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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A035544 Number of partitions of n with equal number of parts congruent to each of 1 and 3 (mod 4).

Original entry on oeis.org

1, 0, 1, 0, 3, 0, 4, 0, 10, 0, 13, 0, 28, 0, 37, 0, 72, 0, 96, 0, 172, 0, 230, 0, 391, 0, 522, 0, 846, 0, 1129, 0, 1766, 0, 2348, 0, 3564, 0, 4722, 0, 6992, 0, 9226, 0, 13371, 0, 17568, 0, 25006, 0, 32708, 0, 45817, 0, 59668, 0, 82430, 0, 106874, 0, 145830, 0, 188260, 0
Offset: 0

Views

Author

Keywords

Comments

From Gus Wiseman, Oct 12 2022: (Start)
Also the number of integer partitions of n whose skew-alternating sum is 0, where we define the skew-alternating sum of a sequence (A, B, C, D, E, F, G, ...) to be A - B - C + D + E - F - G + ... These are the conjugates of the partitions described in the name. For example, the a(0) = 1 through a(8) = 10 partitions are:
() . (11) . (22) . (33) . (44)
(211) (321) (422)
(1111) (2211) (431)
(111111) (2222)
(3221)
(3311)
(22211)
(221111)
(2111111)
(11111111)
The ordered version (compositions) is A138364
These partitions are ranked by A357636, reverse A357632.
The reverse version is A357640 (aerated).
(End)

Examples

			From _Gus Wiseman_, Oct 12 2022: (Start)
The a(0) = 1 through a(8) = 10 partitions:
  ()  .  (2)  .  (4)   .  (6)    .  (8)
                 (22)     (42)      (44)
                 (31)     (222)     (53)
                          (321)     (62)
                                    (71)
                                    (422)
                                    (431)
                                    (2222)
                                    (3221)
                                    (3311)
(End)
		

Crossrefs

The case with at least one odd part is A035550.
Removing zeros gives A035594.
Central column k=0 of A357638.
These partitions are ranked by A357707.
A000041 counts integer partitions.
A344651 counts partitions by alternating sum, ordered A097805.

Programs

  • Mathematica
    skats[f_]:=Sum[f[[i]]*(-1)^(1+Ceiling[(i+1)/2]),{i,Length[f]}];
    Table[Length[Select[IntegerPartitions[n],skats[#]==0&]],{n,0,30}] (* Gus Wiseman,Oct 12 2022 *)

Extensions

More terms from David W. Wilson

A357627 Numbers k such that the k-th composition in standard order has skew-alternating sum 0.

Original entry on oeis.org

0, 3, 10, 11, 15, 36, 37, 38, 43, 45, 54, 55, 58, 59, 63, 136, 137, 138, 140, 147, 149, 153, 166, 167, 170, 171, 175, 178, 179, 183, 190, 191, 204, 205, 206, 212, 213, 214, 219, 221, 228, 229, 230, 235, 237, 246, 247, 250, 251, 255, 528, 529, 530, 532, 536
Offset: 1

Views

Author

Gus Wiseman, Oct 08 2022

Keywords

Comments

We define the skew-alternating sum of a sequence (A, B, C, D, E, F, G, ...) to be A - B - C + D + E - F - G + ....
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 sequence together with the corresponding compositions begins:
    0: ()
    3: (1,1)
   10: (2,2)
   11: (2,1,1)
   15: (1,1,1,1)
   36: (3,3)
   37: (3,2,1)
   38: (3,1,2)
   43: (2,2,1,1)
   45: (2,1,2,1)
   54: (1,2,1,2)
   55: (1,2,1,1,1)
   58: (1,1,2,2)
   59: (1,1,2,1,1)
   63: (1,1,1,1,1,1)
		

Crossrefs

See link for sequences related to standard compositions.
The alternating form is A344619.
Positions of zeros in A357623.
The half-alternating form is A357625, reverse A357626.
The reverse version is A357628.
The version for prime indices is A357632.
The version for Heinz numbers of partitions is A357636.
A124754 gives alternating sum of standard compositions, reverse A344618.
A357637 counts partitions by half-alternating sum, skew A357638.
A357641 counts comps w/ half-alt sum 0, partitions A357639, even A357642.

Programs

  • Mathematica
    stc[n_]:=Differences[Prepend[Join @@ Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    skats[f_]:=Sum[f[[i]]*(-1)^(1+Ceiling[(i+1)/2]),{i,Length[f]}];
    Select[Range[0,100],skats[stc[#]]==0&]

A357628 Numbers k such that the reversed k-th composition in standard order has skew-alternating sum 0.

Original entry on oeis.org

0, 3, 10, 14, 15, 36, 43, 44, 45, 52, 54, 58, 59, 61, 63, 136, 147, 149, 152, 153, 166, 168, 170, 175, 178, 179, 181, 183, 185, 190, 200, 204, 211, 212, 213, 217, 219, 221, 228, 230, 234, 235, 237, 239, 242, 246, 247, 250, 254, 255, 528, 547, 549, 553, 560
Offset: 1

Views

Author

Gus Wiseman, Oct 08 2022

Keywords

Comments

We define the skew-alternating sum of a sequence (A, B, C, D, E, F, G, ...) to be A - B - C + D + E - F - G + ....
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 sequence together with the corresponding compositions begins:
    0: ()
    3: (1,1)
   10: (2,2)
   14: (1,1,2)
   15: (1,1,1,1)
   36: (3,3)
   43: (2,2,1,1)
   44: (2,1,3)
   45: (2,1,2,1)
   52: (1,2,3)
   54: (1,2,1,2)
   58: (1,1,2,2)
   59: (1,1,2,1,1)
   61: (1,1,1,2,1)
   63: (1,1,1,1,1,1)
		

Crossrefs

See link for sequences related to standard compositions.
The alternating form is A344619.
Positions of zeros are A357624, non-reverse A357623.
The half-alternating form is A357626, non-reverse A357625.
The non-reverse version is A357627.
The version for prime indices is A357632.
The version for Heinz numbers of partitions is A357636.
A124754 gives alternating sum of standard compositions, reverse A344618.
A357637 counts partitions by half-alternating sum, skew A357638.
A357641 counts comps w/ half-alt sum 0, partitions A357639, even A357642.

Programs

  • Mathematica
    stc[n_]:=Differences[Prepend[Join @@ Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    skats[f_]:=Sum[f[[i]]*(-1)^(1+Ceiling[(i+1)/2]),{i,Length[f]}];
    Select[Range[0,100],skats[Reverse[stc[#]]]==0&]

A357624 Skew-alternating sum of the reversed n-th composition in standard order.

Original entry on oeis.org

0, 1, 2, 0, 3, -1, 1, -1, 4, -2, 0, -2, 2, -2, 0, 0, 5, -3, -1, -3, 1, -3, -1, 1, 3, -3, -1, -1, 1, -1, 1, 1, 6, -4, -2, -4, 0, -4, -2, 2, 2, -4, -2, 0, 0, 0, 2, 2, 4, -4, -2, -2, 0, -2, 0, 2, 2, -2, 0, 0, 2, 0, 2, 0, 7, -5, -3, -5, -1, -5, -3, 3, 1, -5, -3, 1
Offset: 0

Views

Author

Gus Wiseman, Oct 08 2022

Keywords

Comments

We define the skew-alternating sum of a sequence (A, B, C, D, E, F, G, ...) to be A - B - C + D + E - F - G + ....
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 357-th composition is (2,1,3,2,1) so a(357) = 1 - 2 - 3 + 2 + 1 = -1.
The 358-th composition is (2,1,3,1,2) so a(358) = 2 - 1 - 3 + 1 + 2 = 1.
		

Crossrefs

See link for sequences related to standard compositions.
The half-alternating form is A357622, non-reverse A357621.
The reverse version is A357623.
Positions of zeros are A357628, non-reverse A357627.
The version for prime indices is A357630.
The version for Heinz numbers of partitions is A357634.
A124754 gives alternating sum of standard compositions, reverse A344618.
A357637 counts partitions by half-alternating sum, skew A357638.
A357641 counts comps w/ half-alt sum 0, partitions A357639, even A357642.

Programs

  • Mathematica
    stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    skats[f_]:=Sum[f[[i]]*(-1)^(1+Ceiling[(i+1)/2]),{i,Length[f]}];
    Table[skats[Reverse[stc[n]]],{n,0,100}]

A357635 Numbers k such that the half-alternating sum of the partition having Heinz number k is 1.

Original entry on oeis.org

2, 8, 24, 32, 54, 128, 135, 162, 375, 384, 512, 648, 864, 875, 1250, 1715, 1944, 2048, 2160, 2592, 3773, 4374, 4802, 5000, 6000, 6144, 8192, 9317, 10368, 10935, 13122, 13824, 14000, 15000, 17303, 19208, 20000, 24167, 27440, 29282, 30375, 31104, 32768, 33750
Offset: 1

Views

Author

Gus Wiseman, Oct 28 2022

Keywords

Comments

We define the half-alternating sum of a sequence (A, B, C, D, E, F, G, ...) to be A + B - C - D + E + F - G - ...
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:
    2: {1}
    8: {1,1,1}
   24: {1,1,1,2}
   32: {1,1,1,1,1}
   54: {1,2,2,2}
  128: {1,1,1,1,1,1,1}
  135: {2,2,2,3}
  162: {1,2,2,2,2}
  375: {2,3,3,3}
  384: {1,1,1,1,1,1,1,2}
  512: {1,1,1,1,1,1,1,1,1}
  648: {1,1,1,2,2,2,2}
  864: {1,1,1,1,1,2,2,2}
  875: {3,3,3,4}
		

Crossrefs

The version for k = 0 is A000583, standard compositions A357625-A357626.
The version for original alternating sum is A345958.
Positions of ones in A357633, non-reverse A357629.
The skew version for k = 0 is A357636, non-reverse A357632.
These partitions are counted by A035444, skew A035544.
The non-reverse version is A357851, k = 0 version A357631.
A056239 adds up prime indices, row sums of A112798.
A316524 gives alternating sum of prime indices, reverse A344616.
A351005 = alternately equal and unequal partitions, compositions A357643.
A351006 = alternately unequal and equal partitions, compositions A357644.
A357641 counts comps w/ half-alt sum 0, even-length A357642.

Programs

  • Mathematica
    primeMS[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
    halfats[f_]:=Sum[f[[i]]*(-1)^(1+Ceiling[i/2]),{i,Length[f]}];
    Select[Range[1000],halfats[Reverse[primeMS[#]]]==1&]

A357704 Triangle read by rows where T(n,k) is the number of reversed integer partitions of n with half-alternating sum k, where k ranges from -n to n in steps of 2.

Original entry on oeis.org

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

Views

Author

Gus Wiseman, Oct 10 2022

Keywords

Comments

We define the half-alternating sum of a sequence (A, B, C, D, E, F, G, ...) to be A + B - C - D + E + F - G - ...

Examples

			Triangle begins:
  1
  0  1
  0  0  2
  0  0  1  2
  0  0  2  0  3
  0  0  2  2  0  3
  0  0  3  1  3  0  4
  0  0  3  2  4  2  0  4
  0  0  4  2  6  2  3  0  5
  0  0  4  3  5  7  3  3  0  5
  0  0  5  3  8  4 10  2  4  0  6
  0  0  5  4  8  6 11  9  3  4  0  6
  0  0  6  4 11  5 15  8 13  3  5  0  7
  0  0  6  5 11  8 13 19 10 13  4  5  0  7
  0  0  7  5 14  8 19 13 25  9 17  4  6  0  8
  0  0  7  6 14 11 19 17 29 23 13 18  5  6  0  8
Row n = 7 counts the following reversed partitions:
  .  .  (115)   (124)   (133)      (11113)   .  (7)
        (1114)  (1222)  (223)      (111112)     (16)
        (1123)          (11122)                 (25)
                        (1111111)               (34)
		

Crossrefs

Row sums are A000041.
Last entry of row n is A008619(n).
The central column in the non-reverse case is A035363, skew A035544.
For original reverse-alternating sum we have A344612.
For original alternating sum we have A344651, ordered A097805.
The non-reverse version is A357637, skew A357638.
The central column is A357639, skew A357640.
The non-reverse ordered version (compositions) is A357645, skew A357646.
The skew-alternating version is A357705.
A351005 = alternately equal and unequal partitions, compositions A357643.
A351006 = alternately unequal and equal partitions, compositions A357644.
A357621 gives half-alternating sum of standard compositions, skew A357623.
A357629 gives half-alternating sum of prime indices, skew A357630.
A357633 gives half-alternating sum of Heinz partition, skew A357634.

Programs

  • Mathematica
    halfats[f_]:=Sum[f[[i]]*(-1)^(1+Ceiling[i/2]),{i,Length[f]}];
    Table[Length[Select[Reverse/@IntegerPartitions[n],halfats[#]==k&]],{n,0,15},{k,-n,n,2}]

A357705 Triangle read by rows where T(n,k) is the number of reversed integer partitions of n with skew-alternating sum k, where k ranges from -n to n in steps of 2.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 2, 0, 1, 0, 2, 2, 0, 1, 0, 3, 1, 2, 0, 1, 0, 3, 2, 3, 2, 0, 1, 0, 4, 2, 4, 1, 3, 0, 1, 0, 4, 3, 3, 6, 2, 3, 0, 1, 0, 5, 3, 5, 3, 7, 2, 4, 0, 1, 0, 5, 4, 5, 4, 9, 7, 3, 4, 0, 1, 0, 6, 4, 7, 3, 12, 5, 10, 3, 5, 0, 1
Offset: 0

Views

Author

Gus Wiseman, Oct 10 2022

Keywords

Comments

We define the skew-alternating sum of a sequence (A, B, C, D, E, F, G, ...) to be A - B - C + D + E - F - G + ...

Examples

			Triangle begins:
  1
  0  1
  0  1  1
  0  2  0  1
  0  2  2  0  1
  0  3  1  2  0  1
  0  3  2  3  2  0  1
  0  4  2  4  1  3  0  1
  0  4  3  3  6  2  3  0  1
  0  5  3  5  3  7  2  4  0  1
  0  5  4  5  4  9  7  3  4  0  1
  0  6  4  7  3 12  5 10  3  5  0  1
  0  6  5  7  5 10 16  7 11  4  5  0  1
  0  7  5  9  5 14 11 18  7 14  4  6  0  1
Row n = 7 counts the following reversed partitions:
  .  (16)   (25)   (34)       (1123)  (1114)   .  (7)
     (115)  (223)  (1222)             (11113)
     (124)         (111112)           (11122)
     (133)         (1111111)
		

Crossrefs

Row sums are A000041.
First nonzero entry of each row is A004526.
The central column is A357640, half A357639.
For original alternating sum we have A344651, ordered A097805.
The half-alternating version is A357704.
The ordered non-reverse version (compositions) is A357646, half A357645.
The non-reverse version is A357638, half A357637.
A351005 = alternately equal and unequal partitions, compositions A357643.
A351006 = alternately unequal and equal partitions, compositions A357644.
A357621 gives half-alternating sum of standard compositions, skew A357623.
A357629 gives half-alternating sum of prime indices, skew A357630.
A357633 gives half-alternating sum of Heinz partition, skew A357634.

Programs

  • Mathematica
    skats[f_]:=Sum[f[[i]]*(-1)^(1+Ceiling[(i+1)/2]),{i,Length[f]}];
    Table[Length[Select[Reverse/@IntegerPartitions[n],skats[#]==k&]],{n,0,11},{k,-n,n,2}]

A228248 Number of 2n-step lattice paths from (0,0) to (0,0) using steps in {N, S, E, W} starting with East, then always moving straight ahead or turning left.

Original entry on oeis.org

1, 0, 1, 3, 9, 30, 103, 357, 1257, 4494, 16246, 59246, 217719, 805389, 2996113, 11200113, 42047593, 158452138, 599113966, 2272065638, 8639763574, 32933685102, 125817012366, 481631387438, 1847110931703, 7095928565405, 27302745922817, 105204285608025
Offset: 0

Views

Author

David Scambler and Alois P. Heinz, Aug 18 2013

Keywords

Comments

From Gus Wiseman, Oct 13 2022: (Start)
Also the number of integer compositions of 2n whose half-alternating and skew-alternating sums are both 0, where we define the half-alternating sum of a sequence (A, B, C, D, E, F, G, ...) to be A + B - C - D + E + F - G - ..., and the skew-alternating sum to be A - B - C + D + E - F - G + ... For example, the a(0) = 1 through a(4) = 9 compositions are:
() . (1111) (1212) (1313)
(2121) (2222)
(11211) (3131)
(11312)
(12221)
(21311)
(112211)
(1112111)
(11111111)
For skew-alternating only: A001700, ranked by A357627, reverse A357628.
For partitions: A035363, half only A357639, skew only A357640.
For half-alternating only: A357641, ranked by A357625, reverse A357626.
These compositions are ranked by A357706.
(End)

Examples

			a(0) = 1: [], the empty path.
a(1) = 0.
a(2) = 1: ENWS.
a(3) = 3: EENWWS, ENNWSS, ENWWSE.
		

Crossrefs

Cf. A004006 (same rules, but self-avoiding).

Programs

  • Maple
    b:= proc(x, y, n) option remember; `if`(abs(x)+abs(y)>n, 0,
          `if`(n=0, 1, b(x+1, y, n-1) +b(y+1, -x, n-1)))
        end:
    a:= n-> ceil(b(0, 0, 2*n)/2):
    seq(a(n), n=0..40);
    # second Maple program:
    a:= proc(n) option remember; `if`(n<5, [1, 0, 1, 3, 9][n+1],
         ((n-1)*(414288-1901580*n+186029*n^6-869551*n^5+2393807*n^4
         -3938624*n^3+3753546*n^2+1050*n^8-21605*n^7)*a(n-1)
         +(-17751540*n-12215020*n^5+3494038*n^6+3777840+27478070*n^4
         -39711374*n^3+35488098*n^2-2700*n^9+62370*n^8-621126*n^7)*a(n-2)
         +(-18193248+77490792*n-9138800*n^6+35323128*n^5-88122332*n^4
         +141370392*n^3-140075264*n^2+5400*n^9-135540*n^8+1476432*n^7)*a(n-3)
         +(-192473328*n+48577536+17091500*n^6-70036368*n^5+184890672*n^4
         -313388816*n^3+328043052*n^2-8400*n^9+224440*n^8-2600032*n^7)*a(n-4)
         +8*(n-5)*(150*n^6-2015*n^5+10852*n^4-29867*n^3+44208*n^2-33540*n
         +10416)*(-9+2*n)^2*a(n-5)) / (n^2*(396988*n-487261*n^2+150*n^7
         -3065*n^6+26092*n^5-119602*n^4+317746*n^3-131048)))
        end:
    seq(a(n), n=0..40);
  • Mathematica
    b[x_, y_, n_] := b[x, y, n] = If[Abs[x] + Abs[y] > n, 0, If[n == 0, 1, b[x + 1, y, n - 1] + b[y + 1, -x, n - 1]]];
    a[n_] := Ceiling[b[0, 0, 2n]/2];
    a /@ Range[0, 40] (* Jean-François Alcover, May 14 2020, after Maple *)
    halfats[f_]:=Sum[f[[i]]*(-1)^(1+Ceiling[i/2]),{i,Length[f]}];
    skats[f_]:=Sum[f[[i]]*(-1)^(1+Ceiling[(i+1)/2]),{i,Length[f]}];
    Table[Length[Select[Join@@Permutations/@IntegerPartitions[2n],halfats[#]==0&&skats[#]==0&]],{n,0,7}] (* Gus Wiseman, Oct 12 2022 *)

Formula

a(n) ~ 2^(2n-1)/(Pi*n). - Vaclav Kotesovec, Jul 16 2014

A357706 Numbers k such that the k-th composition in standard order has half-alternating sum and skew-alternating sum both 0.

Original entry on oeis.org

0, 15, 45, 54, 59, 153, 170, 179, 204, 213, 230, 235, 247, 255, 561, 594, 611, 660, 677, 710, 715, 727, 735, 750, 765, 792, 809, 842, 851, 871, 879, 894, 908, 917, 934, 939, 951, 959, 973, 982, 987, 1005, 1014, 1019
Offset: 1

Views

Author

Gus Wiseman, Oct 13 2022

Keywords

Comments

We define the half-alternating sum of a sequence (A, B, C, D, E, F, G, ...) to be A + B - C - D + E + F - G - ..., and the skew-alternating sum to be A - B - C + D + E - F - G + ...
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

For partitions and half only (or both): A000583, counted by A035363.
These compositions are counted by A228248.
For half-alternating only: A357625, reverse A357626, counted by A357641.
For skew-alternating only: A357627, reverse A357628, counted by A001700.
For reversed partitions and half only: A357631, counted by A357639.
For reversed partitions and skew only A357632, counted by A357640.
For partitions and skew only: A357636, counted by A035594.

Programs

  • Mathematica
    stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    halfats[f_]:=Sum[f[[i]]*(-1)^(1+Ceiling[i/2]),{i,Length[f]}];
    skats[f_]:=Sum[f[[i]]*(-1)^(1+Ceiling[(i+1)/2]),{i,Length[f]}];
    Select[Range[0,1000],halfats[stc[#]]==0&&skats[stc[#]]==0&]

Formula

Intersection of A357625 and A357627.

A357707 Numbers whose prime indices have equal number of parts congruent to each of 1 and 3 (mod 4).

Original entry on oeis.org

1, 3, 7, 9, 10, 13, 19, 21, 27, 29, 30, 34, 37, 39, 43, 49, 53, 55, 57, 61, 62, 63, 70, 71, 79, 81, 87, 89, 90, 91, 94, 100, 101, 102, 107, 111, 113, 115, 117, 129, 130, 131, 133, 134, 139, 147, 151, 159, 163, 165, 166, 169, 171, 173, 181, 183, 186, 187, 189
Offset: 1

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Author

Gus Wiseman, Oct 12 2022

Keywords

Comments

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.

Examples

			The terms together with their prime indices begin:
     1: {}
     3: {2}
     7: {4}
     9: {2,2}
    10: {1,3}
    13: {6}
    19: {8}
    21: {2,4}
    27: {2,2,2}
    29: {10}
    30: {1,2,3}
		

Crossrefs

These partitions are counted by A035544.
Includes A066207 = products of primes of even index.
The conjugate partitions are ranked by A357636, reverse A357632.
The conjugate reverse version is A357640 (aerated).
A056239 adds up prime indices, row sums of A112798.
A316524 gives alternating sum of prime indices, reverse A344616.
A344651 counts partitions by alternating sum, ordered A097805.
A357705 counts reversed partitions by skew-alternating sum, half A357704.

Programs

  • Mathematica
    primeMS[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
    Select[Range[100],Count[primeMS[#],?(Mod[#,4]==1&)]==Count[primeMS[#],?(Mod[#,4]==3&)]&]
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