A248434 Number of length three 0..n arrays with the sum of two elements equal to twice the third.

2, 9, 16, 29, 42, 61, 80, 105, 130, 161, 192, 229, 266, 309, 352, 401, 450, 505, 560, 621, 682, 749, 816, 889, 962, 1041, 1120, 1205, 1290, 1381, 1472, 1569, 1666, 1769, 1872, 1981, 2090, 2205, 2320, 2441, 2562, 2689, 2816, 2949, 3082, 3221, 3360, 3505, 3650

Offset: 1

Author

R. H. Hardin, Oct 06 2014

Keywords

• nonn

Comments

Number of length three 0..n vectors that contain their arithmetic mean. - Hywel Normington, Aug 15 2020

Examples

			Some solutions for n=6:
..2....3....6....1....3....4....3....1....6....2....4....0....4....5....4....3
..6....1....2....0....2....3....3....2....5....3....0....1....3....6....4....5
..4....5....4....2....1....5....3....3....4....1....2....2....2....4....4....4
		

Links

• David A. Corneth, Table of n, a(n) for n = 1..10000 (first 210 terms from R. H. Hardin)
• Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

Crossrefs

Row 1 of A248433.

Cf. A168328, A319127, A319127. First differences A168301.

Programs

PARI

a(n) = {my(res = 2); if(n % 2 == 0, res+=(1 + 6*floor(n/2))); n = (n-1)>>1; res+=6*n^2 + 8*n; res} \\ David A. Corneth, Aug 26 2020

PARI

first(n) = {my(res = vector(n), inc = 7); res[1] = 2; for(i = 2, n, res[i] = res[i-1] + inc; inc += 6 * (i%2 == 1)); res} \\ David A. Corneth, Aug 26 2020

Formula

Empirical: a(n) = 2*a(n-1) -2*a(n-3) +a(n-4).

Empirical for n mod 2 = 0: a(n) = (3/2)*n^2 + n + 1.

Empirical for n mod 2 = 1: a(n) = (3/2)*n^2 + n - (1/2).

From Hywel Normington, Aug 21 2020: (Start)

a(n) = a(n-1) + 1 + 6*floor(n/2)

a(n) = A319127(n+1) + n + 1 = 6*floor((n+1)/2)*floor(n/2) + n + 1.

(End)

From Colin Barker, Aug 28 2020: (Start)

G.f.: x*(2 + 5*x - 2*x^2 + x^3) / ((1 - x)^3*(1 + x)).

a(n) = (1 + 3*(-1)^n + 4*n + 6*n^2) / 4 for n>0.

(End)

Extensions

Name simplified by Andrew Howroyd, Aug 14 2020