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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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A113647 Triangle of numbers related to the generalized Catalan sequence C(2;n+1)=A064062(n+1), n>=0.

Original entry on oeis.org

1, 1, 3, 1, 7, 13, 1, 15, 41, 67, 1, 31, 113, 247, 381, 1, 63, 289, 783, 1545, 2307, 1, 127, 705, 2271, 5361, 9975, 14589, 1, 255, 1665, 6207, 16929, 36879, 66057, 95235, 1, 511, 3841, 16255, 50113, 123871, 255985, 446455, 636925, 1, 1023, 8705, 41215, 141441
Offset: 0

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Author

Wolfdieter Lang, Jan 13 2006

Keywords

Comments

This triangle, called Y(2,1), appears in the totally asymmetric exclusion process for the (unphysical) values alpha=2, beta=1. See the Derrida et al. refs. given under A064094, where the triangle entries are called Y_{N,K} for given alpha and beta.
The main diagonal (M=1) gives the generalized Catalan sequence C(2,n):=A064062(n).
The diagonal sequences give A064062(n+1), A115137, A115150-A115153, for n+1>= M=1,..,6.

Examples

			Triangle begins:
  1;
  1,3;
  1,7,13;
  1,15,41,67;
  1,31,113,247,381;
  ...
113=a(4,3)= a(4,2) + 2*a(3,3)= 31 + 2*41.

Links

Crossrefs

Row sums give A115136.

Formula

a(n, n+1)=A064062(n+1) (main diagonal with M=1); a(n, n-M+2)= a(n, n-M+1) + 2*a(n-1, n-M+2), M>=2; a(n, 1)=1; n>=0.
G.f. for diagonal sequence M=1: GY(1, x):=(2*c(2*x)-1)/(1+x) with c(x) g.f. of A000108 (Catalan); for M=2: GY(2, x)=(1-2*x)*GY(1, x)-1; for M>=3: GY(M, x)= GY(M-1, x) -2*x*GY(M-2, x) + x^(M-2).
G.f. for diagonal sequence M (solution to the above given recurrence): GY(M, x)= (x^(M-1)/(1+x))*( 2^(M+1)*x*(p(M, 2*x)-(2*x)*p(M+1, 2*x)*c(2*x))+1), with c(x) g.f. of A000108 (Catalan) and p(n, x):= -((1/sqrt(x))^(n+1))*S(n-1, 1/sqrt(x)) with Chebyshev's S(n, x) polynomials given in A049310.

A333565 O.g.f.: (1 + 4*x)/((1 + x)*sqrt(1 - 8*x)).

Original entry on oeis.org

1, 7, 33, 223, 1537, 11007, 80385, 595455, 4456449, 33615871, 255148033, 1946337279, 14908784641, 114597822463, 883479412737, 6828492980223, 52895475040257, 410544577183743, 3191929428770817, 24855137310736383, 193811815161921537, 1513167009951514623, 11827298001565515777
Offset: 0

Views

Author

Peter Bala, Apr 11 2020

Keywords

Comments

This 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, since the power series E(x) := exp( Sum_{n >= 1} a(n)*x^n/n ) has integer coefficients. See Stanley, Ex. 5.2 (a), p. 72, and its solution on p. 104.
We conjecture that this sequence satisfies the stronger congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 3 and positive integers n and k. The particular case when n = k = 1 follows from the corresponding result for A333564. Some examples of these congruences are given below.

Examples

			Examples of congruences:
a(11) - a(1) = 1946337279 - 7 = (2^3)*(11^3)*182789 == 0 ( mod 11^3 ).
a(2*11) - a(2) = 11827298001565515777 - 33 = (2^5)*(3^2)*(11^3)*107* 288357478039 == 0 ( mod 11^3 ).
a(5^2) - a(5) = 5680983691406772011007 - 11007 = (2^8)*(3^3)*(5^6)*7* 19*1123*352183001 == 0 ( mod 5^6 ).

References

  • R. P. Stanley. Enumerative combinatorics. Vol. 2, (volume 62 of Cambridge Studies in Advanced Mathematics). Cambridge University Press, Cambridge, 1999.

Crossrefs

Cf. A000984, A113647, A115137, A119259, A333564.

Programs

  • Maple
    a := proc (n) option remember; `if`(n = 0, 1, `if`(n = 1, 7, `if`(n = 2, 33, ((3*n+4)*a(n-1)+(36*n-76)*a(n-2)+(32*n-80)*a(n-3))/n)))
end proc:
seq(a(n), n = 0..25);
  • Mathematica
    a[n_] := (-1)^n - 2^(n+2) Binomial[2n, n-1] Hypergeometric2F1[1, 2n +1, n + 2, 2];
Table[Simplify[a[n]], {n, 0, 22}] (* Peter Luschny, Apr 13 2020 *)
CoefficientList[Series[(1+4x)/((1+x)Sqrt[1-8x]),{x,0,30}],x] (* Harvey P. Dale, Jan 24 2021 *)

Formula

a(n) = (2^n)*binomial(2*n,n) + 3*sum_{k = 0..n-1} (-1)^(n+k+1)*2^k* binomial(2*k,k).
a(n) = 4*A333564(n) + (-1)^n for n >= 1.
a(n) = 2*A119259(n) - (-1)^n.
a(n) = (-1)^n + 4*Sum_{k = 1..n} (3*k-1)*2^(k-1)*A000108(k-1).
a(n) ~ 8^n * 4/(3*sqrt(Pi*n)).
Congruences: a(p) == 7 ( mod p^3 ) for all prime p >= 3.
O.g.f. A(x) = 1 + 7*x + 33*x^2 + ... satisfies the differential equation (x + 1)*(4*x + 1)*(8*x - 1)*A'(x) + (16*x^2 - 4*x + 7)*A(x) = 0. Cf. A333564.
P-recursive: n*(3*n - 4)*a(n) = (21*n^2 - 40*n + 12)*a(n-1) + 4*(3*n - 1)*(2*n - 3)*a(n-2) with a(0) = 1 and a(1) = 7.
Alternative form: (a(n) + a(n-1))/(a(n) - a(n-2)) = P(n)/Q(n), where P(n) = 4*(3*n - 1)*(2*n - 3) and Q(n) = (21*n^2 - 40*n + 12).
Also, n*a(n) = (3*n + 4)*a(n-1) + 4*(9*n - 19)*a(n-2) + 16*(2*n - 5)*a(n-3) with a(0) = 1, a(1) = 7 and a(2) = 33.
exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 7*x + 41*x^2 + 247*x^3 + ... is the o.g.f. of the second diagonal of triangle A113647. See also A115137.
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