1 - x^2) )^(-n). Cf. A234839.
a(n) = Sum_{k = 0..floor(n/2)} binomial(2*n-2*k-1,n-2*k)*binomial(n+k-1,k).
exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + x + 3*x^2 + 9*x^3 + 32*x^4 + 119*x^5 + ... is the g.f. of A063020.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p and positive integers n and k.
Conjecture: the supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) hold for all primes p >= 5 and positive integers n and k.
The o.g.f. A(x) is the diagonal of the bivariate rational function 1/(1 - t/((1-x)*(1-x^2))) and hence is an algebraic function over Q(x) by Stanley 1999, Theorem 6.33, p. 197.
Let F(x) = (1/x)*Series_Reversion( x*(1-x)*(1-x^2) ). Then A(x) = 1 + x*d/dx (log(F(x))). (End)
a(n) = Sum_{k = 0..n} (-1)^(n+k)*binomial(2*n+k-1, k)*binomial(2*n-k-1, n-k). Cf. A352373. - Peter Bala, Jun 05 2024

More terms from Alois P. Heinz, Oct 17 2021

A351857 Number of nonnegative integer solutions to n = x_1 + x_2 + ... + x_(2n) + 2y_1 + 2y_2 + ... + 2y_(2*n).

Original entry on oeis.org

2, 14, 92, 654, 4752, 35204, 264112, 2000526, 15264866, 117161264, 903533380, 6995547780, 54343476072, 423360920528, 3306313730592, 25876855432846, 202909132368942, 1593755466338030, 12537009118650016, 98753463725849904, 778825917274945408, 6149069826564738780

Offset: 1

Peter Bala, Feb 22 2022

This is a companion sequence to A348410.

Suppose n identical objects are distributed in 4*n labeled baskets, 2*n colored white and 2*n colored black. White baskets can contain any number of objects (or be empty), while black baskets must contain an even number of objects (or be empty). a(n) is the number of distinct possible distributions.

			n = 2: 14 distributions of 2 identical objects in 4 white and 4 black baskets
             White             Black
   1)   (0) (0) (0) (0)   [2] [0] [0] [0]
   2)   (0) (0) (0) (0)   [0] [2] [0] [0]
   3)   (0) (0) (0) (0)   [0] [0] [2] [0]
   4)   (0) (0) (0) (0)   [0] [0] [0] [2]
   5)   (2) (0) (0) (0)   [0] [0] [0] [0]
   6)   (0) (2) (0) (0)   [0] [0] [0] [0]
   7)   (0) (0) (2) (0)   [0] [0] [0] [0]
   8)   (0) (0) (0) (2)   [0] [0] [0] [0]
   9)   (1) (1) (0) (0)   [0] [0] [0] [0]
  10)   (1) (0) (1) (0)   [0] [0] [0] [0]
  11)   (1) (0) (0) (1)   [0] [0] [0] [0]
  12)   (0) (1) (1) (0)   [0] [0] [0] [0]
  13)   (0) (1) (0) (1)   [0] [0] [0] [0]
  14)   (0) (0) (1) (1)   [0] [0] [0] [0]

R. P. Stanley, Enumerative Combinatorics Volume 2, Cambridge Univ. Press, 1999, Theorem 6.33, p. 197.

Cf. A234839, A348410, A351856, A365855.

seq(add( binomial(3*n-2*k-1,n-2*k)*binomial(2*n+k-1,k), k = 0..floor(n/2) ), n = 0..20);

a(n) = sum(k = 0, n\2, binomial(3*n-2*k-1, n-2*k)*binomial(2*n+k-1, k)); \\ Michel Marcus, Feb 27 2022

a(n) = [x^n] ( 1/((1 - x)*(1 - x^2)) )^(2*n).
a(n) = Sum_{k = 0..floor(n/2)} C(3*n-2*k-1,n-2*k)*C(2*n+k-1,k).
a(n) = Sum_{k = 0..n} (-1)^k*C(5*n-k-1,n-k)*C(2*n+k-1,k).
1024*n*(n-1)*(2*n-1)*(2*n-3)*(4*n-1)*(4*n-3)*P(n-1)*a(n) = 8*(n-1)*(2*n-3)*Q(n)*a(n-1) + 7*(7*n-8)*(7*n-9)*(7*n-10)*(7*n-11)*(7*n-12)*(7*n-13)*P(n)*a(n-2), with a(1) = 2, a(2) = 14, P(n) = 1744*n^4-3815*n^3+ 2920*n^2-912*n+96 and Q(n) = 46599680*n^8-381534880*n^7+1306363456*n^6- 2428492279*n^5+2661904813*n^4 -1747232452*n^3+664205312*n^2- 132046848*n+10321920.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p and positive integers n and k.
Conjecture: the supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) hold for all primes p >= 5 and positive integers n and k.
The o.g.f. A(x) = 2*x + 14*x^2 + 92*x^3 + ... is the diagonal of the bivariate rational function x*t/(1 - t/((1 - x)*(1 - x^2))^2 ) and hence is an algebraic function over Q(x) by Stanley 1999, Theorem 6.33, p. 197.
Let F(x) = (1/x)*Series_Reversion( x*(1 - x)^2*(1 - x^2)^2 ) = A365855(x). Then A(x) = x*(d/dx log(F(x))). [corrected by Jason Yuen, Mar 22 2025]

A352373 a(n) = [x^n] ( 1/((1 - x)^2*(1 - x^2)) )^n for n >= 1.

Original entry on oeis.org

2, 12, 74, 484, 3252, 22260, 154352, 1080612, 7621526, 54071512, 385454940, 2758690636, 19810063392, 142662737376, 1029931873824, 7451492628260, 54013574117106, 392188079586468, 2851934621212598, 20766924805302984, 151403389181347160, 1105047483656041080

Offset: 1

Peter Bala, Mar 14 2022

Suppose n identical objects are distributed in 3*n labeled baskets, 2*n colored white and n colored black. White baskets can contain any number of objects (or be empty), while black baskets must contain an even number of objects (or be empty). a(n) is the number of distinct possible distributions.
Number of nonnegative integer solutions to n = x_1 + x_2 + ... + x_(2*n) + 2*y_1 + 2*y_2 + ... + 2*y_n.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p and positive integers n and k.
Calculation suggests that, in fact, stronger congruences may hold.
Conjecture: the supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) hold for all primes p >= 5 and positive integers n and k.
More generally, let r and s be integers and define a sequence (a(r,s;n))n>=1 by a(r,s;n) = [x^n] ( (1 + x)^r * (1 - x)^s )^n.
Conjecture: for each r and s the above supercongruences hold for the sequence (a(r,s;n))n>=1.
The present sequence is the case r = -1 and s = -3. Other cases include A000984 (r = 2, s = 0), A001700 with offset 1 (r = 0, s = -1), A002003 (r = 1, s = -1), A091527 (r = 3, s = -1), A119259 (r = 2, s = -1), A156894 (r = 1, s = -2), A165817 (r = 0, s = -2), A234839 (r = 1, s = 2), A348410 (r = -1, s = -2) and A351857 (r = -2, s = -4).

			n = 2: 12 distributions of 2 identical objects in 4 white and 2 black baskets
             White         Black
   1)   (0) (0) (0) (0)   [2] [0]
   2)   (0) (0) (0) (0)   [0] [2]
   3)   (2) (0) (0) (0)   [0] [0]
   4)   (0) (2) (0) (0)   [0] [0]
   5)   (0) (0) (2) (0)   [0] [0]
   6)   (0) (0) (0) (2)   [0] [0]
   7)   (1) (1) (0) (0)   [0] [0]
   8)   (1) (0) (1) (0)   [0] [0]
   9)   (1) (0) (0) (1)   [0] [0]
  10)   (0) (1) (1) (0)   [0] [0]
  11)   (0) (1) (0) (1)   [0] [0]
  12)   (0) (0) (1) (1)   [0] [0]
Examples of supercongruences:
a(7) - a(1) = 154352 - 2 = 2*(3^2)*(5^2)*(7^3) == 0 (mod 7^3);
a(2*11) - a(2) = 1105047483656041080 - 12 = (2^2)*3*(11^3)*13*101*103*2441* 209581 == 0 (mod 11^3).

R. P. Stanley, Enumerative Combinatorics Volume 2, Cambridge Univ. Press, 1999, Theorem 6.33, p. 197.

Paolo Xausa, Table of n, a(n) for n = 1..1000

Cf. A000984, A001448, A001700, A002003, A091527, A119259, A156894, A165817, A211419, A211421, A234839, A262733, A276098, A348410, A351856, A351857.

seq(add( binomial(3*n-2*k-1,n-2*k)*binomial(n+k-1,k), k = 0..floor(n/2)), n = 1..25);

nterms=25;Table[Sum[Binomial[3n-2k-1,n-2k]Binomial[n+k-1,k],{k,0,Floor[n/2]}],{n,nterms}] (* Paolo Xausa, Apr 10 2022 *)

a(n) = Sum_{k = 0..floor(n/2)} binomial(3*n-2*k-1,n-2*k)*binomial(n+k-1,k).
a(n) = Sum_{k = 0..n} (-1)^k*binomial(4*n-k-1,n-k)*binomial(n+k-1,k).
a(n) = binomial(4*n-1,n)*hypergeom([n, -n], [1-4*n], -1).
48*n*(n-1)*(3*n-1)*(3*n-2)*(93*n^3-434*n^2+668*n-339)*a(n) = 12*(n-1)*(21762*n^6-134199*n^5+323805*n^4-386685*n^3+237728*n^2-70336*n+7680)*a(n-1) + 5*(5*n-9)*(5*n-8)*(5*n-7)*(5*n-6)*(93*n^3-155*n^2+79*n-12)*a(n-2) with a(1) = 2 and a(2) = 12.
The o.g.f. A(x) = 2*x + 12*x^2 + 74*x^3 + ... is the diagonal of the bivariate rational function x*t/(1 - t/((1 - x)^2*(1 - x^2))) and hence is an algebraic function over Q(x) by Stanley 1999, Theorem 6.33, p. 197.
A(x) = x*d/dx(log(F(x))), where F(x) = (1/x)*Series_Reversion( x*(1 - x)^2*(1 - x^2) ).
a(n) ~ sqrt(4 + sqrt(6)) * (13/4 + 31*sqrt(6)/18)^n / (2*sqrt(5*Pi*n)). - Vaclav Kotesovec, Mar 15 2022

A351856 Number of nonnegative integer solutions to 2n = x_1 + x_2 + ... + x_n + 2y_1 + 2y_2 + ... + 2y_n.

Original entry on oeis.org

2, 14, 119, 1086, 10252, 98735, 963832, 9502014, 94386908, 943206264, 9471346755, 95491466655, 966026045376, 9800968460024, 99685873633744, 1016118049037630, 10377363759903252, 106161722891946356, 1087696666197827374, 11159365823946907336, 114631982782490824420

Offset: 1

Peter Bala, Feb 22 2022

This is a companion sequence to A348410.

Suppose 2*n identical objects are distributed in 2*n labeled baskets, n colored white and n colored black. White baskets can contain any number of objects (or be empty), while black baskets must contain an even number of objects (or be empty). a(n) is the number of distinct possible distributions.

			n = 2: 14 distributions of 4 identical objects in 2 white and 2 black baskets
        White     Black
   1)  (0) (0)   [4] [0]
   2)  (0) (0)   [0] [4]
   3)  (0) (0)   [2] [2]
   4)  (2) (0)   [2] [0]
   5)  (0) (2)   [2] [0]
   6)  (1) (1)   [2] [0]
   7)  (2) (0)   [0] [2]
   8)  (0) (2)   [0] [2]
   9)  (1) (1)   [0] [2]
  10)  (4) (0)   [0] [0]
  11)  (0) (4)   [0] [0]
  12)  (3) (1)   [0] [0]
  13)  (1) (3)   [0] [0]
  14)  (2) (2)   [0] [0]

R. P. Stanley, Enumerative Combinatorics Volume 2, Cambridge Univ. Press, 1999, Theorem 6.33, p. 197.

Cf. A234839, A348410, A351857.

seq( add(binomial(3*n-2*k-1,2*n-2*k)*binomial(n+k-1,k), k = 0..n), n = 1..20);

a(n) = [x^(2*n)] ( 1/((1 - x)*(1 - x^2)) )^n.
a(n) = Sum_{k = 0..n} C(3*n-2*k-1,2*n-2*k)*C(n+k-1,k).
a(n) = Sum_{k = 0..2*n} (-1)^k*C(4*n-k-1,2*n-k)*C(n+k-1,k).
32*n*(n-1)*(2*n-1)*(2*n-3)*(41*n^2-126*n+93)*a(n) = 2*(n-1)*(2*n-3)*(16851*n^4-68637*n^3+93680*n^2-49024*n+7680)*a(n-1) - 5*(5*n-9)*(5*n-8)*(5*n-7)*(5*n-6)*(41*n^2-44*n+8)*a(n-2) with a(1) = 2 and a(2) = 14.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p and positive integers n and k.
Conjecture: the supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) hold for all primes p >= 5 and positive integers n and k.
The o.g.f. A(x) = 2*x + 14*x^2 + 119*x^3 + ... is the diagonal of the bivariate rational function x*t*(x - 1)*((x - 1)^2 - t)/((x - 1)^3 - t*(2*x + t - 2)) and hence is an algebraic function over Q(x) by Stanley 1999, Theorem 6.33, p. 197.
Let F(x) = (1/x)*Series_Reversion( x*sqrt((1-x)*(1-x^2)) ) and put G(x) = x*(d/dx)(log(F(x))). Then A(x^2) = (G(x) + G(-x))/2.

A252355 a(n) = sum_{k = 0..n-1} (-1)^kC(2n-1,k)*C(n-1,k), n>0.

Original entry on oeis.org

1, -2, 1, 8, -29, 34, 92, -512, 919, 818, -9151, 22472, -2924, -156872, 513736, -443392, -2457281, 11094658, -16502221, -31859752, 226433243, -475853006, -217535264, 4333621888, -12126499804, 5346234424

Offset: 1

L. Edson Jeffery, Dec 17 2014

For each n > 0, a(n) is an integer such that A234839(p-n) == 2^(2 - 3*n)*a(n) (mod p), for all primes p >= 2*n+1 [Chamberland, et al., Thm. 2.3].

Marc Chamberland and Karl Dilcher, A Binomial Sum Related to Wolstenholme's Theorem, J. Number Theory, Vol. 171, Issue 11 (Nov. 2009), pp. 2659-2672.

Cf. A000040 (primes), A234839, A045721.

a[n_] := Sum[(-1)^k*Binomial[2*n - 1, k]*Binomial[n - 1, k], {k, 0, n - 1}]; Table[a[n], {n, 26}]

a(n) = sum(k=0, n, (-1)^k*binomial(2*n-1,k)*binomial(n-1,k)); \\ Michel Marcus, Jan 13 2016

a(n) = _2F_1(1-2*n,1-n;1;-1), n>0.
Recurrence: 2*(n-1)*(2*n-1)*(7*n-11)*a(n) = -(91*n^3 - 325*n^2 + 368*n - 128)*a(n-1) - 16*(n-2)*(2*n-3)*(7*n-4)*a(n-2). - Vaclav Kotesovec, Dec 17 2014
Lim sup n->infinity |a(n)|^(1/n) = 2*sqrt(2). - Vaclav Kotesovec, Dec 17 2014
exp( Sum_{n >= 1} 2*a(n)*x^n/n ) = 1 + 2*x - 2*x^3 + 4*x^4 - 2*x^5 - 12*x^6 + 40*x^7 - 44*x^8 - 98*x^9 + 520*x^10 - 882*x^11 - 640*x^12 + ... appears to have integer coefficients. - Peter Bala, Jan 04 2016

A368467 a(n) = Sum_{k=0..n} (-1)^k * binomial(2n,k) binomial(4*n,n-k).

Original entry on oeis.org

1, 2, 2, -16, -126, -498, -880, 3432, 37762, 175916, 411502, -710752, -12482928, -66911830, -190616760, 70959984, 4208145282, 26042918836, 86794308524, 50521487200, -1397839172626, -10176550581570, -38838971577536, -51156092490048, 443929768322704

Offset: 0

Seiichi Manyama, Feb 10 2024

sign

Cf. A234839, A370106.

Cf. A369190.

a(n) = sum(k=0, n, (-1)^k * binomial(2*n, k)*binomial(4*n, n-k));

a(n) = [x^n] ( (1-x)^2 * (1+x)^4 )^n.

The g.f. exp( Sum_{k>=1} a(k) * x^k/k ) has integer coefficients and equals (1/x) * Series_Reversion( x/((1-x)^2*(1+x)^4) ). See A369190.

A370106 a(n) = Sum_{k=0..n} (-1)^k * binomial(2n,k) binomial(3*n,n-k).

Original entry on oeis.org

1, 1, -3, -17, -19, 126, 591, 344, -5907, -22373, 2122, 280842, 854063, -810692, -13254552, -31693392, 67250413, 615932985, 1101123015, -4368359919, -28043889894, -33371056204, 254637122506, 1245324193704, 693586015791, -13913192640499

Offset: 0

Seiichi Manyama, Feb 10 2024

sign

Cf. A234839, A368467.

Cf. A370107.

a(n) = sum(k=0, n, (-1)^k * binomial(2*n, k)*binomial(3*n, n-k));

a(n) = [x^n] ( (1-x)^2 * (1+x)^3 )^n.

The g.f. exp( Sum_{k>=1} a(k) * x^k/k ) has integer coefficients and equals (1/x) * Series_Reversion( x/((1-x)^2*(1+x)^3) ). See A370107.

A252501 Triangle T read by rows: T(n,k) = binomial(2n+1,k)binomial(n,k), n>=0, 0<=k<=n.

Original entry on oeis.org

1, 1, 3, 1, 10, 10, 1, 21, 63, 35, 1, 36, 216, 336, 126, 1, 55, 550, 1650, 1650, 462, 1, 78, 1170, 5720, 10725, 7722, 1716, 1, 105, 2205, 15925, 47775, 63063, 35035, 6435, 1, 136, 3808, 38080, 166600, 346528, 346528, 155584, 24310

Offset: 0

L. Edson Jeffery, Dec 17 2014

			Triangle T begins:
.1
.1.....3
.1....10.....10
.1....21.....63......35
.1....36....216.....336......126
.1....55....550....1650.....1650......462
.1....78...1170....5720....10725.....7722.....1716
.1...105...2205...15925....47775....63063....35035.....6435
.1...136...3808...38080...166600...346528...346528...155584...24310

Marc Chamberland and Karl Dilcher, A Binomial Sum Related to Wolstenholme's Theorem, J. Number Theory, Vol. 171, Issue 11 (Nov. 2009), pp. 2659-2672 (see Thm. 2.3).

Cf. A000012 (col. 1), A014105 (col. 2), A001700 (diag), A045721 (row sums).

Cf. A234839, A252355.

Flatten[Table[Binomial[2*n + 1, k]*Binomial[n, k], {n, 0, 8}, {k, 0, n}]] (* Replace Flatten[] with Grid[] to get the triangle. *)

A268304 Odd numbers n such that binomial(6n, 2n) == -1 (mod 8).

Original entry on oeis.org

1, 5, 21, 73, 85, 273, 293, 297, 329, 341, 529, 545, 1041, 1057, 1089, 1093, 1105, 1173, 1189, 1193, 1297, 1317, 1321, 1353, 1365, 2065, 2081, 2113, 2117, 2129, 2177, 2181, 2209, 2577, 2593, 4113, 4129, 4161, 4165, 4177, 4225, 4229, 4257, 4353, 4357, 4373, 4417, 4421, 4433

Offset: 1

Michel Marcus, Jan 31 2016

nonn

The primes p of this sequence are those that give the even semiprimes of A268303.

Chai Wah Wu, Table of n, a(n) for n = 1..10000
Marc Chamberland and Karl Dilcher, A Binomial Sum Related to Wolstenholme's Theorem, J. Number Theory, Vol. 171, Issue 11 (Nov. 2009), pp. 2659-2672. See Table 2 p. 2669.

Cf. A234839, A268303.

Select[Range[1, 5000, 2], Mod[Binomial[6 #, 2 #], 8] == 7 &] (* Michael De Vlieger, Feb 07 2016 *)

isok(n) = (n%2) && Mod(binomial(6*n, 2*n), 8) == Mod(-1, 8);

from _future_ import division
A268304_list, b, m1, m2 = [], 15, [21941965946880, -54854914867200, 49244258396160, -19011472727040, 2933960577120, -126898662960, 771887070, 385943535, 385945560],  [10569646080, -25763512320, 22419210240, -8309145600, 1209116160, -46992960, 415800, 311850, 311850]
for n in range(10**3):
    if b % 8 == 7:
        A268304_list.append(2*n+1)
    b = b*m1[-1]//m2[-1]
    for i in range(8):
        m1[i+1] += m1[i]
        m2[i+1] += m2[i] # Chai Wah Wu, Feb 05 2016

cp's OEIS Frontend

A268303 Composite numbers n such that Sum_{k = 0..n} (-1)^k * C(n,k) * C(2*n,k) == -1 (mod n^3) (see A234839).

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A348410 Number of nonnegative integer solutions to n = Sum_{i=1..n} (a_i + b_i), with b_i even.

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A351857 Number of nonnegative integer solutions to n = x_1 + x_2 + ... + x_(2*n) + 2*y_1 + 2*y_2 + ... + 2*y_(2*n).

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A352373 a(n) = [x^n] ( 1/((1 - x)^2*(1 - x^2)) )^n for n >= 1.

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A351856 Number of nonnegative integer solutions to 2*n = x_1 + x_2 + ... + x_n + 2*y_1 + 2*y_2 + ... + 2*y_n.

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A252355 a(n) = sum_{k = 0..n-1} (-1)^k*C(2*n-1,k)*C(n-1,k), n>0.

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A368467 a(n) = Sum_{k=0..n} (-1)^k * binomial(2*n,k) * binomial(4*n,n-k).

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A370106 a(n) = Sum_{k=0..n} (-1)^k * binomial(2*n,k) * binomial(3*n,n-k).

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A351857 Number of nonnegative integer solutions to n = x_1 + x_2 + ... + x_(2n) + 2y_1 + 2y_2 + ... + 2y_(2*n).

A351856 Number of nonnegative integer solutions to 2n = x_1 + x_2 + ... + x_n + 2y_1 + 2y_2 + ... + 2y_n.

A252355 a(n) = sum_{k = 0..n-1} (-1)^kC(2n-1,k)*C(n-1,k), n>0.

A368467 a(n) = Sum_{k=0..n} (-1)^k * binomial(2n,k) binomial(4*n,n-k).

A370106 a(n) = Sum_{k=0..n} (-1)^k * binomial(2n,k) binomial(3*n,n-k).

A252501 Triangle T read by rows: T(n,k) = binomial(2n+1,k)binomial(n,k), n>=0, 0<=k<=n.

A268304 Odd numbers n such that binomial(6n, 2n) == -1 (mod 8).