WORKED DEFAULT
Check the calculation with the default inputs
At order n = 5 and ascent index k = 2, the second-order Eulerian recurrence gives 328.
- Set rown=5
- Apply recurrencetwo predecessors
- Read indexA(5,2)=328
STIRLING PERMUTATIONS / ASCENTS
Count Stirling permutations of order n having exactly k ascents under the stated second-order Eulerian convention.
METHOD / WORKED EXAMPLE
Count Stirling permutations of order n having exactly k ascents under the stated second-order Eulerian convention. The result panel keeps the defining recurrence or counting identity visible so the output can be checked independently.
WORKED DEFAULT
At order n = 5 and ascent index k = 2, the second-order Eulerian recurrence gives 328.
READ THE RESULT
This ascent distribution belongs to Stirling permutations and differs from first-order Eulerian numbers.
ASSUMPTIONS AND LIMITS
Indexing conventions vary in references; this page states and tests its zero-based k convention.
COMMON QUESTIONS
Start from the one-entry base row and build each bounded row with the second-order Eulerian two-predecessor recurrence. The indexing, equivalence relation, and counted objects are stated in the method and worked example. Inputs must be whole numbers inside the displayed domain; the page never rounds a decimal into an accepted index. This ties the answer to one explicit convention instead of silently mixing sequence offsets or combinatorial interpretations.
This ascent distribution belongs to Stirling permutations and differs from first-order Eulerian numbers. Small boundary cases and the displayed recurrence or identity provide useful independent checks. Recompute the displayed recurrence or closed form with the same inputs and compare its previous terms or counting factors. That check supports this bounded result, but it does not transfer the interpretation to a different sequence or counting object.
Indexing conventions vary in references; this page states and tests its zero-based k convention. The implementation uses integer arithmetic internally and refuses results beyond the safe display boundary. Combinatorial and recurrence values can grow rapidly even when the inputs look small. The conservative cap prevents browser stalls and avoids presenting an unsafe floating-point integer as exact; larger work needs arbitrary-precision software and independent resource controls.
RELATED TOOLS
Use boundary
Start from the one-entry base row and build each bounded row with the second-order Eulerian two-predecessor recurrence.
A(n,k)=(k+1)A(n-1,k)+(2n-k-1)A(n-1,k-1).