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A241200
For the n in A241199, the index of the first of 4 terms in binomial(n,k) that satisfy a quadratic relation.
4
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.
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
KEYWORD
nonn,easy
AUTHOR
T. D. Noe, Apr 17 2014
STATUS
approved