cp's OEIS Frontend

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.

%I A341092 #48 Apr 16 2022 05:28:18
%S A341092 7,12,14,21,23,32,34,45,47,60,62,77,79,96,98,117,119,140,142,165,167,
%T A341092 192,194,221,223,252,254,285,287,320,322,357,359,396,398,437,439,480,
%U A341092 482,525,527,572,574,621,623,672,674,725,727,780,782,837,839,896,898,957,959
%N 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.
%C A341092 Also, a(2k-1)=(k+2)^2-2; a(2k)=(k+3)^2-4, k>=1.
%C A341092 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."
%C A341092 Proof: Let (n)_(k) denote the falling factorial. With any integer i>=3:
%C A341092 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)}.
%C A341092 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)}.
%C A341092 Although row 19 contains a 3-term arithmetic progression it doesn't fit the pattern found here, so 19 is not in this sequence.
%C A341092 Conjecture 1: Row 19 is the only row that contains a 3-term AP that doesn't fit the pattern found here.
%C A341092 Conjecture 2: No row contains an AP of more than three coefficients.
%C A341092 A brute-force search of n<=1100 found no counterexample of either conjecture above.
%H A341092 Ralf Stephan, <a href="https://arxiv.org/abs/math/0409509">Prove or Disprove. 100 Conjectures from the OEIS</a>, arXiv:math/0409509 [math.CO], 2004.
%H A341092 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,2,-2,-1,1).
%e A341092 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.
%o A341092 (Python)
%o A341092 seq=[]
%o A341092 for n in range(2,101):
%o A341092     k=int(((n)*(n+1))/2)
%o A341092     m=int(((n+1)*(n+2))/2)-4+k
%o A341092     if n==2:
%o A341092         seq.append(m+2)
%o A341092     else:
%o A341092         seq.append(m)
%o A341092         seq.append(m+2)
%o A341092 print(seq)
%o A341092 (PARI) a(n) = (2*n^2 + 22*n + 37 + (2*n + 3)*(-1)^n)/8 \\ _Charles R Greathouse IV_, Apr 02 2022
%Y A341092 Cf. A000217, A000292, A007318, A096338, A006857.
%K A341092 nonn,easy
%O A341092 1,1
%A A341092 _J. Stauduhar_, Feb 13 2022