A241199 Numbers n such that 4 consecutive terms of binomial(n,k) satisfy a quadratic relation for 0 <= k <= n/2.

14, 19, 31, 38, 54, 63, 83, 94, 118, 131, 159, 174, 206, 223, 259, 278, 318, 339, 383, 406, 454, 479, 531, 558, 614, 643, 703, 734, 798, 831, 899, 934, 1006, 1043, 1119, 1158, 1238, 1279, 1363, 1406, 1494, 1539, 1631, 1678, 1774, 1823, 1923, 1974, 2078, 2131

Offset: 1

Author

T. D. Noe, Apr 17 2014

Keywords

• nonn
• easy

Comments

From Robert Israel, Apr 28 2015: (Start)

Numbers n >= 14 such that 3*n + 7 is a square.

This is because

C(n,i+3) - 3*C(n,i+2) + 3*C(n,i+1) - C(n,i) = n!/((n-i)!*(i+3)!) * g(n,i)

where g(n,i) = (n-3-2*i) * ((n-3-2*i)^2 - 3*n - 7). (End)

Examples

			Binomial(14,k) = (1, 14, 91, 364, 1001, 2002, 3003, 3432) for k = 0..7. The 4 terms beginning with 91 equal 182 - 273*x + 182*x^2 for x = 1..4.
		

Links

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

Crossrefs

Sequence A241200 gives the position of the first of the 4 terms. Sequence A008865 gives the terms greater than 2 for which 3 consecutive terms satisfy a linear relation.

A014206 is a related sequence. - Avi Friedlich, Apr 28 2015

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

Programs

Maple

map(k -> (3*k^2+8*k+3,3*k^2+10*k+6),[$1..100]); # Robert Israel, Apr 28 2015

Mathematica

Select[Range[2500], MemberQ[Differences[Binomial[#, Range[0, #/2]], 3], 0] &]
LinearRecurrence[{1,2,-2,-1,1},{14,19,31,38,54},50] (* Harvey P. Dale, Oct 29 2017 *)

PARI

Vec(-x*(6*x^4-3*x^3-16*x^2+5*x+14)/((x-1)^3*(x+1)^2) + O(x^100)) \\ Colin Barker, Apr 29 2015

PARI

a(n)=(6*n^2+42*n+55-(-1)^n*(2*n+7))/8 \\ Charles R Greathouse IV, Apr 15 2016

Formula

a(n) = (55-7*(-1)^n-2*(-21+(-1)^n)*n+6*n^2)/8. G.f.: -x*(6*x^4-3*x^3-16*x^2+5*x+14) / ((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 14, 19, 31, 38, 54. - T. D. Noe, Apr 18 2014

Numbers are of the form A200182(3n+1) and A200182(3n-1). - Avi Friedlich, Apr 25 2015

a(2*k-1) = 3*k^2 + 8*k + 3, a(2*k) = 3*k^2 + 10*k + 6. - Robert Israel, Apr 28 2015