A344619 -id:A344619 - cp's OEIS frontend

A039004 Numbers whose base-4 representation has the same number of 1's and 2's.

0, 3, 6, 9, 12, 15, 18, 24, 27, 30, 33, 36, 39, 45, 48, 51, 54, 57, 60, 63, 66, 72, 75, 78, 90, 96, 99, 102, 105, 108, 111, 114, 120, 123, 126, 129, 132, 135, 141, 144, 147, 150, 153, 156, 159, 165, 177, 180, 183, 189, 192, 195, 198, 201, 204, 207, 210, 216, 219

Offset: 1

Olivier Gérard

Numbers such that sum (-1)^k*b(k) = 0 where b(k)=k-th binary digit of n (see A065359). - Benoit Cloitre, Nov 18 2003
Conjecture: a(C(2n,n)-1) = 4^n - 1. (A000984 is C(2n,n)). - Gerald McGarvey, Nov 18 2007
From Russell Jay Hendel, Jun 23 2015: (Start)
We prove the McGarvey conjecture (A) a(e(n,n)-1) = 4^n-1, with e(n,m) = A034870(n,m) = binomial(2n,m), the even rows of Pascal's triangle. By the comment from Hendel in A034870, we have the function s(n,k) = #{n-digit, base-4 numbers with n-k more 1-digits than 2-digits}. As shown in A034870, (B) #s(n,k)= e(n,k) with # indicating cardinality, that is, e(n,k) = binomial(2n,k) gives the number of n-digit, base-4 numbers with n-k more 1-digits than 2-digits.
We now show that (B) implies (A). By definition, s(n,n) contains the e(n,n) = binomial(2n,n) numbers with an equal number of 1-digits and 2-digits. The biggest n-digit, base-4 number is 333...3 (n copies of 3). Since 333...33 has zero 1-digits and zero 2-digits it follows that 333...333 is a member of s(n,n) and hence it is the biggest member of s(n,n). But 333...333 (n copies of 3) in base 4 has value 4^n-1. Since A039004 starts with index 0 (that is, 0 is the 0th member of A039004), it immediately follows that 4^n-1 is the (e(n,n)-1)st member of A039004, proving the McGarvey conjecture. (End)
Also numbers whose alternating sum of binary expansion is 0, i.e., positions of zeros in A345927. These are numbers whose binary expansion has the same number of 1's at even positions as at odd positions. - Gus Wiseman, Jul 28 2021

Alois P. Heinz, Table of n, a(n) for n = 1..10000

Cf. A139370, A139371, A139372, A139373.
Cf. A139351, A139352, A139353, A139354, A139355.
A subset of A001969 (evil numbers).
A base-2 version is A031443 (digitally balanced numbers).
Positions of 0's in A065359 and A345927.
Positions of first appearances are A086893.
The version for standard compositions is A344619.
A000120 and A080791 count binary digits, with difference A145037.
A003714 lists numbers with no successive binary indices.
A011782 counts compositions.
A030190 gives the binary expansion of each nonnegative integer.
A070939 gives the length of an integer's binary expansion.
A097805 counts compositions by alternating (or reverse-alternating) sum.
A101211 lists run-lengths in binary expansion:
- row-lengths: A069010
- reverse: A227736
- ones only: A245563
A138364 counts compositions with alternating sum 0:
- bisection: A001700/A088218
- complement: A058622
A328594 lists numbers whose binary expansion is aperiodic.
A345197 counts compositions by length and alternating sum.
Cf. A000302, A000346, A003242, A029837, A053738, A053754, A071321, A103919, A191232, A328596.

Fortran
```
c See link in A139351.
```

N:= 1000: # to get all terms up to N, which should be divisible by 4
B:= Array(0..N-1):
d:= ceil(log[4](N));
S:= Array(0..N-1,[seq(op([0,1,-1,0]),i=1..N/4)]):
for i from 1 to d do
  B:= B + S;
  S:= Array(0..N-1,i-> S[floor(i/4)]);
od:
select(t -> B[t]=0, [$0..N-1]); # Robert Israel, Jun 24 2015

ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}];
Select[Range[0,100],ats[IntegerDigits[#,2]]==0&] (* Gus Wiseman, Jul 28 2021 *)

for(n=0,219,if(sum(i=1,length(binary(n)),(-1)^i*component(binary(n),i))==0,print1(n,",")))

Conjecture: there is a constant c around 5 such that a(n) is asymptotic to c*n. - Benoit Cloitre, Nov 24 2002

That conjecture is false. The number of members of the sequence from 0 to 4^d-1 is binomial(2d,d) which by Stirling's formula is asymptotic to 4^d/sqrt(Pi*d). If Cloitre's conjecture were true we would have 4^d-1 asymptotic to c*4^d/sqrt(Pi*d), a contradiction. - Robert Israel, Jun 24 2015

A357623 Skew-alternating sum of the n-th composition in standard order.

Original entry on oeis.org

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

Offset: 0

Gus Wiseman, Oct 08 2022

sign

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.

			The 358-th composition is (2,1,3,1,2) so a(358) = 2 - 1 - 3 + 1 + 2 = 1.

Gus Wiseman, Statistics, classes, and transformations of standard compositions

See link for sequences related to standard compositions.
Positions of positive firsts appear to be A029744.
The half-alternating form is A357621, reverse A357622.
The reverse version is A357624.
Positions of zeros are A357627, reverse A357628.
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.
Cf. A001700, A001511, A053251, A344619, A357136, A357182, A357183, A357184, A357185, A357625, A357626.

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[stc[n]],{n,0,100}]

A357625 Numbers k such that the k-th composition in standard order has half-alternating sum 0.

Original entry on oeis.org

0, 14, 15, 44, 45, 46, 52, 53, 54, 59, 61, 152, 153, 154, 156, 168, 169, 170, 172, 179, 181, 185, 200, 201, 202, 204, 211, 213, 217, 230, 231, 234, 235, 239, 242, 243, 247, 254, 255, 560, 561, 562, 564, 568, 592, 593, 594, 596, 600, 611, 613, 617, 625, 656

Offset: 1

Gus Wiseman, Oct 08 2022

nonn

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.

			The sequence together with the corresponding compositions begins:
    0: ()
   14: (1,1,2)
   15: (1,1,1,1)
   44: (2,1,3)
   45: (2,1,2,1)
   46: (2,1,1,2)
   52: (1,2,3)
   53: (1,2,2,1)
   54: (1,2,1,2)
   59: (1,1,2,1,1)
   61: (1,1,1,2,1)

John Tyler Rascoe, Table of n, a(n) for n = 1..7762
Gus Wiseman, Statistics, classes, and transformations of standard compositions

See link for sequences related to standard compositions.
The version for full alternating sum is A344619.
Positions of zeros in A357621.
The reverse version is A357626.
The skew-alternating form is A357627, reverse A357628.
The version for prime indices is A357631.
The version for Heinz numbers of partitions is A357635.
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.
Cf. A001511, A053251, A357136, A357182, A357183, A357184, A357185, A357622, A357623, A357629, A357633.

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]}];
Select[Range[0,100],halfats[stc[#]]==0&]

from itertools import count, islice
def comp(n): #row n of A066099 after Franklin T. Adams-Watters
    v,k = [],0
    while n > 0:
        k += 1
        if n%2 == 1:
            v.append(k)
            k = 0
        n = n//2
    return(v[::-1])
def a_gen():
    for n in count(0):
        c = comp(n)
        x = sum(c[i]*(-1)**(i//2) for i in range(len(c)))
        if x == 0:
            yield(n)
A357625_list = list(islice(a_gen(), 60)) # John Tyler Rascoe, Jun 01 2024

A357642 Number of even-length integer compositions of 2n whose half-alternating sum is 0.

Original entry on oeis.org

1, 0, 1, 4, 13, 48, 186, 712, 2717, 10432, 40222, 155384, 601426, 2332640, 9063380, 35269392, 137438685, 536257280, 2094786870, 8191506136, 32063203590, 125613386912, 492516592620, 1932569186288, 7588478653938, 29816630378368, 117226929901676, 461151757861552

Offset: 0

Gus Wiseman, Oct 12 2022

nonn

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 a(0) = 1 through a(4) = 13 compositions:
  ()  .  (1111)  (1212)  (1313)
                 (1221)  (1322)
                 (2112)  (1331)
                 (2121)  (2213)
                         (2222)
                         (2231)
                         (3113)
                         (3122)
                         (3131)
                         (111311)
                         (112211)
                         (113111)
                         (11111111)

David A. Corneth, Table of n, a(n) for n = 0..1665

The skew-alternating version appears to be A000984.
For original alternating sum we have A001700/A088218.
The version for partitions of any length is A357639, ranked by A357631.
For length multiple of 4 we have A110145.
These compositions of any length are ranked by A357625, reverse A357626.
A124754 gives alternating sum of standard compositions, reverse A344618.
A357621 = half-alternating sum of standard compositions, reverse A357622.
A357637 counts partitions by half-alternating sum, skew A357638.
Cf. A000583, A001511, A035363, A053251, A344619, A357136, A357182, A357627, A357628, A357629, A357633.

Table[Length[Select[Join @@ Permutations/@IntegerPartitions[2n],EvenQ[Length[#]]&&halfats[#]==0&]],{n,0,9}]

a(n) = {my(v, res); if(n < 3, return(1 - bitand(n,1))); res = 0; v = vector(2*n, i, binomial(n-1,i-1)); forstep(i = 4, 2*n, 2, lp = i\4 * 2; rp = i - lp; res += v[lp] * v[rp]; ); res } \\ David A. Corneth, Oct 13 2022

More terms from Alois P. Heinz, Oct 12 2022

A357626 Numbers k such that the reversed k-th composition in standard order has half-alternating sum 0.

Original entry on oeis.org

0, 11, 15, 37, 38, 45, 46, 53, 54, 55, 59, 137, 138, 140, 153, 154, 156, 167, 169, 170, 171, 172, 179, 191, 201, 202, 204, 205, 206, 213, 214, 229, 230, 231, 235, 243, 247, 251, 255, 529, 530, 532, 536, 561, 562, 564, 568, 583, 587, 593, 594, 595, 596, 600

Offset: 1

Gus Wiseman, Oct 08 2022

nonn

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.

			The sequence together with the corresponding compositions begins:
    0: ()
   11: (2,1,1)
   15: (1,1,1,1)
   37: (3,2,1)
   38: (3,1,2)
   45: (2,1,2,1)
   46: (2,1,1,2)
   53: (1,2,2,1)
   54: (1,2,1,2)
   55: (1,2,1,1,1)
   59: (1,1,2,1,1)

Gus Wiseman, Statistics, classes, and transformations of standard compositions

See link for sequences related to standard compositions.
The alternating form is A344619.
Positions of zeros in A357622.
The non-reverse version is A357625.
The skew-alternating form is A357628, reverse A357627.
The version for prime indices is A357631.
The version for Heinz numbers of partitions is A357635.
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.
Cf. A001511, A053251, A357136, A357182, A357183, A357184, A357185, A357621, A357623, A357629, A357633.

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]}];
Select[Range[0,100],halfats[Reverse[stc[#]]]==0&]

A345908 Traces of the matrices (A345197) counting integer compositions by length and alternating sum.

Original entry on oeis.org

1, 1, 0, 1, 3, 3, 6, 15, 24, 43, 92, 171, 315, 629, 1218, 2313, 4523, 8835, 17076, 33299, 65169

Offset: 0

Gus Wiseman, Jul 26 2021

The matrices (A345197) count the integer compositions of n of length k with alternating sum i, where 1 <= k <= n, and i ranges from -n + 2 to n in steps of 2. Here, the alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i. So a(n) is the number of compositions of n of length (n + s)/2, where s is the alternating sum of the composition.

			The a(0) = 1 through a(7) = 15 compositions of n = 0..7 of length (n + s)/2 where s = alternating sum (empty column indicated by dot):
  ()  (1)  .  (2,1)  (2,2)    (2,3)    (2,4)      (2,5)
                     (1,1,2)  (1,2,2)  (1,3,2)    (1,4,2)
                     (2,1,1)  (2,2,1)  (2,3,1)    (2,4,1)
                                       (1,1,3,1)  (1,1,3,2)
                                       (2,1,2,1)  (1,2,3,1)
                                       (3,1,1,1)  (2,1,2,2)
                                                  (2,2,2,1)
                                                  (3,1,1,2)
                                                  (3,2,1,1)
                                                  (1,1,1,1,3)
                                                  (1,1,2,1,2)
                                                  (1,1,3,1,1)
                                                  (2,1,1,1,2)
                                                  (2,1,2,1,1)
                                                  (3,1,1,1,1)

Traces of the matrices given by A345197.
Diagonals and antidiagonals of the same matrices are A346632 and A345907.
Row sums of A346632.
A011782 counts compositions.
A097805 counts compositions by alternating (or reverse-alternating) sum.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A316524 gives the alternating sum of prime indices (reverse: A344616).
Other diagonals are A008277 of A318393 and A055884 of A320808.
Compositions of n, 2n, or 2n+1 with alternating/reverse-alternating sum k:
- k = 0:  counted by A088218, ranked by A344619/A344619.
- k = 1:  counted by A000984, ranked by A345909/A345911.
- k = -1: counted by A001791, ranked by A345910/A345912.
- k = 2:  counted by A088218, ranked by A345925/A345922.
- k = -2: counted by A002054, ranked by A345924/A345923.
- k >= 0: counted by A116406, ranked by A345913/A345914.
- k <= 0: counted by A058622(n-1), ranked by A345915/A345916.
- k > 0:  counted by A027306, ranked by A345917/A345918.
- k < 0:  counted by A294175, ranked by A345919/A345920.
- k != 0: counted by A058622, ranked by A345921/A345921.
- k even: counted by A081294, ranked by A053754/A053754.
- k odd:  counted by A000302, ranked by A053738/A053738.
Cf. A000070, A000346, A001405, A007318, A008549, A025047, A114121, A163493.

ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}];
Table[Length[Select[Join@@Permutations/@IntegerPartitions[n],Length[#]==(n+ats[#])/2&]],{n,0,15}]

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

Gus Wiseman, Oct 08 2022

nonn

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.

			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)

Gus Wiseman, Statistics, classes, and transformations of standard compositions

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.
Cf. A001700, A001511, A053251, A357136, A357182-A357185, A357621, A357624, A357630, A357634, A357640.

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

Gus Wiseman, Oct 08 2022

nonn

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.

			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)

Gus Wiseman, Statistics, classes, and transformations of standard compositions

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.
Cf. A001700, A001511, A053251, A357136, A357182, A357183, A357184, A357185, A357622, A357635, A357640.

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&]

A348614 Numbers k such that the k-th composition in standard order has sum equal to twice its alternating sum.

Original entry on oeis.org

0, 9, 11, 14, 130, 133, 135, 138, 141, 143, 148, 153, 155, 158, 168, 177, 179, 182, 188, 208, 225, 227, 230, 236, 248, 2052, 2057, 2059, 2062, 2066, 2069, 2071, 2074, 2077, 2079, 2084, 2089, 2091, 2094, 2098, 2101, 2103, 2106, 2109, 2111, 2120, 2129, 2131

Offset: 1

Gus Wiseman, Oct 29 2021

nonn

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 alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i.

			The terms together with their binary indices begin:
    0: ()
    9: (3,1)
   11: (2,1,1)
   14: (1,1,2)
  130: (6,2)
  133: (5,2,1)
  135: (5,1,1,1)
  138: (4,2,2)
  141: (4,1,2,1)
  143: (4,1,1,1,1)
  148: (3,2,3)
  153: (3,1,3,1)
  155: (3,1,2,1,1)
  158: (3,1,1,1,2)

Gus Wiseman, Statistics, classes, and transformations of standard compositions

The unordered case (partitions) is counted by A000712, reverse A006330.
These compositions are counted by A262977.
Except for 0, a subset of A345917 (which is itself a subset of A345913).
A000346 = even-length compositions with alt sum != 0, complement A001700.
A011782 counts compositions.
A025047 counts wiggly compositions, ranked by A345167.
A034871 counts compositions of 2n with alternating sum 2k.
A097805 counts compositions by alternating (or reverse-alternating) sum.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A116406 counts compositions with alternating sum >=0, ranked by A345913.
A138364 counts compositions with alternating sum 0, ranked by A344619.
A345197 counts compositions by length and alternating sum.
Cf. A000984, A002458, A004331, A005810, A224274.
Cf. A008549, A013777, A027306, A058622, A088218, A114121, A120452, A126869, A163493, A294175, A344604.

ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}];
stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
Select[Range[0,1000],Total[stc[#]]==2*ats[stc[#]]&]

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

Gus Wiseman, Oct 08 2022

sign

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.

			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.

Gus Wiseman, Statistics, classes, and transformations of standard compositions

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.
Cf. A001700, A001511, A053251, A344619, A357136, A357182, A357183, A357184, A357185, A357625, A357626, A357629, A357640.

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}]

cp's OEIS Frontend

A039004 Numbers whose base-4 representation has the same number of 1's and 2's.

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A357623 Skew-alternating sum of the n-th composition in standard order.

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A357625 Numbers k such that the k-th composition in standard order has half-alternating sum 0.

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A357642 Number of even-length integer compositions of 2n whose half-alternating sum is 0.

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A357626 Numbers k such that the reversed k-th composition in standard order has half-alternating sum 0.

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A345908 Traces of the matrices (A345197) counting integer compositions by length and alternating sum.

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

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

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