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A220691 Table A(i,j) read by antidiagonals in order A(1,1), A(1,2), A(2,1), A(1,3), A(2,2), A(3,1), ..., where A(i,j) is the number of ways in which we can add 2 distinct integers from the range 1..i in such a way that the sum is divisible by j.

%I A220691 #23 Jun 12 2021 23:33:02
%S A220691 0,0,1,0,0,3,0,1,1,6,0,0,1,2,10,0,0,1,2,4,15,0,0,1,1,4,6,21,0,0,0,2,2,
%T A220691 5,9,28,0,0,0,1,2,3,7,12,36,0,0,0,1,2,3,5,10,16,45,0,0,0,0,2,2,4,6,12,
%U A220691 20,55,0,0,0,0,1,3,3,6,8,15,25,66,0,0,0,0
%N A220691 Table A(i,j) read by antidiagonals in order A(1,1), A(1,2), A(2,1), A(1,3), A(2,2), A(3,1), ..., where A(i,j) is the number of ways in which we can add 2 distinct integers from the range 1..i in such a way that the sum is divisible by j.
%H A220691 A. Karttunen, <a href="/A220691/b220691.txt">The first 150 antidiagonals of the square array, flattened</a>
%H A220691 Stackexchange, <a href="http://math.stackexchange.com/questions/142323/sequence-generation/142364">Question 142323</a>
%H A220691 <a href="/index/Su#subsetsums">Index entries for sequences related to subset sums modulo m</a>
%F A220691 See _Robert Israel_'s formula at A061857.
%e A220691 The upper left corner of this square array starts as:
%e A220691    0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...
%e A220691    1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, ...
%e A220691    3, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, ...
%e A220691    6, 2, 2, 1, 2, 1, 1, 0, 0, 0, 0, ...
%e A220691   10, 4, 4, 2, 2, 2, 2, 1, 1, 0, 0, ...
%e A220691   15, 6, 5, 3, 3, 2, 3, 2, 2, 1, 1, ...
%e A220691 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.
%e A220691 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}.
%t A220691 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 *)
%o A220691 (Scheme function, written after _Robert Israel_'s formula given at A061857):
%o A220691 (define (A220691 n) (A220691bi (A002260 n) (A004736 n)))
%o A220691 (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)))))))
%Y A220691 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.
%Y A220691 The left column of the table: A000217. The following cases should be checked: the second column: A002620, the third column: A058212 (after the first two terms), the fourth column: A001971.
%K A220691 nonn,tabl
%O A220691 1,6
%A A220691 _Antti Karttunen_, Feb 18 2013