A241200 - OEIS

%I #17 Dec 18 2022 13:25:18

%S 2,4,9,12,19,23,32,37,48,54,67,74,89,97,114,123,142,152,173,184,207,

%T 219,244,257,284,298,327,342,373,389,422,439,474,492,529,548,587,607,

%U 648,669,712,734,779,802,849,873,922,947,998,1024,1077,1104,1159,1187

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

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

%H Colin Barker, <a href="/A241200/b241200.txt">Table of n, a(n) for n = 1..10000</a>

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,2,-2,-1,1).

%F 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

%F 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

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

%t 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

%t LinearRecurrence[{1,2,-2,-1,1},{2,4,9,12,19},60] (* _Harvey P. Dale_, Dec 18 2022 *)

%o (PARI) Vec(x*(x^2-2)*(x^2+x+1)/((x-1)^3*(x+1)^2) + O(x^100)) \\ _Colin Barker_, Apr 29 2015

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

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

%Y Cf. A241199 (the values of n).

%K nonn,easy

%O 1,1

%A _T. D. Noe_, Apr 17 2014