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Let

  • $T(n,k)$ be A375837, i.e., triangle read by rows: $T(n,k)$ is the number of rooted chains starting with the cycle $(1)(2)(3)\dotsc(n)$ of length $k$ of permutation poset of $n$ letters.
  • $a(n)$ be A375838, i.e., number of rooted chains starting with the cycle $(1)(2)(3)\dotsc(n)$ in the permutation poset of $[n]$. Here $$a(n) = \sum\limits_{k=0}^{n} T(n,k). $$
  • $R(n,k)$ be an integer coefficients such that $$ R(n,k) = (k+1)\left(\sum\limits_{i=0}^{n-1} R(n-1,i) + \sum\limits_{j=0}^{k-1} R(n-1,j)\right), \\ R(n,n) = 1. $$
  • $P(n,k)$ be the family of polynomials such that $$ P(n,k) = (k+1)\left(\sum\limits_{i=0}^{n-1} P(n-1,i) + x\sum\limits_{j=0}^{k-1} P(n-1,j)\right), \\ P(n,n) = x^n. $$

I conjecture that $R(n,0) = a(n)$. I also conjecture that $$ P(n,0) = \sum\limits_{k=0}^{n} T(n,k) x^{n-k}. $$

Here is the PARI/GP program to generate $R(n,0)$:

upto(n) = {my(A, v1, v2, v3);
v1 = vector(n+1, i, 0); v1[1] = 1; 
v2 = vector(n+1, i, 0); v2[1] = 1; 
for(i=1, n, v3 = v2; A = vecsum(v3); v2[1] = A; 
for(j=1, i-1, v2[j+1] = (j+1)*(A += v3[j])); 
v2[i+1] = 1; v1[i+1] = v2[1]); 
v1}

Note that this algorithm is pretty simple and very fast.

Is there a way to prove it?

Improve this question
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    $\begingroup$ The notes on the sequence mention that it's the row sum of oeis.org/A375837 . Have you checked your T(n,k) against the values there to see if, e.g., you have $T(n,k)=\sum_{i=0}^{n-k} A_{375837}(n, n-i)$? $\endgroup$ Commented Jun 17 at 16:37
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    $\begingroup$ What is "the permutation poset of [n]"? There is more than one poset of permutations. $\endgroup$ Commented Jun 17 at 16:41
  • $\begingroup$ @IraGessel, thank you for comment! I just copied the name of the sequence. If you have any corrections, feel free to add them. $\endgroup$ Commented Jun 18 at 4:18
  • $\begingroup$ @StevenStadnicki, thank you for comment! Done. $\endgroup$ Commented Jun 18 at 7:14

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