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.

A273261 Irregular triangle read by rows: T(n,k) = sum of the elements of the k-th row of the difference table of the divisors of n.

Original entry on oeis.org

1, 3, 1, 4, 2, 7, 3, 1, 6, 4, 12, 5, 2, 2, 8, 6, 15, 7, 3, 1, 13, 8, 4, 18, 9, 4, 0, 12, 10, 28, 11, 5, 4, 3, 1, 14, 12, 24, 13, 6, -2, 24, 14, 8, 8, 31, 15, 7, 3, 1, 18, 16, 39, 17, 8, 6, 4, 12, 20, 18, 42, 19, 9, 4, 3, -11, 32, 20, 12, 8, 36, 21, 10, -6, 24, 22, 60, 23, 11, 8, 6, 3, 4, -12, 31, 24, 16, 42, 25, 12, -8
Offset: 1

Views

Author

Omar E. Pol, May 20 2016

Keywords

Comments

Row 2^k gives the first k+1 positive terms of A000225 in decreasing order, k >= 0.
If n is prime then row n contains only two terms: n+1 and n-1.
First differs from A274531 at a(41).
For n = p^k, T(n, 1) = n - 1, T(n, n) = (p - 1)^k. a(A006218(n - 1) + 1) = T(n, 0), a(A006218(n)) = T(n, t-1) where t is the number of divisors of n. - David A. Corneth, Jun 18 2016
Let D_n(m, c) be the k-th element in row m. The divisors of n are in row m = 0. Let t be the number of divisors of n. Then T(n, k) = D_n(k - 1, t-1) - D_n(k - 1, 0). - David A. Corneth, Jun 25 2016
For n in A187204, the last term of the n-th row is 0. - Michel Marcus, Apr 02 2017

Examples

			Triangle begins:
1;
3, 1;
4, 2;
7, 3, 1;
6, 4;
12, 5, 2, 2;
8, 6;
15, 7, 3, 1;
13, 8, 4;
18, 9, 4, 0;
12, 10;
28, 11, 5, 4, 3, 1;
14, 12;
24, 13, 6, -2;
24, 14, 8, 8;
31, 15, 7, 3, 1;
18, 16;
39, 17, 8, 6, 4, 12;
20, 18;
42, 19, 9, 4, 3, -11;
32, 20, 12, 8;
36, 21, 10, -6;
24, 22;
60, 23, 11, 8, 6, 3, 4, -12;
31, 24, 16;
42, 25, 12, -8;
...
For n = 14 the divisors of 14 are 1, 2, 7, 14, and the difference triangle of the divisors is
1, 2, 7, 14;
1, 5, 7;
4, 2;
-2;
The row sums give [24, 13, 6, -2] which is also the 14th row of the irregular triangle.
In the first row, the last element is 14, the first is 1. So the sum of the second row is 14 - 1 is 13. Similarly, the sum of the third row is 7 - 1 = 6, and of the last row, 2 - 4 = -2. - _David A. Corneth_, Jun 25 2016

Crossrefs

Row lengths give A000005. Column 1 is A000203.
Right border gives A187202. Row sums give A273103.
Cf. A000225, A006218, A187204, A273102, A273262, A273263, A274531.

Programs

  • Mathematica
    Map[Total, Table[NestWhileList[Differences, Divisors@ n, Length@ # > 1 &], {n, 26}], {2}] // Flatten (* Michael De Vlieger, Jun 26 2016 *)
  • PARI
    row(n) = {my(d = divisors(n));my(nd = #d); my(m = matrix(#d, #d)); for (j=1, nd, m[1,j] = d[j];); for (i=2, nd, for (j=1, nd - i +1, m[i,j] = m[i-1,j+1] - m[i-1,j];);); vector(nd, i, sum(j=1, nd, m[i, j]));}
tabf(nn) = for (n=1, nn, print(row(n)););
lista(nn) = for (n=1, nn, v = row(n); for (j=1, #v, print1(v[j], ", "));); \\ Michel Marcus, Jun 25 2016

A274532 Irregular triangle read by rows: T(n,k) = sum of the elements of the k-th antidiagonal of the absolute difference table of the divisors of n.

Original entry on oeis.org

1, 1, 3, 1, 5, 1, 3, 7, 1, 9, 1, 3, 4, 13, 1, 13, 1, 3, 7, 15, 1, 5, 19, 1, 3, 10, 17, 1, 21, 1, 3, 4, 5, 11, 28, 1, 25, 1, 3, 16, 25, 1, 5, 7, 41, 1, 3, 7, 15, 31, 1, 33, 1, 3, 4, 13, 14, 47, 1, 37, 1, 3, 7, 7, 25, 39, 1, 5, 13, 53, 1, 3, 28, 41, 1, 45, 1, 3, 4, 5, 11, 12, 22, 61, 1, 9, 61, 1, 3, 34, 49, 1, 5, 19, 65
Offset: 1

Views

Author

Omar E. Pol, Jun 27 2016

Keywords

Comments

If n is prime then row n contains only two terms: 1 and 2*n-1.
Row 2^k gives the first k+1 positive terms of A000225, k >= 0.
Note that this sequence is not the absolute values of A273262.
First differs from A273262 at a(41).

Examples

			Triangle begins:
1;
1, 3;
1, 5;
1, 3, 7;
1, 9;
1, 3, 4, 13;
1, 13;
1, 3, 7, 15;
1, 5, 19;
1, 3, 10, 17;
1, 21;
1, 3, 4, 5, 11, 28;
1, 25;
1, 3, 16, 25;
1, 5, 7, 41;
1, 3, 7, 15, 31;
1, 33;
1, 3, 4, 13, 14, 47;
1, 37;
1, 3, 7, 7, 25, 39;
1, 5, 13, 53;
1, 3, 28, 41;
1, 45;
1, 3, 4, 5, 11, 12, 22, 61;
1, 9, 61;
1, 3, 34, 49;
1, 5, 19, 65;
...
For n = 18 the divisors of 18 are 1, 2, 3, 6, 9, 18, and the absolute difference triangle of the divisors is
1, 2, 3, 6, 9, 18;
1, 1, 3, 3, 9;
0, 2, 0, 6;
2, 2, 6;
0, 4;
4;
The antidiagonal sums give [1, 3, 4, 13, 14, 47] which is also the 18th row of the irregular triangle.

Crossrefs

Row lengths give A000005. Column 1 is A000012. Row sums give A187215.
Cf. A000225, A187203, A272121, A273132, A273135, A273262, A274531.

Programs

  • Mathematica
    Table[Map[Total, Table[#[[m - k + 1, k]], {m, Length@ #}, {k, m}], {1}] &@ NestWhileList[Abs@ Differences@ # &, Divisors@ n, Length@ # > 1 &], {n, 27}] // Flatten (* Michael De Vlieger, Jun 27 2016 *)

A274533 Irregular triangle read by rows: T(n,k) = sum of the elements of the k-th column of the absolute difference table of the divisors of n.

Original entry on oeis.org

1, 2, 2, 3, 3, 3, 4, 4, 5, 5, 4, 5, 6, 6, 7, 7, 4, 6, 8, 8, 7, 9, 9, 4, 7, 10, 10, 11, 11, 4, 6, 8, 10, 12, 12, 13, 13, 8, 9, 14, 14, 11, 13, 15, 15, 5, 8, 12, 16, 16, 17, 17, 8, 11, 12, 15, 18, 18, 19, 19, 7, 10, 10, 15, 20, 20, 13, 17, 21, 21, 16, 13, 22, 22, 23, 23, 6, 7, 10, 12, 16, 20, 24, 24, 21, 25, 25
Offset: 1

Views

Author

Omar E. Pol, Jun 29 2016

Keywords

Comments

If n is prime then row n is [n, n].
It appears that the last two terms of the n-th row are [n, n], n > 1.
Note that this sequence is not the absolute values of A273263.
First differs from A273263 at a(38).

Examples

			Triangle begins:
   1;
   2,  2;
   3,  3;
   3,  4,  4;
   5,  5;
   4,  5,  6,  6;
   7,  7;
   4,  6,  8,  8;
   7,  9,  9;
   4,  7, 10, 10;
  11, 11;
   4,  6,  8, 10, 12, 12;
  13, 13;
   8,  9, 14, 14;
  11, 13, 15, 15;
   5,  8, 12, 16, 16;
  17, 17;
   8, 11, 12, 15, 18, 18;
  19, 19;
   7, 10, 10, 15, 20, 20;
  13, 17, 21, 21;
  16, 13, 22, 22;
  23, 23;
   6,  7, 10, 12, 16, 20, 24, 24;
  21, 25, 25;
  20, 15, 26, 26;
  ...
For n = 18 the divisors of 18 are 1, 2, 3, 6, 9, 18, and the absolute difference triangle of the divisors is
  1,  2,  3,  6,  9, 18;
  1,  1,  3,  3,  9;
  0,  2,  0,  6;
  2,  2,  6;
  0,  4;
  4;
The column sums give [8, 11, 12, 15, 18, 18] which is also the 18th row of the irregular triangle.

Crossrefs

Row lengths give A000005. Right border gives A000027. Row sums give A187215.
Cf. A187203, A272121, A273132, A273104, A273137, A273263, A274531, A274532.

Programs

  • Mathematica
    Table[Total /@ Table[#[[m - k + 1, -k]], {m, Length@ #, 1, -1}, {k, m}] &@ NestWhileList[Abs@ Differences@ # &, Divisors@ n, Length@ # > 1 &], {n, 25}] // Flatten (* Michael De Vlieger, Jun 29 2016 *)
Showing 1-3 of 3 results.

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