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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A099837 Expansion of (1 - x^2) / (1 + x + x^2) in powers of x.

Original entry on oeis.org

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

Views

Author

Paul Barry, Oct 27 2004

Keywords

Comments

A transform of (-1)^n.
Row sums of Riordan array ((1-x)/(1+x), x/(1+x)^2), A110162.
Let b(n) = Sum_{k=0..floor(n/2)} binomial(n-k,k)(-1)^(n-2k). Then a(n) = b(n) - b(n-2) = A049347(n) - A049347(n-2) (n > 0). The g.f. 1/(1+x) of (-1)^n is transformed to (1-x^2)/(1+x+x^2) under the mapping G(x)->((1-x^2)/(1+x^2))G(x/(1+x^2)). Partial sums of A099838.
A(n) = a(n+3) (or a(n) if a(0) is replaced by 2) appears, together with B(n) = A049347(n) in the formula 2*exp(2*Pi*n*i/3) = A(n) + B(n)*sqrt(3)*i, n >= 0, with i = sqrt(-1). See A164116 for the case N=5. - Wolfdieter Lang, Feb 27 2014

Examples

			G.f. = 1 - x - x^2 + 2*x^3 - x^4 - x^5 + 2*x^6 - x^7 - x^8 + 2*x^9 - x^10 + ...

Links

Crossrefs

Cf. A061347, A100051, A100063, A098554, A110162.

Programs

  • Maple
    A099837 := proc(n)
    option remember;
    if n <=2 then
        op(n+1,[1,-1,-1]) ;
    else
        -procname(n-1)-procname(n-2) ;
    end if;
end proc:
seq(A099837(n),n=0..80) ; # R. J. Mathar, Apr 26 2022
  • Mathematica
    a[0] = 1; a[n_] := Mod[n+2, 3] - Mod[n, 3]; A099837 = Table[a[n], {n, 0, 71}](* Jean-François Alcover, Feb 15 2012, after Michael Somos *)
LinearRecurrence[{-1, -1}, {1, -1, -1}, 50] (* G. C. Greubel, Aug 08 2017 *)
  • Maxima
    A099837(n) := block(
        if n = 0 then 1 else [2,-1,-1][1+mod(n,3)]
)$ /* R. J. Mathar, Mar 19 2012 */
  • PARI
    {a(n) = [2, -1, -1][n%3 + 1] - (n == 0)}; /* Michael Somos, Jan 19 2012 */
  • PARI
    Vec((1-x^2)/(1+x+x^2) + O(x^20)) \\ Felix Fröhlich, Aug 08 2017

Formula

G.f.: (1-x^2)/(1+x+x^2).
Euler transform of length 3 sequence [-1, -1, 1]. - Michael Somos, Mar 21 2011
Moebius transform is length 3 sequence [-1, 0, 3]. - Michael Somos, Mar 22 2011
a(n) = -b(n) where b(n) = A061347(n) is multiplicative with b(3^e) = -2 if e > 0, b(p^e) = 1 otherwise. - Michael Somos, Jan 19 2012
a(n) = a(-n). a(n) = c_3(n) if n > 1 where c_k(n) is Ramanujan's sum. - Michael Somos, Mar 21 2011
G.f.: (1 - x) * (1 - x^2) / (1 - x^3). a(n) = -a(n-1) - a(n-2) unless n = 0, 1, 2. - Michael Somos, Jan 19 2012
Dirichlet g.f.: Sum_{n>=1} a(n)/n^s = zeta(s)*(3^(1-s)-1). - R. J. Mathar, Apr 11 2011
a(n+3) = R(n,-1) for n >= 0, with the monic Chebyshev T-polynomials R with coefficient table A127672. - Wolfdieter Lang, Feb 27 2014
For n > 0, a(n) = 2*cos(n*Pi/3)*cos(n*Pi). - Wesley Ivan Hurt, Sep 25 2017
From Peter Bala, Apr 20 2024: (Start)
a(n) is equal to the n-th order Taylor polynomial (centered at 0) of 1/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. Cf. A333093.
Row sums of the Riordan array A110162. (End)

A333090 a(n) is equal to the n-th order Taylor polynomial (centered at 0) of S(x)^n evaluated at x = 1, where S(x) = (1 - x - sqrt(1 - 6*x + x^2))/(2*x) is the o.g.f. of the Schröder numbers A006318.

Original entry on oeis.org

1, 3, 21, 183, 1729, 17003, 171237, 1752047, 18130433, 189218451, 1987916021, 20996253479, 222730436161, 2371369720827, 25325636818629, 271189884041183, 2910628489408513, 31302328583021091, 337241582882175189, 3639109029230457751, 39324814984207649729
Offset: 0

Views

Author

Peter Bala, Mar 22 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 all primes 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 S(x)^(m*n) evaluated at x = 1, satisfies the same supercongruences. See A333091 for m = 2 and A333092 for m = 3. For similarly defined sequences see A333093 through A333097.

Examples

			n-th order Taylor polynomial of S(x)^n:
  n = 0: S(x)^0 = 1 + O(x)
  n = 1: S(x)^1 = 1 + 2*x + O(x^2)
  n = 2: S(x)^2 = 1 + 4*x + 16*x^2 + O(x^3)
  n = 3: S(x)^3 = 1 + 6*x + 30*x^2 + 146*x^3 + O(x^4)
  n = 4: S(x)^4 = 1 + 8*x + 48*x^2 + 264*x^3 + 1408*x^4 + O(x^5)
Setting x = 1 gives a(0) = 1, a(1) = 1 + 2 = 3, a(2) = 1 + 4 + 16 = 21, a(3) = 1 + 6 + 30 + 146 = 183 and a(4) = 1 + 8 + 48 + 264 + 1408 = 1729.
The triangle of coefficients of the n-th order Taylor polynomial of S(x)^n, n >= 0, in descending powers of x begins
                                          row sums
  n = 0 |    1                                1
  n = 1 |    2    1                           3
  n = 2 |   16    4    1                     21
  n = 3 |  146   30    6   1                183
  n = 4 | 1408  264   48   8   1           1729
   ...
This is a Riordan array belonging to the Hitting time subgroup of the Riordan group. The first column sequence is [x^n]S(x)^n = A103885(n).
Examples of supercongruences:
a(13) - a(1) = 2371369720827 - 3 = (2^3)*(3^2)*(13^3)*83*180617 == 0 ( mod 13^3 ).
a(3*7) - a(3) = 425495386400395896971 - 183 = (2^2)*(7^3*)*19*47* 347287606554703 == 0 ( mod 7^3 ).
a(5^2) - a(5) = 5894174066435445232142003 - 17003 = (2^3)*(3^4)*(5^6)*17* 41*101*5081*1627513421 == 0 ( mod 5^6 ).

Crossrefs

Cf. A006318, A027307, A103885, A333091 through A333097, A372214, A372215.

Programs

  • Maple
    S:= x -> (1/2)*(1-x-sqrt(1-6*x+x^2))/x:
G := (x,n) -> series(S(x)^n, x, 51):
seq(add(coeff(G(x, n), x, k), k = 0..n), n = 0..25);
  • Mathematica
    Table[SeriesCoefficient[((1 + x)*(1 - Sqrt[1 - 4*x - 4*x^2])/(2*x))^n, {x, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Mar 28 2020 *)

Formula

a(n) = [x^n] ( (1 + x)*S(x/(1 + x)) )^n.
O.g.f.: ( 1 + x*f'(x)/f(x) )/( 1 - x*f(x) ), where f(x) = 1 + 2*x + 10*x^2 + 66*x^3 + 498*x^4 + ... = (1/x)*Revert( x/S(x) ) is the o.g.f. of A027307.
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) ~ phi^(5*n+2) / (2*5^(3/4)*sqrt(Pi*n)), where phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - Vaclav Kotesovec, Mar 28 2020

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

Cf. A000108, A233834, A333090 through A333097.

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

Cf. A000108, A069271, A333090 through A333097.

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

Cf. A000108, A118970, A333090 through A333097.

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

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), 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

Cf. A000108, A212073, A333090 through A333097.

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

A372215 a(n) is equal to the n-th order Taylor polynomial (centered at 0) of G(x)^n evaluated at x = 1, where G(x) = (1 - 3*x - sqrt(1 - 10*x + 9*x^2))/(2*x) is the g.f. of A082298.

Original entry on oeis.org

1, 5, 65, 1013, 16897, 292005, 5157569, 92456341, 1675300865, 30604622405, 562675808065, 10398428960309, 192983418877441, 3594314403564773, 67146652988725697, 1257656071818605013, 23609209650223284225, 444081728926392461445, 8367715805572617168449
Offset: 0

Views

Author

Peter Bala, Apr 23 2024

Keywords

Comments

The sequence satisfies the Gauss congruences: a(n*p^k) == a(n*p^(k-1)) ( mod p^k ) for all primes 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 all primes p >= 5 and positive integers n and k. Examples of these supercongruences 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 G(x)^(m*n) evaluated at x = 1, satisfies the same supercongruences.

Examples

			n-th order Taylor polynomial of G(x)^n:
  n = 0: G(x)^0 = 1 + O(x)
  n = 1: G(x)^1 = 1 + 4*x + O(x^2)
  n = 2: G(x)^2 = 1 + 8*x + 56*x^2 + O(x^3)
  n = 3: G(x)^3 = 1 + 12*x + 108*x^2 + 892 *x^3 + O(x^4)
  n = 4: G(x)^4 = 1 + 16*x + 176*x^2 + 1680*x^3 + 15024*x^4 + O(x^5)
Setting x = 1 gives a(0) = 1, a(1) = 1 + 4 = 5, a(2) = 1 + 8 + 56 = 65, a(3) = 1 + 12 +  108 + 892 = 1013 and a(4) = 1 + 16 + 176 + 1680 + 15024 = 16897.
The triangle of coefficients of the n-th order Taylor polynomial of G(x)^n, n >= 0, in descending powers of x begins
                                            row sums
  n = 0 |     1                                 1
  n = 1 |     4      1                          5
  n = 2 |    56      8     1                   65
  n = 3 |   892    108    12    1            1013
  n = 4 | 15024   1680   176   16   1       16897
   ...
This is a Riordan array belonging to the Hitting time subgroup of the Riordan group.
Examples of supercongruences:
a(13) - a(1) = 3594314403564773 - 5 = (2^5)*(3^3)*(13^3)*29*7643*8543 == 0 (mod 13^3).
a(2*7) - a(2) = 67146652988725697 - 65 = (2^7)*3*(7^4)*23*3943*803057 = 0 (mod 7^4).

Crossrefs

Cf. A082298, A182959, A333090, A333093, A372214.

Programs

  • Maple
    G := x -> (1/2)*(1 - 3*x - sqrt(1 - 10*x + 9*x^2))/x:
H := (x, n) -> series(G(x)^n, x, 21):
seq(add(coeff(H(x, n), x, k), k = 0..n), n = 0..20);
  • Mathematica
    Table[SeriesCoefficient[(2*(1 + x)^2/(1 - 2*x + Sqrt[1 - 8*x]))^n, {x, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, May 02 2024 *)
  • PARI
    G(x) = (1 - 3*x - sqrt(1 - 10*x + 9*x^2))/(2*x);
a(n) = my(x='x+O('x^(n+2))); subst(Pol(Vec(G(x)^n)), 'x, 1); \\ Michel Marcus, May 07 2024

Formula

a(n) = [x^n] ( (1 + x)*G(x/(1 + x)) )^n.
O.g.f.: ( 1 + x*F'(x)/F(x) )/( 1 - x*F(x) ), where F(x) = (1/x)*Revert( x/G(x) ) = = 1 + 4*x + 36*x^2 + 420*x^3 + 5572*x^4 + ....
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/2 + 3/4) * (1 + sqrt(3))^(2*n-1) / (sqrt(Pi*n) * 2^(n+1)). - Vaclav Kotesovec, May 02 2024
a(n) = [x^n] H(x)^n, where H(x) = 2*(1 + x)^2/(1 - 2*x + sqrt(1 - 8*x)), the g.f. of A182959, satisfies [x^(n)] H(x)^(3*n) = binomial(6*n, 2*n). - Peter Bala, Nov 07 2024

A372214 a(n) is equal to the n-th order Taylor polynomial (centered at 0) of G(x)^n evaluated at x = 1, where G(x) = (1 - 2*x - sqrt(1 - 8*x + 4*x^2))/(2*x).

Original entry on oeis.org

1, 4, 40, 487, 6376, 86629, 1203823, 16984678, 242274280, 3484593028, 50444222665, 734066291974, 10728052396111, 157349171819155, 2314894133906086, 34145661019248487, 504810905195542504, 7478066502444399439, 110972913533524676080, 1649407167353221551706, 24549982881130265421001
Offset: 0

Views

Author

Peter Bala, Apr 23 2024

Keywords

Comments

x*G(x) = (1 - 2*x - sqrt(1 - 8*x + 4*x^2))/2 is the o.g.f. of A047891.
The sequence satisfies the Gauss congruences: a(n*p^k) == a(n*p^(k-1)) ( mod p^k ) for all primes 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 all primes p >= 5 and positive integers n and k. Examples of these supercongruences 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 G(x)^(m*n) evaluated at x = 1, satisfies the same supercongruences.

Examples

			n-th order Taylor polynomial of G(x)^n:
  n = 0: G(x)^0 = 1 + O(x)
  n = 1: G(x)^1 = 1 + 3*x + O(x^2)
  n = 2: G(x)^2 = 1 + 6*x + 33*x^2 + O(x^3)
  n = 3: G(x)^3 = 1 + 9*x + 63*x^2 + 414*x^3 + O(x^4)
  n = 4: G(x)^4 = 1 + 12*x + 102*x^2 + 768*x^3 + 5493*x^4 + O(x^5)
Setting x = 1 gives a(0) = 1, a(1) = 1 + 3 = 4, a(2) = 1 + 6 + 33 = 40, a(3) = 1 + 9 + 63 + 414 = 487 and a(4) = 1 + 12 + 102 + 768 + 5493 = 6376.
The triangle of coefficients of the n-th order Taylor polynomial of G(x)^n, n >= 0, in descending powers of x begins
                                             row sums
  n = 0 |    1                                   1
  n = 1 |    3    1                              4
  n = 2 |   33    6     1                       40
  n = 3 |  414   63     9    1                 487
  n = 4 | 5493  768   102   12   1            6376
   ...
This is a Riordan array belonging to the Hitting time subgroup of the Riordan group.
Examples of supercongruences:
a(13) - a(1) = 157349171819155 - 4 = (3^3)*(13^3)*269*9860941 == 0 (mod 13^3).
a(2*7) - a(2) = 2314894133906086 - 40 = 2*(3^4)*(7^3)*11*12119*312509 == 0 (mod 7^3).

Crossrefs

Cf. A047891, A219535, A333090, A333093, A372215.

Programs

  • Maple
    G := x -> (1/2)*(1 - 2*x - sqrt(1 - 8*x + 4*x^2))/x:
H := (x, n) -> series(G(x)^n, x, 41):
seq(add(coeff(H(x, n), x, k), k = 0..n), n = 0..20);
  • Mathematica
    Table[SeriesCoefficient[(2*(1 + x)^2/(1 - x + Sqrt[1 - 6*x - 3*x^2]))^n, {x, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, May 02 2024 *)
  • PARI
    G(x) = (1 - 2*x - sqrt(1 - 8*x + 4*x^2))/(2*x);
a(n) = my(x='x+O('x^(n+2))); subst(Pol(Vec(G(x)^n)), 'x, 1); \\ Michel Marcus, May 07 2024

Formula

a(n) = [x^n] ( (1 + x)*G(x/(1 + x)) )^n.
O.g.f.: ( 1 + x*F'(x)/F(x) )/( 1 - x*F(x) ), where F(x) = (1/x)*Revert( x/G(x) ) = = 1 + 3*x + 21*x^2 + 192*x^3 + 2001*x^4 + ... is the o.g.f. of A219535.
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) ~ sqrt(1 + 17/sqrt(33)) * (59 + 11*sqrt(33))^n / (sqrt(3*Pi*n) * 2^(3*n + 3/2)). - Vaclav Kotesovec, May 02 2024

A333091 a(n) is the n-th order Taylor polynomial (centered at 0) of S(x)^(2*n ) evaluated at x = 1, where S(x) = (1 - x - sqrt(1 - 6*x + x^2))/(2*x) is the o.g.f. of the Schröder numbers A006318.

Original entry on oeis.org

1, 5, 57, 761, 10817, 159005, 2386857, 36348401, 559362561, 8676917429, 135445348057, 2125030235113, 33479772021953, 529326516063181, 8393856020704841, 133449301759137761, 2126391547960594433, 33948765589280671589, 542950968447834265209, 8697032976559212532953
Offset: 0

Views

Author

Peter Bala, Mar 22 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 S(x)^(m*n) evaluated at x = 1 satisfies the same congruences. For cases see A333090 (m = 1) and A333092 (m = 3). For similarly defined sequences see A333093 through A333097.

Examples

			n-th order Taylor polynomial of S(x)^(2*n):
  n = 0: S(x)^0 = 1 + O(x)
  n = 1: S(x)^2 = 1 + 4*x + O(x^2)
  n = 2: S(x)^4 = 1 + 8*x + 48*x^2 + O(x^3)
  n = 3: S(x)^6 = 1 + 12*x + 96*x^2 + 652*x^3 + O(x^4)
  n = 4: S(x)^8 = 1 + 16*x + 160*x^2 + 1296*x^3 + 9344*x^4 + O(x^5)
Setting x = 1 gives a(0) = 1, a(1) = 1 + 4 = 5, a(2) = 1 + 8 + 48 = 57, a(3) = 1 + 12 + 96 + 652 = 761 and a(4) = 1 + 16 + 160 + 1296 + 9344 = 10817.
The triangle of coefficients of the n-th order Taylor polynomial of S(x)^(2*n), n >= 0, in descending powers of x begins
                                          row sums
  n = 0 |    1                                1
  n = 1 |    4    1                           5
  n = 2 |   48    8    1                     57
  n = 3 |  652   96   12   1                761
  n = 4 | 9344 1296  160  16   1          10817
   ...
This is a Riordan array belonging to the Hitting time subgroup of the Riordan group. The first column sequence [1, 4, 48, 652, 9344, 138004, ...]  = [x^n] S(x)^(2*n), and may also satisfy the above congruences.
Examples of congruences:
a(13) - a(1) = 529326516063181 - 5 = (2^3)*(13^3)*30116438101 == 0 ( mod 13^3 ).
a(3*7) - a(3) = 2240508640665255893197949 - 761 = (2^2)*3*(7^3)*11* 49485569411283149863 == 0 ( mod 7^3 ).
a(5^2) - a(5) = 150633078429259494145205034005 - 159005 = (2^3)*(3^3)*(5^6)*11*23*61*2663*28097*119633*323083 == 0 ( mod 5^6 ).

Crossrefs

A006318, A333090 through A333097.

Programs

  • Maple
    S:= x -> (1/2)*(1-x-sqrt(1-6*x+x^2))/x:
G := (x,n) -> series(S(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) * (1 - 2*x*(1+x) - Sqrt[1 - 4*x*(1+x)]) / (2*x^2))^n, {x, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Mar 28 2020 *)

Formula

a(n) = [x^n] ( (1 + x)*S^2(x/(1 + x)) )^n
O.g.f.: ( 1 + x*f'(x)/f(x) )/( 1 - x*f(x) ), where f(x) = 1 + 4*x + 32*x^2 + 324*x^3 + 3696*x^4 + ... = (1/x)*Revert( x/S^2(x) ).
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) ~ sqrt(120 + 39*sqrt(10)) * (223 + 70*sqrt(10))^n / (30*sqrt(Pi*n) * 3^(3*n)). - Vaclav Kotesovec, Mar 28 2020

A333092 a(n) is the n-th order Taylor polynomial (centered at 0) of S(x)^(3*n) evaluated at x = 1, where S(x) = (1 - x - sqrt(1 - 6*x + x^2))/(2*x) is the o.g.f. of the Schröder numbers A006318.

Original entry on oeis.org

1, 7, 109, 1951, 36993, 724007, 14457421, 292732671, 5987886081, 123440423047, 2560421160109, 53373725431583, 1117198199782785, 23465732683090471, 494330214846965389, 10440064992542621951, 220978578227187097601, 4686426367646858888711, 99559270036968523118317
Offset: 0

Views

Author

Peter Bala, Mar 22 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 S(x)^(m*n) evaluated at x = 1 satisfies the same congruences. For cases, see A333090 (m = 1) and A333091 (m = 2). For similarly defined sequences see A333093 through A333097.

Examples

			n-th order Taylor polynomial of S(x)^(3*n):
  n = 0: S(x)^0 = 1 + O(x)
  n = 1: S(x)^3 = 1 + 6*x + O(x^2)
  n = 2: S(x)^6 = 1 + 12*x + 96*x^2 + O(x^3)
  n = 3: S(x)^9 = 1 + 18*x + 198*x^2 + 1734*x^3 + O(x^4)
  n = 4: S(x)^12 = 1 + 24*x + 336*x^2 + 3608*x^3 + 33024*x^4 + O(x^5)
Setting x = 1 gives a(0) = 1, a(1) = 1 + 6 = 7, a(2) = 1 + 12 + 96 = 109, a(3) = 1 + 18 + 198 + 1734 = 1951 and a(4) = 1 + 24 + 336 + 3608 + 33024 = 36993.
The triangle of coefficients of the n-th order Taylor polynomial of S(x)^(2*n), n >= 0, in descending powers of x begins
                                           row sums
  n = 0 |     1                                1
  n = 1 |     6    1                           7
  n = 2 |    96   12    1                    109
  n = 3 |  1734  198   18   1               1951
  n = 4 | 33024 3608  336  24   1          36993
   ...
This is a Riordan array belonging to the Hitting time subgroup of the Riordan group. The first column sequence [1, 6, 96, 1734, 33024, 648006, ...]  = [x^n] S(x)^(3*n), and may also satisfy the above congruences.
Examples of congruences:
a(13) - a(1) = 23465732683090471 - 7 = (2^5)*(3^4)*(13^3)*83*911*54497 == 0 ( mod 13^3 ).
a(3*7) - a(3) = 962815680123979633351467303 - 1951 = (2^3)*(7^3)*29*41* 1832861*161008076794727  == 0 ( mod 7^3 ).
a(5^2) - a(5) = 201479167004032422703424646224007 - 724007 = (2^5)*(5^6)* 402958334008064845406849291 == 0 ( mod 5^6 ).

Crossrefs

Cf. A006318, A333090 through A333097.

Programs

  • Maple
    S:= x -> (1/2)*(1-x-sqrt(1-6*x+x^2))/x:
G := (x,n) -> series(S(x)^(3*n), x, 101):
seq(add(coeff(G(x, n), x, n-k), k = 0..n), n = 0..25);
  • Mathematica
    Table[SeriesCoefficient[((1+x)*(1 - 3*x*(1+x) + (x^2 + x - 1)*Sqrt[1 - 4*x*(1+x)]) / (2*x^3))^n, {x, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Mar 28 2020 *)

Formula

a(n) = [x^n] ( (1 + x)*S^3(x/(1 + x)) )^n.
O.g.f.: ( 1 + x*f'(x)/f(x) )/( 1 - x*f(x) ), where f(x) = 1 + 6*x + 66*x^2 + 902*x^3 + 13794*x^4 + ... = (1/x) * series reversion of ( x/S^3(x) ).
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*sqrt(85 + 21*sqrt(17)) * (349 + 85*sqrt(17))^n / (68 * sqrt(Pi*n) * 2^(5*n)). - Vaclav Kotesovec, Mar 28 2020
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