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A394082
Diagonal of the 4-dimensional geode array: a(n) = G(n,n,n,n).
0
1, 12344, 2408941884, 894971463204720, 446324644841317281200, 263656050352833337510832640, 173882340006327290808417397911384, 123903876312804417119950974199179053760, 93504544544138169501999410471960172058628640, 73755525269360717394343255792950241790850509604000
OFFSET
0,2
COMMENTS
The geode G is a 4-dimensional array of positive integers introduced by Wildberger and Rubine (2025) via operator inversion of the hyper-Catalan generating series. If S denotes the hyper-Catalan series in variables t_2, t_3, t_4, t_5, then S - 1 = (t_2 + t_3 + t_4 + t_5) * G. This sequence records the main diagonal a(n) = G(n,n,n,n).
Every geode entry G(m2,m3,m4,m5) can be expressed as a closed-form alternating multinomial sum of hyper-Catalan numbers (Kotlarz, 2026). The proof follows by formally inverting the geode recurrence via a Neumann series and expanding with the multinomial theorem.
The hyper-Catalan number C(m2,m3,m4,m5) counts subdivisions of a roofed convex polygon into m2 triangles, m3 quadrilaterals, m4 pentagons, and m5 hexagons (Erdelyi and Etherington, 1940). Its closed form is C(m2,m3,m4,m5) = (S+L-1)! / (m2! * m3! * m4! * m5! * L!) where S = m2+m3+m4+m5 and L = m2 + 2*m3 + 3*m4 + 4*m5 + 1.
Combinatorial interpretation (Kotlarz, 2026): G(m2,m3,m4,m5) counts exactly the ballot sequences (Lukasiewicz words) of type (m2+1,m3,m4,m5) in which the first negative step is -1. Equivalently, no step of value -k (k >= 2) precedes the first step of value -1. Equivalently, in the corresponding ordered tree (via the Lukasiewicz bijection), no node of arity r >= 3 whose first r-2 children are all leaves precedes the first binary node in pre-order.
The value G(1000,1000,1000,1000) (a 6303-digit integer) was computed by Rubine in December 2025, claiming a $100 prize offered by Amdeberhan, Kauers, and Zeilberger in arXiv:2508.10245.
Amdeberhan, Kauers, and Zeilberger (arXiv:2508.10245) found a second-order holonomic recurrence for the 3-dimensional diagonal G(n,n,n) (with degree-35 polynomial coefficients) but were unable to find one for the 4-dimensional case. By Lipshitz's theorem the diagonal is D-finite, so such a recurrence exists but likely has very high order and degree.
Empirical: Numerical fitting of 120 terms suggests a(n) ~ K * alpha^n * n^beta where beta = -3.000 to at least 3 decimal places. The exponential base alpha has not fully converged at n=119 (successive ratios a(n+1)/a(n) ~ 1083000 and increasing); it is conjectured to be 14^14/10^10 = 1111200.6826... based on saddle-point heuristics.
LINKS
T. Amdeberhan, M. Kauers, and D. Zeilberger, The Challenge of Computing Geode Numbers, arXiv:2508.10245 [math.CO], 2025; Palestine J. Math. 14(4) (2025), 709-713.
T. Amdeberhan and D. Zeilberger, Proofs Of Three Geode Conjectures, arXiv:2506.17862 [math.CO], 2025.
I. M. Gessel, Lattice Paths and the Geode, arXiv:2507.09405 [math.CO], 2025.
M. Gossow, Ordered Trees and the Geode, arXiv:2507.18097 [math.CO], 2025.
D. Rubine, Computing the 4D Geode, arXiv:2512.21785 [math.CO], 2025.
N. J. Wildberger and D. Rubine, A Hyper-Catalan Series Solution to Polynomial Equations, and the Geode, Amer. Math. Monthly, 132(5) (2025), 420-437.
FORMULA
a(n) = G(n,n,n,n) where G(m2,m3,m4,m5) = Sum_{i=0..m3} Sum_{j=0..m4} Sum_{k=0..m5} (-1)^{i+j+k} * ((i+j+k)!/(i!*j!*k!)) * C(m2+1+i+j+k, m3-i, m4-j, m5-k), where C(m2,m3,m4,m5) = (S+L-1)!/(m2!*m3!*m4!*m5!*L!) with S = m2+m3+m4+m5, L = m2+2*m3+3*m4+4*m5+1.
Equivalently, G satisfies the recurrence G(m2,m3,m4,m5) = C(m2+1,m3,m4,m5) - G(m2+1,m3-1,m4,m5) - G(m2+1,m3,m4-1,m5) - G(m2+1,m3,m4,m5-1) (nonnegative indices only).
EXAMPLE
For n=1, a(1) = G(1,1,1,1) = 12344. This counts the ballot sequences of type (2,1,1,1) whose first negative step is -1.
PROG
(Python) # Direct sum formula
from math import factorial
def C(m2, m3, m4, m5):
S, L = m2+m3+m4+m5, m2+2*m3+3*m4+4*m5+1
return factorial(S+L-1) // (factorial(m2)*factorial(m3)*factorial(m4)*factorial(m5)*factorial(L))
def a(n):
res = 0
for i in range(n+1):
for j in range(n+1):
for k in range(n+1):
t = factorial(i+j+k)//(factorial(i)*factorial(j)*factorial(k)) * C(n+1+i+j+k, n-i, n-j, n-k)
res += t if (i+j+k)%2==0 else -t
return res
print([a(n) for n in range(8)])
CROSSREFS
Cf. A000108 (first column of 2-dimensional geode), A001764 (first row of 2-dimensional geode), A104978, A383439 (diagonal of 2-dimensional geode), A383450, A383451.
Sequence in context: A371833 A350153 A251949 * A178475 A104972 A193493
KEYWORD
nonn
AUTHOR
Steven Kotlarz, Mar 09 2026
STATUS
approved