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.

Previous Showing 11-17 of 17 results.

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}]

A253566 Permutation of natural numbers: a(n) = A243071(A122111(n)).

Original entry on oeis.org

0, 1, 2, 3, 4, 6, 8, 7, 5, 12, 16, 14, 32, 24, 10, 15, 64, 13, 128, 28, 20, 48, 256, 30, 9, 96, 11, 56, 512, 26, 1024, 31, 40, 192, 18, 29, 2048, 384, 80, 60, 4096, 52, 8192, 112, 22, 768, 16384, 62, 17, 25, 160, 224, 32768, 27, 36, 120, 320, 1536, 65536, 58, 131072, 3072, 44, 63, 72, 104, 262144, 448, 640, 50, 524288, 61, 1048576, 6144, 21
Offset: 1

Views

Author

Antti Karttunen, Jan 03 2015

Keywords

Comments

Note the indexing: domain starts from one, while the range includes also zero. See also comments in A253564.
The a(n)-th composition in standard order (graded reverse-lexicographic, A066099) is one plus the first differences of the weakly increasing sequence of prime indices of n with 1 prepended. See formula for a simplification. The triangular form is A358169. The inverse is A253565. Not prepending 1 gives A358171. For Heinz numbers instead of standard compositions we have A325351 (without prepending A325352). - Gus Wiseman, Dec 23 2022

Examples

			From _Gus Wiseman_, Dec 23 2022: (Start)
This represents the following bijection between partitions and compositions. The reversed prime indices of n together with the a(n)-th composition in standard order are:
   1:        () -> ()
   2:       (1) -> (1)
   3:       (2) -> (2)
   4:     (1,1) -> (1,1)
   5:       (3) -> (3)
   6:     (2,1) -> (1,2)
   7:       (4) -> (4)
   8:   (1,1,1) -> (1,1,1)
   9:     (2,2) -> (2,1)
  10:     (3,1) -> (1,3)
  11:       (5) -> (5)
  12:   (2,1,1) -> (1,1,2)
  13:       (6) -> (6)
  14:     (4,1) -> (1,4)
  15:     (3,2) -> (2,2)
  16: (1,1,1,1) -> (1,1,1,1)
(End)
		

Crossrefs

Inverse: A253565.
Applying A000120 gives A001222.
A reverse version is A156552, inverse essentially A005940.
The inverse is A253565, triangular form A242628.
The triangular form is A358169.
A048793 gives partial sums of reversed standard comps, Heinz number A019565.
A066099 lists standard compositions, lengths A000120, sums A070939.
A112798 list prime indices, sum A056239.
A358134 gives partial sums of standard compositions, Heinz number A358170.

Programs

  • Mathematica
    primeMS[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
    stcinv[q_]:=Total[2^(Accumulate[Reverse[q]])]/2;
    stcinv/@Table[Differences[Prepend[primeMS[n],1]]+1,{n,100}] (* Gus Wiseman, Dec 23 2022 *)
  • Scheme
    (define (A253566 n) (A243071 (A122111 n)))

Formula

a(n) = A243071(A122111(n)).
As a composition of other permutations:
a(n) = A054429(A253564(n)).
a(n) = A336120(n) + A336125(n). - Antti Karttunen, Jul 18 2020
If 2n = Product_{i=1..k} prime(x_i) then a(n) = Sum_{i=1..k-1} 2^(x_k-x_{k-i}+i-1). - Gus Wiseman, Dec 23 2022

A359402 Numbers whose binary expansion and reversed binary expansion have the same sum of positions of 1's, where positions in a sequence are read starting with 1 from the left.

Original entry on oeis.org

0, 1, 3, 5, 7, 9, 15, 17, 21, 27, 31, 33, 45, 51, 63, 65, 70, 73, 78, 85, 93, 99, 107, 119, 127, 129, 150, 153, 165, 189, 195, 219, 231, 255, 257, 266, 273, 282, 294, 297, 310, 313, 325, 334, 341, 350, 355, 365, 371, 381, 387, 397, 403, 413, 427, 443, 455, 471
Offset: 1

Views

Author

Gus Wiseman, Jan 05 2023

Keywords

Comments

Also numbers whose binary expansion and reversed binary expansion have the same sum of partial sums.
Also numbers whose average position of a 1 in their binary expansion is (c+1)/2, where c is the number of digits.
Conjecture: Also numbers whose binary expansion has as least squares fit a line of zero slope, counted by A222955.

Examples

			The binary expansion of 70 is (1,0,0,0,1,1,0), with positions of 1's {1,5,6}, while the reverse positions are {2,3,7}. Both sum to 12, so 70 is in the sequence.
		

Crossrefs

Binary words of this type appear to be counted by A222955.
For greater instead of equal sums we have A359401.
These are the indices of 0's in A359495.
A030190 gives binary expansion, reverse A030308.
A048793 lists partial sums of reversed standard compositions, sums A029931.
A070939 counts binary digits, 1's A000120.
A326669 lists numbers with integer mean position of a 1 in binary expansion.

Programs

  • Mathematica
    Select[Range[0,100],#==0||Mean[Join@@Position[IntegerDigits[#,2],1]]==(IntegerLength[#,2]+1)/2&]
  • Python
    from functools import reduce
    from itertools import count, islice
    def A359402_gen(startvalue=0): # generator of terms
        return filter(lambda n:(r:=reduce(lambda c, d:(c[0]+d[0]*(e:=int(d[1])),c[1]+e),enumerate(bin(n)[2:],start=1),(0,0)))[0]<<1==(n.bit_length()+1)*r[1],count(max(startvalue,0)))
    A359402_list = list(islice(A359402_gen(),30)) # Chai Wah Wu, Jan 08 2023

Formula

A230877(a(n)) = A029931(a(n)).

A358135 Difference of first and last parts of the n-th composition in standard order.

Original entry on oeis.org

0, 0, 0, 0, -1, 1, 0, 0, -2, 0, -1, 2, 0, 1, 0, 0, -3, -1, -2, 1, -1, 0, -1, 3, 0, 1, 0, 2, 0, 1, 0, 0, -4, -2, -3, 0, -2, -1, -2, 2, -1, 0, -1, 1, -1, 0, -1, 4, 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 0, -5, -3, -4, -1, -3, -2, -3, 1, -2, -1, -2, 0, -2
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.
This is the first minus last part of row n of A066099.
The version for Heinz numbers of partitions is A243055.
Row sums of A358133.
The partial sums of standard compositions are A358134, adjusted A242628.
A011782 counts compositions.
A333766 and A333768 give max and min in standard compositions, diff A358138.
A351014 counts distinct runs in standard compositions.

Programs

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

Formula

a(n) = A001511(n) - A065120(n).

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).
Previous Showing 11-17 of 17 results.