THE FOLD / GRIND / THE CRON JOB / THE WORPITZKY
THE WORPITZKY
powers rebuilt from Eulerian numbers
1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION
Worpitzky’s identity rebuilds any power from binomial coefficients, weighted by the Eulerian numbers. The Eulerian number A(n,k) counts the permutations of n elements with exactly k ascents. Worpitzky proved that xn = ∑k A(n,k)·C(x+k, n) — the monomial xn is a fixed integer combination of the ‘binomial staircase’ C(x+k, n), with the Eulerian numbers as the exact coefficients. It is the bridge between powers, binomial coefficients, and the ascent statistic on permutations, and it is what makes Eulerian numbers appear whenever you sum kn.
LIT verified live with exact integer arithmetic: for n = 1..12 and x = 0..20, the sum ∑k A(n,k)·C(x+k, n) equals xn exactly, with the Eulerian numbers generated independently by their own recurrence — A(3,·)=[1,4,1] (window.__worpitzky). FIG no framing; the Eulerian recurrence, the binomial staircase, and the power xn all run in-browser and agree exactly.
LIT verified live with exact integer arithmetic: for n = 1..12 and x = 0..20, the sum ∑k A(n,k)·C(x+k, n) equals xn exactly, with the Eulerian numbers generated independently by their own recurrence — A(3,·)=[1,4,1] (window.__worpitzky). FIG no framing; the Eulerian recurrence, the binomial staircase, and the power xn all run in-browser and agree exactly.
2 HOW IT WAS WEAVED · AI + HUMAN
David (human) seated this at the-cron-job — the grind that reruns every tick: for each x, rebuild xn mechanically from the same Eulerian coefficients and the binomial staircase. AVAN (AI) built the instrument: the Eulerian-number recurrence, the binomial coefficients, and their exact reconstruction of xn.
Credit as content: Julius Worpitzky (1883); Eulerian numbers from Leonhard Euler. The weave: David names the recurring job; I confirm xn equals the Eulerian-weighted binomial sum, exactly.
Credit as content: Julius Worpitzky (1883); Eulerian numbers from Leonhard Euler. The weave: David names the recurring job; I confirm xn equals the Eulerian-weighted binomial sum, exactly.
3 ONE DIMENSION
The Eulerian triangle A(n,k) — permutations of n with k ascents — the coefficients in Worpitzky's identity.
4 TWO DIMENSIONS · INTERACTIVE
Cycle n and x; the Eulerian-weighted binomial sum is compared, term by term, to x^n.
5 THREE DIMENSIONS + AVAN’S INVERSE
The green forward object: the power x^n, rebuilt exactly.
AVAN’s addition (the inverse-companion): don’t exponentiate — sum a staircase. The inverse of ‘the power xn’ is ‘the Eulerian-weighted sum ∑A(n,k)C(x+k,n) of binomial coefficients’, tying powers to the ascent statistic on permutations. Magenta are the Eulerian-weighted binomial pieces; green is the power xn they rebuild. A monomial as a staircase sum.
LIT Genuine Worpitzky's identity (Julius Worpitzky, 1883; Eulerian numbers from Euler). Verified live with exact BigInt: for n=1..12 and x=0..20, Σ_k A(n,k)·C(x+k,n) equals x^n exactly, where A(n,k) are the Eulerian numbers from their recurrence A(n,k)=(k+1)A(n−1,k)+(n−k)A(n−1,k−1); A(3,·)=[1,4,1] (window.__worpitzky.ok).
FIG No framing; the Eulerian recurrence, the binomial staircase, and the power x^n all run in-browser and agree exactly. The AVAN inverse is honest — instead of exponentiating, sum a staircase: the inverse of 'the power x^n' is 'the Eulerian-weighted sum ΣA(n,k)C(x+k,n)', tying powers to the ascent statistic on permutations. Magenta are the Eulerian-weighted binomial pieces; green is the power x^n they rebuild. A monomial as a staircase sum.
FIG No framing; the Eulerian recurrence, the binomial staircase, and the power x^n all run in-browser and agree exactly. The AVAN inverse is honest — instead of exponentiating, sum a staircase: the inverse of 'the power x^n' is 'the Eulerian-weighted sum ΣA(n,k)C(x+k,n)', tying powers to the ascent statistic on permutations. Magenta are the Eulerian-weighted binomial pieces; green is the power x^n they rebuild. A monomial as a staircase sum.
◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of THE CRON JOB · David Lee Wise (ROOT0), with AVAN