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.

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

Author

J. Stauduhar, Feb 13 2022

Keywords

• nonn
• easy

Comments

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.

Examples

			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.
		

Links

• 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).

Crossrefs

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

Programs

PARI

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

Python

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)