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User: Noel B. Lacpao

Noel B. Lacpao has authored 2 sequences.

A385108 Triangle a(n,k) read by antidiagonals: a(n,k) is the number of dots in the k-augmented centered triangle of order n, k>=0, n>=1.

1, 1, 4, 1, 10, 10, 1, 31, 31, 19, 1, 109, 109, 64, 31, 1, 409, 409, 235, 109, 46, 1, 1585, 1585, 901, 409, 166, 64, 1, 6241, 6241, 3529, 1585, 631, 235, 85, 1, 24769, 24769, 13969, 6241, 2461, 901, 316, 109, 1, 98689, 98689, 55585, 24769, 9721, 3529, 1219, 409, 136

Offset: 1

Noel B. Lacpao, Jun 18 2025

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.

			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, ...

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

Cf. A005448, A085473, A381424.

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);

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}]]

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)

# 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))

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.

A382251 a(n) = 7n^3 - 6n^2.

Original entry on oeis.org

1, 32, 135, 352, 725, 1296, 2107, 3200, 4617, 6400, 8591, 11232, 14365, 18032, 22275, 27136, 32657, 38880, 45847, 53600, 62181, 71632, 81995, 93312, 105625, 118976, 133407, 148960, 165677, 183600, 202771, 223232, 245025, 268192, 292775, 318816, 346357, 375440, 406107, 438400, 472361

Offset: 1

Noel B. Lacpao, May 17 2025

Consider a figurate cubic number of the form a(n)=n^3. n is interpreted as the number of dots or nodes in each edge of the cube. Refer this cube as "central cube". Suppose one identical cube is attached to each of its six faces of the central cube. The resulting geometric structure consists of a total of seven arranged cubes so that each of the six surrounding cubes shares an entire face with the central cube. The overlapping dots along these shared faces are counted once. The number of dots in this configuration is given by the formula: a(n) = 7*n^3-6*n^2 for n>=1.

			For n=2, a(2) = 7*(2^3) - 6*(2^2) = 32.
For n=5, a(5) = 7*(5^3) - 6*(5^2) = 725.

Jejemae S. Maque, "Augmented Cubic Numbers," Undergraduate Thesis, Bukidnon State University, 2024.

Noel B. Lacpao, Table of n, a(n) for n = 1..1000
Noel B. Lacpao, Illustration of the augmented cubic number structure for n=2
Noel B. Lacpao, Illustration of the augmented cubic number structure for n=3
Noel B. Lacpao, Colab notebook for generating augmented cubic numbers
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Maple
```
seq(7*n^3 - 6*n^2, n=1..20);
```
Mathematica
```
Table[7 n^3 - 6 n^2, {n, 1, 20}]
```
Python
```
[7*n**3 - 6*n**2 for n in range(1, 21)]
```

a(n) = 7*n^3 - 6*n^2.

G.f.: x*(1 + 28*x + 13*x^2) / (1-x)^4.

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User: Noel B. Lacpao

A385108 Triangle a(n,k) read by antidiagonals: a(n,k) is the number of dots in the k-augmented centered triangle of order n, k>=0, n>=1.

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A382251 a(n) = 7n^3 - 6n^2.

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cp's OEIS Frontend

User: Noel B. Lacpao

A385108 Triangle a(n,k) read by antidiagonals: a(n,k) is the number of dots in the k-augmented centered triangle of order n, k>=0, n>=1.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Python

Python

Formula

A382251 a(n) = 7*n^3 - 6*n^2.

Views

Author

Keywords

Comments

Examples

References

Links

Programs

Maple

Mathematica

Python

Formula

A382251 a(n) = 7n^3 - 6n^2.