A333094 -id:A333094 - cp's OEIS frontend

A333093 a(n) is equal to the n-th order Taylor polynomial (centered at 0) of c(x)^n evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4x))/(2x) is the o.g.f. of the Catalan numbers A000108.

1, 2, 8, 41, 232, 1377, 8399, 52138, 327656, 2077934, 13270633, 85226594, 549837391, 3560702069, 23132584742, 150695482041, 984021596136, 6438849555963, 42208999230224, 277144740254566, 1822379123910857, 11998811140766701, 79095365076843134

Offset: 0

Peter Bala, Mar 07 2020

The 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.
We conjecture that the sequence satisfies the stronger supercongruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k. Examples of these congruences are given below.
More generally, for each integer m, we conjecture that the sequence
  {a_m(n) : n >= 0}, defined by setting a_m(n) = the n-th order Taylor polynomial of c(x)^(m*n) evaluated at x = 1, satisfies the same supercongruences. For cases, see A099837 (m = -2), A100219 (m = -1), A000012 (m = 0), A333094 (m = 2), A333095 (m = 3), A333096 (m = 4), A333097 (m = 5).

			n-th order Taylor polynomial of c(x)^n:
  n = 0: c(x)^0 = 1 + O(x)
  n = 1: c(x)^1 = 1 + x + O(x^2)
  n = 2: c(x)^2 = 1 + 2*x + 5*x^2 + O(x^3)
  n = 3: c(x)^3 = 1 + 3*x + 9*x^2 + 28*x^3 + O(x^4)
  n = 4: c(x)^4 = 1 + 4*x + 14*x^2 + 48*x^3 + 165*x^4 + O(x^5)
Setting x = 1 gives a(0) = 1, a(1) = 1 + 1 = 2, a(2) = 1 + 2 + 5 = 8, a(3) = 1 + 3 + 9 + 28 = 41 and a(4) = 1 + 4 + 14 + 48 + 165 = 232.
The triangle of coefficients of the n-th order Taylor polynomial of c(x)^n, n >= 0, in descending powers of x begins
                                        row sums
  n = 0 |   1                               1
  n = 1 |   1   1                           2
  n = 2 |   5   2    1                      8
  n = 3 |  28   9    3   1                 41
  n = 4 | 165  48   14   4   1            232
   ...
This is a Riordan array belonging to the Hitting time subgroup of the Riordan group. The first column sequence [1, 1, 5, 28, 165, ...] = [x^n] c(x)^n = A025174(n).
Examples of supercongruences:
a(13) - a(1) = 3560702069 - 2 = (3^2)*(13^3)*31*37*157 == 0 ( mod 13^3 ).
a(3*7) - a(3) = 11998811140766701 - 41 = (2^2)*5*(7^4)*32213*7756841 == 0 ( mod 7^3 ).
a(5^2) - a(5) = 22794614296746579502 - 1377 = (5^6)*7*53*6491*605796421 == 0 ( mod 5^6 ).

Peter Bala, Notes on A333093

Cf. A000108, A001006, A001764, A005554, A025174, A333090 through A333097, A372214, A372215.

seq(add(n/(n+k)*binomial(n+2*k-1,k), k = 0..n), n = 1..25);
#alternative program
c:= x -> (1/2)*(1-sqrt(1-4*x))/x:
G := (x,n) -> series(c(x)^n, x, 51):
seq(add(coeff(G(x, n), x, k), k = 0..n), n = 0..25);

Table[SeriesCoefficient[((1 + x)^2 * (1 - Sqrt[(1 - 3*x)/(1 + x)]) / (2*x))^n, {x, 0, n}], {n, 0, 25}] (* Vaclav Kotesovec, Mar 28 2020 *)

a(n) = Sum_{k = 0..n} n/(n+k)*binomial(n+2*k-1,k) for n >= 1.
a(n) = [x^n] ( (1 + x)*c(x/(1 + x)) )^n = [x^n] ( (1 + x)*(1 + x*M(x)) )^n, where M(x) = ( 1 - x - sqrt(1 - 2*x - 3*x^2) ) / (2*x^2) is the o.g.f. of the Motzkin numbers A001006.
O.g.f.: ( 1 + x*f'(x)/f(x) )/( 1 - x*f(x) ), where f(x) = 1 + x + 3*x^2 + 12*x^3 + 55*x^4 + ... = (1/x)*Revert( x/c(x) ) is the o.g.f. of A001764.
Row sums of the Riordan array ( 1 + x*f'(x)/f(x), x*f(x) ) belonging to the Hitting time subgroup of the Riordan group.
a(n) ~ 3^(3*n + 3/2) / (7 * sqrt(Pi*n) * 2^(2*n+1)). - Vaclav Kotesovec, Mar 28 2020
a(n) = Sum_{k = 0..n} n/(n+2*k)*binomial(n+2*k, k) for n >= 1. - Peter Bala, Apr 20 2024
D-finite with recurrence 2*n*(2*n-1)*(3991*n -21664)*a(n) +(-1329757*n^3 +9119565*n^2 -18270518*n +10657440)*a(n-1) +10*(947050*n^3 -6943257*n^2 +15944396*n -11260008)*a(n-2) +12*(-787878*n^3 +5778161*n^2 -13283386*n +9383340)*a(n-3) +9*(3*n-10)*(3*n-8)*(100503*n -141587)*a(n-4)=0, n>=5. - R. J. Mathar, Nov 22 2024

A333097 a(n) = the n-th order Taylor polynomial (centered at 0) of c(x)^(5n) evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4x))/(2*x) is the o.g.f. of the sequence of Catalan numbers A000108.

Original entry on oeis.org

1, 6, 76, 1101, 16876, 266881, 4305247, 70414133, 1163355884, 19369868385, 324486751951, 5462851474614, 92346622131103, 1566455916243068, 26649562889363259, 454528917186429226, 7769463895152496364, 133064720735632286722, 2282869928179537263601, 39225214245206751480102

Offset: 0

Peter Bala, Mar 15 2020

The 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.
We conjecture that the sequence satisfies the stronger supercongruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k. Examples of these congruences are given below.
More generally, for each integer m, we conjecture that the sequence a_m(n) := the n-th order Taylor polynomial of c(x)^(m*n) evaluated at x = 1 satisfies the same supercongruences. For cases see A099837 (m = -2), A100219 (m = -1), A000012 (m = 0), A333093 (m = 1), A333094 (m = 2), A333095 (m = 3), A333096 (m = 4).
In general, for m > 0 and c(x)^(m*n) is a(n) ~ m * (m+2)^((m+2)*n + 3/2) / (((m+1)*(m+2)+1) * sqrt(2*Pi*n) * (m+1)^((m+1)*n + 1/2)). - Vaclav Kotesovec, Mar 28 2020

			n-th order Taylor polynomial of c(x)^(5*n):
  n = 0: c(x)^0 = 1 + O(x)
  n = 1: c(x)^5 = 1 + 5*x + O(x^2)
  n = 2: c(x)^10 = 1 + 10*x + 65*x^2 + O(x^3)
  n = 3: c(x)^15 = 1 + 15*x + 135*x^2 + 950*x^3 + O(x^4)
  n = 4: c(x)^20 = 1 + 20*x + 230*x^2 + 2000*x^3 + 14625*x^4 + O(x^5)
Setting x = 1 gives a(0) = 1, a(1) = 1 + 5 = 6, a(2) = 1 + 10 + 65 = 76, a(3) = 1 + 15 + 135 + 950 = 1101 and a(4) = 1 + 20 + 230 + 2000 + 14625 = 16876.
The triangle of coefficients of the n-th order Taylor polynomial of c(x)^(5*n), n >= 0, in descending powers of x begins
                                                row sums
  n = 0 |     1                                     1
  n = 1 |     5        1                            6
  n = 2 |    65       10      1                    76
  n = 3 |   950      135     15    1             1101
  n = 4 | 14625     2000    230   20    1       16876
   ...
This is a Riordan array belonging to the Hitting time subgroup of the Riordan group.
Examples of supercongruences:
a(13) - a(1) = 1566455916243068 - 6 = 2*(13^3)*104701*3404923 == 0 ( mod 13^3 ).
a(3*7) - a(3) = 11627033261887689372357353 - 1101 = (2^2)*(7^4)*19*29* 2197177609353575713 == 0 ( mod 7^3 ).
a(5^2) - a(5) = 1034770243516278817426081673131 - 266881 = 2*3*(5^7)*31* 13305359*5351978496238483 == 0 ( mod 5^6 ).

Cf. A000108, A233834, A333090 through A333097.

seq(add(5*n/(5*n+k)*binomial(5*n+2*k-1,k), k = 0..n), n = 1..25);
#alternative program
c:= x → (1/2)*(1-sqrt(1-4*x))/x:
G := (x,n) → series(c(x)^(5*n), x, 151):
seq(add(coeff(G(x, n), x, n-k), k = 0..n), n = 0..25);

Join[{1}, Table[5*Binomial[7*n-1, n] * HypergeometricPFQ[{1, -6*n, -n}, {1/2 - 7*n/2, 1 - 7*n/2}, 1/4]/6, {n, 1, 20}]] (* Vaclav Kotesovec, Mar 28 2020 *)

a(n) = Sum_{k = 0..n} 5*n/(5*n+k)*binomial(5*n+2*k-1,k) for n >= 1.
a(n) = [x^n] ( (1 + x)*c^5(x/(1 + x)) )^n.
O.g.f.: ( 1 + x*f'(x)/f(x) )/( 1 - x*f(x) ), where f(x) = 1 + 5*x + 45*x^2 + 500*x^3 + 6200*x^4 + ... = (1/x)*Revert( x/c^5(x) ) is the o.g.f. of A233834.
Row sums of the Riordan array ( 1 + x*f'(x)/f(x), f(x) ) belonging to the Hitting time subgroup of the Riordan group.
a(n) ~ 5 * 7^(7*n + 3/2) / (43 * sqrt(Pi*n) * 2^(6*n + 1) * 3^(6*n + 1/2)). - Vaclav Kotesovec, Mar 28 2020
a(n) = Sum_{k = 0..n} 5*n/(5*n+2*k)*binomial(5*n+2*k, k) for n >= 1. - Peter Bala, Apr 20 2024

A333095 a(n) = the n-th order Taylor polynomial (centered at 0) of c(x)^(3n) evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4x))/(2*x) is the o.g.f. of the sequence of Catalan numbers A000108.

Original entry on oeis.org

1, 4, 34, 337, 3554, 38754, 431521, 4874377, 55639010, 640177033, 7412165034, 86256322816, 1007980394849, 11820510331777, 139032549536551, 1639506780365337, 19376785465043938, 229458302589724067, 2721958273545613513, 32339465512495259708, 384758834631081248554

Offset: 0

Peter Bala, Mar 15 2020

The 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.
We conjecture that the sequence satisfies the stronger supercongruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k. Examples of these congruences are given below.
More generally, for each integer m, we conjecture that the sequence a_m(n) := the n-th order Taylor polynomial of c(x)^(m*n) evaluated at x = 1 satisfies the same supercongruences. For cases see A099837 (m = -2), A100219 (m = -1), A000012 (m = 0), A333093 (m = 1), A333094 (m = 2), A333096 (m = 4), A333097 (m = 5).

			n-th order Taylor polynomial of c(x)^(3*n):
  n = 0: c(x)^0 = 1 + O(x)
  n = 1: c(x)^3 = 1 + 3*x + O(x^2)
  n = 2: c(x)^6 = 1 + 6*x + 27*x^2 + O(x^3)
  n = 3: c(x)^9 = 1 + 9*x + 54*x^2 + 273*x^3 + O(x^4)
  n = 4: c(x)^12 = 1 + 12*x + 90*x^2 + 544*x^3 + 2907*x^4 + O(x^5)
Setting x = 1 gives a(0) = 1, a(1) = 1 + 3 = 4, a(2) = 1 + 6 + 27 = 34, a(3) = 1 + 9 + 54 + 273 = 337 and a(4) = 1 + 12 + 90 + 544 + 2907 = 3554.
The triangle of coefficients of the n-th order Taylor polynomial of c(x)^n, n >= 0, in descending powers of x begins
                                              row sums
  n = 0 |    1                                    1
  n = 1 |    3       1                            4
  n = 2 |   27       6    1                      34
  n = 3 |  273      54    9   1                 337
  n = 4 | 2907     544   90  12   1            3554
   ...
This is a Riordan array belonging to the Hitting time subgroup of the Riordan group.
Examples of supercongruences:
a(13) - a(1) = 11820510331777 - 4 = 3*11*(13^3)*(43^2)*88177 == 0 ( mod 13^3 ).
a(3*7) - a(3) = 4583419703934987639046 - 337 = (3^2)*(7^4)*2441* 86893477573061 == 0 ( mod 7^3 ).
a(5^2) - a(5) = 93266278848727959965820004 - 38754 = 2*(5^7)*19* 31416009717466260199 == 0 ( mod 5^6 ).

Cf. A000108, A118970, A333090 through A333097.

seq(add(3*n/(3*n+k)*binomial(3*n+2*k-1,k), k = 0..n), n = 1..25);
#alternative program
c:= x → (1/2)*(1-sqrt(1-4*x))/x:
G := (x,n) → series(c(x)^(3*n), x, 101):
seq(add(coeff(G(x, n), x, n-k), k = 0..n), n = 0..25);

Join[{1}, Table[3*Binomial[5*n-1, n] * HypergeometricPFQ[{1, -4*n, -n}, {1/2 - 5*n/2, 1 - 5*n/2}, 1/4]/4, {n, 1, 20}]] (* Vaclav Kotesovec, Mar 28 2020 *)

a(n) = Sum_{k = 0..n} 3*n/(3*n+k)*binomial(3*n+2*k-1, k) for n >= 1.
a(n) = [x^n] ( (1 + x)*c^3(x/(1 + x)) )^n.
O.g.f.: ( 1 + x*f'(x)/f(x) )/( 1 - x*f(x) ), where f(x) = 1 + 3*x + 18*x^2 + 136*x^3 + 1155*x^4 + ... = (1/x)*Revert( x/c^3(x) ) is the o.g.f. of A118970.
Row sums of the Riordan array ( 1 + x*f'(x)/f(x), f(x) ) belonging to the Hitting time subgroup of the Riordan group.
a(n) ~ 5^(5*n + 3/2) / (7 * 2^(8*n + 3/2) * sqrt(Pi*n)). - Vaclav Kotesovec, Mar 28 2020
a(n) = Sum_{k = 0..n} 3*n/(3*n+2*k)*binomial(3*n+2*k, k) for n >= 1. - Peter Bala, May 03 2024

A333096 a(n) = the n-th order Taylor polynomial (centered at 0) of c(x)^(4n) evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4x))/(2*x) is the o.g.f. of the sequence of Catalan numbers A000108.

Original entry on oeis.org

1, 5, 53, 647, 8373, 111880, 1525511, 21093476, 294663349, 4148593604, 58770091928, 836722722951, 11961868391175, 171601856667701, 2469036254872996, 35615467194043147, 514888180699419829, 7458193213805231529, 108219144962546395364, 1572690742149983040857

Offset: 0

Peter Bala, Mar 15 2020

The 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.
We conjecture that the sequence satisfies the stronger supercongruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k. Examples of these congruences are given below.
More generally, for each integer m, we conjecture that the sequence a_m(n) := the n-th order Taylor polynomial of c(x)^(m*n) evaluated at x = 1 satisfies the same supercongruences. For cases see A099837 (m = -2), A100219 (m = -1), A000012 (m = 0), A333093 (m = 1), A333094 (m = 2), A333095 (m = 3), A333097 (m = 5).

			n-th order Taylor polynomial of c(x)^(4*n):
  n = 0: c(x)^0 = 1 + O(x)
  n = 1: c(x)^4 = 1 + 4*x + O(x^2)
  n = 2: c(x)^8 = 1 + 8*x + 44*x^2 + O(x^3)
  n = 3: c(x)^12 = 1 + 12*x + 90*x^2 + 544*x^3 + O(x^4)
  n = 4: c(x)^16 = 1 + 16*x + 152*x^2 + 1120*x^3 + 7084*x^4 + O(x^5)
Setting x = 1 gives a(0) = 1, a(1) = 1 + 4 = 5, a(2) = 1 + 8 + 44 = 53, a(3) = 1 + 12 + 90 + 544 = 647 and a(4) = 1 + 16 + 152 + 1120 + 7084 = 8373.
The triangle of coefficients of the n-th order Taylor polynomial of c(x)^(4*n), n >= 0, in descending powers of x begins
                                         row sums
  n = 0 |    1                               1
  n = 1 |    4     1                         5
  n = 2 |   44     8     1                  53
  n = 3 |  544    90    12     1           647
  n = 4 | 7084  1120   152    16   1      8373
   ...
This is a Riordan array belonging to the Hitting time subgroup of the Riordan group.
Examples of congruences:
a(13) - a(1) = 171601856667701 - 5 = (2^4)*3*(7^2)*(13^3)*33208909 == 0 ( mod 13^3 ).
a(3*7) - a(3) = 333475516822140871773101 - 647 = 2*(3^2)*(7^3)* 54012879303877692221 == 0 ( mod 7^3 ).
a(5^2) - a(5) = 15187725485911657497382846255 - 111880 = (3^3)*(5^7)*29* 248279548173268475053 == 0 ( mod 5^6 ).

Cf. A000108, A212073, A333090 through A333097.

seq(add(4*n/(4*n+k)*binomial(4*n+2*k-1,k), k = 0..n), n = 1..25);
#alternative program
c:= x → (1/2)*(1-sqrt(1-4*x))/x:
G := (x,n) → series(c(x)^(4*n), x, 126):
seq(add(coeff(G(x, n), x, n-k), k = 0..n), n = 0..25);

Join[{1}, Table[4*Binomial[6*n-1, n] * HypergeometricPFQ[{1, -5*n, -n}, {1/2 - 3*n, 1 - 3*n}, 1/4]/5, {n, 1, 20}]] (* Vaclav Kotesovec, Mar 28 2020 *)

a(n) = Sum_{k = 0..n} 4*n/(4*n+k)*binomial(4*n+2*k-1, k) for n >= 1.
a(n) = [x^n] ( (1 + x)*c^4(x/(1 + x)) )^n.
O.g.f.: ( 1 + x*f'(x)/f(x) )/( 1 - x*f(x) ), where f(x) = 1 + 4*x + 30*x^2 + 280*x^3 + 2925*x^4 + ... = (1/x)*Revert( x/c^4(x) ) is the o.g.f. of A212073.
Row sums of the Riordan array ( 1 + x*f'(x)/f(x), f(x) ) belonging to the Hitting time subgroup of the Riordan group.
a(n) ~ 2^(6*n + 3) * 3^(6*n + 3/2) / (31 * sqrt(Pi*n) * 5^(5*n + 1/2)). - Vaclav Kotesovec, Mar 28 2020
a(n) = Sum_{k = 0..n} 4*n/(4*n+2*k)*binomial(4*n+2*k, k) for n >= 1. - Peter Bala, May 03 2024

cp's OEIS Frontend

A333093 a(n) is equal to the n-th order Taylor polynomial (centered at 0) of c(x)^n evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. of the Catalan numbers A000108.

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A333097 a(n) = the n-th order Taylor polynomial (centered at 0) of c(x)^(5*n) evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. of the sequence of Catalan numbers A000108.

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A333095 a(n) = the n-th order Taylor polynomial (centered at 0) of c(x)^(3*n) evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. of the sequence of Catalan numbers A000108.

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A333096 a(n) = the n-th order Taylor polynomial (centered at 0) of c(x)^(4*n) evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. of the sequence of Catalan numbers A000108.

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Maple

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A333093 a(n) is equal to the n-th order Taylor polynomial (centered at 0) of c(x)^n evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4x))/(2x) is the o.g.f. of the Catalan numbers A000108.

A333097 a(n) = the n-th order Taylor polynomial (centered at 0) of c(x)^(5n) evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4x))/(2*x) is the o.g.f. of the sequence of Catalan numbers A000108.

A333095 a(n) = the n-th order Taylor polynomial (centered at 0) of c(x)^(3n) evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4x))/(2*x) is the o.g.f. of the sequence of Catalan numbers A000108.

A333096 a(n) = the n-th order Taylor polynomial (centered at 0) of c(x)^(4n) evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4x))/(2*x) is the o.g.f. of the sequence of Catalan numbers A000108.