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 31-40 of 40 results.

A344090 Flattened tetrangle of strict integer partitions, sorted first by sum, then by length, then lexicographically.

Original entry on oeis.org

1, 2, 3, 2, 1, 4, 3, 1, 5, 3, 2, 4, 1, 6, 4, 2, 5, 1, 3, 2, 1, 7, 4, 3, 5, 2, 6, 1, 4, 2, 1, 8, 5, 3, 6, 2, 7, 1, 4, 3, 1, 5, 2, 1, 9, 5, 4, 6, 3, 7, 2, 8, 1, 4, 3, 2, 5, 3, 1, 6, 2, 1, 10, 6, 4, 7, 3, 8, 2, 9, 1, 5, 3, 2, 5, 4, 1, 6, 3, 1, 7, 2, 1, 4, 3, 2, 1
Offset: 0

Views

Author

Gus Wiseman, May 12 2021

Keywords

Comments

The zeroth row contains only the empty partition.
A tetrangle is a sequence of finite triangles.

Examples

			Tetrangle begins:
  0: ()
  1: (1)
  2: (2)
  3: (3)(21)
  4: (4)(31)
  5: (5)(32)(41)
  6: (6)(42)(51)(321)
  7: (7)(43)(52)(61)(421)
  8: (8)(53)(62)(71)(431)(521)
  9: (9)(54)(63)(72)(81)(432)(531)(621)
		

Crossrefs

Starting with reversed partitions gives A026793.
The version for compositions is A124734.
Showing partitions as Heinz numbers gives A246867.
The non-strict version is A334301 (reversed: A036036).
Ignoring length gives A344086 (reversed: A246688).
Same as A344089 with partitions reversed.
The version for revlex instead of lex is A344092.
A026791 reads off lexicographically ordered reversed partitions.
A080577 reads off reverse-lexicographically ordered partitions.
A112798 reads off reversed partitions by Heinz number.
A296150 reads off partitions by Heinz number.

Programs

  • Mathematica
    Table[Sort[Select[IntegerPartitions[n],UnsameQ@@#&]],{n,0,10}]

A344087 Flattened tetrangle of strict integer partitions sorted first by sum, then colexicographically.

Original entry on oeis.org

1, 2, 2, 1, 3, 3, 1, 4, 4, 1, 3, 2, 5, 3, 2, 1, 5, 1, 4, 2, 6, 4, 2, 1, 6, 1, 5, 2, 4, 3, 7, 5, 2, 1, 4, 3, 1, 7, 1, 6, 2, 5, 3, 8, 6, 2, 1, 5, 3, 1, 8, 1, 4, 3, 2, 7, 2, 6, 3, 5, 4, 9, 4, 3, 2, 1, 7, 2, 1, 6, 3, 1, 5, 4, 1, 9, 1, 5, 3, 2, 8, 2, 7, 3, 6, 4, 10
Offset: 0

Views

Author

Gus Wiseman, May 11 2021

Keywords

Comments

The zeroth row contains only the empty partition.
A tetrangle is a sequence of finite triangles.

Examples

			Tetrangle begins:
  0: ()
  1: (1)
  2: (2)
  3: (21)(3)
  4: (31)(4)
  5: (41)(32)(5)
  6: (321)(51)(42)(6)
  7: (421)(61)(52)(43)(7)
  8: (521)(431)(71)(62)(53)(8)
  9: (621)(531)(81)(432)(72)(63)(54)(9)
		

Crossrefs

Positions of first appearances are A015724.
Triangle sums are A066189.
Taking revlex instead of colex gives A118457.
The not necessarily strict version is A211992.
Taking lex instead of colex gives A344086.
A026793 gives reversed strict partitions in A-S order (sum/length/lex).
A319247 sorts strict partitions by Heinz number.
A329631 sorts reversed strict partitions by Heinz number.
A344090 gives strict partitions in A-S order (sum/length/lex).

Programs

  • Mathematica
    colex[f_,c_]:=OrderedQ[PadRight[{Reverse[f],Reverse[c]}]];
    Table[Sort[Select[IntegerPartitions[n],UnsameQ@@#&],colex],{n,0,10}]

A344088 Flattened tetrangle of reversed strict integer partitions sorted first by sum, then colexicographically.

Original entry on oeis.org

1, 2, 1, 2, 3, 1, 3, 4, 2, 3, 1, 4, 5, 1, 2, 3, 2, 4, 1, 5, 6, 1, 2, 4, 3, 4, 2, 5, 1, 6, 7, 1, 3, 4, 1, 2, 5, 3, 5, 2, 6, 1, 7, 8, 2, 3, 4, 1, 3, 5, 4, 5, 1, 2, 6, 3, 6, 2, 7, 1, 8, 9, 1, 2, 3, 4, 2, 3, 5, 1, 4, 5, 1, 3, 6, 4, 6, 1, 2, 7, 3, 7, 2, 8, 1, 9, 10
Offset: 0

Views

Author

Gus Wiseman, May 12 2021

Keywords

Comments

The zeroth row contains only the empty partition.
A tetrangle is a sequence of finite triangles.

Examples

			Tetrangle begins:
  0: ()
  1: (1)
  2: (2)
  3: (12)(3)
  4: (13)(4)
  5: (23)(14)(5)
  6: (123)(24)(15)(6)
  7: (124)(34)(25)(16)(7)
  8: (134)(125)(35)(26)(17)(8)
  9: (234)(135)(45)(126)(36)(27)(18)(9)
		

Crossrefs

Positions of first appearances are A015724.
Triangle sums are A066189.
The non-strict version is A080576.
Taking lex instead of colex gives A246688 (non-reversed: A344086).
The non-reversed version is A344087.
Taking revlex instead of colex gives A344089 (non-reversed: A118457).
A026793 gives reversed strict partitions in A-S order (sum/length/lex).
A319247 sorts strict partitions by Heinz number.
A329631 sorts reversed strict partitions by Heinz number.
A344090 gives strict partitions in A-S order (sum/length/lex).

Programs

  • Mathematica
    colex[f_,c_]:=OrderedQ[PadRight[{Reverse[f],Reverse[c]}]];
    Table[Sort[Reverse/@Select[IntegerPartitions[n],UnsameQ@@#&],colex],{n,0,10}]

A344091 Flattened tetrangle of all finite multisets of positive integers sorted first by sum, then by length, then colexicographically.

Original entry on oeis.org

1, 2, 1, 1, 3, 1, 2, 1, 1, 1, 4, 2, 2, 1, 3, 1, 1, 2, 1, 1, 1, 1, 5, 2, 3, 1, 4, 1, 2, 2, 1, 1, 3, 1, 1, 1, 2, 1, 1, 1, 1, 1, 6, 3, 3, 2, 4, 1, 5, 2, 2, 2, 1, 2, 3, 1, 1, 4, 1, 1, 2, 2, 1, 1, 1, 3, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1
Offset: 0

Views

Author

Gus Wiseman, May 12 2021

Keywords

Comments

First differs from A334302 for partitions of 9.
The zeroth row contains only the empty partition.
A tetrangle is a sequence of finite triangles.

Examples

			Tetrangle begins:
  0: ()
  1: (1)
  2: (2)(11)
  3: (3)(12)(111)
  4: (4)(22)(13)(112)(1111)
  5: (5)(23)(14)(122)(113)(1112)(11111)
  6: (6)(33)(24)(15)(222)(123)(114)(1122)(1113)(11112)(111111)
		

Crossrefs

The version for lex instead of colex is A036036.
Starting with reversed partitions gives A036037.
Ignoring length gives A211992 (reversed: A080576).
Same as A334301 with partitions reversed.
The version for revlex instead of colex is A334302.
The Heinz numbers of these partitions are A334433.
The strict case is A344089.
A026791 reads off lexicographically ordered reversed partitions.
A080577 reads off reverse-lexicographically ordered partitions.
A112798 reads off reversed partitions by Heinz number.
A193073 reads off lexicographically ordered partitions.
A296150 reads off partitions by Heinz number.

Programs

  • Mathematica
    Table[Reverse/@Sort[IntegerPartitions[n]],{n,0,9}]

A382255 Heinz number of the partition corresponding to run lengths in the bits of n.

Original entry on oeis.org

1, 2, 4, 3, 6, 8, 6, 5, 10, 12, 16, 12, 9, 12, 10, 7, 14, 20, 24, 18, 24, 32, 24, 20, 15, 18, 24, 18, 15, 20, 14, 11, 22, 28, 40, 30, 36, 48, 36, 30, 40, 48, 64, 48, 36, 48, 40, 28, 21, 30, 36, 27, 36, 48, 36, 30, 25, 30, 40, 30, 21, 28, 22, 13, 26, 44, 56, 42
Offset: 0

Views

Author

M. F. Hasler and Ali Sada, Mar 19 2025

Keywords

Comments

The run lengths (number of consecutive bits that are equal) in the binary numbers in [2^(L-1), 2^L-1], i.e., of bit length L, yield all possible compositions of L, i.e., the partitions with any possible order of the parts.
Associated to any composition (p1, ..., pK) is their Heinz number prime(p1)*...*prime(pK) which depends only on the partition, i.e., not on the order of the parts.
The sequence can also be read as a table with row lengths 1, 1, 2, 4, 8, 16, 32, ... (= A011782), where row L = 0, 1, 2, 3, ... lists the 2^(L-1) compositions of L through their Heinz numbers (which will appear more than once if they contain at least two distinct parts).

Examples

			   n | binary | partition | a(n) = Heinz number
  ---+--------+-----------+--------------------
   0 |   (0)  | empty sum | 1 = empty product
   1 |     1  |     1     | 2 = prime(1)
   2 |    10  |    1+1    | 4 = prime(1) * prime(1)
   3 |    11  |     2     | 3 = prime(2)
   4 |   100  |    1+2    | 6 = prime(1) * prime(2)
   5 |   101  |   1+1+1   | 8 = 2^3 = prime(1) * prime(1) * prime(1)
   6 |   110  |    2+1    | 6 = prime(2) * prime(1)
   7 |   111  |     3     | 5 = prime(3)
   8 |  1000  |    1+3    | 10 = 2*5 = prime(1) * prime(3)
   9 |  1001  |   1+2+1   | 12 = 2^2*3 = prime(1) * prime(2) * prime(1)
  ...|   ...  |    ...    | ...
For example, n = 4 = 100[2] (in binary) has run lengths (1, 2), namely: one bit 1 followed by two bits 0. This gives a(4) = prime(1)*prime(2) = 6.
Next, n = 5 = 101[2] (in binary) has run lengths (1, 1, 1): one bit 1, followed by one bit 0, followed by one bit 1. This gives a(4) = prime(1)^3 = 8.
Then, n = 6 = 110[2] (in binary) has run lengths (2, 1): first two bits 1, then one bit 0. This is the same as for 4, just in reverse order, so it yields the same Heinz number a(6) = prime(2)*prime(1) = 6.
Then, n = 7 = 111[2] (in binary) has run lengths (3), namely: three bits 1. This gives a(5) = prime(3) = 5.
Sequence written as irregular triangle:
   1;
   2;
   4,  3;
   6,  8,  6,  5;
  10, 12, 16, 12,  9, 12, 10,  7;
  14, 20, 24, 18, 24, 32, 24, 20, 15, 18, 24, 18, 15, 20, 14, 11;
  ...
		

Crossrefs

Cf. A112798 and A296150 (partitions sorted by Heinz number).
Cf. A185974, A334433, A334435, A334438, A334434, A129129, A334436 (partitions given as Heinz numbers, in Abramowitz-Stegun, Maple, Mathematica order).
For "constructive" lists of partitions see A036036 (Abramowitz and Stegun order), A036036 (reversed), A080576 (Maple order), A080577 (Mathematica order).
Row sums of triangle give A030017(n+1).
Cf. A007088 (the binary numbers).
Cf. A101211 (the run lengths as rows of a table).

Programs

  • Maple
    a:= proc(n) option remember; `if`(n<2, 1+n, (p->
          a(iquo(n, 2^p))*ithprime(p))(padic[ordp](n+(n mod 2), 2)))
        end:
    seq(a(n), n=0..100);  # Alois P. Heinz, Mar 20 2025
  • PARI
    Heinz(p)=vecprod([ prime(k) | k <- p ])
    RL(v) = if(#v, v=Vec(select(t->t,concat([1,v[^1]-v[^-1],1]),1)); v[^1]-v[^-1])
    apply( {A382255(n) = Heinz(RL(binary(n)))}, [0..99] )

Formula

a(2^n) = A001747(n+1).
a(2^n-1) = A008578(n+1).
a(2^n+1) = A001749(n-1) for n>=2.

A339677 Partition array: T(n, k) is the number of aperiodic necklaces (Lyndon words) on a multiset of colored beads (of size n) whose color multiplicities form the k-th partition of n in Abramowitz-Stegun order.

Original entry on oeis.org

1, 0, 1, 0, 1, 2, 0, 1, 1, 3, 6, 0, 1, 2, 4, 6, 12, 24, 0, 1, 2, 3, 5, 10, 14, 20, 30, 60, 120, 0, 1, 3, 5, 6, 15, 20, 30, 30, 60, 90, 120, 180, 360, 720, 0, 1, 3, 7, 8, 7, 21, 35, 51, 70, 42, 105, 140, 210, 312, 210, 420, 630, 840, 1260, 2520, 5040, 0, 1, 4, 9, 14, 8, 28, 56, 70, 84, 140
Offset: 1

Views

Author

Álvar Ibeas, Dec 12 2020

Keywords

Comments

As in A212359, A072605, and A261600, for each partition, the base set of beads is fixed.
Abuse of notation: we write T(n, L) for T(n, k), where L is the k-th partition of n in A-St order. We do accordingly for A036038 and A212359.

Examples

			Array begins:
  k:  1 2 3 4 5  6  7  8  9 10  11  12  13  14  15
      --------------------------------------------
n=1:  1
n=2:  0 1
n=3:  0 1 2
n=4:  0 1 1 3 6
n=5:  0 1 2 4 6 12 24
n=6:  0 1 2 3 5 10 14 20 30 60 120
n=7:  0 1 3 5 6 15 20 30 30 60  90 120 180 360 720
Consider partition L = (4, 2). There are 3 = A212359(6, L) necklaces on the bead set {a^4, b^2}: (aaaabb), (aaabab), and (aabaab). The latter has a period smaller than its size (3 < 6), whereas the other two are aperiodic. Hence, T(6, L) = 2.
T(n, (1,...,1)) = A212359(n, (1,...,1)) = (n-1)!, counting necklaces with n beads, each in a different color.
		

Crossrefs

Programs

  • PARI
    C(sig)={my(n=vecsum(sig)); sumdiv(gcd(sig), d, moebius(d)*(n/d)!/prod(i=1, #sig, (sig[i]/d)!))/n}
    Row(n)=[C(Vec(p)) | p<-partitions(n)]
    for(n=1, 7, print(Row(n))) \\ Andrew Howroyd, Dec 14 2020

Formula

Let L be a partition of n and d be the gcd of its parts. Then,
T(n, L) = n^(-1) * Sum_{v|d} mu(v) * A036038(n/v, L/v), where L/v is the partition obtained from L after dividing each part by v.
T(n, L) = Sum_{v|d} mu(v) * A212359(n/v, L/v).
T(n, L) = n^(-1) * A036038(n, L) - Sum_{1
T(n,k) = A298941(A036035(n,k)) = A318808(A185974(n,k)). - Andrew Howroyd, Dec 14 2020

A344084 Concatenated list of all finite nonempty sets of positive integers sorted first by maximum, then by length, and finally lexicographically.

Original entry on oeis.org

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

Author

Gus Wiseman, May 11 2021

Keywords

Examples

			The sets are the columns below:
  1 2 1 3 1 2 1 4 1 2 3 1 1 2 1 5 1 2 3 4 1 1 1 2 2 3 1
      2   3 3 2   4 4 4 2 3 3 2   5 5 5 5 2 3 4 3 4 4 2
              3         4 4 4 3           5 5 5 5 5 5 3
                              4                       5
As a tetrangle, the first four triangles are:
  {1}
  {2},{1,2}
  {3},{1,3},{2,3},{1,2,3}
  {4},{1,4},{2,4},{3,4},{1,2,4},{1,3,4},{2,3,4},{1,2,3,4}
		

Crossrefs

Triangle lengths are A000079.
Triangle sums are A001793.
Positions of first appearances are A005183.
Set maxima are A070939.
Set lengths are A124736.

Programs

  • Mathematica
    SortBy[Rest[Subsets[Range[5]]],Last]

A344092 Flattened tetrangle of strict integer partitions, sorted first by sum, then by length, and finally reverse-lexicographically.

Original entry on oeis.org

1, 2, 3, 2, 1, 4, 3, 1, 5, 4, 1, 3, 2, 6, 5, 1, 4, 2, 3, 2, 1, 7, 6, 1, 5, 2, 4, 3, 4, 2, 1, 8, 7, 1, 6, 2, 5, 3, 5, 2, 1, 4, 3, 1, 9, 8, 1, 7, 2, 6, 3, 5, 4, 6, 2, 1, 5, 3, 1, 4, 3, 2, 10, 9, 1, 8, 2, 7, 3, 6, 4, 7, 2, 1, 6, 3, 1, 5, 4, 1, 5, 3, 2, 4, 3, 2, 1
Offset: 0

Author

Gus Wiseman, May 14 2021

Keywords

Comments

First differs from A118457 at a(53) = 4, A118457(53) = 2.
The zeroth row contains only the empty partition.
A tetrangle is a sequence of finite triangles.

Examples

			Tetrangle begins:
   0: ()
   1: (1)
   2: (2)
   3: (3)(21)
   4: (4)(31)
   5: (5)(41)(32)
   6: (6)(51)(42)(321)
   7: (7)(61)(52)(43)(421)
   8: (8)(71)(62)(53)(521)(431)
   9: (9)(81)(72)(63)(54)(621)(531)(432)
		

Crossrefs

Same as A026793 with rows reversed.
Ignoring length gives A118457.
The non-strict version is A334439 (reversed: A036036/A334302).
The version for lex instead of revlex is A344090.
A026791 reads off lexicographically ordered reversed partitions.
A080577 reads off reverse-lexicographically ordered partitions.
A112798 reads off reversed partitions by Heinz number.
A193073 reads off lexicographically ordered partitions.
A296150 reads off partitions by Heinz number.

Programs

  • Mathematica
    revlensort[f_,c_]:=If[Length[f]!=Length[c],Length[f]
    				

A185975 Prime number factorization of n mapped to a(n)-th partition in A-St order.

Original entry on oeis.org

1, 2, 3, 4, 5, 7, 6, 9, 8, 12, 10, 19, 13, 14, 11, 30, 16, 45, 15, 21, 20, 67, 17, 22, 31, 25, 23, 97, 24, 139, 18, 32, 46, 33, 27, 195, 68, 47, 26, 272, 35, 373, 34, 37, 98, 508, 28, 49, 36, 69, 50, 684, 40, 48, 38, 99, 140, 915, 39, 1212, 196, 53, 29, 70, 51, 1597, 72, 141, 52, 2087, 42, 2714, 273, 54, 103, 71, 73, 3506, 41, 59, 374, 4508, 56, 100, 509, 197, 55, 5763, 58, 101, 145, 274, 685, 142, 43, 7338, 75, 76, 57
Offset: 2

Author

Wolfdieter Lang, Feb 11 2011

Keywords

Comments

This is the inverse of the map n->A185974(n), n>=1.
The prime number factorization
n = p(1)^e(1)*p(2)^e(2)*...*p(M)^e(M), with e(M)>=1, and e(j)>=0, j=1,..,M-1, with the prime numbers p(j):=A000040(j), is mapped to the partition 1^e(1),2^e(2),...,M^e(M), with M=M(n) = A061395(n). Note that j^0 means that j does not show up in the partition, it is not 1. a(n) is the position of this partition of N=N(n):=sum(j*e(j), j=1..M(n)) in the A-St (Abramowitz-Stegun) list of all partitions. See A036036 and the reference for this order.
In order to obtain an exponent list of length N=N(n), appropriate for a partition of N(n), one has to append N(n)-M(n)>=0 zeros to the list [e(1),e(2),...,e(M(n))].
E.g., n=10, M(10)=3, N(10)=4, from the partition 1^2,3^1; N(10)-M(10)= 1, hence the complete exponent list for this partition is [2,0,1,0].
In order to have offset 1 one could add a(1):=0.

Examples

			a(20)=15 because 20=p(1)^2 p(3)^1 which maps to the partition 1^2,3^1 (of N(20)=5) which appears at position 15 in the A-St list. M(20)=3, hence the exponent list for this partition is [2,0,1,0,0] (2=5-3 zeros added to the list [2,0,1] from the prime number factorization exponent list).
		

Crossrefs

Cf. A185974 (inverse map).

Formula

a(n) gives the a(n)-th position of the partition in A-St order (see A036036 and the reference) obtained from the prime number factorization of n read as partition. This mapping is explained in the comment above.

A187448 One half of the smallest number with prime signature of the multiset defining partition, taken in Abramowitz-Stegun order.

Original entry on oeis.org

1, 2, 3, 4, 6, 8, 12, 16, 15, 18, 24, 32, 30, 36, 48, 64, 60, 72, 96, 128, 90, 120, 108, 144, 192, 256, 105, 180, 240, 216, 288, 384, 512, 210, 360, 480, 432, 576, 768, 1024, 420, 450, 540, 720, 648, 960, 864, 1152, 1536, 2048
Offset: 1

Author

Wolfdieter Lang, Mar 15 2011

Keywords

Comments

For a list of the multiset repetition class defining partitions in Abramowitz-Stegun (A-St)order see the links under A176725 and A187447.
For the A-St ordering of all partitions see A036036.
The actual sequence is 2*a(n): 2, 4, 6, 8, 12, 16, 24, 32, 30, 36, 48, 64, 60, 72, 96, 128, 120, 144, 192, 256,... This is similar to A025487 without the leading 1 (products of primorial numbers A002110, ordered increasingly, which is not the case here).
The analog sequence for all partitions in A-St order is A185974.

Examples

			2*a(11)=2*24=48 =2^4*3^1, the smallest number with prime signature e[1]=4, e[2]=1, read as multiset defining partition 1^4,2^1, which is the 11th one in Abramowitz-Stegun order. The corresponding 5-multiset is {1,1,1,1,2}.
		

Formula

a(n)=((p(1)^e[1])*(p(2)^e^[2])*...*(p(M)^e[M]))/2 with the prime numbers p(j):=A000040(j), and the n-th multiset defining partition with positive integer exponents e[1]>=e[2]>=...>=e[M]>=1; M=M(n)=A176725(n), read as sequence. These partitions are taken in A-St order. See the links to A176725 and A187447 for this partition list.
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