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%I #14 Jul 25 2017 02:27:37
%S 1,2,3,3,1,6,6,1,4,9,2,12,10,5,1,4,12,1,18,4,20,15,1,2,6,15,3,8,24,2,
%T 30,6,30,21,1,7,1,3,7,18,30,1,5,14,40,3,45,9,42,28,1,3,8,1,4,21,1,3,7,
%U 14,36,50,2,8,21,60,5,63,12,56,36
%N Sum_{k=1..n} floor(k*p/q), where (p,q) are either coprime positive integers or q=1 or p=1, n*p>=q, ordered by (n + p + q) ascending, then n ascending, then p ascending.
%C Since this is a sequence with 3 indexes (n,p,q), then the order proposed is an ordering by planes of 3D-discrete points (similar to a diagonal ordering of 2D-discrete points). It is not possible to order by rows, columns since n, p, q are boundless.
%C This sequence generalizes other sequences like A130518, A001840, A058937, A130519, A001972 and maybe others (most of those sequences are replica of each other up to an offset), by providing a closed formula (see formulas).
%F a(n, p, q) = Sum_{k=1..n} floor(k*p/q) defines the sequence.
%F a(n, p, q) = n*(n+1)*p/q/2 - floor(n/q) * (q-1)/2 - Sum_{k=1...(n mod q)} (k*p mod q)/q (the remaining sum has at most q-1 terms, and can assume at most q values when n varies, i.e., that sum for n is equal to the sum for n+q, so the computation of a(n, p, q) requires adding at most (q+1) terms). [_Renzo Benedetti_, Jul 27 2012]
%e a(n, 1, 3) = n*(n+1)/ 6 - floor(n/3) - Sum_{k=1..(n mod 3)} (k mod 3) = n*(n+1)/ 6 - floor(n/3) - (4 mod 3)/3 = A130518(n).
%e Example of the ordering (n,p,q): (1,1,1), (1,1,2), (1,2,1), (2,1,1), (1,1,3), (1,3,1), (2,1,2), (2,2,1), (3,1,1), (1,1,4), ...
%K nonn
%O 1,2
%A _Renzo Benedetti_, Jul 27 2012
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