OFFSET
1,3
COMMENTS
The k-augmented centered triangular numbers a(n,k) count the number of dots in the k-augmented centered triangle of order n. The order, n, refers to the number of exact dots along each side of the base equilateral triangle in the initial unaugmented centered triangle. For k=0, the configuration is simply this base triangle. It would be a single dot for n=1 or for n>=2, a central dot surrounded by dots so that each side contains exactly n dots. For k>=1, the k-augmented centered triangle of order n is recursively constructed by taking a (k-1)-augmented centered triangle of order n and attaching one congruent copy on each side. Each copy shares a whole side with the central triangle. Any overlapping dots that occur along the shared sides and vertices are counted only once. This recursive construction generalizes the classic centered triangular numbers. As k increases, it produces more complex and symmetric triangular patterns.
For k=0, a(n,0) gives the centered triangular numbers (A005448).
For k=1, a(n,1) matches A085473.
When k=2, a(n,1) matches the truncated hex numbers A381424.
The rows appear to be new for all k>=3.
For geometric illustrations, see the linked images.
LINKS
Noel B. Lacpao, Table of n, a(n) for n = 1..1000
Noel B. Lacpao, Summary Table for n=1..50,k=0..3
Noel B. Lacpao, Geometric_Illustration for n=3,k=2
Noel B. Lacpao, Geometric_Illustration for n=3,k=3
FORMULA
a(n,k) = 4^k * (3*n^2-3*n+2) / 2 - 3*n * Sum_{j=1..k} 4^(k-j) * 2^(j-1) + 3 * Sum_{j=1..k} 4^(k-j) * (2^(j-1)-1).
a(n,k) = 4 * a(n,k-1) - 3 * (2^(k-1) * n - (2^(k-1)-1)).
G.f. for fixed k: G_k(x) = (A_kx(x+1) + B_kx(1-x) + C_kx(1-x)^2) / (1-x)^3, where a(n,k)=A_kn^2 + B_kn + C_k.
EXAMPLE
For n=1, any k:
a(1,k) = 4^k*(3*1^2-3*1+2)/2 - 3*1*Sum_{j=1..k} 4^(k-j)*2^(j-1) + 3*Sum_{j=1..k} 4^{k-j}*(2^{j-1}-1)
= 4^k*1 - 3*Sum_{j=1..k} 4^(k-j)*2^(j-1) + 3*Sum_{j=1..k} 4^(k-j)*2^(j-1) - 3*Sum_{j=1..k} 4^(k-j)
= 4^k-3*Sum_{j=1..k} 4^(k-j)
= 4^k-3*(4^k-1)/3
= 4^k-4^k+1
= 1.
For n=2,k=0:
a(2,0) = 4^0*(3*2^2-3*2+2)/2
= 8/2
= 4.
For n=3, k=2:
a(3,2) = 4^2*(3*3^2-3*3+2)/2 - 3*3*Sum_{j=1..2} 4^(2-j)*2^(j-1) + 3*Sum_{j=1..22} 4^(2-j)*(2^(j-1)-1)
= 16*20/2-9*6+3*1
= 109.
Square array begins:
1, 1, 1, 1, 1, 1, 1, 1, ...
4, 10, 31, 109, 409, 1585, 6241, 24769, ...
10, 31, 109, 409, 1585, 6241, 24769, 98689, ...
19, 64, 235, 901, 3529, 13969, 55585, 221761, ...
31, 109, 409, 1585, 6241, 24769, 98689, 393985, ...
46, 166, 631, 2461, 9721, 38641, 154081, 615361, ...
64, 235, 901, 3529, 13969, 55585, 221761, 885889, ...
85, 316, 1219, 4789, 18985, 75601, 301729, 1205569, ...
MAPLE
a := proc(n, k)
local S1, S2, j;
S1 := add(4^(k-j)*2^(j-1), j=1..k);
S2 := add(4^(k-j)*(2^(j-1)-1), j=1..k);
return 4^k*(3*n^2 - 3*n + 2)/2 - 3*n*S1 + 3*S2;
end proc:
seq(seq(a(n, d-n), n=1..d), d=1..10);
MATHEMATICA
a[n_, k_] := Module[{S1, S2},
S1 = Sum[4^(k - j) * 2^(j - 1), {j, 1, k}];
S2 = Sum[4^(k - j) * (2^(j - 1) - 1), {j, 1, k}];
4^k * (3 n^2 - 3 n + 2)/2 - 3 n * S1 + 3 * S2
];
Table[a[n, k], {n, 1, 5}, {k, 0, 5}]
TableForm[Table[a[n, k], {n, 1, 5}, {k, 0, 5}]]
PROG
(Python)
def a(n, k):
return 4**k*(3*n**2-3*n+2)//2-3*n*sum(4**(k-j)*2**(j-1) for j in range(1, k+1))+ 3*sum(4**(k-j)*(2**(j-1)-1) for j in range(1, k+1))
for n in range(1, 10):
row = [a(n, k)
for k in range(0, 10)]
print(row)
(Python)
# For the antidiagonal terms.
def a(n, k):
term1 = 4**k * (3*n**2 - 3*n + 2) // 2
if k == 0:
return term1
return term1 - 3*n*sum([4**(k-j) * 2**(j-1) for j in range(1, k+1)]) + 3*sum([4**(k-j) * (2**(j-1) - 1) for j in range(1, k+1)])
def antidiagonal_sequence(num_terms):
terms = []
diag = 0
while len(terms) < num_terms:
for n in range(1, diag+2):
k = diag - (n-1)
if k < 0:
continue
terms.append(a(n, k))
if len(terms) == num_terms:
break
diag += 1
return terms
N = 55
seq = antidiagonal_sequence(N)
print(', '.join(str(x) for x in seq))
CROSSREFS
KEYWORD
AUTHOR
Noel B. Lacpao, Jun 18 2025
STATUS
approved