cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A220692 Square array A220691 transposed.

Original entry on oeis.org

0, 1, 0, 3, 0, 0, 6, 1, 1, 0, 10, 2, 1, 0, 0, 15, 4, 2, 1, 0, 0, 21, 6, 4, 1, 1, 0, 0, 28, 9, 5, 2, 2, 0, 0, 0, 36, 12, 7, 3, 2, 1, 0, 0, 0, 45, 16, 10, 5, 3, 2, 1, 0, 0, 0, 55, 20, 12, 6, 4, 2, 2, 0, 0, 0, 0, 66, 25, 15, 8, 6, 3, 3, 1, 0, 0, 0, 0
Offset: 1

Views

Author

Antti Karttunen, Feb 18 2013

Keywords

Links

Crossrefs

Cf. A220691, A220693, A061857.

Programs

  • 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, 1, n}] // Flatten (* Jean-François Alcover, Mar 04 2014 *)
  • Scheme
    (define (A220692 n) (A220691bi (A004736 n) (A002260 n)))

Formula

a(n) = A220691bi(A004736(n),A002260(n)). (As a sequence. As an array this is the transpose of the square array given in A220691).

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.

Original entry on oeis.org

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

Views

Author

Antti Karttunen, May 11 2001

Keywords

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

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

A220693 Irregular triangle where the k-th item in the n-th row (both starting from 1) tells in how many ways we can add 2 distinct integers from 1 to n in such a way that the sum is divisible by k. Row n has 2n-1 terms.

Original entry on oeis.org

0, 1, 0, 1, 3, 1, 1, 1, 1, 6, 2, 2, 1, 2, 1, 1, 10, 4, 4, 2, 2, 2, 2, 1, 1, 15, 6, 5, 3, 3, 2, 3, 2, 2, 1, 1, 21, 9, 7, 5, 4, 3, 3, 3, 3, 2, 2, 1, 1, 28, 12, 10, 6, 6, 4, 4, 3, 4, 3, 3, 2, 2, 1, 1, 36, 16, 12, 8, 8, 5, 5, 4, 4, 4, 4, 3, 3, 2, 2, 1, 1
Offset: 1

Views

Author

Antti Karttunen, Feb 18 2013. Proposed by Robert Israel, May 07 2012

Keywords

Comments

After the first two rows, this irregular table gives all the nonzero terms from the beginning of each row of A220691. See the comments there and at A061857.

Examples

			Row n (starting from row 1) has 2n-1 terms in this irregular table:
   0;
   1, 0, 1;
   3, 1, 1, 1, 1;
   6, 2, 2, 1, 2, 1, 1;
  10, 4, 4, 2, 2, 2, 2, 1, 1;
  15, 6, 5, 3, 3, 2, 3, 2, 2, 1, 1;
  etc.
See A220691 and A061857 for the meaning of each term.

Links

Programs

Formula

See Robert Israel's formula at A061857.
Showing 1-3 of 3 results.

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