|
%I #24 Dec 21 2016 02:44:18
%S 1,2,4,7,10,14,19,25,31,38,46,55,65,75,86,98,111,125,140,155,171,188,
%T 206,225,245,266,287,309,332,356,381,407,434,462,490,519,549,580,612,
%U 645,679,714,750,786,823,861,900,940,981,1023,1066,1110,1155
%N Convolution of natural numbers with A023532.
%C From _Vladimir Letsko_, Dec 18 2016: (Start)
%C Also, a(n) is the number of possible values for the number of diagonals in a convex polyhedron with n+3 vertices.
%C Let v>4 denote the number of vertices of convex polyhedra. The set of possible numbers of diagonals is the union of sets {(k-1)(v-k-4), ..., (k-1)(v-(k+6)/2)}, where 1 <= k <= floor((sqrt(8v-15)-5)/2), and the set {(k-1)(v-k-4), ..., (v-3)(v-4)/2}, where k = floor((sqrt(8v-15)-3)/2). Note that cardinalities of all sets of this union excluding the last one are consecutive triangular numbers. (End)
%H Vladimir Letsko, <a href="/A023536/b023536.txt">Table of n, a(n) for n = 1..500</a>
%F a(n) = (n(n + 5) - 4 )/2 - Sum_{k=2..n} floor(1/2 + sqrt(2(k + 2))). - Jan Hagberg (jan.hagberg(AT)stat.su.se), Oct 16 2002
%F From _Paul Barry_, May 24 2004: (Start)
%F a(n) = (n+1)(n+2)/2 - Sum_{k=1..n+1} floor((sqrt(8k+1)-1)/2);
%F a(n) = Sum_{k=1..n+1} k-floor((sqrt(8k+1)-1)/2). (End)
%Y Cf. A005230, A279681.
%K nonn
%O 1,2
%A _Clark Kimberling_
%E Corrected by Jan Hagberg (jan.hagberg(AT)stat.su.se), Oct 16 2002
|
