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a(n) is the maximum number of triangles in planar connected graphs of triangles with n+3 nodes. - Ya-Ping Lu, Jun 25 2024

a(k*m) = k*a(m) - (k-1). - Ya-Ping Lu, Jun 25 2024

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From Klaus Purath, Apr 16 2024: (Start)

This sequence is closed with respect to multiplication, which is valid for all recursive sequences (t) with constant coefficients (2,-1) and initial value t(0) = 1, t(n) = bn+1. Proof: (bn+1)*(b(n+1)+1) = b*(bn^2 + bn + 2n + 1) + 1, which can be verified by expanding both sides of the equation.

a(2n^2-1) = A000466(n) is the product of two consecutive terms a(n-1)*a(n).

For any two consecutive terms (a(n), a(n+1)) = (x,y): x^2 - 2xy + y^2 = 2^2 = A000290(2).

In general, the following applies to all recursive sequences (t) with constant

coefficients (2,-1) and two consecutive terms (x,y): x^2 - 2xy + y^2 = A000290(t(1)-t(0)).

By analogy to this, for three consecutive terms (x,y,z) of any recursive sequence

(t) of the form (2,-1): y^2 - xz = A000290(t(1)-t(0)).

If (t) is a recursive sequence with the constant coefficients (3,-3,1) or (2,-1) including this sequence itself, then a(n) = [t(k+2n+1) - t(k)]/[t(k+n+1) - t(k+n)] always applies, k and n = integer. This algorithm works not only with the integer sequences mentioned, but also with the functions f(x) = ax^2 + bx + c or f(x) = bx + c. Then a(n) = [f(x+(2n+1)d) - f(x)]/[f(x+(n+1)d) - f(x+nd)] with a, b, c, d, x in R. And it works with the first differences of all them too. (End)

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This sequence is closed with respect to multiplication, which is valid for all recursive sequences (t) with constant coefficients (2,-1) and initial value t(0) = 1, t(n) = bn+1. Proof: (bn+1)*(b(n+1)+1) = b*(bn^2 + bn + 2n + 1) + 1, which can be verified by expanding both sides of the equation.

a(2n^2-1) = A000466(n) is valid for all recursive sequences (t) with constant coefficients the product of two consecutive terms a(2,n-1) and*a(n).

initial value t(0) = 1, t(n) = bn+1. Proof: (bn+1)*(b(n+1)+1) = b*(bn^2 +bn +

2n +1) + 1, which can be verified by expanding both sides of the equation.

a(2n^2-1) = A000466(n) is the product of For any two consecutive terms (a(n-1)*, a(n+1)) = (x,y): x^2 - 2xy + y^2 = 2^2 = A000290(2).

For

any two consecutive terms (a(n), a(n+1)) = (x,y): x^2 - 2xy + y^2 = 2^2 = A000290(2).

By analogy to this, for three consecutive terms (x,y,z) of any recursive sequence

analogy to this, for three consecutive terms (x,y,z) of any recursive sequence

If (t) is a recursive sequence with the constant coefficients (3,-3,1) or (2,-1) including this sequence itself, then a(n) = [t(k+2n+1) - t(k)]/[t(k+n+1) - t(k+n)] always applies, k and n = integer. This algorithm works not only with the integer sequences mentioned, but also with the functions f(x) = ax^2 + bx + c or f(x) = bx + c. Then a(n) = [f(x+(2n+1)d) - f(x)]/[f(x+(n+1)d) - f(x+nd)] with a, b, c, d, x in R. And it works with the first differences of all them too. (End)

with the integer sequences mentioned, but also with the functions f(x) = ax^2 +

bx + c or f(x) = bx + c. Then a(n) = [f(x+(2n+1)d) - f(x)]/[f(x+(n+1)d) - f(x+nd)]

with a, b, c, x in R. And it works with the first differences of all them too. (End)

This sequence is closed with respect to multiplication, which

is valid for all recursive sequences (t) with constant coefficients (2,-1) and

initial value t(0) = 1, t(n) = bn+1. Proof: (bn+1)*(b(n+1)+1) = b*(bn^2 +bn +

2n +1) + 1, which can be verified by expanding both sides of the equation.

This sequence is closed with respect to multiplication. a(2n^2-1) = A000466(n) is the product of two consecutive terms a(n-1)*a(n).

For

any two consecutive terms (a(n), a(n+1)) = (x,y): x^2 - 2xy + y^2 = 2^2 = A000290(2).

In general, the following applies to all recursive sequences (t) with constant

coefficients (2,-1) and two consecutive terms (x,y): x^2 - 2xy + y^2 = A000290(t(1)-t(0)).

By

analogy to this, for three consecutive terms (x,y,z) of any recursive sequence

(t) of the form (2,-1): y^2 - xz = A000290(t(1)-t(0)).

The sum of an odd number of any terms belongs to If (t) is a recursive sequence with the constant coefficients (3,-3,1) or (2,-1) including this sequence. For odd k, itself, then a(n) = [t(k^2+2kn-2n+1)/2 - t(k) is the sum of ]/[t(k consecutive terms starting from a+n+1) - t(k+n)] always applies, k and n = integer. This algorithm works not only

If (t) is a recursive sequence with the constant coefficients (3,-3,1) or (2,-1) including this sequence itself, then a(n) = [t(k+2n+1) - t(k)]/[t(k+n+1) - t(k+n)] always applies. (End)

with the integer sequences mentioned, but also with the functions f(x) = ax^2 +

bx + c or f(x) = bx + c. Then a(n) = [f(x+(2n+1)d) - f(x)]/[f(x+(n+1)d) - f(x+nd)]

with a, b, c, x in R. And it works with the first differences of all them too. (End)

a(n) = A055112(n)/oblong(n) = A193218(n+1)/Hex number(n). Compare to the Sep 27 2008 comment by Pierre CAMI. - Klaus Purath, Apr 23 2024