proposed

approved

editing

proposed

allocated for Samuel J. Bevins

Triangle read by rows: T(n,k) is the k-th Lie-Betti number of the Fibonacci trees of order n >= 2.

1, 2, 2, 1, 1, 4, 11, 16, 16, 11, 4, 1, 1, 7, 33, 95, 212, 344, 444, 444, 344, 212, 95, 33, 7, 1, 1, 12, 90, 454, 1780, 5489, 14036, 29804, 54007, 83404, 111361, 128378, 128378, 111361, 83404, 54007, 29804, 14036, 5489, 1780, 454, 90, 12, 1

2,2

Marco Aldi and Samuel Bevins, <a href="https://arxiv.org/abs/2212.13608">2-step Nilpotent L_oo-algebras and Hypergraphs</a>, arXiv:2212.13608 [math.CO], 2023. See page 9.

Meera Mainkar, <a href="https://arxiv.org/abs/1310.3414">Graphs and two step nilpotent Lie algebras</a>, arXiv:1310.3414 [math.DG], 2013. See page 1.

SageMath Graph Theory, <a href="https://doc.sagemath.org/html/en/reference/graphs/sage/graphs/generators/families.html">Various families of graphs</a>, see FibonacciTree().

Triangle begins:

k=0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

n=2: 1 2 2 1

n=3: 1 4 11 16 16 11 4 1

n=4: 1 7 33 95 212 344 444 444 344 212 95 33 7 1

n=5: 1 12 90 454 1780 5489 14036 29804 54007 83404 111361 128378 128378 111361 83404 54007 ...

(SageMath)

from sage.algebras.lie_algebras.lie_algebra import LieAlgebra, LieAlgebras

def BettiNumbers(graph):

D = {}

for edge in graph.edges():

e = "x" + str(edge[0])

f = "x" + str(edge[1])

D[(e, f)] = {e + f : 1}

C = (LieAlgebras(QQ).WithBasis().Graded().FiniteDimensional().

Stratified().Nilpotent())

L = LieAlgebra(QQ, D, nilpotent=True, category=C)

H = L.cohomology()

d = L.dimension() + 1

return [H[n].dimension() for n in range(d)]

# Example usage:

n = 5

X = BettiNumbers(graphs.FibonacciTree(n))

Cf. A360572 (cycle graph), A088459 (star graph), A360625 (complete graph), A360938 (ladder graph), A360937 (wheel graph).

allocated

nonn,tabf

Samuel J. Bevins, Jan 11 2024

approved

editing

allocated for Samuel J. Bevins

recycled

allocated

reviewed

approved

proposed

reviewed

editing

proposed

Numbers k such that k^(1/digsum(k)) is integer and k is not a multiple of 10.

1, 32, 1180591620717411303424

1,2

There are no other such numbers up to 10^50.

32 is in the sequence because the sum of the digits of 32 is 5, and the fifth root of 32 is 2, which is an integer.

(Python)

# Generates all that numbers up to 10^m

def digsum(x):

r = 0

for i in str(x): r+=int(i)

return r

def check(x, k):

n = int(x**(1/k))

if (n**k == x): print(x, k)

def recur(x, k, path, s):

if (k == 0) and (x == 0): check(path, s)

if (k == 0) and (x > 0): return

if (x < 0): return

n = min(9, k)

for i in range(n+1): recur(x-1, k-i, path*10+i, s)

# Setting the limit

m = 50

# Checking exponents 4 and 5

for k in range(4, 6): recur(m, k, 0, k)

# Checking all exponents bigger than 6

i = 2

lim = 10**m

j = i**7

while (j <= lim):

j2 = j

s = 7

while (j2 <= lim):

if (digsum(j2)==s): print(j2, s)

s+=1

j2*=i

i+=1

j = i**7

(PARI) isok(k) = (k%10) && (type(k^sumdigits(k)) == "t_INT"); \\ Michel Marcus, Dec 14 2023

Cf. A007953.

nonn,hard,base

recycled

Ivan Borysiuk, Dec 13 2023

32 is in the sequence because the sum of the digits of 32 is 5, and the fifth root of 32 = 2^(3+is 2), which is an integer.

if (i%1000==0): print(i, " \r", end='')