A006857 -id:A006857 - cp's OEIS frontend

A240444 Triangle T(n, k) = Number of ways to arrange k indistinguishable points on an n X n square grid so that no four of them are vertices of a square of any orientation.

1, 1, 1, 4, 6, 4, 1, 9, 36, 84, 120, 96, 32, 1, 16, 120, 560, 1800, 4128, 6726, 7492, 5238, 1924, 232, 1, 25, 300, 2300, 12600, 52080, 166702, 416622, 808488, 1196196, 1306464, 1001364, 497940, 141336, 18208, 636, 1, 36, 630, 7140, 58800, 373632, 1895938, 7835492

Offset: 1

Heinrich Ludwig, May 07 2014

The triangle is irregularly shaped: 0 <= k <= A240443(n). The first row corresponds to n = 1.

The maximal number of points that can be placed on an n X n square grid so that no four points are vertices of a square is A240443(n).

			The triangle begins:
  1,  1;
  1,  4,   6,   4;
  1,  9,  36,  84,  120,   96,   32;
  1, 16, 120, 560, 1800, 4128, 6726, 7492, 5238, 1924, 232;
...

Heinrich Ludwig, Table of n, a(n) for n = 1..101

Cf. A240443, A000290 (column 2), A083374 (column 3), A178208 (column 4), A006857 (column 5 divided by 120), A240445 (column 6), A240446 (column 7).

A341092 Rows of Pascal's triangle which contain a 3-term arithmetic progression of a certain form: a(n) = (2n^2 + 22n + 37 + (2n + 3)*(-1)^n)/8.

Original entry on oeis.org

7, 12, 14, 21, 23, 32, 34, 45, 47, 60, 62, 77, 79, 96, 98, 117, 119, 140, 142, 165, 167, 192, 194, 221, 223, 252, 254, 285, 287, 320, 322, 357, 359, 396, 398, 437, 439, 480, 482, 525, 527, 572, 574, 621, 623, 672, 674, 725, 727, 780, 782, 837, 839, 896, 898, 957, 959

Offset: 1

J. Stauduhar, Feb 13 2022

Also, a(2k-1)=(k+2)^2-2; a(2k)=(k+3)^2-4, k>=1.
Conjecture (67) in Ralf Stephan's paper, "Prove or Disprove. 100 Conjectures from the OEIS" asks if it is true that: "The numbers n such that the n-th row of Pascal's triangle contains an arithmetic progresion are n = 19 ∨ n = (1/8)*[2*k^2 + 22k + 37 + (2k + 3)*(-1)^k], k > 0."
Proof: Let (n)_(k) denote the falling factorial. With any integer i>=3:
For a(n) = i^2-2, if we set x=binomial(i,2), and y=binomial(i-1,2), we can calculate three integers in arithmetic progression, {a,b,c}, such that a=[(x+y-2)(y-2)*(y*(y-1))]/y!, b=[(x+y-2)(y-2)*(x*y)]/y!, c =[(x+y-2)_(y-2)*(x*(x-1))]/y!; {a,b,c}={C(i^2-2,y-2), C(i^2-2,y-1), C(i^2-2, y)}.
For a(n) = (i+1)^2-4, if we set x=binomial(i+1,2), and y=binomial(i,2), we can calculate three integers in arithmetic progression, {a,b,c}, such that a=[(x+y-4)(y-4)*(y)(y-4)]/y!, b=[(x+y-4)(y-4)*(x)(2)*(y)(2)]/y!, c =[(x+y-4)(y-4)*(x)_(x-4)]/y!; {a,b,c}={C((i+1)^2-4,y-4), C((i+1)^2-4,y-2 ), C((i+1)^2-4,y)}.
Although row 19 contains a 3-term arithmetic progression it doesn't fit the pattern found here, so 19 is not in this sequence.
Conjecture 1: Row 19 is the only row that contains a 3-term AP that doesn't fit the pattern found here.
Conjecture 2: No row contains an AP of more than three coefficients.
A brute-force search of n<=1100 found no counterexample of either conjecture above.

			With n=2, k=binomial(n+2=4,2)=6. m=binomial(n+3=5,2)-4+k=12. [C(m,k-4), C(m,k-2), C(m,k)] = [66,495,924], and [C(m+2,k-2), C(m+2,k-1), C(m+2,k)] = [1001,2002,3003], so a(2)=m=12 and a(3)=m+2=14.

Ralf Stephan, Prove or Disprove. 100 Conjectures from the OEIS, arXiv:math/0409509 [math.CO], 2004.
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. A000217, A000292, A007318, A096338, A006857.

a(n) = (2*n^2 + 22*n + 37 + (2*n + 3)*(-1)^n)/8 \\ Charles R Greathouse IV, Apr 02 2022

seq=[]
for n in range(2,101):
    k=int(((n)*(n+1))/2)
    m=int(((n+1)*(n+2))/2)-4+k
    if n==2:
        seq.append(m+2)
    else:
        seq.append(m)
        seq.append(m+2)
print(seq)

cp's OEIS Frontend

A240444 Triangle T(n, k) = Number of ways to arrange k indistinguishable points on an n X n square grid so that no four of them are vertices of a square of any orientation.

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