OFFSET
1,6
LINKS
FORMULA
See Robert Israel's formula at A061857.
EXAMPLE
The upper left corner of this square array starts as:
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...
1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, ...
3, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, ...
6, 2, 2, 1, 2, 1, 1, 0, 0, 0, 0, ...
10, 4, 4, 2, 2, 2, 2, 1, 1, 0, 0, ...
15, 6, 5, 3, 3, 2, 3, 2, 2, 1, 1, ...
Row 1 is all zeros, because it's impossible to choose two distinct integers from range [1]. A(2,1) = 1, as there is only one possibility to choose a pair of distinct numbers from the range [1,2] such that it is divisible by 1, namely 1+2. Also A(2,3) = 1, as 1+2 is divisible by 3.
A(4,1) = 2, as from [1,2,3,4] one can choose two pairs of distinct numbers whose sum is even: {1+3} and {2+4}.
MATHEMATICA
a[n_, 1] := n*(n-1)/2; a[n_, k_] := Module[{r}, r = Reduce[1 <= i < j <= n && Mod[i + j, k] == 0, {i, j}, Integers]; Which[Head[r] === Or, Length[r], Head[r] === And, 1, r === False, 0, True, Print[r, " not parsed"]]]; Table[a[n-k+1, k], {n, 1, 13} , {k, n, 1, -1}] // Flatten (* Jean-François Alcover, Mar 04 2014 *)
PROG
(Scheme function, written after Robert Israel's formula given at A061857):
(define (A220691bi n k) (let* ((b (modulo (+ 1 n) k)) (q (/ (- (+ 1 n) b) k)) (c (modulo k 2))) (cond ((< b 2) (+ (* q q k (/ 1 2)) (* q b) (* -2 q) (* -1 b) 1 (* c q (/ 1 2)))) ((>= b (/ (+ k 3) 2)) (+ (* q q k (/ 1 2)) (* q b) (* -2 q) b -1 (* (/ k -2)) (* c (+ 1 q) (/ 1 2)))) (else (+ (* q q k (/ 1 2)) (* q b) (* -2 q) (* c q (/ 1 2)))))))
CROSSREFS
Transpose: A220692. The lower triangular region of this square array is given by A061857, which leaves out about half of the nonzero terms. A220693 is another variant giving 2n-1 terms from the beginning of each row, thus containing all the nonzero terms of this array.
KEYWORD
nonn,tabl
AUTHOR
Antti Karttunen, Feb 18 2013
STATUS
approved