A087161 -id:A087161 - cp's OEIS frontend

A087159 Satisfies a(1)=1, a(A087160(n+1)) = a(n)+1, with a(m)=2 for all m not found in A087160, where A087160(n+2)=A087160(n+1)+a(n)+1.

1, 2, 2, 3, 2, 2, 3, 2, 2, 4, 2, 2, 2, 3, 2, 2, 3, 2, 2, 4, 2, 2, 2, 3, 2, 2, 3, 2, 2, 5, 2, 2, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 4, 2, 2, 2, 3, 2, 2, 3, 2, 2, 4, 2, 2, 2, 3, 2, 2, 3, 2, 2, 5, 2, 2, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 4, 2, 2, 2, 3, 2, 2, 3, 2, 2, 4, 2, 2, 2, 3, 2, 2, 3, 2, 2, 6, 2, 2, 2, 2, 2, 3, 2

Offset: 1

Paul D. Hanna, Aug 22 2003

nonn

Records form A087161: a(A087161(n))=n and satisfy recurrence: A087161(n+3) = 5*A087161(n+2) - 6*A087161(n+1) + 2*A087161(n).

			Initialize all terms to 2. Set a(1)=1, go one term forward,
set a(2)=a(1)+1=2, go 2 terms forward,
set a(4)=a(2)+1=3, go 3 terms forward,
set a(7)=a(3)+1=3, go 3 terms forward,
set a(10)=a(4)+1=4, go 4 terms forward,
set a(14)=a(5)+1=3, etc.
The indices 1,2,4,7,10,14,... form A087160.

Cf. A087160, A087161.

A100631 Triangle read by rows: T(n,k) = 2*(T(n-1,k-1) - T(n-2,k-1) + T(n-1,k)) for 0 < k < n, T(n,0) = T(n,n) = 1.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 4, 4, 1, 1, 8, 12, 8, 1, 1, 16, 32, 32, 16, 1, 1, 32, 80, 104, 80, 32, 1, 1, 64, 192, 304, 304, 192, 64, 1, 1, 128, 448, 832, 1008, 832, 448, 128, 1, 1, 256, 1024, 2176, 3072, 3072, 2176, 1024, 256, 1, 1, 512, 2304, 5504, 8832, 10272, 8832, 5504, 2304, 512, 1

Offset: 0

Reinhard Zumkeller, Dec 03 2004

From Petros Hadjicostas, Feb 09 2021: (Start)
The rectangular version (R(n,k): n,k >= 0) of this symmetric triangular array (T(n,k): 0 <= k <= n) is given by R(n,k) = T(n+k,k) for n,k >= 0. Conversely, T(n,k) = R(n-k, k) for 0 <= k <= n.
Note that [o.g.f of R](x,y) = [o.g.f. of T](x, y/x) and [o.g.f of T](x,y) = [o.g.f of R](x,x*y). (End)
From Petros Hadjicostas, Feb 10 2021: (Start)
All the conjectures below are true because one has to prove only one of them, and the rest follow from the proved one.
As Peter Luschny pointed out, one has to show only that the function S(n,k) = 2^n*hypergeom([-k + 1, n], [1], -1) satisfies the recurrence S(n,k) = 2*(S(n,k-1) - S(n-1,k-1) + S(n-1,k)) for n, k > 0 and the initial conditions S(n,0) = S(0,n) = 1 for n >= 0.
This is quite easy to achieve because S(n,k) = 2^n*Sum_{s=0}^{k-1} binomial(k-1,s)*binomial(n+s-1,s) for n >= 0 and k >= 1. The proof of the recurrence relies on the identity binomial(m,n) = binomial(m-1, n) + binomial(m-1,n-1).
Note that without the 2^n in the formula R(n,k) = 2^n*hypergeom([-k + 1, n], [1], -1), we essentially get array A049600.
In addition, note that without the 2^(n-k-1) in the formula T(n,k+1) = 2^(n-k-1)*hypergeom([-k, n-k+1], [1], -1), we essentially get A208341 (without the first column and the main diagonal of T). (End)

			From _Petros Hadjicostas_, Feb 09 2021: (Start)
Triangle T(n,k) (with rows n >= 0 and columns 0 <= k <= n) begins:
  1,
  1,   1,
  1,   2,    1,
  1,   4,    4,    1,
  1,   8,   12,    8,    1,
  1,  16,   32,   32,   16,    1,
  1,  32,   80,  104,   80,   32,    1,
  1,  64,  192,  304,  304,  192,   64,    1,
  1, 128,  448,  832, 1008,  832,  448,  128,   1,
  1, 256, 1024, 2176, 3072, 3072, 2176, 1024, 256, 1,
  ...
Rectangular array R(n,k) (with rows n >= 0 and columns k >= 0) begins:
  1,   1,    1,    1,     1,     1,      1,       1, ...
  1,   2,    4,    8,    16,    32,     64,     128, ...
  1,   4,   12,   32,    80,   192,    448,    1024, ...
  1,   8,   32,  104,   304,   832,   2176,    5504, ...
  1,  16,   80,  304,  1008,  3072,   8832,   24320, ...
  1,  32,  192,  832,  3072, 10272,  32064,   95104, ...
  1,  64,  448, 2176,  8832, 32064, 107712,  341504, ...
  1, 128, 1024, 5504, 24320, 95104, 341504, 1150592, ...
  ... (End)

Index entries for triangles and arrays related to Pascal's triangle

Cf. A000079, A001787, A049600, A084773, A087161, A100312, A152254, A208341, A341344.

T[, 0] = T[n, n_] = 1;
T[n_, k_] /; 0, ] = 0;
Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Oct 03 2021 *)

From Petros Hadjicostas, Feb 09 2021: (Start)
Formulas for the triangular array (T(n,k): 0 <= k <= n):
T(n,k) = T(n,n-k) for 0 <= k <= n.
Sum_{k=0..n} T(n,k) = A087161(n+1).
T(n,1) = T(n,n-1) = 2^(n-1) = A000079(n-1) for n >= 1.
T(n,2) = T(n,n-2) = (n-1)*2^(n-2) = A001787(n-1) for n >= 2.
T(n,3) = T(n,n-3) = (n^2-n-4)*2^(n-4) = A100312(n-3) for n >= 3.
T(n,floor(n/2)) = T(n,ceiling(n/2)) = A341344(n).
Bivariate o.g.f.: Sum_{n,k >= 0} T(n,k)*x^n*y^k = (3*x^2*y - 2*x*y - 2*x + 1)/((1 - x)*(-x*y + 1)*(2*x^2*y - 2*x*y - 2*x + 1)).
Conjecture based on Peter Luschny's formulas in other sequences: T(n,k) = 2^(n-k)*hypergeom([-k + 1, n-k], [1], -1) = 2^k*hypergeom([-(n-k) + 1, k], [1], -1).
Formulas for the rectangular array (R(n,k): n,k >= 0):
R(n,k) = 2*(R(n,k-1) - R(n-1,k-1) + R(n-1,k)) for n,k > 0 with R(n,0) = R(0,n) = 1 for n >= 0.
R(n,k) = R(k, n) for n,k >= 0.
R(1,n) = R(n,1) = 2^n = A000079(n).
R(2,n) = R(n,2) = (n+1)*2^n = A001787(n+1).
R(3,n) = R(n,3) = (n^2+5*n+2)*2^(n-1) = A100312(n).
R(n,n) = A152254(n-1) = 2*A084773(n-1) for n >= 1.
Bivariate o.g.f.: Sum_{n,k >= 0} R(n,k)*x^n*y^k = (3*x*y - 2*x - 2*y - 1)/((1 - x)*(1 - y)*(2*x*y - 2*x - 2*y - 1)).
Conjecture based on Peter Luschny's formulas in other sequences: R(n,k) = 2^n*hypergeom([-k + 1, n], [1], -1) = 2^k*hypergeom([-n + 1, k], [1], -1). (End)
From Petros Hadjicostas, Feb 10 2021: (Start)
The above conjecture is true (see the comments).
R(n,k) = 2^k*Sum_{s=0}^{n-1} binomial(n-1,s)*binomial(k+s-1,s) = 2^n*Sum_{s=0}^{k-1} binomial(k-1,s)*binomial(n+s-1,s) for n, k >= 1.
To get two binomial formulas for T(n,k), use the equation T(n,k) = R(n-k, k) for 1 <= k <= n and the above formulas for R(n,k). (End)

Offset changed by Petros Hadjicostas, Feb 09 2021

cp's OEIS Frontend

A087159 Satisfies a(1)=1, a(A087160(n+1)) = a(n)+1, with a(m)=2 for all m not found in A087160, where A087160(n+2)=A087160(n+1)+a(n)+1.

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A087160 Satisfies a(1)=1, a(2)=2, a(n+2)=a(n+1)+A087159(n)+1.

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A100631 Triangle read by rows: T(n,k) = 2*(T(n-1,k-1) - T(n-2,k-1) + T(n-1,k)) for 0 < k < n, T(n,0) = T(n,n) = 1.

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