A241200 For the n in A241199, the index of the first of 4 terms in binomial(n,k) that satisfy a quadratic relation.

2, 4, 9, 12, 19, 23, 32, 37, 48, 54, 67, 74, 89, 97, 114, 123, 142, 152, 173, 184, 207, 219, 244, 257, 284, 298, 327, 342, 373, 389, 422, 439, 474, 492, 529, 548, 587, 607, 648, 669, 712, 734, 779, 802, 849, 873, 922, 947, 998, 1024, 1077, 1104, 1159, 1187

Offset: 1

Author

T. D. Noe, Apr 17 2014

Keywords

• nonn
• easy

Comments

This value of k appears to approach n/2 as n grows larger.

Examples

			Binomial(14,k) = (1, 14, 91, 364, 1001, 2002, 3003, 3432) for k = 0..7. The 4 quadratic terms begin at k = 2.
		

Links

• Colin Barker, Table of n, a(n) for n = 1..10000
• Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Crossrefs

Cf. A008865 (binomial(n,k) has 3 consecutive terms in a linear relation).

Cf. A062730 (3 terms in arithmetic progression in Pascal's triangle).

Cf. A241199 (the values of n).

Programs

Mathematica

t = {}; Do[b = Binomial[n, Range[0, n/2]]; d = Differences[b, 3]; If[MemberQ[d, 0], AppendTo[t, Position[d, 0, 1, 1][[1, 1]] - 1]], {n, 3000}]; t
LinearRecurrence[{1,2,-2,-1,1},{2,4,9,12,19},60] (* Harvey P. Dale, Dec 18 2022 *)

PARI

Vec(x*(x^2-2)*(x^2+x+1)/((x-1)^3*(x+1)^2) + O(x^100)) \\ Colin Barker, Apr 29 2015

Formula

a(n) = (-11-5*(-1)^n-2*(-15+(-1)^n)*n+6*n^2)/16. G.f.: x*(x^2-2)*(x^2+x+1) / ((x-1)^3*(x+1)^2). - Colin Barker, Apr 18 2014 and Apr 29 2015

The terms appear to satisfy a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5), with initial terms 2, 4, 9, 12, 19. - T. D. Noe, Apr 18 2014