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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A357633 Half-alternating sum of the partition having Heinz number n.

Original entry on oeis.org

0, 1, 2, 2, 3, 3, 4, 1, 4, 4, 5, 2, 6, 5, 5, 0, 7, 3, 8, 3, 6, 6, 9, 1, 6, 7, 2, 4, 10, 4, 11, 1, 7, 8, 7, 2, 12, 9, 8, 2, 13, 5, 14, 5, 3, 10, 15, 2, 8, 5, 9, 6, 16, 1, 8, 3, 10, 11, 17, 3, 18, 12, 4, 2, 9, 6, 19, 7, 11, 6, 20, 3, 21, 13, 4, 8, 9, 7, 22, 3, 0
Offset: 1

Views

Author

Gus Wiseman, Oct 09 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 partition with Heinz number 525 is (4,3,3,2) so a(525) = 4 + 3 - 3 - 2 = 2.
		

Crossrefs

The original alternating sum is A316524, reverse A344616.
The version for standard compositions is A357622, non-reverse A357621.
The skew-alternating form is A357634, non-reverse A357630.
Positions of zeros are A000583, non-reverse A357631.
The reverse version is A357629.
These partitions are counted by A357637, skew A357638.
A056239 adds up prime indices, row sums of A112798.
A351005 = alternately equal and unequal partitions, compositions A357643.
A351006 = alternately unequal and equal partitions, compositions A357644.
A357641 counts comps w/ half-alt sum 0, partitions A357639, even 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]}];
    Table[halfats[Reverse[primeMS[n]]],{n,30}]

A053253 Coefficients of the '3rd-order' mock theta function omega(q).

Original entry on oeis.org

1, 2, 3, 4, 6, 8, 10, 14, 18, 22, 29, 36, 44, 56, 68, 82, 101, 122, 146, 176, 210, 248, 296, 350, 410, 484, 566, 660, 772, 896, 1038, 1204, 1391, 1602, 1846, 2120, 2428, 2784, 3182, 3628, 4138, 4708, 5347, 6072, 6880, 7784, 8804, 9940, 11208, 12630
Offset: 0

Views

Author

Dean Hickerson, Dec 19 1999

Keywords

Comments

Empirical: a(n) is the number of integer partitions mu of 2n+1 such that the diagram of mu has an odd number of cells in each row and in each column. - John M. Campbell, Apr 24 2020
From Gus Wiseman, Jun 26 2022: (Start)
By Campbell's conjecture above that a(n) is the number of partitions of 2n+1 with all odd parts and all odd conjugate parts, the a(0) = 1 through a(5) = 8 partitions are (B = 11):
(1) (3) (5) (7) (9) (B)
(111) (311) (511) (333) (533)
(11111) (31111) (711) (911)
(1111111) (51111) (33311)
(3111111) (71111)
(111111111) (5111111)
(311111111)
(11111111111)
These partitions are ranked by A352143. (End)

References

  • Srinivasa Ramanujan, The Lost Notebook and Other Unpublished Papers, Narosa Publishing House, New Delhi, 1988, pp. 15, 17, 31.

Crossrefs

Other '3rd-order' mock theta functions are at A000025, A053250, A053251, A053252, A053254, A053255, A261401.
Cf. A095913(n)=a(n-3).
Cf. A259094.
Conjectured to count the partitions ranked by A352143.
A069911 = strict partitions w/ all odd parts, ranked by A258116.
A078408 = partitions w/ all odd parts, ranked by A066208.
A117958 = partitions w/ all odd parts and multiplicities, ranked by A352142.

Programs

  • Mathematica
    Series[Sum[q^(2n(n+1))/Product[1-q^(2k+1), {k, 0, n}]^2, {n, 0, 6}], {q, 0, 100}]
  • PARI
    {a(n)=local(A); if(n<0, 0, A=1+x*O(x^n); polcoeff( sum(k=0, (sqrtint(2*n+1)-1)\2, A*=(x^(4*k)/(1-x^(2*k+1))^2 +x*O(x^(n-2*(k^2-k))))), n))} /* Michael Somos, Aug 18 2006 */
    
  • PARI
    {a(n)=local(A); if(n<0, 0, n++; A=1+x*O(x^n); polcoeff( sum(k=0, n-1, A*=(x/(1-x^(2*k+1)) +x*O(x^(n-k)))), n))} /* Michael Somos, Aug 18 2006 */

Formula

G.f.: omega(q) = Sum_{n>=0} q^(2*n*(n+1))/((1-q)*(1-q^3)*...*(1-q^(2*n+1)))^2.
G.f.: Sum_{k>=0} x^k/((1-x)(1-x^3)...(1-x^(2k+1))). - Michael Somos, Aug 18 2006
G.f.: (1 - G(0))/(1-x) where G(k) = 1 - 1/(1-x^(2*k+1))/(1-x/(x-1/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jan 18 2013
a(n) ~ exp(Pi*sqrt(n/3)) / (4*sqrt(n)). - Vaclav Kotesovec, Jun 10 2019
Conjectural g.f.: 1/(1 - x)*( 1 + Sum_{n >= 0} x^(3*n+1) /((1 - x)*(1 - x^3)*...*(1 - x^(2*n+1))) ). - Peter Bala, Nov 18 2024

A351007 Number of even-length integer partitions of n into parts that are alternately unequal and equal.

Original entry on oeis.org

1, 0, 0, 1, 1, 2, 2, 3, 4, 5, 5, 7, 8, 9, 10, 13, 14, 16, 18, 20, 23, 27, 28, 32, 37, 40, 44, 51, 54, 60, 67, 73, 81, 90, 96, 107, 118, 127, 139, 154, 166, 181, 198, 213, 232, 256, 273, 297, 325, 348, 377, 411, 440, 476, 516, 555, 598, 647, 692, 746, 807
Offset: 0

Views

Author

Gus Wiseman, Jan 31 2022

Keywords

Comments

These are partitions whose multiplicities begin with a 1, are followed by any number of 2's, and end with another 1.

Examples

			The a(3) = 1 through a(15) = 13 partitions (A..E = 10..14):
  21  31  32  42  43  53    54    64    65    75    76    86    87
          41  51  52  62    63    73    74    84    85    95    96
                  61  71    72    82    83    93    94    A4    A5
                      3221  81    91    92    A2    A3    B3    B4
                            4221  5221  A1    B1    B2    C2    C3
                                        4331  4332  C1    D1    D2
                                        6221  5331  5332  5441  E1
                                              7221  6331  6332  5442
                                                    8221  7331  6441
                                                          9221  7332
                                                                8331
                                                                A221
                                                                433221
		

Crossrefs

The alternately equal and unequal version is A035457, any length A351005.
This is the even-length case of A351006, odd-length A053251.
Without equalities we have A351008, any length A122129, opposite A122135.
Without inequalities we have A351012, any length A351003, opposite A351004.

Programs

  • Mathematica
    Table[Length[Select[IntegerPartitions[n],EvenQ[Length[#]]&&And@@Table[#[[i]]==#[[i+1]],{i,2,Length[#]-1,2}]&&And@@Table[#[[i]]!=#[[i+1]],{i,1,Length[#]-1,2}]&]],{n,0,30}]

A357636 Numbers k such that the skew-alternating sum of the partition having Heinz number k is 0.

Original entry on oeis.org

1, 4, 9, 12, 16, 25, 30, 36, 49, 63, 64, 70, 81, 90, 100, 108, 121, 144, 154, 165, 169, 192, 196, 210, 225, 256, 273, 286, 289, 300, 324, 325, 360, 361, 400, 441, 442, 462, 480, 484, 525, 529, 550, 561, 576, 588, 595, 625, 646, 676, 700, 729, 741, 750, 784
Offset: 1

Views

Author

Gus Wiseman, Oct 09 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 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: {}
    4: {1,1}
    9: {2,2}
   12: {1,1,2}
   16: {1,1,1,1}
   25: {3,3}
   30: {1,2,3}
   36: {1,1,2,2}
   49: {4,4}
   63: {2,2,4}
   64: {1,1,1,1,1,1}
   70: {1,3,4}
   81: {2,2,2,2}
   90: {1,2,2,3}
  100: {1,1,3,3}
  108: {1,1,2,2,2}
  121: {5,5}
  144: {1,1,1,1,2,2}
		

Crossrefs

The version for original alternating sum is A000290.
The half-alternating form is A000583, non-reverse A357631.
The version for standard compositions is A357628, non-reverse A357627.
The non-reverse version is A357632.
Positions of zeros in A357634, non-reverse A357630.
These partitions are counted by A357640, half A357639.
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 A357642.

Programs

  • Mathematica
    primeMS[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
    skats[f_]:=Sum[f[[i]]*(-1)^(1+Ceiling[(i+1)/2]),{i,Length[f]}];
    Select[Range[1000],skats[Reverse[primeMS[#]]]==0&]

A356232 Numbers whose prime indices are all odd and cover an initial interval of odd positive integers.

Original entry on oeis.org

1, 2, 4, 8, 10, 16, 20, 32, 40, 50, 64, 80, 100, 110, 128, 160, 200, 220, 250, 256, 320, 400, 440, 500, 512, 550, 640, 800, 880, 1000, 1024, 1100, 1210, 1250, 1280, 1600, 1760, 1870, 2000, 2048, 2200, 2420, 2500, 2560, 2750, 3200, 3520, 3740, 4000, 4096, 4400
Offset: 1

Views

Author

Gus Wiseman, Aug 20 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.
Also positions of first appearances of rows in A356226.

Examples

			The terms together with their prime indices begin:
      1: {}
      2: {1}
      4: {1,1}
      8: {1,1,1}
     10: {1,3}
     16: {1,1,1,1}
     20: {1,1,3}
     32: {1,1,1,1,1}
     40: {1,1,1,3}
     50: {1,3,3}
     64: {1,1,1,1,1,1}
     80: {1,1,1,1,3}
    100: {1,1,3,3}
    110: {1,3,5}
    128: {1,1,1,1,1,1,1}
    160: {1,1,1,1,1,3}
    200: {1,1,1,3,3}
    220: {1,1,3,5}
    250: {1,3,3,3}
    256: {1,1,1,1,1,1,1,1}
    320: {1,1,1,1,1,1,3}
    400: {1,1,1,1,3,3}
		

Crossrefs

The partitions with these Heinz numbers are counted by A053251.
This is the odd restriction of A055932.
A subset of A066208 (numbers with all odd prime indices).
This is the sorted version of A356603.
These are the positions of first appearances of rows in A356226. Other statistics are:
- length: A287170, firsts A066205
- minimum: A356227
- maximum: A356228
- bisected length: A356229
- standard composition: A356230
- Heinz number: A356231
- positions of first appearances: A356232 (this sequence)
A001221 counts distinct prime factors, with sum A001414.
A001223 lists the prime gaps, reduced A028334.
A003963 multiplies together the prime indices.
A056239 adds up the prime indices, row sums of A112798.
A073491 lists numbers with gapless prime indices, complement A073492.

Programs

  • Mathematica
    primeMS[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
    normQ[m_]:=Or[m=={},Union[m]==Range[Max[m]]];
    Select[Range[1000],normQ[(primeMS[#]+1)/2]&]

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

A053254 Coefficients of the '3rd-order' mock theta function nu(q).

Original entry on oeis.org

1, -1, 2, -2, 2, -3, 4, -4, 5, -6, 6, -8, 10, -10, 12, -14, 15, -18, 20, -22, 26, -29, 32, -36, 40, -44, 50, -56, 60, -68, 76, -82, 92, -101, 110, -122, 134, -146, 160, -176, 191, -210, 230, -248, 272, -296, 320, -350, 380, -410, 446, -484, 522, -566, 612, -660, 715, -772, 830, -896, 966, -1038
Offset: 0

Views

Author

Dean Hickerson, Dec 19 1999

Keywords

Comments

In Watson 1936 the function is denoted by upsilon(q). - Michael Somos, Jul 25 2015

Examples

			G.f. = 1 - x + 2*x^2 - 2*x^3 + 2*x^4 - 3*x^5 + 4*x^6 - 4*x^7 + 5*x^8 + ...
		

References

  • George E. Andrews, The Theory of Partitions, Addison-Wesley, 1976, (Example 6, p. 29).
  • Srinivasa Ramanujan, The Lost Notebook and Other Unpublished Papers, Narosa Publishing House, New Delhi, 1988, p. 31.

Crossrefs

Other '3rd-order' mock theta functions are at A000025, A053250, A053251, A053252, A053253, A053255.

Programs

  • Mathematica
    Series[Sum[q^(n(n+1))/Product[1+q^(2k+1), {k, 0, n}], {n, 0, 9}], {q, 0, 100}]
  • PARI
    /* Continued Fraction Expansion: */
    {a(n)=local(CF); CF=1+x; for(k=0, n, CF=1/(1 + x^(n-k+1)*(1 - x^(n-k+1))*CF+x*O(x^n))); polcoeff(CF, n)} \\ Paul D. Hanna, Jul 09 2013

Formula

G.f.: nu(q) = Sum_{n >= 0} q^(n*(n+1))/((1+q)*(1+q^3)*...*(1+q^(2*n+1)))
(-1)^n*a(n) = number of partitions of n in which even parts are distinct and if k occurs then so does every positive even number less than k.
G.f.: 1/(1 + x*(1-x)/(1 + x^2*(1-x^2)/(1 + x^3*(1-x^3)/(1 + x^4*(1-x^4)/(1 + x^5*(1-x^5)/(1 + ...)))))), a continued fraction. - Paul D. Hanna, Jul 09 2013
a(2*n) = A085140(n). a(2*n + 1) = - A053253(n). - Michael Somos, Jul 25 2015
a(n) ~ (-1)^n * exp(Pi*sqrt(n/6)) / (2^(3/2)*sqrt(n)). - Vaclav Kotesovec, Jun 15 2019
From Peter Bala, Jan 03 2025: (Start)
a(n) = (-1)^n * A067357(n).
nu(-q) = Sum_{n >= 0} q^n * (1 + q)*(1 + q^3)*...*(1 + q^(2*n-1)) (Andrews, p. 29: in Example 6 take x = q and y = -q).
Conjecture: a(n) = (-1)^n * (A376628(n) + A376628(n+1)), or equivalently, (1 + q * nu(-q))/(1 + q) = Sum_{n >= 0} q^(n*(n+1))/((1 - q)*(1 - q^3)*...*(1 - q^(2*n-1))), the g.f. of A376628. (End)

A357622 Half-alternating sum of the reversed n-th composition in standard order.

Original entry on oeis.org

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

Views

Author

Gus Wiseman, Oct 08 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 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 - 1 + 2 = 1.
		

Crossrefs

See link for sequences related to standard compositions.
This is the reverse version of A357621.
The skew-alternating form is A357624, non-reverse A357623.
Positions of zeros are A357626, reverse A357625.
The version for prime indices is A357629.
The version for Heinz numbers of partitions is A357633.
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;
    halfats[f_]:=Sum[f[[i]]*(-1)^(1+Ceiling[i/2]),{i,Length[f]}];
    Table[halfats[Reverse[stc[n]]],{n,0,100}]

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&]
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