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• Let $a(n)$ be A069999 (i.e., number of possible dimensions for commutators of $n \times n$ matrices; it is independent of the field). OEIS states that no generating function is known.
• Let $P(n,k)$ be a triangle read by rows such that $$ P(n, k) = \begin{cases} 0 & \textrm{if } k > n \\ 1 & \textrm{if } k = 0 \\ z^{n-k+1}P(n, k-1) + P(k-1, k-1) & \textrm{otherwise} \end{cases} $$

I conjecture that number of positive terms in $P(n-1, n-1)$ equals $a(n)$.

Here is the PARI/GP program to generate number of positive terms of $P(n-1, n-1)$:

upto1(n) = my(v1); v1 = vector(n, i, 1); for(i=1, n-1, for(j=i+1, n, v1[j] = v1[i] + z^(j-i)*v1[j])); v1 = vector(n, i, #select(x->x>0, Vec(v1[i])))

Is there a way to prove it? Is it possible to get a closed-form or generating function after some manipulations with $P(n,k)$?

• Let $a(n)$ be A069999 (i.e., number of possible dimensions for commutators of $n \times n$ matrices; it is independent of the field). OEIS states that no generating function is known.
• Let $P(n,k)$ be a triangle read by rows such that $$ P(n, k) = \begin{cases} 0 & \textrm{if } k > n \\ 1 & \textrm{if } k = 0 \\ z^{n-k+1}P(n, k-1) + P(k-1, k-1) & \textrm{otherwise} \end{cases} $$

I conjecture that number of positive terms in $P(n-1, n-1)$ equals $a(n)$.

Here is the PARI/GP program to generate number of positive terms of $P(n-1, n-1)$:

upto1(n) = my(v1); v1 = vector(n, i, 1); for(i=1, n-1, for(j=i+1, n, v1[j] = v1[i] + z^(j-i)*v1[j])); v1 = vector(n, i, #select(x->x>0, Vec(v1[i])))

Is there a way to prove it? Is it possible to get a closed-form or generating function after some manipulations with $P(n,k)$?

• Let $a(n)$ be A069999 (i.e., number of possible dimensions for commutators of $n \times n$ matrices; it is independent of the field). OEIS states that no generating function is known.
• Let $P(n,k)$ be a triangle read by rows such that $$ P(n, k) = \begin{cases} 0 & \textrm{if } k > n \\ 1 & \textrm{if } k = 0 \\ z^{n-k+1}P(n, k-1) + P(k-1, k-1) & \textrm{otherwise} \end{cases} $$

I conjecture that number of positive terms in $P(n-1, n-1)$ equals $a(n)$.

Here is the PARI/GP program to generate number of positive terms of $P(n-1, n-1)$:

upto1(n) = my(v1); v1 = vector(n, i, 1); for(i=1, n-1, for(j=i+1, n, v1[j] = v1[i] + z^(j-i)*v1[j])); v1 = vector(n, i, #select(x->x>0, Vec(v1[i])))

Is there a way to prove it? Is it possible to get a closed-form or generating function after some manipulations with $P(n-1,n-1)$?

• Let $a(n)$ be A069999 (i.e., number of possible dimensions for commutators of $n \times n$ matrices; it is independent of the field). OEIS states that no generating function is known.
• Let $P(n,k)$ be a triangle read by rows such that $$ P(n, k) = \begin{cases} 0 & \textrm{if } k > n \\ 1 & \textrm{if } k = 0 \\ z^{n-k+1}P(n, k-1) + P(k-1, k-1) & \textrm{otherwise} \end{cases} $$

I conjecture that number of positive terms in $P(n-1, n-1)$ equals $a(n)$.

Here is the PARI/GP program to generate number of positive terms of $P(n-1, n-1)$:

upto1(n) = my(v1); v1 = vector(n, i, 1); for(i=1, n-1, for(j=i+1, n, v1[j] = v1[i] + z^(j-i)*v1[j])); v1 = vector(n, i, #select(x->x>0, Vec(v1[i])))

Is there a way to prove it? Is it possible to get a closed-form or generating function after some manipulations with $P(n,k)$?

• Let $a(n)$ be A069999 (i.e., number of possible dimensions for commutators of $n \times n$ matrices; it is independent of the field). OEIS states that no generating function is known.
• Let $P(n,k)$ be a triangle read by rows such that $$ P(n, k) = \begin{cases} 0 & \textrm{if } k > n \\ 1 & \textrm{if } k = 0 \\ z^{n-k+1}P(n, k-1) + P(k-1, k-1) & \textrm{otherwise} \end{cases} $$

I conjecture that number of positive terms in $P(n-1, n-1)$ equals $a(n)$.

Here is the PARI/GP program to generate number of positive terms of $P(n-1, n-1)$:

upto1(n) = my(v1); v1 = vector(n, i, 1); for(i=1, n-1, for(j=i+1, n, v1[j] = v1[i] + z^(j-i)*v1[j])); v1 = vector(n, i, #select(x->x>0, Vec(v1[i])))

Is there a way to prove it?

• Let $a(n)$ be A069999 (i.e., number of possible dimensions for commutators of $n \times n$ matrices; it is independent of the field). OEIS states that no generating function is known.
• Let $P(n,k)$ be a triangle read by rows such that $$ P(n, k) = \begin{cases} 0 & \textrm{if } k > n \\ 1 & \textrm{if } k = 0 \\ z^{n-k+1}P(n, k-1) + P(k-1, k-1) & \textrm{otherwise} \end{cases} $$

I conjecture that number of positive terms in $P(n-1, n-1)$ equals $a(n)$.

Here is the PARI/GP program to generate number of positive terms of $P(n-1, n-1)$:

upto1(n) = my(v1); v1 = vector(n, i, 1); for(i=1, n-1, for(j=i+1, n, v1[j] = v1[i] + z^(j-i)*v1[j])); v1 = vector(n, i, #select(x->x>0, Vec(v1[i])))

Is there a way to prove it? Is it possible to get a closed-form or generating function after some manipulations with $P(n-1,n-1)$?