A112880 -id:A112880 - cp's OEIS frontend

A112879 Define a(0)=3; then a(n)=k*a(n-1)^2-1=least prime first of twin primes.

3, 107, 206081, 20894934252011

Offset: 0

Pierre CAMI, Oct 01 2005

k sequence in A112880

			12*3^2-1=107, 107 and 109 twin primes so a(1)=107
18*107^2-1=206081, 206081 and 206083 twin primes so a(2)=206081
a(3)=492*(18*(12*3^2-1)^2-1)^2-1
a(4)=702*(492*(18*(12*3^2-1)^2-1)^2-1)^2-1
a(5)=2310*(702*(492*(18*(12*3^2-1)^2-1)^2-1)^2-1)^2-1
a(6)=2370*(2310*(702*(492*(18*(12*3^2-1)^2-1)^2-1)^2-1)^2-1)^2-1
a(7)=130650*(2370*(2310*(702*(492*(18*(12*3^2-1)^2-1)^2-1)^2-1)^2-1)^2-1)^2-1
a(8)=980910*(130650*(2370*(2310*(702*(492*(18*(12*3^2-1)^2-1)^2-1)^2-1)^2-1)^2-1)^2-1)^2-1
a(9)=5528418*(980910*(130650*(2370*(2310*(702*(492*(18*(12*3^2-1)^2-1)^2-1)^2-1)^2-1)^2-1)^2-1)^2-1)^2-1
a(10)=21357012*(5528418*(980910*(130650*(2370*(2310*(702*(492*(18*(12*3^2-1)^2-1)^2-1)^2-1)^2-1)^2-1)^2-1)^2-1)^2-1)^2-1

Cf. A112880.

A122881 Triangle read by rows: number of Catalan paths of 2n steps of all values less than or equal to m.

Original entry on oeis.org

1, 1, 2, 1, 2, 5, 1, 2, 5, 13, 1, 2, 5, 14, 34, 1, 2, 5, 14, 42, 89, 1, 2, 5, 14, 42, 131, 233, 1, 2, 5, 14, 42, 132, 417, 610, 1, 2, 5, 14, 42, 132, 429, 1341, 1597, 1, 2, 5, 14, 42, 132, 429, 1429, 4334, 4181

Offset: 1

Gary W. Adamson, Sep 16 2006

Convergents of k-th diagonals relate to (2k+3)-polygons; e.g., right border relates to the pentagon (N=5), next border relates to the heptagon (N=7). Convergents of the diagonals are 2 + 2*cos(2*Pi/N) and are roots to Morgan-Voyce polynomials. k2 diagonal = A080937, number of Catalan paths of 2n steps of all values less than or equal to 5. k3 diagonal = A080938, number of Catalan paths of 2n steps of all values less than or equal to 7.

			For the right border, odd-indexed Fibonacci numbers (1, 2, 5, 13, 34...), we begin with (M2) = [1, 1; 1, 0], then P2 = [1, -1; -1, 2] = 1/(M2)^2. Performing (P2)^n * [1,0] we extract the left vector (1, 2, 5, 13, ...), making it the right border of the triangle, k1 diagonal.
For the next diagonal going to the left, we begin with the Heptagonal matrix M3 = [1, 1, 1; 1, 1, 0; 1, 0, 0], take the inverse square (P3) and then perform the analogous operation getting 1, 2, 5, 14, 42, ...
First few rows of the triangle are:
  1;
  1, 2;
  1, 2, 5;
  1, 2, 5, 13;
  1, 2, 5, 14, 34;
  1, 2, 5, 14, 42, 89;
  1, 2, 5, 14, 42, 131, 233;
  1, 2, 5, 14, 42, 132, 417, 610;
  ...

Cf. A112880, A001519, A000108, A080937, A080938.

Begin with polygonal matrices of the form (exemplified by the Heptagonal matrix M3: [1, 1, 1; 1, 1, 0; 1, 0, 0]). Let matrix P3 = 1 / M3^2; then for n X n matrices P2, P3, P4...perform P^n * [1, 0, 0] letting this vector = k-th diagonal of the triangle.

cp's OEIS Frontend

A112879 Define a(0)=3; then a(n)=k*a(n-1)^2-1=least prime first of twin primes.

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