

A061857


Triangle where the kth item at nth row (both starting from 1) tells in how many ways we can add 2 distinct integers from 1 to n in such way that the sum is divisible by k.


8



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
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OFFSET

1,4


COMMENTS

Since the sum of two distinct integers from 1 to n can be as much as 2n1, 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.


LINKS

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


FORMULA

Contribution from Robert Israel, May 08 2012: (Start)
Let n+1 = b mod k with 0<=b<k, q = (n+1b)/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)


EXAMPLE

E.g. 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}
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;


MAPLE

[seq(DivSumChoose2Triangle(j), j=1..120)]; DivSumChoose2Triangle := (n) > nops(DivSumChoose2(trinv(n1), (n((trinv(n1)*(trinv(n1)1))/2))));
DivSumChoose2 := proc(n, k) local a, i, j; a := []; for i from 1 to (n1) 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*(n1)/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 (* JeanFrançois Alcover, Mar 04 2014 *)


PROG

(Haskell)
a061857 n k = length [() i < [2..n], j < [1..i1], mod (i + j) k == 0]
a061857_row n = map (a061857 n) [1..n]
a061857_tabl = map a061857_row [1..]
 Reinhard Zumkeller, May 08 2012
(Scheme): (define (A061857 n) (A220691bi (A002024 n) (A002260 n)))  Antti Karttunen, Feb 18 2013. Needs A220691bi from A220691.


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.
Sequence in context: A162315 A109446 A088441 * A067433 A133567 A184049
Adjacent sequences: A061854 A061855 A061856 * A061858 A061859 A061860


KEYWORD

nonn,tabl


AUTHOR

Antti Karttunen, May 11 2001


EXTENSIONS

Offset corrected by Reinhard Zumkeller, May 08 2012


STATUS

approved



