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.

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A363622 Irregular triangle read by rows where T(n,k) is the number of integer partitions of n with weighted alternating sum k (leading and trailing 0's omitted).

Original entry on oeis.org

1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 2, 0, 0, 1, 0, 1, 1, 2, 0, 1, 1, 1, 1, 1, 3, 0, 0, 2, 0, 1, 1, 2, 1, 1, 3, 0, 2, 2, 1, 1, 2, 2, 1, 1, 5, 0, 0, 3, 0, 2, 2, 2, 1, 3, 2, 1, 1, 5, 0, 3, 3, 2, 2, 3, 2, 2, 4, 2, 1, 1, 7, 0, 0, 5, 0, 3, 3, 4, 2, 4, 2, 4, 4, 2, 1, 1
Offset: 0

Views

Author

Gus Wiseman, Jun 15 2023

Keywords

Comments

We define the weighted alternating sum of a sequence (y_1,...,y_k) to be Sum_{i=1..k} (-1)^(i-1) i * y_i. For example:
- (3,3,2,1,1) has weighted alternating sum 1*3 - 2*3 + 3*2 - 4*1 + 5*1 = 4.
- (1,2,2,3) has weighted alternating sum 1*1 - 2*2 + 3*2 - 4*3 = -9.

Examples

			Triangle begins:
  1
  1
  1  0  0  1
  1  0  1  1
  2  0  0  1  0  1  1
  2  0  1  1  1  1  1
  3  0  0  2  0  1  1  2  1  1
  3  0  2  2  1  1  2  2  1  1
  5  0  0  3  0  2  2  2  1  3  2  1  1
  5  0  3  3  2  2  3  2  2  4  2  1  1
  7  0  0  5  0  3  3  4  2  4  2  4  4  2  1  1
  7  0  5  5  3  3  5  4  3  5  3  5  4  2  1  1
Row n = 6 counts the following partitions:
  k=-3            k=0        k=2    k=3   k=4      k=5    k=6
  -----------------------------------------------------------
  (33)      .  .  (42)    .  (321)  (51)  (222)    (411)  (6)
  (2211)          (3111)                  (21111)
  (111111)
		

Crossrefs

Row sums are A000041.
The unweighted version is A103919 with leading zeros removed.
Row-lengths appear to be A168233.
Central column T(n,0) is A363532, ranks A363621.
The corresponding rank statistic is A363619, reverse A363620.
The reverse version is A363623.
A053632 counts compositions by weighted sum.
A264034 counts partitions by weighted sum, reverse A358194.
A316524 gives alternating sum of prime indices, reverse A344616.
A363624 gives weighted alternating sum of Heinz partition, reverse A363625.

Programs

  • Mathematica
    altwtsum[y_]:=Sum[(-1)^(k-1)*k*y[[k]],{k,1,Length[y]}];
    Table[Length[Select[IntegerPartitions[n],altwtsum[#]==k&]],{n,0,15},{k,Min[altwtsum/@IntegerPartitions[n]], Max[altwtsum/@IntegerPartitions[n]]}]

A363623 Irregular triangle read by rows where T(n,k) is the number of integer partitions of n with reverse-weighted alternating sum k (leading and trailing 0's omitted).

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 2, 0, 1, 2, 2, 1, 1, 1, 1, 1, 3, 1, 0, 3, 0, 1, 1, 1, 1, 3, 2, 0, 3, 1, 2, 0, 1, 0, 1, 2, 5, 1, 0, 3, 1, 2, 2, 2, 1, 1, 0, 1, 0, 1, 2, 5, 3, 0, 4, 2, 2, 0, 3, 2, 1, 3, 0, 0, 1, 0, 1, 1, 1, 1, 7, 2, 0, 4, 1, 5, 2, 3, 1, 3, 0, 2, 3, 1, 2, 1, 0, 0, 1, 0, 1, 1, 1, 1
Offset: 0

Views

Author

Gus Wiseman, Jun 15 2023

Keywords

Comments

We define the reverse-weighted alternating sum of a sequence (y_1,...,y_k) to be Sum_{i=1..k} (-1)^(k-i) i * y_{k-i+1}. For example:
- (3,3,2,1,1) has reverse-weighted alternating sum 1*1 - 2*1 + 3*2 - 4*3 + 5*3 = 8.
- (1,2,2,3) has reverse-weighted alternating sum -1*3 + 2*2 - 3*2 + 4*1 = -1.

Examples

			Triangle begins:
  1
  1
  1  1
  1  2
  2  0  1  2
  2  1  1  1  1  1
  3  1  0  3  0  1  1  1  1
  3  2  0  3  1  2  0  1  0  1  2
  5  1  0  3  1  2  2  2  1  1  0  1  0  1  2
  5  3  0  4  2  2  0  3  2  1  3  0  0  1  0  1  1  1  1
Row n = 6 counts the following partitions:
  k=3       k=4       k=6       k=8      k=9   k=10    k=11
--------------------------------------------------------------
  (33)      (222)  .  (6)    .  (21111)  (51)  (3111)  (411)
  (2211)              (42)
  (111111)            (321)
		

Crossrefs

Row sums are A000041.
Column k = floor((n+1)/2) is A119620.
The unweighted version is A344612 aerated, reverse A103919.
The corresponding rank statistic is A363620, reverse A363619.
The reverse version is A363622.
A053632 counts compositions by weighted sum.
A264034 counts partitions by weighted sum, reverse A358194.
A316524 gives alternating sum of prime indices, reverse A344616.
A363624 gives weighted alternating sum of Heinz partition, reverse A363625.

Programs

  • Mathematica
    revaltwtsum[y_]:=Sum[(-1)^(Length[y]-k)*k*y[[-k]],{k,1,Length[y]}];
    Table[Length[Select[IntegerPartitions[n],revaltwtsum[#]==k&]],{n,0,15},{k,Floor[(n+1)/2],Ceiling[n*(n+1)/4]}]

A363624 Weighted alternating sum of the integer partition with Heinz number n.

Original entry on oeis.org

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

Views

Author

Gus Wiseman, Jun 13 2023

Keywords

Comments

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.
We define the weighted alternating sum of a sequence (y_1,...,y_k) to be Sum_{i=1..k} (-1)^(i - 1) * i * y_i.

Examples

			The partition with Heinz number 600 is (3,3,2,1,1,1), with weighted alternating sum 1*3 - 2*3 + 3*2 - 4*1 + 5*1 - 6*1 = -2, so a(600) = -2.
		

Crossrefs

The non-alternating version is A318283, reverse A304818.
The unweighted version is A344616, reverse A316524.
For multisets instead of partitions we have A363619.
Positions of zeros are A363621, counted by A363532.
The triangle for this rank statistic is A363622, reverse A363623.
The reverse version is A363625, for multisets A363620.
A055396 gives minimum prime index, maximum A061395.
A112798 lists prime indices, length A001222, sum A056239.
A264034 counts partitions by weighted sum, reverse A358194.
A320387 counts multisets by weighted sum, reverse A007294.
A359677 gives zero-based weighted sum of prime indices, reverse A359674.
A363626 counts compositions with reverse-weighted alternating sum 0.

Programs

  • Mathematica
    prix[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n], {p_,k_}:>Table[PrimePi[p],{k}]]]];
    altwtsum[y_]:=Sum[(-1)^(k-1)*k*y[[k]],{k,1,Length[y]}];
    Table[altwtsum[Reverse[prix[n]]],{n,100}]

A363625 Reverse-weighted alternating sum of the integer partition with Heinz number n.

Original entry on oeis.org

0, 1, 2, 1, 3, 3, 4, 2, 2, 5, 5, 5, 6, 7, 4, 2, 7, 3, 8, 8, 6, 9, 9, 6, 3, 11, 4, 11, 10, 6, 11, 3, 8, 13, 5, 3, 12, 15, 10, 10, 13, 9, 14, 14, 7, 17, 15, 8, 4, 4, 12, 17, 16, 5, 7, 14, 14, 19, 17, 7, 18, 21, 10, 3, 9, 12, 19, 20, 16, 7, 20, 4, 21, 23, 5, 23
Offset: 1

Views

Author

Gus Wiseman, Jun 15 2023

Keywords

Comments

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.
We define the reverse-weighted alternating sum of a sequence (y_1,...,y_k) to be Sum_{i=1..k} (-1)^(k-i) * i * y_{k-i+1}.

Examples

			The partition with Heinz number 600 is (3,3,2,1,1,1), so a(600) = -1*1 + 2*1 - 3*1 + 4*2 - 5*3 + 6*3 = 9.
		

Crossrefs

The non-alternating version is A304818, reverse A318283.
The unweighted version is A316524, reverse A344616.
For multisets instead of partitions we have A363620.
The triangle for this rank statistic is A363623, reverse A363622.
The reverse version is A363624, for multisets A363619.
A055396 gives minimum prime index, maximum A061395.
A112798 lists prime indices, length A001222, sum A056239.
A264034 counts partitions by weighted sum, reverse A358194.
A320387 counts multisets by weighted sum, reverse A007294.

Programs

  • Mathematica
    prix[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
    revaltwtsum[y_]:=Sum[(-1)^(Length[y]-k)*k*y[[-k]],{k,1,Length[y]}];
    Table[revaltwtsum[Reverse[prix[n]]],{n,100}]

A320385 Number of partitions of n into distinct parts such that the successive differences of consecutive parts are decreasing, and first difference < first part.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 2, 2, 2, 2, 3, 2, 3, 4, 3, 3, 5, 3, 5, 6, 4, 5, 7, 5, 7, 8, 6, 7, 10, 8, 9, 11, 8, 11, 13, 9, 13, 15, 12, 14, 17, 13, 16, 20, 15, 18, 22, 18, 21, 25, 20, 23, 27, 23, 28, 30, 26, 30, 34, 30, 33, 38, 31, 38, 43, 36, 42, 46, 42, 47, 50, 45, 50, 58, 51, 55
Offset: 0

Views

Author

Seiichi Manyama, Oct 12 2018

Keywords

Examples

			There are a(29) = 10 such partitions of 29:
01: [29]
02: [10, 19]
03: [11, 18]
04: [12, 17]
05: [13, 16]
06: [14, 15]
07: [6, 10, 13]
08: [6, 11, 12]
09: [7, 10, 12]
10: [8, 10, 11]
There are a(30) = 8 such partitions of 30:
01: [30]
02: [11, 19]
03: [12, 18]
04: [13, 17]
05: [14, 16]
06: [6, 11, 13]
07: [7, 11, 12]
08: [4, 7, 9, 10]
		

Crossrefs

Programs

  • Ruby
    def partition(n, min, max)
      return [[]] if n == 0
      [max, n].min.downto(min).flat_map{|i| partition(n - i, min, i - 1).map{|rest| [i, *rest]}}
    end
    def f(n)
      return 1 if n == 0
      cnt = 0
      partition(n, 1, n).each{|ary|
        ary << 0
        ary0 = (1..ary.size - 1).map{|i| ary[i - 1] - ary[i]}
        cnt += 1 if ary0.sort == ary0 && ary0.uniq == ary0
      }
      cnt
    end
    def A320385(n)
      (0..n).map{|i| f(i)}
    end
    p A320385(50)

A359756 First position of n in the sequence of zero-based weighted sums of standard compositions (A124757), if we start with position 0.

Original entry on oeis.org

0, 3, 6, 7, 13, 14, 15, 27, 29, 30, 31, 55, 59, 61, 62, 63, 111, 119, 123, 125, 126
Offset: 0

Views

Author

Gus Wiseman, Jan 17 2023

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 zero-based weighted sum of a sequence (y_1,...,y_k) is Sum_{i=1..k} (i-1)*y_i.
Is this sequence strictly increasing?

Examples

			The terms together with their standard compositions begin:
    0: ()
    3: (1,1)
    6: (1,2)
    7: (1,1,1)
   13: (1,2,1)
   14: (1,1,2)
   15: (1,1,1,1)
   27: (1,2,1,1)
   29: (1,1,2,1)
   30: (1,1,1,2)
   31: (1,1,1,1,1)
		

Crossrefs

The one-based version is A089633, for prime indices A359682.
First index of n in A124757, reverse A231204.
The version for prime indices is A359676, reverse A359681.
A053632 counts compositions by zero-based weighted sum.
A066099 lists standard compositions.
A304818 gives weighted sums of prime indices, reverse A318283.
A320387 counts multisets by weighted sum, zero-based A359678.

Programs

  • Mathematica
    nn=10;
    stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    wts[y_]:=Sum[(i-1)*y[[i]],{i,Length[y]}];
    seq=Table[wts[stc[n]],{n,0,2^(nn-1)}];
    Table[Position[seq,k][[1,1]]-1,{k,0,nn}]

Formula

Appears to be the complement of A083329 in A089633.

A363531 Heinz numbers of integer partitions such that 3*(sum) = (reverse-weighted sum).

Original entry on oeis.org

1, 32, 144, 216, 243, 672, 1008, 1350, 2176, 2250, 2520, 2673, 3125, 3969, 4160, 4200, 5940, 6240, 6615, 7344, 7424, 7744, 8262, 9261, 9800, 9900, 10400, 11616, 12250, 12312, 12375, 13104, 13720, 14720, 14742, 16767, 16807, 17150, 19360, 21840, 22080, 23100
Offset: 1

Views

Author

Gus Wiseman, Jun 12 2023

Keywords

Comments

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.
The (one-based) weighted sum of a sequence (y_1,...,y_k) is Sum_{i=1..k} i*y_i. The reverse-weighted sum is the weighted sum of the reverse, also the sum of partial sums. For example, the weighted sum of (4,2,2,1) is 1*4 + 2*2 + 3*2 + 4*1 = 18 and the reverse-weighted sum is 4*4 + 3*2 + 2*2 + 1*1 = 27.

Examples

			The terms together with their prime indices begin:
      1: {}
     32: {1,1,1,1,1}
    144: {1,1,1,1,2,2}
    216: {1,1,1,2,2,2}
    243: {2,2,2,2,2}
    672: {1,1,1,1,1,2,4}
   1008: {1,1,1,1,2,2,4}
   1350: {1,2,2,2,3,3}
   2176: {1,1,1,1,1,1,1,7}
   2250: {1,2,2,3,3,3}
   2520: {1,1,1,2,2,3,4}
   2673: {2,2,2,2,2,5}
   3125: {3,3,3,3,3}
   3969: {2,2,2,2,4,4}
   4160: {1,1,1,1,1,1,3,6}
		

Crossrefs

These partitions are counted by A363526.
The non-reverse version is A363530, counted by A363527.
A053632 counts compositions by weighted sum.
A055396 gives minimum prime index, maximum A061395.
A112798 lists prime indices, length A001222, sum A056239.
A264034 counts partitions by weighted sum, reverse A358194.
A304818 gives weighted sum of prime indices, row-sums of A359361.
A318283 gives weighted sum of reversed prime indices, row-sums of A358136.
A320387 counts multisets by weighted sum, zero-based A359678.

Programs

  • Mathematica
    prix[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
    Select[Range[1000],3*Total[prix[#]]==Total[Accumulate[prix[#]]]&]

Formula

A056239(a(n)) = A318283(a(n))/3.

A363526 Number of integer partitions of n with reverse-weighted sum 3*n.

Original entry on oeis.org

1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 1, 2, 2, 2, 4, 3, 2, 4, 4, 4, 5, 5, 4, 7, 7, 5, 8, 7, 6, 11, 9, 8, 11, 10, 10, 13, 12, 11, 15, 15, 12, 17, 16, 14, 20, 18, 16, 22, 20, 19, 24, 22, 20, 27, 26, 23, 29, 27, 25, 33, 30, 28, 35, 33, 31, 38, 36, 33, 41, 40
Offset: 0

Views

Author

Gus Wiseman, Jun 10 2023

Keywords

Comments

Are the partitions counted all of length 4 or 5?
The (one-based) weighted sum of a sequence (y_1,...,y_k) is Sum_{i=1..k} i*y_i. The reverse-weighted sum is the weighted sum of the reverse, also the sum of partial sums. For example, the weighted sum of (4,2,2,1) is 1*4 + 2*2 + 3*2 + 4*1 = 18 and the reverse-weighted sum is 4*4 + 3*2 + 2*2 + 1*1 = 27.

Examples

			The partition (6,4,4,1) has sum 15 and reverse-weighted sum 45 so is counted under a(15).
The a(n) partitions for n = {5, 10, 15, 16, 21, 24}:
  (1,1,1,1,1)  (4,3,2,1)    (6,4,4,1)    (6,5,4,1)  (8,6,6,1)   (9,7,7,1)
               (2,2,2,2,2)  (6,5,2,2)    (6,6,2,2)  (8,7,4,2)   (9,8,5,2)
                            (7,3,3,2)    (7,4,3,2)  (9,5,5,2)   (9,9,3,3)
                            (3,3,3,3,3)             (9,6,3,3)   (10,6,6,2)
                                                    (10,4,4,3)  (10,7,4,3)
                                                                (11,5,5,3)
                                                                (12,4,4,4)
		

Crossrefs

Positions of terms with omega > 4 appear to be A079998.
The version for compositions is A231429.
The non-reverse version is A363527.
These partitions have ranks A363530, reverse A363531.
A000041 counts integer partitions, strict A000009.
A053632 counts compositions by weighted sum, rank statistic A029931/A359042.
A264034 counts partitions by weighted sum, reverse A358194.
A304818 gives weighted sum of prime indices, row-sums of A359361.
A318283 gives weighted sum of reversed prime indices, row-sums of A358136.
A320387 counts multisets by weighted sum, zero-based A359678.

Programs

  • Mathematica
    Table[Length[Select[IntegerPartitions[n],Total[Accumulate[#]]==3n&]],{n,0,30}]

A363527 Number of integer partitions of n with weighted sum 3*n.

Original entry on oeis.org

1, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 1, 3, 4, 4, 6, 8, 7, 10, 13, 13, 21, 25, 24, 37, 39, 40, 58, 63, 72, 94, 106, 118, 144, 165, 181, 224, 256, 277, 341, 387, 417, 504, 560, 615, 743, 818, 899, 1066, 1171, 1285, 1502, 1655, 1819, 2108, 2315, 2547, 2915
Offset: 0

Views

Author

Gus Wiseman, Jun 11 2023

Keywords

Comments

Are the partitions counted all of length > 4?
The (one-based) weighted sum of a sequence (y_1,...,y_k) is Sum_{i=1..k} i*y_i. The reverse-weighted sum is the weighted sum of the reverse, also the sum of partial sums. For example, the weighted sum of (4,2,2,1) is 1*4 + 2*2 + 3*2 + 4*1 = 18 and the reverse-weighted sum is 4*4 + 3*2 + 2*2 + 1*1 = 27.

Examples

			The partition (2,2,1,1,1,1) has sum 8 and weighted sum 24 so is counted under a(8).
The a(13) = 1 through a(18) = 8 partitions:
  (332221)  (333221)    (33333)     (442222)    (443222)    (443331)
            (4322111)   (522222)    (5322211)   (4433111)   (444222)
            (71111111)  (4332111)   (55111111)  (5332211)   (533322)
                        (63111111)  (63211111)  (55211111)  (4443111)
                                                (63311111)  (7222221)
                                                (72221111)  (55311111)
                                                            (64221111)
                                                            (A11111111)
		

Crossrefs

The version for compositions is A231429.
The reverse version is A363526.
These partitions have ranks A363531.
A000041 counts integer partitions, strict A000009.
A053632 counts compositions by weighted sum, rank statistic A029931/A359042.
A264034 counts partitions by weighted sum, reverse A358194.
A304818 gives weighted sum of prime indices, row-sums of A359361.
A318283 gives weighted sum of reversed prime indices, row-sums of A358136.
A320387 counts multisets by weighted sum, zero-based A359678.

Programs

  • Mathematica
    Table[Length[Select[IntegerPartitions[n],Total[Accumulate[Reverse[#]]]==3n&]],{n,0,30}]

A320388 Number of partitions of n into distinct parts such that the successive differences of consecutive parts are decreasing.

Original entry on oeis.org

1, 1, 1, 2, 2, 3, 3, 4, 5, 5, 6, 8, 7, 9, 11, 10, 12, 15, 14, 16, 19, 18, 21, 25, 23, 26, 31, 29, 33, 38, 36, 40, 46, 44, 49, 56, 53, 58, 66, 64, 70, 77, 76, 82, 92, 89, 96, 106, 104, 113, 123, 120, 130, 142, 141, 149, 162, 160, 172, 186, 184, 195, 211, 210, 223, 238
Offset: 0

Views

Author

Seiichi Manyama, Oct 12 2018

Keywords

Comments

Partitions are usually written with parts in descending order, but the conditions are easier to check "visually" if written in ascending order.
Partitions into distinct parts (p(1), p(2), ..., p(m)) such that p(k-1) - p(k-2) > p(k) - p(k-1) for all k >= 3.

Examples

			There are a(17) = 15 such partitions of 17:
  01: [17]
  02: [1, 16]
  03: [2, 15]
  04: [3, 14]
  05: [4, 13]
  06: [5, 12]
  07: [6, 11]
  08: [7, 10]
  09: [1, 6, 10]
  10: [8, 9]
  11: [1, 7, 9]
  12: [2, 6, 9]
  13: [2, 7, 8]
  14: [3, 6, 8]
  15: [4, 6, 7]
There are a(18) = 14 such partitions of 18:
  01: [18]
  02: [1, 17]
  03: [2, 16]
  04: [3, 15]
  05: [4, 14]
  06: [5, 13]
  07: [6, 12]
  08: [7, 11]
  09: [8, 10]
  10: [1, 7, 10]
  11: [1, 8, 9]
  12: [2, 7, 9]
  13: [3, 7, 8]
  14: [1, 4, 6, 7]
		

Crossrefs

Programs

  • Ruby
    def partition(n, min, max)
      return [[]] if n == 0
      [max, n].min.downto(min).flat_map{|i| partition(n - i, min, i - 1).map{|rest| [i, *rest]}}
    end
    def f(n)
      return 1 if n == 0
      cnt = 0
      partition(n, 1, n).each{|ary|
        ary0 = (1..ary.size - 1).map{|i| ary[i - 1] - ary[i]}
        cnt += 1 if ary0.sort == ary0 && ary0.uniq == ary0
      }
      cnt
    end
    def A320388(n)
      (0..n).map{|i| f(i)}
    end
    p A320388(50)
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