Peter Bala - cp's OEIS frontend

A387248 a(n) = 3/(n + 1) * Catalan(2*n).

3, 3, 14, 99, 858, 8398, 89148, 1002915, 11785890, 143291610, 1790214660, 22870640910, 297670187844, 3935861372604, 52749590350072, 715309969142307, 9800129095949682, 135490673691621794, 1888389218820071604, 26510079418051005210, 374589577468070301260, 5324240442532424176260, 76082624294738699098440

Offset: 0

Peter Bala, Aug 24 2025

Bisection of A007054.
Compare with Catalan(n) = 1/(n + 1) * binomial(2*n, n).
For r >= 2, there is a constant K_r such that K_r/(n + 1) * Catalan(r*n) is integral for all n.

Cf. A000108, A000984, A001448, A007054, A387249, A387250.

seq( 6/((2*n+1)*(2*n+2)) * binomial(4*n, 2*n), n = 0..22);

a(n) = 6/((2*n + 1)*(2*n + 2)) * binomial(4*n, 2*n).
a(n) = 4*Catalan(2*n) - Catalan(2*n+1) (showing a(n) to be an integer)
G.f.: A(x) = ((2 - f(x))*sqrt(2 + 2*f(x)) - 2)/(4*x), where f(x) = sqrt(1 - 16*x).
a(n) = 2*(4*n - 1)*(4*n - 3)/((n + 1)*(2*n + 1)) * a(n-1) with a(0) = 3.
a(n) ~ 3/(2*sqrt(2*Pi)) * 16^n/n^(5/2).
a(n) is odd iff n = 2^k - 1 for some k, so a(n) has the same parity as Catalan(n).
E.g.f.: 3*hypergeom([1/4, 3/4], [3/2, 2], 16*x). - Stefano Spezia, Aug 27 2025

A387250 a(n) = 105/(n + 1) * Catalan(4*n).

Original entry on oeis.org

105, 735, 50050, 5460315, 742511070, 114872107350, 19348562209860, 3461691866723475, 647897423565562310, 125577883051534761666, 25029394494457424675100, 5103876046438721064520350, 1060725331955983336553011500, 224018752093294694626068131340, 47967198494914114482847609250184

Offset: 0

Peter Bala, Aug 25 2025

For r >= 2, there is a constant K_r such that K_r/(n + 1) * Catalan(r*n) is integral for all n.

Cf. A000108, A387248, A387249.

seq( 105/((n + 1)*(4*n + 1)) * binomial(8*n, 4*n), n = 0..20);

a[n_]:=105/(n+1)*CatalanNumber[4n];Array[a,15,0] (* James C. McMahon, Aug 29 2025 *)

a(n) = 105/((n + 1)*(4*n + 1)) * binomial(8*n, 4*n).
a(n) = 2*(8*n - 1)*(8*n - 3)*(8*n - 5)*(8*n - 7)/((n + 1)*(2*n - 1)*(4*n + 1)*(4*n - 1)) * a(n-1) with a(0) = 105.
a(n) ~ 105/(8*sqrt(Pi)) * 256^n/n^(5/2).
E.g.f.: 105*hypergeom([1/8, 3/8, 5/8, 7/8], [1/2, 3/4, 5/4, 2], 256*x). - Stefano Spezia, Aug 27 2025

A387249 a(n) = 10/(n + 1) * Catalan(3*n).

Original entry on oeis.org

10, 25, 440, 12155, 416024, 16158075, 682341000, 30582833775, 1433226830360, 69533550916004, 3468169547356640, 176946775343535925, 9199844912200348840, 486018122664268428850, 26029619941269629306160, 1410698658798280045783575, 77251704848334920869407000, 4269325372507953547350453420

Offset: 0

Peter Bala, Aug 24 2025

Compare with Catalan(n) = 1/(n + 1) * binomial(2*n, n).

For r >= 2, there is a constant K_r such that K_r/(n + 1) * Catalan(r*n) is integral for all n.

Cf. A000108, A000984, A007272, A066802, A387248, A387250.

seq( 10/((n+1)*(3*n+1)) * binomial(6*n, 3*n), n = 0..20);

a(n) = 10/((n + 1)*(3*n + 1)) * binomial(6*n, 3*n).
a(n) = (3*n + 2)/2 * (16*Catalan(3*n) - 8*Catalan(3*n+1) + Catalan(3n+2)) (shows a(n) to be an integer since Catalan(n) is odd iff n = 2^k - 1 for some k, so Catalan(3*n+2) is always even).
a(n) = (3*n + 2)/2 * A007272(3*n).
a(n) = 8*(2*n - 1)*(6*n - 1)*(6*n - 5)/((n + 1)*(3*n + 1)*(3*n - 1)) * a(n-1) with a(0) = 10.
a(n) ~ 10/(sqrt(27*Pi)) * 64^n/n^(5/2).
E.g.f.: 10*hypergeom([1/6, 1/2, 5/6], [2/3, 4/3, 2], 64*x). - Stefano Spezia, Aug 27 2025

A386877 Triangle read by rows: T(n, k) = n! / (k! * (n/k)!) if k divides n otherwise 0; T(n, 0) = 0^n.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 6, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 60, 60, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 840, 0, 840, 0, 0, 0, 1, 0, 1, 0, 10080, 0, 0, 0, 0, 0, 1, 0, 1, 15120, 0, 0, 15120, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1

Offset: 0

Peter Bala and Peter Luschny, Aug 09 2025

			Triangle starts:
  [ 0] [1]
  [ 1] [0, 1]
  [ 2] [0, 1,     1]
  [ 3] [0, 1,     0,     1]
  [ 4] [0, 1,     6,     0,   1]
  [ 5] [0, 1,     0,     0,   0,     1]
  [ 6] [0, 1,    60,    60,   0,     0, 1]
  [ 7] [0, 1,     0,     0,   0,     0, 0, 1]
  [ 8] [0, 1,   840,     0, 840,     0, 0, 0, 1]
  [ 9] [0, 1,     0, 10080,   0,     0, 0, 0, 0, 1]
  [10] [0, 1, 15120,     0,   0, 15120, 0, 0, 0, 0, 1]
  [11] [0, 1,     0,     0,   0,     0, 0, 0, 0, 0, 0, 1]

Paolo Xausa, Table of n, a(n) for n = 0..11475 (rows 0..150 of triangle, flattened).

Cf. A121860 (row sums), A113704 (sign).

A386877[n_, k_] := Which[k == 0, Boole[n == 0], Divisible[n, k], n!/(k!*(n/k)!), True, 0];
Table[A386877[n, k], {n, 0, 12}, {k, 0, n}] (* Paolo Xausa, Aug 09 2025 *)

F = factorial
def T(n, k):
    if k == 0: return 0**n
    return F(n)/(F(k)*F(n//k)) if k.divides(n) else 0
for n in range(33): print([T(n,k) for k in srange(n+1)])

sign(T(n, k)) = A113704(n, k).

A386548 a(n) = [x^n] ((1 - x)/(1 - x + x^2))^n.

Original entry on oeis.org

1, 0, -2, -3, 6, 25, 1, -147, -218, 591, 2223, -484, -14871, -18759, 68353, 222697, -116058, -1629671, -1656989, 8275203, 23266031, -20154144, -184550412, -141418628, 1019061001, 2468408775, -3122976521, -21213927840, -10837119735, 126256071125, 262294667301, -456407675223

Offset: 0

Peter Bala, Jul 25 2025

The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p and all positive integers n and k.

Conjecture: the stronger supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(2*k)) hold for all primes p >= 5 and all positive integers n and k.

Cf. A104507, A246437, A364374, A370616.

a := proc(n) option remember; if n = 0 then 1 elif n = 1 then 0 elif n = 2 then -2 else
( 2*(n-1)*(2*n-3)*(19*n^2-60*n+36)*a(n-1) - 2*(190*n^4-1170*n^3+2519*n^2-2229*n+666)*a(n-2) - 2*(n-3)*(2*n-3)*(19*n^2-41*n+18)*a(n-3) )/(3*n*(n-1)*(19*n^2-79*n+78)) fi; end:
seq(a(n), n = 0..30);

a[n_]:=SeriesCoefficient[((1 - x)/(1 - x + x^2))^n,{x,0,n}]; Array[a,32,0] (* Stefano Spezia, Jul 29 2025 *)

a(n) = my(x='x+O('x^(n+1))); polcoef(((1 - x)/(1 - x + x^2))^n, n); \\ Michel Marcus, Aug 03 2025

a(n) = Sum_{k = 0..floor(n/2)} binomial(-n, k)*binomial(n-k-1, n-2*k) = Sum_{k = 0
..floor(n/2)} (-1)^k*binomial(n+k-1, k)*binomial(n-k-1, n-2*k). Cf. A246437.
a(n) = -n*hypergeom([n+1, 1 - (1/2)*n, 3/2 - (1/2)*n], [2, 2 - n], 4) for n >= 3.
P-recursive: 3*n*(n - 1)*(19*n^2 - 79*n + 78)*a(n) = 2*(n - 1)*(2*n - 3)*(19*n^2 - 60*n + 36)*a(n-1) - 2*(190*n^4 - 1170*n^3 + 2519*n^2 - 2229*n + 666)*a(n-2) - 2*(n - 3)*(2*n - 3)*(19*n^2 - 41*n + 18)*a(n-3) with a(0) = 1, a(1) = 0 and a(2) = -2.
exp( Sum_{n >= 1} a(n)*(-x)^n/n ) = 1 - x^2 + x^3 + 2*x^4 - 6*x^5 - x^6 +  ... is the g.f. of A364374.

A382527 a(n) = Sum_{j = 1..n} (-1)^(n+j) * j^(2n+4) binomial(2*n, n-j).

Original entry on oeis.org

1, 252, 52920, 12640320, 3632428800, 1264085222400, 529085049292800, 263564384219136000, 154550100069421056000, 105562401683780321280000, 83178863857362412339200000, 74951718050379657373286400000, 76628603945744083606044672000000, 88258468221509704910254374912000000

Offset: 1

Peter Bala, Mar 30 2025

Compare with the identities Sum_{j = 1..n} (-1)^(n+j) * j^n * binomial(n, n-j) = n! and Sum_{j = 1..n} (-1)^(n+j) * j^(2*n+2) * binomial(2*n, n-j) = n*(n+1)*(2*n+1)/6 * (2*n)!/2 = A000330(n) * (2*n)!/2. (Campbell, Eq. 17).

Paolo Xausa, Table of n, a(n) for n = 1..200
John Campbell, Applications of Gosper’s nonlocal derangement identity, Maple Transactions, 5, 1, Article 16724, Feb. 2025.

Cf. A002674, A060493.

seq(add((-1)^(n+j) * j^(2*n+4) * binomial(2*n, n-j), j = 1..n), n = 1..20);

A382527[n_] := n*(5*n - 1)*(2*n + 4)!/2880; Array[A382527, 15] (* Paolo Xausa, Apr 03 2025 *)

a(n) = n*(n+1)*(n+2)*(2*n+3)*(5*n-1)*(2*n+1)!/6!.
a(n) = (2*n)!/2 * (Sum_{1 <= i <= j <= n} i^2*j^2) = (2*n)!/2 * A060493(n).
a(n) = 2*n*(n+2)*(2*n+3)*(5*n-1)/((n-1)*(5*n-6)) * a(n-1) with a(1) = 1.

A380549 List of numbers of the form i + 3j + 4i*j for i, j >= 1.

Original entry on oeis.org

8, 13, 15, 18, 22, 23, 24, 28, 29, 33, 35, 36, 38, 42, 43, 46, 48, 50, 51, 53, 57, 58, 60, 61, 63, 64, 68, 69, 71, 73, 74, 78, 79, 80, 83, 85, 87, 88, 90, 92, 93, 96, 97, 98, 99, 100, 101, 103, 105, 106, 108, 112, 113, 114, 118, 120, 123, 126, 127, 128, 131, 132, 133, 134, 137, 138, 139, 141, 143, 145, 148, 150

Offset: 1

Peter Bala, Jan 26 2025

This is a companion sequence to A380509. If N != 6 is a positive integer not in this list then 4*N + 3 is either a prime or three times a prime. See A380550.

Compare with A072668, numbers of the form i + j + i*j, and A047845, numbers of the form i + j + 2*i*j.

Cf. A047845, A072668, A380509, A380550.

L := 150:  S := {}:
for i from 1 to L do
  for j from 1 to L do
    if i + 3*j + 4*i*j <= L then S := `union`(S, {i+3*j+4*i*j}) end if
  end do;
end do:
S;

A380550 List of numbers not of the form i + 3j + 4i*j for i, j >= 1.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 12, 14, 16, 17, 19, 20, 21, 25, 26, 27, 30, 31, 32, 34, 37, 39, 40, 41, 44, 45, 47, 49, 52, 54, 55, 56, 59, 62, 65, 66, 67, 70, 72, 75, 76, 77, 81, 82, 84, 86, 89, 91, 94, 95, 102, 104, 107, 109, 110, 111, 115, 116, 117, 119, 121, 122, 124, 125, 129, 130, 135, 136, 140, 142, 144, 146, 147, 149

Offset: 1

Peter Bala, Jan 26 2025

This is a companion sequence to A380572. It is the complementary sequence to A380549.
Compare with A006093, numbers not of the form i + j + i*j, and A005097, numbers not of the form i + j + 2*i*j.
Apart from a(6) = 6, this sequence consists of all the positive integers N such that 4*N + 3 is either a prime or three times a prime.

			Factorization of 4*a(n) + 3 for n = 1..78:
[7,  11, 3*5, 19, 23, 3^3, 31, 3*13, 43, 47, 3*17, 59, 67, 71, 79, 83, 3*29, 103, 107, 3*37, 3*41, 127, 131, 139, 151, 3*53, 163, 167, 179, 3*61, 191, 199, 211, 3*73, 223, 227, 239, 251, 263, 3*89, 271, 283, 3*97, 3*101, 307, 311, 3*109, 331, 3*113, 347, 359, 367, 379, 383, 3*137, 419, 431, 439, 443, 3*149, 463, 467, 3*157, 479, 487, 491, 499, 503, 3*173, 523, 3*181, 547, 563, 571, 3*193, 587, 3*197, 599]

Peter Bala, 4*A380550(n) + 3 is either a prime or three times a prime except when n = 6

Cf. A005097, A006093, A380549, A380572.

L := 150: N := {seq(n, n= 1..L)}: S := {}:
for i from 1 to L do
  for j from 1 to L do
    if i + 3*j + 4*i*j <= L then S := `union`(S, {i+3*j+4*i*j}) end if
  end do;
end do:
N minus S;

A380583 a(n) = [x^n] G(x)^n, where G(x) = Product_{k >= 1} ((1 + x^(2*k))/(1 - x^k))^(k^2).

Original entry on oeis.org

1, 1, 13, 82, 665, 5026, 40180, 319677, 2583401, 20965150, 171276238, 1405008925, 11571476120, 95601033542, 792038546739, 6577523807332, 54737967873385, 456368114019558, 3811136362823056, 31873576059000827, 266919720010452190, 2237944814420991135, 18784073017650350445

Offset: 0

Peter Bala, Jan 27 2025

The sequence satisfies the Gauss congruences: a(n*p^r) == a(n*p^(r-1)) (mod p^r) for all primes p and for all positive integers n and r. In fact, stronger congruences appear to hold for this sequence.
We conjecture that a(p) == 1 (mod p^3) for all primes p >= 7.
More generally, we conjecture that the supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) holds for all primes p >= 7 and positive integers n and r. Some examples are given below.
Let A and B be integers. Let C be a positive integer. Define u(n) = [x^(C*n)] Product_{k >= 1} ((1 + - x^(2*k))^A * (1 + - x^k)^B)^(k^2). The present sequence is the case A = 1, B = -1 and C = 1, with the appropriate choice of signs. We conjecture that the above supercongruences also hold for the sequence {u(n)} for all primes p >= 7.

			Examples of supercongruences:
a(7) - a(1) = 319677 - 1 = (2^2)*(7^3)*233 == 0 (mod 7^3)
a(11) - a(1) = 1405008925 - 1 = (2^2)*3*(11^5)*727 == 0 (mod 11^5)
a(22) - a(2) = 18784073017650350445 - 13 = (2^5)*(11^3)*222773*1979699077 == 0 (mod 11^3)

Cf. A380290, A380581, A380582.

G(x) := series(mul( ( (1 + x^(2*k))/(1 - x^k) )^(k^2), k = 1..22), x, 23):
a:= n-> coeftayl(G(x)^n, x = 0, n):
seq(a(n), n = 0..22);

A380582 a(n) = [x^n] G(x)^n, where G(x) = Product_{k >= 1} ((1 + x^k)/(1 - x^k))^(k^2) is the g.f. of A206622.

Original entry on oeis.org

1, 2, 24, 236, 2432, 25752, 277152, 3019088, 33186816, 367378814, 4089875024, 45741207228, 513537853952, 5784253405192, 65332622356032, 739706089046736, 8392732289277952, 95401363286044260, 1086232605119042424, 12386037358495697292, 141422619808922418432, 1616691574828234720352

Offset: 0

Peter Bala, Jan 27 2025

Given an integer sequence {f(n) : n >= 0} with f(0) = 1, there is a unique power series F(x) with rational coefficients, where F(0) = 1, such that f(n) = [x^n] F(x)^n. F(x) is given by F(x) = series_reversion(x/E(x)), where E(x) = exp(Sum_{n >= 1} f(n)*x^n/n). Furthermore, if the series E(x) has integer coefficients then the series F(x) also has integer coefficients and the sequence {f(n)} satisfies the Gauss congruences: f(n*p^r) == f(n*p^(r-1)) (mod p^r) for all primes p and positive integers n and r (by Stanley, Ch. 5, Ex. 5.2(a), p. 72 and the Lagrange inversion formula).
Thus the present sequence satisfies the Gauss congruences. In fact, stronger congruences appear to hold for this sequence.
We conjecture that a(p) == 1 (mod p^3) for all primes p >= 5 (checked up to p = 61).
More generally, we conjecture that the supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) holds for all primes p >= 5 and positive integers n and r. Some examples are given below.
Let A, B be integers and let C be a positive integer. Define u(n) = [x^(C*n)] Product_{k >= 1} ((1 + x^k)^A * (1 - x^k)^B)^(k^2). The present sequence is the case A = 1, B = -1 and C = 1. We conjecture that the above supercongruences also hold for the sequence {u(n)} for all primes p >= 7.

			Examples of supercongruences:
a(7) - a(1) = 3019088 - 2 = 2*(3^3)*(7^3)*163 == 0 (mod 7^3)
a(13) - a(1) = 5784253405192 - 2 = 2*5*(13^4)*20252279 == 0 (mod 13^4)
a(2*11) - a(2) = 18501616629347623668448 - 24 = (2^3)*(11^3)*17*1951*4243*9817*1257719 == 0 (mod 11^3)
a(5^2) - a(5) = 1884578634304981694792832319004 - 256504 = (2^2)*(5^6)*193381* 155926684363405438573 == 0 (mod 5^6)

Cf. A001158, A015128, A206622, A380290, A380581, A380583.

with(numtheory):
G(x) := series(exp(add( (1/4)*(sigma[3](2*k) - sigma[3](k))*x^k/k, k = 1..23 )),x,24):
seq(coeftayl(G(x)^n, x = 0, n), n = 0..23);

a(n) = [x^n] exp( n*Sum_{k >= 1} (sigma_3(2*k) - sigma_3(k))/4 * x^k/k ).

cp's OEIS Frontend

User: Peter Bala

A387248 a(n) = 3/(n + 1) * Catalan(2*n).

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Maple

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A387250 a(n) = 105/(n + 1) * Catalan(4*n).

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A387249 a(n) = 10/(n + 1) * Catalan(3*n).

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A386877 Triangle read by rows: T(n, k) = n! / (k! * (n/k)!) if k divides n otherwise 0; T(n, 0) = 0^n.

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A386548 a(n) = [x^n] ((1 - x)/(1 - x + x^2))^n.

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A382527 a(n) = Sum_{j = 1..n} (-1)^(n+j) * j^(2*n+4) * binomial(2*n, n-j).

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A380549 List of numbers of the form i + 3*j + 4*i*j for i, j >= 1.

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Maple

A380550 List of numbers not of the form i + 3*j + 4*i*j for i, j >= 1.

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A382527 a(n) = Sum_{j = 1..n} (-1)^(n+j) * j^(2n+4) binomial(2*n, n-j).

A380549 List of numbers of the form i + 3j + 4i*j for i, j >= 1.

A380550 List of numbers not of the form i + 3j + 4i*j for i, j >= 1.