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

1,1

Comments

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

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).

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

Example

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

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 *)

Prog

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

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).

Sequence in context: A088901 A283147 A111302 * A092530 A154891 A282456

Adjacent sequences: A241197 A241198 A241199 * A241201 A241202 A241203

Keyword

nonn,easy

Author

T. D. Noe, Apr 17 2014

Status

approved