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 51-56 of 56 results.

A371738 Numbers with non-quanimous binary indices. Numbers whose binary indices have only one set partition with all equal block-sums.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 8, 9, 10, 11, 12, 14, 16, 17, 18, 19, 20, 21, 23, 24, 26, 28, 29, 32, 33, 34, 35, 36, 37, 38, 40, 41, 43, 44, 46, 48, 50, 52, 53, 55, 56, 57, 58, 61, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 77, 78, 79, 80, 81, 83, 84, 86, 88, 89, 91, 92
Offset: 1

Views

Author

Gus Wiseman, Apr 14 2024

Keywords

Comments

A binary index of n is any position of a 1 in its reversed binary expansion. The binary indices of n are row n of A048793.

Examples

			The binary indices of 165 are {1,3,6,8}, with qualifying set partitions {{1,8},{3,6}}, and {{1,3,6,8}}, so 165 is not in the sequence.
The terms together with their binary expansions and binary indices begin:
   1:     1 ~ {1}
   2:    10 ~ {2}
   3:    11 ~ {1,2}
   4:   100 ~ {3}
   5:   101 ~ {1,3}
   6:   110 ~ {2,3}
   8:  1000 ~ {4}
   9:  1001 ~ {1,4}
  10:  1010 ~ {2,4}
  11:  1011 ~ {1,2,4}
  12:  1100 ~ {3,4}
  14:  1110 ~ {2,3,4}
  16: 10000 ~ {5}
  17: 10001 ~ {1,5}
  18: 10010 ~ {2,5}
  19: 10011 ~ {1,2,5}
  20: 10100 ~ {3,5}
  21: 10101 ~ {1,3,5}
  23: 10111 ~ {1,2,3,5}
		

Crossrefs

Set partitions with all equal block-sums are counted by A035470.
Positions of 1's in A336137 and A371735.
The complement is A371784.
A000110 counts set partitions.
A002219 (aerated) counts biquanimous partitions, ranks A357976.
A048793 lists binary indices, length A000120, reverse A272020, sum A029931.
A070939 gives length of binary expansion.
A321451 counts non-quanimous partitions, ranks A321453.
A321452 counts quanimous partitions, ranks A321454.
A371789 counts non-quanimous sets, differences A371790.
A371796 counts quanimous sets, differences A371797.

Programs

  • Mathematica
    bix[n_]:=Join@@Position[Reverse[IntegerDigits[n,2]],1];
    sps[{}]:={{}};sps[set:{i_,_}]:=Join@@Function[s,Prepend[#,s]& /@ sps[Complement[set,s]]]/@Cases[Subsets[set],{i,_}];
    Select[Range[100],Length[Select[sps[bix[#]],SameQ@@Total/@#&]]==1&]

A371784 Numbers with quanimous binary indices. Numbers whose binary indices can be partitioned in more than one way into blocks with the same sum.

Original entry on oeis.org

7, 13, 15, 22, 25, 27, 30, 31, 39, 42, 45, 47, 49, 51, 54, 59, 60, 62, 63, 75, 76, 82, 85, 87, 90, 93, 94, 95, 97, 99, 102, 107, 108, 109, 110, 115, 117, 119, 120, 122, 125, 126, 127, 141, 143, 147, 148, 153, 155, 158, 162, 165, 167, 170, 173, 175, 179, 180
Offset: 1

Views

Author

Gus Wiseman, Apr 16 2024

Keywords

Comments

A binary index of n is any position of a 1 in its reversed binary expansion. The binary indices of n are row n of A048793.

Examples

			The binary indices of 165 are {1,3,6,8}, with qualifying set partitions {{1,8},{3,6}}, and {{1,3,6,8}}, so 165 is in the sequence.
The terms together with their binary expansions and binary indices begin:
   7:     111 ~ {1,2,3}
  13:    1101 ~ {1,3,4}
  15:    1111 ~ {1,2,3,4}
  22:   10110 ~ {2,3,5}
  25:   11001 ~ {1,4,5}
  27:   11011 ~ {1,2,4,5}
  30:   11110 ~ {2,3,4,5}
  31:   11111 ~ {1,2,3,4,5}
  39:  100111 ~ {1,2,3,6}
  42:  101010 ~ {2,4,6}
  45:  101101 ~ {1,3,4,6}
  47:  101111 ~ {1,2,3,4,6}
  49:  110001 ~ {1,5,6}
  51:  110011 ~ {1,2,5,6}
  54:  110110 ~ {2,3,5,6}
  59:  111011 ~ {1,2,4,5,6}
  60:  111100 ~ {3,4,5,6}
  62:  111110 ~ {2,3,4,5,6}
  63:  111111 ~ {1,2,3,4,5,6}
		

Crossrefs

Set partitions with all equal block-sums are counted by A035470.
Positions of terms > 1 in A336137 and A371735.
The complement is A371738.
A000110 counts set partitions.
A002219 (aerated) counts biquanimous partitions, ranks A357976.
A048793 lists binary indices, length A000120, reverse A272020, sum A029931.
A070939 gives length of binary expansion.
A321451 counts non-quanimous partitions, ranks A321453.
A321452 counts quanimous partitions, ranks A321454.
A371789 counts non-quanimous sets, differences A371790.
A371796 counts quanimous sets, differences A371797.

Programs

  • Mathematica
    bix[n_]:=Join@@Position[Reverse[IntegerDigits[n,2]],1];
    sps[{}]:={{}};sps[set:{i_,_}]:=Join@@Function[s,Prepend[#,s]& /@ sps[Complement[set,s]]]/@Cases[Subsets[set],{i,_}];
    Select[Range[100],Length[Select[sps[bix[#]],SameQ@@Total/@#&]]>1&]

A359041 Number of finite sets of integer partitions with all equal sums and total sum n.

Original entry on oeis.org

1, 1, 2, 3, 6, 7, 14, 15, 32, 31, 63, 56, 142, 101, 240, 211, 467, 297, 985, 490, 1524, 1247, 2542, 1255, 6371, 1979, 7486, 7070, 14128, 4565, 32953, 6842, 42229, 37863, 56266, 17887, 192914, 21637, 145820, 197835, 371853, 44583, 772740, 63261, 943966, 1124840
Offset: 0

Views

Author

Gus Wiseman, Dec 14 2022

Keywords

Examples

			The a(1) = 1 through a(6) = 14 sets:
  {(1)}  {(2)}   {(3)}    {(4)}       {(5)}      {(6)}
         {(11)}  {(21)}   {(22)}      {(32)}     {(33)}
                 {(111)}  {(31)}      {(41)}     {(42)}
                          {(211)}     {(221)}    {(51)}
                          {(1111)}    {(311)}    {(222)}
                          {(2),(11)}  {(2111)}   {(321)}
                                      {(11111)}  {(411)}
                                                 {(2211)}
                                                 {(3111)}
                                                 {(21111)}
                                                 {(111111)}
                                                 {(3),(21)}
                                                 {(3),(111)}
                                                 {(21),(111)}
		

Crossrefs

This is the constant-sum case of A261049, ordered A358906.
The version for all different sums is A271619, ordered A336342.
Allowing repetition gives A305551, ordered A279787.
The version for compositions instead of partitions is A358904.
A001970 counts multisets of partitions.
A034691 counts multisets of compositions, ordered A133494.
A098407 counts sets of compositions, ordered A358907.

Programs

  • Mathematica
    Table[If[n==0,1,Sum[Binomial[PartitionsP[d],n/d],{d,Divisors[n]}]],{n,0,50}]
  • PARI
    a(n) = if (n, sumdiv(n, d, binomial(numbpart(d), n/d)), 1); \\ Michel Marcus, Dec 14 2022

Formula

a(n) = Sum_{d|n} binomial(A000041(d),n/d).

A381872 Number of multisets that can be obtained by taking the sum of each block of a multiset partition of the prime indices of n into blocks having a common sum.

Original entry on oeis.org

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

Views

Author

Gus Wiseman, Mar 14 2025

Keywords

Comments

First differs from A321455 at a(144) = 4, A321455(144) = 3.
A prime index of n is a number m such that prime(m) divides n. The multiset of prime indices of n is row n of A112798, sum A056239.

Examples

			The prime indices of 144 are {1,1,1,1,2,2}, with the following 4 multiset partitions having common block sum:
  {{1,1,1,1,2,2}}
  {{2,2},{1,1,1,1}}
  {{1,1,2},{1,1,2}}
  {{2},{2},{1,1},{1,1}}
with sums: 8, 4, 4, 2, of which 3 are distinct, so a(144) = 3.
The prime indices of 1296 are {1,1,1,1,2,2,2,2}, with the following 7 multiset partitions having common block sum:
  {{1,1,1,1,2,2,2,2}}
  {{2,2,2},{1,1,1,1,2}}
  {{1,1,2,2},{1,1,2,2}}
  {{2,2},{2,2},{1,1,1,1}}
  {{2,2},{1,1,2},{1,1,2}}
  {{1,2},{1,2},{1,2},{1,2}}
  {{2},{2},{2},{2},{1,1},{1,1}}
with sums: 12, 6, 6, 4, 4, 3, 2, of which 5 are distinct, so a(1296) = 5.
		

Crossrefs

With equal blocks instead of sums we have A089723.
Without equal sums we have A317141, before sums A001055, lower A300383.
Positions of terms > 1 are A321454.
Before taking sums we had A321455.
With distinct instead of equal sums we have A381637, before sums A321469.
A000041 counts integer partitions, strict A000009, constant A000005.
A055396 gives least prime index, greatest A061395.
A056239 adds up prime indices, row sums of A112798.
A265947 counts refinement-ordered pairs of integer partitions.
Other multiset partitions of prime indices:
- For multisets of constant multisets (A000688) see A381455 (upper), A381453 (lower).
- For sets of constant multisets (A050361) see A381715.
- For sets of constant multisets with distinct sums (A381635) see A381716, A381636.

Programs

  • Mathematica
    prix[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
    sps[{}]:={{}};sps[set:{i_,_}]:=Join@@Function[s,Prepend[#,s]&/@sps[Complement[set,s]]]/@Cases[Subsets[set],{i,_}];
    mps[mset_]:=Union[Sort[Sort/@(#/.x_Integer:>mset[[x]])]&/@sps[Range[Length[mset]]]];
    Table[Length[Union[Sort[Total/@#]&/@Select[mps[prix[n]],SameQ@@Total/@#&]]],{n,100}]

A387388 a(n) is the maximum number of ways in which any strict partition of 2n can be partitioned into two disjoint subsets of equal sum.

Original entry on oeis.org

0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 4, 4, 4, 4, 7, 6, 6, 6, 6, 11, 11, 10, 10, 10, 19, 18, 18, 18, 17, 35, 33, 32, 32, 31, 31, 62, 60, 58, 57, 57, 55, 56, 108, 105, 103, 101, 100, 100, 99, 195, 191, 187, 184, 182, 181, 180, 361, 352, 344, 340, 336, 333
Offset: 1

Views

Author

Jesús Bellver Arnau, Aug 28 2025

Keywords

Comments

Finding the number of ways in which a set can be partitioned into two disjoint subsets with equal sum is often referred to as the "partition search problem".
The sequence is defined for partitions of 2n because for odd numbers there are no solutions.

Examples

			a(2) = 0, because strict partitions of 4 are {4} and {3,1}. None of these partitions can be partitioned into two disjoint subsets of equal sum.
a(3) = 1, because strict partitions of 6 are {6}, {5,1}, {4,2} and {3,2,1}. There is one way to partition {3,2,1} into two disjoint subsets of equal sum: {3}={2,1}. For the other partitions, this cannot be done.
a(11) = 2, because among the 89 strict partitions of 22 there is {7, 5, 4, 3, 2, 1}. There are two ways to partition {7, 5, 4, 3, 2, 1} into two disjoint subsets of equal sum: {7,4}={5,3,2,1} and {7,3,1}={5,4,2}. And this cannot be done in three ways for any strict partition of 22.
		

Crossrefs

Programs

  • Python
    def partitions_distinct(n):
        def _build(remaining, max_next):
            if remaining == 0:
                return [[]]
            res = []
            for k in range(min(remaining, max_next), 0, -1):
                for tail in _build(remaining - k, k - 1):
                    res.append([k] + tail)
            return res
        return _build(n, n//2) # The biggest number in the subset can't be bigger than n/2
    def count_half_subsets(partition, n):
        if n % 2:
            return 0
        half = n // 2
        dp = [0] * (half + 1)
        dp[0] = 1
        for x in partition:
            for s in range(half, x - 1, -1):
                dp[s] += dp[s - x]
        return int(dp[half]/2) #-> to not count {X}={Y} and {Y}={X} as two different solutions
    #---- Generate Sequence -----
    sequence = []
    max_n=25  #number of terms
    for N in range(1, max_n):
        parts = partitions_distinct(2*N)
        max_sols = 0
        for p in parts:
            subsets = count_half_subsets(p, 2*N)
            if subsets > max_sols:
                max_sols = subsets
        sequence.append(max_sols)

A387389 a(n) is the smallest positive integer for which there exists a strict partition that can be partitioned into two disjoint subsets with equal sum in n ways.

Original entry on oeis.org

6, 22, 32, 28, 40, 38, 36, 52, 50, 48, 46, 66, 64, 64, 62, 62, 60, 58, 56, 80, 80, 78, 78, 76, 78, 76, 74, 74, 72, 72, 70, 70, 68, 96, 66, 96, 94, 96, 92, 94, 92, 92, 90, 92, 90, 88, 88, 90, 86, 88, 86, 86, 84, 84, 84, 82, 82, 82, 80, 80, 110, 78, 112, 114
Offset: 1

Views

Author

Jesús Bellver Arnau, Aug 28 2025

Keywords

Comments

Finding ways in which a set can be partitioned into two disjoint subsets with equal sum is often referred to as the "partition search problem".
All the numbers in the sequence are even because for odd numbers there is no solution to the partition search problem.

Examples

			a(1) = 6, because S={3,2,1} is a strict partition of 6 and there is a way to partition S into two disjoint subsets of equal sum: {3}={2,1}. It is not possible to do this for any strict partition of integers smaller than 6.
a(2) = 22, because S={7, 5, 4, 3, 2, 1} is a strict partition of 22 and there are two ways to partition S into two disjoint subsets of equal sum: {7,4}={5,3,2,1} and {7,3,1}={5,4,2}. There are no strict partitions of any smaller number for which this can be done.
a(3) = 32, because S={11, 6, 5, 4, 3, 2, 1} is a strict partition of 32 and there are three ways to partition S into two disjoint subsets of equal sum: {11,5}={6,4,3,2,1}, {11,4,1}={6,5,3,2} and {11,3,2}={6,5,4,1}. There are no strict partitions of any smaller number for which this can be done.
		

Crossrefs

Programs

  • Python
    def partitions_distinct(n):
        def _build(remaining, max_next):
            if remaining == 0:
                return [[]]
            res = []
            for k in range(min(remaining, max_next), 0, -1):
                for tail in _build(remaining - k, k - 1):
                    res.append([k] + tail)
            return res
        return _build(n, n//2) # The biggest number in the subset can't be bigger than n/2
    def count_half_subsets(partition, n):
        if n % 2:
            return 0
        half = n // 2
        dp = [0] * (half + 1)
        dp[0] = 1
        for x in partition:
            for s in range(half, x - 1, -1):
                dp[s] += dp[s - x]
        return int(dp[half]/2) #-> to not count {X}={Y} and {Y}={X} as two different solutions
    #---- Generate Sequence -----
    max_n = 15 #number of terms
    sequence = []
    for n in range(1, max_n):
        p_N_exists = False
        N=1
        while p_N_exists==False:
            partes = partitions_distinct(2*N)
            for p in partes:
                subsets = count_half_subsets(p, 2*N)
                if subsets == n:
                    sequence.append(2*N)
                    p_N_exists = True
                    break
            N = N+1
Previous Showing 51-56 of 56 results.