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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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.

Original entry on oeis.org

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

Views

Author

Peter Bala, Mar 07 2020

Keywords

Comments

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).

Examples

			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 ).
		

Crossrefs

Programs

  • Maple
    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);
  • Mathematica
    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 *)

Formula

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

A100218 Riordan array ((1-2*x)/(1-x), (1-x)).

Original entry on oeis.org

1, -1, 1, -1, -2, 1, -1, 0, -3, 1, -1, 0, 2, -4, 1, -1, 0, 0, 5, -5, 1, -1, 0, 0, -2, 9, -6, 1, -1, 0, 0, 0, -7, 14, -7, 1, -1, 0, 0, 0, 2, -16, 20, -8, 1, -1, 0, 0, 0, 0, 9, -30, 27, -9, 1, -1, 0, 0, 0, 0, -2, 25, -50, 35, -10, 1, -1, 0, 0, 0, 0, 0, -11, 55, -77, 44, -11, 1, -1, 0, 0, 0, 0, 0, 2, -36, 105, -112, 54, -12, 1
Offset: 0

Views

Author

Paul Barry, Nov 08 2004

Keywords

Examples

			Triangle begins as:
   1;
  -1,  1;
  -1, -2,  1;
  -1,  0, -3,  1;
  -1,  0,  2, -4,  1;
  -1,  0,  0,  5, -5,   1;
  -1,  0,  0, -2,  9,  -6,   1;
  -1,  0,  0,  0, -7,  14,  -7,  1;
  -1,  0,  0,  0,  2, -16,  20, -8,  1;
  -1,  0,  0,  0,  0,   9, -30, 27, -9,  1;
		

Crossrefs

Row sums are A100219.
Matrix inverse of A100100.
Apart from signs, same as A098599.
Very similar to triangle A111125.

Programs

  • Magma
    A100218:= func< n,k | n eq 0 select 1 else (-1)^(n+k)*(Binomial(k,n-k) + Binomial(k-1,n-k-1)) >;
    [A100218(n,k): k in [0..n], n in [0..13]]; // G. C. Greubel, Mar 28 2024
    
  • Mathematica
    T[0,0]:= 1; T[1,1]:= 1; T[1,0]:= -1; T[n_, k_]:= T[n, k]= If[k<0 || k>n, 0, T[n- 1,k] +T[n-1,k-1] -2*T[n-2,k-1] +T[n-3,k-1]]; Table[T[n,k], {n,0,14}, {k,0,n} ]//Flatten (* G. C. Greubel, Mar 13 2017 *)
  • SageMath
    def A100218(n,k): return 1 if n==0 else (-1)^(n+k)*(binomial(k,n-k) + binomial(k-1,n-k-1))
    flatten([[A100218(n,k) for k in range(n+1)] for n in range(14)]) # G. C. Greubel, Mar 28 2024

Formula

Sum_{k=0..n} T(n, k) = A100219(n) (row sums).
Number triangle T(n, k) = (-1)^(n-k)*(binomial(k, n-k) + binomial(k-1, n-k-1)), with T(0, 0) = 1. - Paul Barry, Nov 09 2004
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)=-1, T(1,1)=1, T(n,k)=0 if k < 0 or if k > n. - Philippe Deléham, Jan 09 2014
From G. C. Greubel, Mar 28 2024: (Start)
T(n, n-1) = A000027(n), n >= 1.
T(n, n-2) = -A080956(n-1), n >= 2.
T(2*n, n) = A280560(n).
T(2*n-1, n) = A157142(n-1), n >= 1.
T(2*n+1, n) = -A000007(n) = A154955(n+2).
T(3*n, n) = T(4*n, n) = A000007(n).
Sum_{k=0..n} (-1)^k*T(n, k) = A355021(n).
Sum_{k=0..floor(n/2)} T(n-k, k) = (-1)^n*A098601(n).
Sum_{k=0..floor(n/2)} (-1)^k*T(n-k, k) = -1 + 2*A077961(n) + A077961(n-2). (End)
From Peter Bala, Apr 28 2024: (Start)
This Riordan array has the form ( x*h'(x)/h(x), h(x) ) with h(x) = x*(1 - x) and hence belongs to the hitting time subgroup of the Riordan group (see Peart and Woan for properties of this subgroup).
T(n,k) = [x^(n-k)] (1/c(x))^n, where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the g.f. of the Catalan numbers A000108. In general the (n, k)-th entry of the hitting time array ( x*h'(x)/h(x), h(x) ) has the form [x^(n-k)] f(x)^n, where f(x) = x/( series reversion of h(x) ). (End)

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.

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

Views

Author

Peter Bala, Mar 15 2020

Keywords

Comments

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

Examples

			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 ).
		

Crossrefs

Programs

  • Maple
    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);
  • Mathematica
    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 *)

Formula

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

A333094 a(n) is the n-th order Taylor polynomial (centered at 0) of c(x)^(2*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.

Original entry on oeis.org

1, 3, 19, 144, 1171, 9878, 85216, 746371, 6609043, 59008563, 530279894, 4790262348, 43458522976, 395683988547, 3613641184739, 33088666355144, 303670285138067, 2792497004892302, 25724693177503987, 237350917999324431, 2193027397174233046, 20288470364637624223
Offset: 0

Views

Author

Peter Bala, Mar 15 2020

Keywords

Comments

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) defined as the n-th order Taylor polynomial of c(x)^(m*n) evaluated at x = 1 satisfies the same supercongruences. For cases A099837(m = -2), A100219 (m = -1), A000012 (m = 0), A333093 (m = 1), A333095 (m = 3), A333096 (m = 4), A333097 (m = 5).

Examples

			n-th order Taylor polynomial of c(x)^(2*n):
  n = 0: c(x)^0 = 1 + O(x)
  n = 1: c(x)^2 = 1 + 2*x + O(x^2)
  n = 2: c(x)^4 = 1 + 4*x + 14*x^2 + O(x^3)
  n = 3: c(x)^6 = 1 + 6*x + 27*x^2 + 110*x^3 + O(x^4)
  n = 4: c(x)^8 = 1 + 8*x + 44*x^2 + 208*x^3 + 910*x^4 + O(x^5)
Setting x = 1 gives a(0) = 1, a(1) = 1 + 2 = 3, a(2) = 1 + 4 + 14 = 19, a(3) = 1 + 6 + 27 + 110 = 144 and a(4) = 1 + 8 + 44 + 208 + 910 = 1171.
The triangle of coefficients of the n-th order Taylor polynomial of c(x)^(2*n), n >= 0, in descending powers of x begins
                                          row sums
  n = 0 |   1                                 1
  n = 1 |   2    1                            3
  n = 2 |  14    4    1                      19
  n = 3 | 110   27    6   1                 144
  n = 4 | 910  208   44   8   1            1171
   ...
This is a Riordan array belonging to the Hitting time subgroup of the Riordan group.
Examples of supercongruences:
a(13) - a(1) = 395683988547 - 3 = (2^6)*(3^2)*(13^3)*312677 == 0 ( mod 13^3 ).
a(3*7) - a(3) = 20288470364637624223 - 144 = (7^3)*17*269*12934629208861 == 0 ( mod 7^3 ).
a(5^2) - a(5) = 150194008594715226556753 - 9878 = (5^6)*2593*5471* 677584325533 == 0 ( mod 5^6 ).
		

Crossrefs

Programs

  • Maple
    seq(add(2*n/(2*n+k)*binomial(2*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)^(2*n), x, 76):
    seq(add(coeff(G(x, n), x, n-k), k = 0..n), n = 0..25);
  • Mathematica
    Table[SeriesCoefficient[((1 + x)^2 * (1 - x - Sqrt[(1 - 3*x)*(1 + x)]) / (2*x^2))^n, {x, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Mar 28 2020 *)

Formula

a(n) = Sum_{k = 0..n} 2*n/(2*n+k)*binomial(2*n+2*k-1, k) for n >= 1.
a(n) = [x^n] ( (1 + x)*c^2(x/(1 + x)) )^n.
O.g.f.: ( 1 + x*f'(x)/f(x) )/( 1 - x*f(x) ), where f(x) = 1 + 2*x + 9*x^2 + 52*x^3 + 340*x^4 + ... = (1/x)*Revert( x/c^2(x) ) is the o.g.f. of A069271.
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) ~ 2^(8*n + 7/2) / (13 * sqrt(Pi*n) * 3^(3*n + 1/2)). - Vaclav Kotesovec, Mar 28 2020
a(n) = Sum_{k = 0..n} n/(2*n+2*k)*binomial(2*n+2*k, k) for n >= 1. - Peter Bala, Apr 19 2024

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.

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

Views

Author

Peter Bala, Mar 15 2020

Keywords

Comments

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).

Examples

			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 ).
		

Crossrefs

Programs

  • Maple
    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);
  • Mathematica
    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 *)

Formula

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)^(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.

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

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Author

Peter Bala, Mar 15 2020

Keywords

Comments

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).

Examples

			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 ).
		

Crossrefs

Programs

  • Maple
    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);
  • Mathematica
    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 *)

Formula

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

A131290 1 followed by period 6: repeat [3, 2, 0, -1, 0, 2].

Original entry on oeis.org

1, 3, 2, 0, -1, 0, 2, 3, 2, 0, -1, 0, 2, 3, 2, 0, -1, 0, 2, 3, 2, 0, -1, 0, 2, 3, 2, 0, -1, 0, 2, 3, 2, 0, -1, 0, 2, 3, 2, 0, -1, 0, 2, 3, 2, 0, -1, 0, 2, 3, 2, 0, -1, 0, 2, 3, 2, 0, -1, 0, 2, 3, 2, 0, -1, 0, 2, 3, 2, 0, -1, 0, 2, 3, 2, 0, -1, 0, 2, 3, 2, 0, -1, 0, 2, 3, 2, 0, -1, 0, 2, 3, 2, 0, -1, 0, 2, 3, 2, 0, -1
Offset: 0

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Author

Paul Curtz, Sep 29 2007

Keywords

Crossrefs

Programs

  • Magma
    [1] cat &cat [[3, 2, 0, -1, 0, 2]^^30]; // Wesley Ivan Hurt, Jun 20 2016
  • Maple
    A131290 := proc(n) if n = 0 then 1; else op(((n-1)mod 6)+1,[3,2,0,-1,0,2]) ; fi ; end: seq(A131290(n),n=0..100) ; # R. J. Mathar, Feb 27 2008
  • Mathematica
    PadRight[{1},110,{2,3,2,0,-1,0}] (* or *) Join[{1},LinearRecurrence[ {2,-2,1},{3,2,0},110]] (* Harvey P. Dale, Jun 22 2012 *)

Formula

G.f.: (1+x-2*x^2+x^3)/((1-x)*(1-x+x^2)). - R. J. Mathar, Feb 27 2008
If n mod 6 = 4 then a(n) = (Fibonacci(n-3)*Fibonacci(n+1)) mod 4 -2, else a(n) = (Fibonacci(n-3)*Fibonacci(n+1)) mod 4, n>0. - Gary Detlefs, Dec 12 2010
From Wesley Ivan Hurt, Jun 20 2016: (Start)
a(n) = 2*a(n-1) - 2*a(n-2) + a(n-3) for n>4.
a(0) = 1, a(n) = 1 + cos(n*Pi/3) + sqrt(3)*sin(n*Pi/3) for n>0. (End)

Extensions

More terms from R. J. Mathar, Feb 27 2008
Showing 1-7 of 7 results.