A337130 a(n) is the sum of all products of pairs of numbers joined by the diagonals of an n-gon when its vertices are numbered from 1 to n in order.

0, 0, 0, 11, 40, 99, 203, 370, 621, 980, 1474, 2133, 2990, 4081, 5445, 7124, 9163, 11610, 14516, 17935, 21924, 26543, 31855, 37926, 44825, 52624, 61398, 71225, 82186, 94365, 107849, 122728, 139095, 157046, 176680, 198099, 221408, 246715, 274131, 303770

Offset: 1

Author

Mohammed Yaseen, Aug 17 2020

Keywords

• nonn
• easy

Comments

For n < 4, no n-gon has a diagonal and thus a(n)=0.

Examples

			The diagonals of 4-gon would be numbered (1,3) and (2,4). So a(4) = 1*3 + 2*4 = 11.
The diagonals of 5-gon would be numbered (1,3), (1,4), (2,4), (2,5) and (3,5). So a(5) = 1*3 + 1*4 + 2*4 + 2*5 + 3*5 = 40.
		

Links

• Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Crossrefs

Partial sums of A117560. Cf. A000914 (products including sides), A007569, A007678.

Programs

PARI

concat([0,0,0],Vec(x^4*(11 - 15*x + 9*x^2 - 2*x^3) / (1 - x)^5 + O(x^40))) \\ Colin Barker, Aug 19 2020

Formula

a(n) = 3*binomial(n+1, 4) - n = (n-2)*(n-1)*n*(n+1)/8 - n for n>=3; a(1) = a(2) = 0.

a(n) = A000914(n-1) - A006527(n).

From Colin Barker, Aug 19 2020: (Start)

G.f.: x^4*(11 - 15*x + 9*x^2 - 2*x^3) / (1 - x)^5.

a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n>7.

(End)

E.g.f.: x + x^2 + exp(x)*x*(-8 + 4*x^2 + x^3)/8. - Stefano Spezia, Aug 19 2020