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 1-8 of 8 results.

A253565 Permutation of natural numbers: a(0) = 1, a(1) = 2; after which, a(2n) = A253550(a(n)), a(2n+1) = A253560(a(n)).

Original entry on oeis.org

1, 2, 3, 4, 5, 9, 6, 8, 7, 25, 15, 27, 10, 18, 12, 16, 11, 49, 35, 125, 21, 75, 45, 81, 14, 50, 30, 54, 20, 36, 24, 32, 13, 121, 77, 343, 55, 245, 175, 625, 33, 147, 105, 375, 63, 225, 135, 243, 22, 98, 70, 250, 42, 150, 90, 162, 28, 100, 60, 108, 40, 72, 48, 64, 17, 169, 143, 1331, 91, 847, 539, 2401, 65, 605, 385, 1715, 275, 1225, 875, 3125, 39
Offset: 0

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Author

Antti Karttunen, Jan 03 2015

Keywords

Comments

This sequence can be represented as a binary tree. Each child to the left is obtained by applying A253550 to the parent, and each child to the right is obtained by applying A253560 to the parent:
1
|
...................2...................
3 4
5......../ \........9 6......../ \........8
/ \ / \ / \ / \
/ \ / \ / \ / \
/ \ / \ / \ / \
7 25 15 27 10 18 12 16
11 49 35 125 21 75 45 81 14 50 30 54 20 36 24 32
etc.
Sequence A253563 is the mirror image of the same tree. Also in binary trees A005940 and A163511 the terms on level of the tree are some permutation of the terms present on the level n of this tree. A252464(n) gives the distance of n from 1 in all these trees. Of these four trees, this is the one where the left child is always smaller than the right child.
Note that the indexing of sequence starts from 0, although its range starts from one.
The term a(n) is the Heinz number of the adjusted partial sums of the n-th composition in standard order, where (1) the k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again, (2) the Heinz number of a partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k), and (3) we define the adjusted partial sums of a composition to be obtained by subtracting one from all parts, taking partial sums, and adding one back to all parts. See formula for a simplification. A triangular form is A242628. The inverse is A253566. The non-adjusted version is A358170. - Gus Wiseman, Dec 17 2022

Examples

			From _Gus Wiseman_, Dec 23 2022: (Start)
This represents the following bijection between compositions and partitions. The n-th composition in standard order together with the reversed prime indices of a(n) are:
   0:        () -> ()
   1:       (1) -> (1)
   2:       (2) -> (2)
   3:     (1,1) -> (1,1)
   4:       (3) -> (3)
   5:     (2,1) -> (2,2)
   6:     (1,2) -> (2,1)
   7:   (1,1,1) -> (1,1,1)
   8:       (4) -> (4)
   9:     (3,1) -> (3,3)
  10:     (2,2) -> (3,2)
  11:   (2,1,1) -> (2,2,2)
  12:     (1,3) -> (3,1)
  13:   (1,2,1) -> (2,2,1)
  14:   (1,1,2) -> (2,1,1)
  15: (1,1,1,1) -> (1,1,1,1)
(End)
		

Crossrefs

Inverse: A253566.
Cf. A252737 (row sums), A252738 (row products).
Applying A001222 gives A000120.
A reverse version is A005940.
These are the Heinz numbers of the rows of A242628.
Sum of prime indices of a(n) is A359043, reverse A161511.
A048793 gives partial sums of reversed standard comps, Heinz number A019565.
A066099 lists standard compositions.
A112798 list prime indices, sum A056239.
A358134 gives partial sums of standard compositions, Heinz number A358170.

Programs

  • Mathematica
    stc[n_]:=Differences[Prepend[Join @@ Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    Times@@Prime/@#&/@Table[Accumulate[stc[n]-1]+1,{n,0,60}] (* Gus Wiseman, Dec 17 2022 *)

Formula

a(0) = 1, a(1) = 2; after which, a(2n) = A253550(a(n)), a(2n+1) = A253560(a(n)).
As a composition of related permutations:
a(n) = A122111(A163511(n)).
a(n) = A253563(A054429(n)).
Other identities and observations. For all n >= 0:
a(2n+1) - a(2n) > 0. [See the comment above.]
If n = 2^(x_1)+...+2^(x_k) then a(n) = Product_{i=1..k} prime(x_k-x_{i-1}-k+i) where x_0 = 0. - Gus Wiseman, Dec 23 2022

A242628 Irregular table enumerating partitions; n-th row has partitions in previous row with each part incremented, followed by partitions in previous row with an additional part of size 1.

Original entry on oeis.org

1, 2, 1, 1, 3, 2, 2, 2, 1, 1, 1, 1, 4, 3, 3, 3, 2, 2, 2, 2, 3, 1, 2, 2, 1, 2, 1, 1, 1, 1, 1, 1, 5, 4, 4, 4, 3, 3, 3, 3, 4, 2, 3, 3, 2, 3, 2, 2, 2, 2, 2, 2, 4, 1, 3, 3, 1, 3, 2, 1, 2, 2, 2, 1, 3, 1, 1, 2, 2, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 6, 5, 5, 5, 4, 4, 4, 4, 5, 3, 4, 4, 3, 4, 3, 3, 3, 3, 3, 3, 5, 2
Offset: 1

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Author

Keywords

Comments

This can be calculated using the binary expansion of n; see the PARI program.
The n-th row consists of all partitions with hook size (maximum + number of parts - 1) equal to n.
The partitions in row n of this sequence are the conjugates of the partitions in row n of A125106 taken in reverse order.
Row n is also the reversed partial sums plus one of the n-th composition in standard order (A066099) minus one. - Gus Wiseman, Nov 07 2022

Examples

			The table starts:
  1;
  2; 1,1;
  3; 2,2; 2,1; 1,1,1;
  4; 3,3; 3,2; 2,2,2; 3,1 2,2,1 2,1,1 1,1,1,1;
  ...
		

Crossrefs

Cf. A241596 (another version of this list of partitions), A125106, A240837, A112531, A241597 (compositions).
For other schemes to list integer partitions, please see for example A227739, A112798, A241918, A114994.
First element in each row is A008687.
Last element in each row is A065120.
Heinz numbers of rows are A253565.
Another version is A358134.

Programs

  • Maple
    b:= proc(n) option remember; `if`(n=1, [[1]],
          [map(x-> map(y-> y+1, x), b(n-1))[],
           map(x-> [x[], 1], b(n-1))[]])
        end:
    T:= n-> map(x-> x[], b(n))[]:
    seq(T(n), n=1..7);  # Alois P. Heinz, Sep 25 2015
  • Mathematica
    T[1] = {{1}};
    T[n_] := T[n] = Join[T[n-1]+1, Append[#, 1]& /@ T[n-1]];
    Array[T, 7] // Flatten (* Jean-François Alcover, Jan 25 2021 *)
  • PARI
    apart(n) = local(r=[1]); while(n>1,if(n%2==0,for(k=1,#r,r[k]++),r=concat(r,[1]));n\=2);r \\ Generates n-th partition.

A359042 Sum of partial sums of the n-th composition in standard order (A066099).

Original entry on oeis.org

0, 1, 2, 3, 3, 5, 4, 6, 4, 7, 6, 9, 5, 8, 7, 10, 5, 9, 8, 12, 7, 11, 10, 14, 6, 10, 9, 13, 8, 12, 11, 15, 6, 11, 10, 15, 9, 14, 13, 18, 8, 13, 12, 17, 11, 16, 15, 20, 7, 12, 11, 16, 10, 15, 14, 19, 9, 14, 13, 18, 12, 17, 16, 21, 7, 13, 12, 18, 11, 17, 16, 22
Offset: 0

Views

Author

Gus Wiseman, Dec 20 2022

Keywords

Comments

The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.

Examples

			The 29th composition in standard order is (1,1,2,1), with partial sums (1,2,4,5), with sum 12, so a(29) = 12.
		

Crossrefs

See link for sequences related to standard compositions.
Each n appears A000009(n) times.
The reverse version is A029931.
Comps counted by this statistic are A053632, ptns A264034, rev ptns A358194.
This is the sum of partial sums of rows of A066099.
The version for Heinz numbers of partitions is A318283, row sums of A358136.
Row sums of A358134.
A011782 counts compositions.
A065120 gives first part of standard compositions, last A001511.
A242628 lists adjusted partial sums, ranked by A253565, row sums A359043.
A358135 gives last minus first of standard compositions.

Programs

  • Mathematica
    stc[n_]:=Differences[Prepend[Join @@ Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    Table[Total[Accumulate[stc[n]]],{n,0,100}]

A359043 Sum of adjusted partial sums of the n-th composition in standard order (A066099). Row sums of A242628.

Original entry on oeis.org

0, 1, 2, 2, 3, 4, 3, 3, 4, 6, 5, 6, 4, 5, 4, 4, 5, 8, 7, 9, 6, 8, 7, 8, 5, 7, 6, 7, 5, 6, 5, 5, 6, 10, 9, 12, 8, 11, 10, 12, 7, 10, 9, 11, 8, 10, 9, 10, 6, 9, 8, 10, 7, 9, 8, 9, 6, 8, 7, 8, 6, 7, 6, 6, 7, 12, 11, 15, 10, 14, 13, 16, 9, 13, 12, 15, 11, 14, 13
Offset: 0

Views

Author

Gus Wiseman, Dec 21 2022

Keywords

Comments

We define the adjusted partial sums of a composition to be obtained by subtracting one from all parts, taking partial sums, and adding one back to all parts.
The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.

Examples

			The 29th composition in standard order is (1,1,2,1), with adjusted partial sums (1,1,2,2), with sum 6, so a(29) = 6.
		

Crossrefs

See link for sequences related to standard compositions.
The unadjusted reverse version is A029931, row sums of A048793.
The reverse version is A161511, row sums of A125106.
Row sums of A242628, ranked by A253565.
The unadjusted version is A359042, row sums of A358134.
A011782 counts compositions.
A066099 lists standard compositions.
A358135 gives last minus first of standard compositions.
A358194 counts partitions by sum and weighted sum.

Programs

  • Mathematica
    stc[n_]:=Differences[Prepend[Join @@ Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    Table[Total[Accumulate[stc[n]-1]+1],{n,0,100}]

A358134 Triangle read by rows whose n-th row lists the partial sums of the n-th composition in standard order (row n of A066099).

Original entry on oeis.org

1, 2, 1, 2, 3, 2, 3, 1, 3, 1, 2, 3, 4, 3, 4, 2, 4, 2, 3, 4, 1, 4, 1, 3, 4, 1, 2, 4, 1, 2, 3, 4, 5, 4, 5, 3, 5, 3, 4, 5, 2, 5, 2, 4, 5, 2, 3, 5, 2, 3, 4, 5, 1, 5, 1, 4, 5, 1, 3, 5, 1, 3, 4, 5, 1, 2, 5, 1, 2, 4, 5, 1, 2, 3, 5, 1, 2, 3, 4, 5, 6, 5, 6, 4, 6, 4, 5
Offset: 1

Views

Author

Gus Wiseman, Oct 31 2022

Keywords

Comments

The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.

Examples

			Triangle begins:
  1
  2
  1 2
  3
  2 3
  1 3
  1 2 3
  4
  3 4
  2 4
  2 3 4
  1 4
  1 3 4
  1 2 4
  1 2 3 4
		

Crossrefs

See link for sequences related to standard compositions.
First element in each row is A065120.
Rows are the partial sums of rows of A066099.
Last element in each row is A070939.
An adjusted version is A242628, ranked by A253565.
The first differences instead of partial sums are A358133.
The version for Heinz numbers of partitions is A358136, ranked by A358137.
Row sums are A359042.
A011782 counts compositions.
A351014 counts distinct runs in standard compositions.
A358135 gives last minus first of standard compositions.

Programs

  • Mathematica
    stc[n_]:=Differences[Prepend[Join @@ Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    Join@@Table[Accumulate[stc[n]],{n,100}]

A358133 Triangle read by rows whose n-th row lists the first differences of the n-th composition in standard order (row n of A066099).

Original entry on oeis.org

0, -1, 1, 0, 0, -2, 0, -1, 0, 2, 1, -1, 0, 1, 0, 0, 0, -3, -1, -2, 0, 1, 0, -1, -1, 1, -1, 0, 0, 3, 2, -2, 1, 0, 1, -1, 0, 0, 2, 0, 1, -1, 0, 0, 1, 0, 0, 0, 0, -4, -2, -3, 0, 0, -1, -1, -2, 1, -2, 0, 0, 2, 1, -2, 0, 0, 0, -1, 0, -1, 2, -1, 1, -1, -1, 0, 1, -1
Offset: 3

Views

Author

Gus Wiseman, Oct 31 2022

Keywords

Comments

The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.

Examples

			Triangle begins (dots indicate empty rows):
   1:   .
   2:   .
   3:   0
   4:   .
   5:  -1
   6:   1
   7:   0  0
   8:   .
   9:  -2
  10:   0
  11:  -1  0
  12:   2
  13:   1 -1
  14:   0  1
  15:   0  0  0
		

Crossrefs

See link for sequences related to standard compositions.
First differences of rows of A066099.
The version for Heinz numbers of partitions is A355536, ranked by A253566.
The partial sums instead of first differences are A358134.
Row sums are A358135.
A011782 counts compositions.
A351014 counts distinct runs in standard compositions.

Programs

  • Mathematica
    stc[n_]:=Differences[Prepend[Join @@ Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    Table[Differences[stc[n]],{n,100}]

A358170 Heinz number of the partial sums of the n-th composition in standard order (A066099).

Original entry on oeis.org

1, 2, 3, 6, 5, 15, 10, 30, 7, 35, 21, 105, 14, 70, 42, 210, 11, 77, 55, 385, 33, 231, 165, 1155, 22, 154, 110, 770, 66, 462, 330, 2310, 13, 143, 91, 1001, 65, 715, 455, 5005, 39, 429, 273, 3003, 195, 2145, 1365, 15015, 26, 286, 182, 2002, 130, 1430, 910, 10010
Offset: 0

Views

Author

Gus Wiseman, Dec 20 2022

Keywords

Comments

The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.
The Heinz number of a partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k). This gives a bijective correspondence between positive integers and integer partitions.

Examples

			The terms together with their prime indices begin:
           1: {}
           2: {1}
           3: {2}
           6: {1,2}
           5: {3}
          15: {2,3}
          10: {1,3}
          30: {1,2,3}
           7: {4}
          35: {3,4}
          21: {2,4}
         105: {2,3,4}
          14: {1,4}
          70: {1,3,4}
          42: {1,2,4}
         210: {1,2,3,4}
		

Crossrefs

See link for sequences related to standard compositions.
Applying A001221 or A001222 gives A000120.
The image is A005117 (squarefree numbers).
The reverse version is A019565, triangular version A048793.
Greatest prime index of a(n) is A029837 or A070939.
Least prime index of a(n) is A065120.
The adjusted version is A253565, inverse A253566, reverse A005940.
These are the Heinz numbers of the rows of A358134.
Sum of prime indices of a(n) is A359042.
A066099 lists standard compositions.
A112798 list prime indices, sum A056239.

Programs

  • Mathematica
    stc[n_]:=Differences[Prepend[Join @@ Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    Times@@Prime/@#&/@Table[Accumulate[stc[n]],{n,0,100}]

A358138 Difference between maximum and minimum part in the n-th composition in standard order.

Original entry on oeis.org

0, 0, 0, 0, 1, 1, 0, 0, 2, 0, 1, 2, 1, 1, 0, 0, 3, 1, 2, 1, 1, 1, 1, 3, 2, 1, 1, 2, 1, 1, 0, 0, 4, 2, 3, 0, 2, 2, 2, 2, 2, 0, 1, 2, 1, 1, 1, 4, 3, 2, 2, 2, 1, 1, 1, 3, 2, 1, 1, 2, 1, 1, 0, 0, 5, 3, 4, 1, 3, 3, 3, 1, 2, 1, 2, 2, 2, 2, 2, 3, 3, 1, 2, 1, 1, 1, 1
Offset: 1

Views

Author

Gus Wiseman, Oct 31 2022

Keywords

Comments

The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.

Crossrefs

See link for sequences related to standard compositions.
The first and last parts are A065120 and A001511, difference A358135.
This is the maximum minus minimum part in row n of A066099.
The version for Heinz numbers of partitions is A243055.
The maximum and minimum parts are A333766 and A333768.
The partial sums of standard compositions are A358134, adjusted A242628.
A011782 counts compositions.
A351014 counts distinct runs in standard compositions.

Programs

  • Mathematica
    stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    Table[Max[stc[n]]-Min[stc[n]],{n,1,100}]

Formula

a(n) = A333766(n) - A333768(n).
Showing 1-8 of 8 results.