A061857 Triangle in which the k-th item in the n-th row (both starting from 1) is the number of ways in which we can add 2 distinct integers from 1 to n in such a way that the sum is divisible by k.

0, 1, 0, 3, 1, 1, 6, 2, 2, 1, 10, 4, 4, 2, 2, 15, 6, 5, 3, 3, 2, 21, 9, 7, 5, 4, 3, 3, 28, 12, 10, 6, 6, 4, 4, 3, 36, 16, 12, 8, 8, 5, 5, 4, 4, 45, 20, 15, 10, 9, 7, 6, 5, 5, 4, 55, 25, 19, 13, 11, 9, 8, 6, 6, 5, 5, 66, 30, 22, 15, 13, 10, 10, 7, 7, 6, 6, 5, 78, 36, 26, 18, 16, 12, 12, 9, 8, 7, 7

Offset: 1

Author

Antti Karttunen, May 11 2001

Keywords

• nonn
• tabl

Comments

Since the sum of two distinct integers from 1 to n can be as much as 2n-1, this triangular table cannot show all the possible cases. For larger triangles showing all solutions, see A220691 and A220693. - Antti Karttunen, Feb 18 2013 [based on Robert Israel's mail, May 07 2012]

Examples

			The second term on the sixth row is 6 because we have 6 solutions: {1+3, 1+5, 2+4, 2+6, 3+5, 4+6} and the third term on the same row is 5 because we have solutions {1+2,1+5,2+4,3+6,4+5}.
Triangle begins:
   0;
   1,  0;
   3,  1,  1;
   6,  2,  2,  1;
  10,  4,  4,  2,  2;
  15,  6,  5,  3,  3,  2;
  21,  9,  7,  5,  4,  3,  3;
  28, 12, 10,  6,  6,  4,  4,  3;
  36, 16, 12,  8,  8,  5,  5,  4,  4;
  45, 20, 15, 10,  9,  7,  6,  5,  5,  4;
		

Links

• Reinhard Zumkeller, Rows n = 1..150 of triangle, flattened
• Stackexchange, Question 142323
• Index entries for sequences related to subset sums modulo m

Crossrefs

This is the lower triangular region of square array A220691. See A220693 for all nonzero solutions.

The left edge (first diagonal) of the triangle: A000217, the second diagonal is given by C(((n+(n mod 2))/2), 2)+C(((n-(n mod 2))/2), 2) = A002620, the third diagonal by A058212, the fourth by A001971, the central column by A042963? trinv is given at A054425. Cf. A061865.

Programs

Haskell

a061857 n k = length [()| i <- [2..n], j <- [1..i-1], mod (i + j) k == 0]
a061857_row n = map (a061857 n) [1..n]
a061857_tabl = map a061857_row [1..]
-- Reinhard Zumkeller, May 08 2012

Maple

[seq(DivSumChoose2Triangle(j),j=1..120)]; DivSumChoose2Triangle := (n) -> nops(DivSumChoose2(trinv(n-1),(n-((trinv(n-1)*(trinv(n-1)-1))/2))));
DivSumChoose2 := proc(n,k) local a,i,j; a := []; for i from 1 to (n-1) do for j from (i+1) to n do if(0 = ((i+j) mod k)) then a := [op(a),[i,j]]; fi; od; od; RETURN(a); end;

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], {n, 1, 13}, {k, 1, n}] // Flatten (* Jean-François Alcover, Mar 04 2014 *)

Scheme

(define (A061857 n) (A220691bi (A002024 n) (A002260 n))) ;; Antti Karttunen, Feb 18 2013. Needs A220691bi from A220691.

Formula

From Robert Israel, May 08 2012: (Start)

Let n+1 = b mod k with 0 <= b < k, q = (n+1-b)/k. Let k = c mod 2, c = 0 or 1.

If b = 0 or 1 then a(n,k) = q^2*k/2 + q*b - 2*q - b + 1 + c*q/2.

If b >= (k+3)/2 then a(n,k) = q^2*k/2 + q*b - 2*q + b - 1 - k/2 + c*(q+1)/2.

Otherwise a(n,k) = q^2*k/2 + q*b - 2*q + c*q/2. (End)

Extensions

Offset corrected by Reinhard Zumkeller, May 08 2012