A048896 -id:A048896 - cp's OEIS frontend

A358135 Difference of first and last parts of the n-th composition in standard order.

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

Offset: 1

Gus Wiseman, Oct 31 2022

sign

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.

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

See link for sequences related to standard compositions.
The first and last parts are A065120 and A001511.
This is the first minus last part of row n of A066099.
The version for Heinz numbers of partitions is A243055.
Row sums of A358133.
The partial sums of standard compositions are A358134, adjusted A242628.
A011782 counts compositions.
A333766 and A333768 give max and min in standard compositions, diff A358138.
A351014 counts distinct runs in standard compositions.
Cf. A000120, A001511, A029837, A029931, A048896, A058891, A070939, A133494, A329395, A357187.

stc[n_]:=Differences[Prepend[Join @@ Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
Table[-First[stc[n]]+Last[stc[n]],{n,1,100}]

a(n) = A001511(n) - A065120(n).

A160468 Triangle of polynomial coefficients related to the o.g.f.s of the RES1 polynomials.

Original entry on oeis.org

1, 1, 2, 1, 17, 26, 2, 62, 192, 60, 1, 1382, 7192, 5097, 502, 2, 21844, 171511, 217186, 55196, 2036, 2, 929569, 10262046, 20376780, 9893440, 1089330, 16356, 4, 6404582, 94582204, 271154544, 215114420, 48673180, 2567568, 16376, 1

Offset: 1

Johannes W. Meijer, May 24 2009

In A160464 we defined the ES1 matrix by ES1[2*m-1,n=1] and in A094665 it was shown that the n-th term of the coefficients of matrix row ES1[1-2*m,n] for m >= 1 can be generated with the RES1(1-2*m,n) polynomials.
We define the o.g.f.s. of these polynomials by GFRES1(z,1-2*m) = sum(RES1(1-2*m,n)*z^(n-1), n=1..infinity) for m >= 1. The general expression of the o.g.f.s. is GFRES1(z,1-2*m) = (-1)*RE(z,1-2*m)/(2*p(m-1)*(z-1)^(m)). The p(m-1), m >= 1, sequence is Gould's sequence A001316.
The coefficients of the RE(z,1-2*m) polynomials lead to the triangle given above.
The E(z,n) = numer(sum((-1)^(n+1)*k^n*z^(k-1), k=1..infinity)) polynomials with n >= 1, see the Maple algorithm, lead to the Eulerian numbers A008292.
Some of our results are conjectures based on numerical evidence.

			The first few rows are:
[1]
[1]
[2, 1]
[17, 26, 2]
[62, 192, 60, 1]
The first few polynomials RE(z,m) are:
RE(z,-1) = 1
RE(z,-3) = 1
RE(z,-5) = 2+z
RE(z,-7) = 17+26*z+2*z^2
The first few GFRES1(z,m) are:
GFRES1(z,-1) = -(1/1)*(1)/(2*(z-1)^1)
GFRES1(z,-3) = -(1/2)*(1)/(2*(z-1)^2)
GFRES1(z,-5) = -(1/2)*(2+z)/(2*(z-1)^3)
GFRES1(z,-7) = -(1/4)*(17+26*z+2*z^2)/(2*(z-1)^4)

Grzegorz Rzadkowski, M Urlinska, A Generalization of the Eulerian Numbers, arXiv preprint arXiv:1612.06635, 2016

Cf. A160464, A094665 and A083061.
For the Eulerian numbers E(n, k) see A008292.
The p(n) sequence equals Gould's sequence A001316.
The first right hand column of the triangle equals A048896.
The first left hand column equals A160469.
The row sums equal the absolute values of A117972.

nmax := 8; mmax := nmax: T(0, x) := 1: for i from 1 to nmax do dgr := degree(T(i-1, x), x): for na from 0 to dgr do c(na) := coeff(T(i-1, x), x, na) od: T(i-1, x+1) := 0: for nb from 0 to dgr do T(i-1, x+1) := T(i-1, x+1) + c(nb)*(x+1)^nb od: for nc from 0 to dgr do ECGP(i-1, nc+1) := coeff(T(i-1, x), x, nc) od: T(i, x) := expand((2*x+1)*(x+1)*T(i-1, x+1) - 2*x^2*T(i-1, x)) od: dgr := degree (T(nmax, x), x): kmax := nmax: for k from 1 to kmax do p := k: for m from 1 to k do E(m, k) := sum((-1)^(m-q)*(q^k)*binomial(k+1, m-q), q=1..m) od: fx(p) := (-1)^(p+1) * (sum(E(r, k)*z^(k-r), r=1..k))/(z-1)^(p+1): GF(-(2*p+1)) := sort(simplify(((-1)^p* 1/2^(p+1)) * sum(ECGP(k-1, k-s)*fx(k-s), s=0..k-1)), ascending): NUMGF(-(2*p+1)) := -numer(GF(-(2*p+1))): for n from 1 to mmax+1 do A(k+1, n) := coeff(NUMGF(-(2*p+1)), z, n-1) od: od: for m from 2 to mmax do A(1, m) := 0 od: A(1, 1) := 1: FT(1) := 1: for n from 1 to nmax do for m from 1 to n do FT((n)*(n-1)/2+m+1) := A(n+1, m) end do end do: a := n-> FT(n): seq(a(n), n = 1..(nmax+1)*(nmax)/2+1);

T[ n_, k_] := Coefficient[a[2 n]/2^IntegerExponent[(2 n)!, 2], x, n + k];
a[0] = a[1] = 1; a[ m_] := a[m] = With[{n = m - 1}, x Sum[ a[k] a[n - k] Binomial[n, k], {k, 0, n}]]; Join[{1}, Flatten@Table[T[n, k], {n, 1, 8}, {k, 0, n - 1}]] (* Michael Somos, Apr 22 2020 *)

Edited by Johannes W. Meijer, Sep 23 2012

A261363 Triangle read by rows: partial row sums of Sierpinski's triangle.

Original entry on oeis.org

1, 1, 2, 1, 1, 2, 1, 2, 3, 4, 1, 1, 1, 1, 2, 1, 2, 2, 2, 3, 4, 1, 1, 2, 2, 3, 3, 4, 1, 2, 3, 4, 5, 6, 7, 8, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 2, 2, 2, 2, 2, 2, 3, 4, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 4, 1, 2, 3, 4, 4, 4, 4, 4, 5, 6, 7, 8, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4

Offset: 0

Reinhard Zumkeller, Aug 16 2015

T(n,n) = number of distinct terms in row n = number of odd terms in row n+1 = A001316(n).

Central terms, for n > 0: T(2*n,n) = A048896(n-1).

			.   n |  Sierpinski: A047999(n,*)  |  Partial row sums: T(n,*)
. ----+----------------------------+----------------------------
.   0 |              1             |              1
.   1 |             1 1            |             1 2
.   2 |            1 0 1           |            1 1 2
.   3 |           1 1 1 1          |           1 2 3 4
.   4 |          1 0 0 0 1         |          1 1 1 1 2
.   5 |         1 1 0 0 1 1        |         1 2 2 2 3 4
.   6 |        1 0 1 0 1 0 1       |        1 1 2 2 3 3 4
.   7 |       1 1 1 1 1 1 1 1      |       1 2 3 4 5 6 7 8
.   8 |      1 0 0 0 0 0 0 0 1     |      1 1 1 1 1 1 1 1 2
.   9 |     1 1 0 0 0 0 0 0 1 1    |     1 2 2 2 2 2 2 2 3 4
.  10 |    1 0 1 0 0 0 0 0 1 0 1   |    1 1 2 2 2 2 2 2 3 3 4
.  11 |   1 1 1 1 0 0 0 0 1 1 1 1  |   1 2 3 4 4 4 4 4 5 6 7 8
.  12 |  1 0 0 0 1 0 0 0 1 0 0 0 1 |  1 1 1 1 2 2 2 2 3 3 3 3 4  .

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened

Cf. A047999, A008949, A048896 (central terms), A001316 (right edge), A261366.

a261363 n k = a261363_tabl !! n !! k
a261363_row n = a261363_tabl !! n
a261363_tabl = map (scanl1 (+)) a047999_tabl

row[n_] := Accumulate[Array[Boole[0 == BitAnd[n-#, #]] &, n + 1, 0]]; Array[row, 13, 0] // Flatten (* Amiram Eldar, May 13 2025 *)

A357187 First differences A357186 = "Take the k-th composition in standard order for each part k of the n-th composition in standard order, then add up everything.".

Original entry on oeis.org

1, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, -1, 1, 0, 0, -1, 1, 0, 0, 0, 1, 0, 0, -1, 0, 1, 0, -1, 1, 0, 0, -2, 1, 1, 0, -1, 1, 0, 0, 0, 0, 1, 0, -1, 1, 0, 0, -2, 1, 0, 0, 0, 1, 0, 0, -1, 0, 1, 0, -1, 1, 0, 0, -3, 1, 1, 0, 0, 1, 0, 0, -1, 0, 1, 0, -1, 1, 0, 0, -1, 1, 0

Offset: 0

Gus Wiseman, Sep 28 2022

sign

Are there any terms > 1?

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.

			We have A357186(5) - A357186(4) = 3 - 2 = 1, so a(4) = 1.

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

See link for sequences related to standard compositions.
Positions of first appearances appear to all belong to A052955.
Differences of A357186 (row-sums of A357135).
The version for partitions is A357458, differences of A325033.
Cf. A000120, A029837, A029931, A048896, A058891, A070939, A133494, A333766, A357134, A357137.

stc[n_]:=Differences[Prepend[Join @@ Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
Differences[Table[stc/@stc[n]/.List->Plus,{n,0,100}]]

a(n) = A357186(n + 1) - A357186(n).

A358133 Triangle read by rows whose n-th row lists the first differences of the n-th composition in standard order (row n of A066099).

Original entry on oeis.org

0, -1, 1, 0, 0, -2, 0, -1, 0, 2, 1, -1, 0, 1, 0, 0, 0, -3, -1, -2, 0, 1, 0, -1, -1, 1, -1, 0, 0, 3, 2, -2, 1, 0, 1, -1, 0, 0, 2, 0, 1, -1, 0, 0, 1, 0, 0, 0, 0, -4, -2, -3, 0, 0, -1, -1, -2, 1, -2, 0, 0, 2, 1, -2, 0, 0, 0, -1, 0, -1, 2, -1, 1, -1, -1, 0, 1, -1

Offset: 3

Gus Wiseman, Oct 31 2022

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.

			Triangle begins (dots indicate empty rows):
   1:   .
   2:   .
   3:   0
   4:   .
   5:  -1
   6:   1
   7:   0  0
   8:   .
   9:  -2
  10:   0
  11:  -1  0
  12:   2
  13:   1 -1
  14:   0  1
  15:   0  0  0

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

See link for sequences related to standard compositions.
First differences of rows of A066099.
The version for Heinz numbers of partitions is A355536, ranked by A253566.
The partial sums instead of first differences are A358134.
Row sums are A358135.
A011782 counts compositions.
A351014 counts distinct runs in standard compositions.
Cf. A000120, A001511, A029837, A029931, A048896, A070939, A133494, A242628, A357135, A357187.

stc[n_]:=Differences[Prepend[Join @@ Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
Table[Differences[stc[n]],{n,100}]

A151780 a(n) = 5^(wt(n) - 1) where wt(n) = A000120(n).

Original entry on oeis.org

1, 1, 5, 1, 5, 5, 25, 1, 5, 5, 25, 5, 25, 25, 125, 1, 5, 5, 25, 5, 25, 25, 125, 5, 25, 25, 125, 25, 125, 125, 625, 1, 5, 5, 25, 5, 25, 25, 125, 5, 25, 25, 125, 25, 125, 125, 625, 5, 25, 25, 125, 25, 125, 125, 625, 25, 125, 125, 625, 125, 625, 625, 3125, 1, 5, 5, 25, 5, 25, 25, 125, 5

Offset: 1

N. J. A. Sloane, Jun 25 2009

nonn

			From _Omar E. Pol_, Jul 21 2009: (Start)
If written as a triangle:
  1;
  1,5;
  1,5,5,25;
  1,5,5,25,5,25,25,125;
  1,5,5,25,5,25,25,125,5,25,25,125,25,125,125,625;
  1,5,5,25,5,25,25,125,5,25,25,125,25,125,125,625,5,25,25,125,25,125,125,625,...
(End)

Essentially A151779/6.
Cf. A000120, A048881, A048896, A147610, A151783.
Cf. A000351. - Omar E. Pol, Jul 21 2009

a(n) = 5^(hammingweight(n)-1); \\ Michel Marcus, Nov 15 2022

A368225 Irregular table of nonnegative integers read by rows: the 1's in the binary expansion of n exactly match the nonzero digits in the balanced ternary expansions of the terms in the n-th row.

Original entry on oeis.org

0, 1, 3, 2, 4, 9, 8, 10, 6, 12, 5, 7, 11, 13, 27, 26, 28, 24, 30, 23, 25, 29, 31, 18, 36, 17, 19, 35, 37, 15, 21, 33, 39, 14, 16, 20, 22, 32, 34, 38, 40, 81, 80, 82, 78, 84, 77, 79, 83, 85, 72, 90, 71, 73, 89, 91, 69, 75, 87, 93, 68, 70, 74, 76, 86, 88, 92, 94

Offset: 0

Rémy Sigrist, Dec 18 2023

As a flat sequence, this is a permutation of the nonnegative integers with inverse A368226 and infinitely many fixed points (see Formula section).
Row 0 has one term, and for n > 0, row n has A048896(n-1) terms.
For any n >= 0, row n ends with A005836(n+1).

			Table T(n, k) begins:
    0;
    1;
    3;
    2, 4;
    9;
    8, 10;
    6, 12;
    5, 7, 11, 13;
    27;
    26, 28;
    24, 30;
    23, 25, 29, 31;
    18, 36;
    17, 19, 35, 37;
    15, 21, 33, 39;
    14, 16, 20, 22, 32, 34, 38, 40;
    81;
    ...

Rémy Sigrist, Table of n, a(n) for n = 0..9841 (rows for n = 0..2^9-1 flattened)
Index entries for sequences that are permutations of the natural numbers

See A368229 and A368239 for similar sequences.

Cf. A003462, A005836, A048896, A343231, A353662, A368226 (inverse).

row(n) = { my (r = [sign(n)], b = binary(n)); for (k = 2, #b, r = [3*v+b[k]|v<-r]; if (b[k], r = concat(r, [v-2|v<-r]););); Set(r); }

A343231(T(n, k)) = n.

a(m) = m for any m in A003462.

A382291 a(n) = A037445(n)/A034444(n).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1

Offset: 1

Amiram Eldar, Mar 21 2025

First differs from A368168 at n = 64, and from A359411, A367516 and A368979 at n = 128.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A000120, A034444, A037445, A138302, A243036, A382290, A382292.

Cf. A359411, A367516, A368168, A368979.

f[p_, e_] := 2^(DigitCount[e, 2, 1] - 1); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 105]

a(n) = vecprod(apply(x -> 1 << (hammingweight(x)-1), factor(n)[, 2]));

a(n) = 2^A382290(n).
Multiplicative with a(p^e) = 2^(A000120(e)-1) = A048896(e-1) (= A243036(e) for e >= 2).
a(n) >= 1, with equality if and only if n is in A138302.
a(n) = 2 if and only if n is in A382292.

A151783 a(n) = 4^(wt(n) - 1) where wt(n) = A000120(n).

Original entry on oeis.org

1, 1, 4, 1, 4, 4, 16, 1, 4, 4, 16, 4, 16, 16, 64, 1, 4, 4, 16, 4, 16, 16, 64, 4, 16, 16, 64, 16, 64, 64, 256, 1, 4, 4, 16, 4, 16, 16, 64, 4, 16, 16, 64, 16, 64, 64, 256, 4, 16, 16, 64, 16, 64, 64, 256, 16, 64, 64, 256, 64, 256, 256, 1024, 1, 4, 4, 16, 4, 16, 16, 64, 4, 16, 16, 64, 16, 64

Offset: 1

N. J. A. Sloane, Jun 25 2009

nonn

			From _Omar E. Pol_, Jul 21 2009: (Start)
If written as a triangle:
  1;
  1,4;
  1,4,4,16;
  1,4,4,16,4,16,16,64;
  1,4,4,16,4,16,16,64,4,16,16,64,16,64,64,256;
  1,4,4,16,4,16,16,64,4,16,16,64,16,64,64,256,4,16,16,64,16,64,64,256,16,64,...
(End)

Cf. A000120, A048881, A048896, A147610, A151780.
Cf. A000302, A102376. [Omar E. Pol, Jul 21 2009]
A102376 is a very similar sequence.

A245195 a(n) = 2^A014081(n).

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 2, 4, 1, 1, 1, 2, 2, 2, 4, 8, 1, 1, 1, 2, 1, 1, 2, 4, 2, 2, 2, 4, 4, 4, 8, 16, 1, 1, 1, 2, 1, 1, 2, 4, 1, 1, 1, 2, 2, 2, 4, 8, 2, 2, 2, 4, 2, 2, 4, 8, 4, 4, 4, 8, 8, 8, 16, 32, 1, 1, 1, 2, 1, 1, 2, 4, 1, 1, 1, 2, 2, 2, 4, 8, 1, 1, 1, 2, 1, 1, 2, 4, 2, 2, 2, 4, 4, 4, 8, 16, 2, 2, 2, 4, 2

Offset: 0

N. J. A. Sloane, Jul 24 2014

This sequence provides a bridge between A245180 (and, presumably, A160239) and A014081.
See A245196 for more about this class of sequences.
Run length transform of A011782: 1,1,2,4,8,16,32,64,... - Chai Wah Wu, Oct 19 2016

Chai Wah Wu and Robert Israel, Table of n, a(n) for n = 0..10000
Robert Israel, Proof that A277560 is the same as A245195 [This will be modified to reflect the fact that the two sequences have now been merged]
Chai Wah Wu, Sums of products of binomial coefficients mod 2 and run length transforms of sequences, arXiv:1610.06166 [math.CO], 2016.

Cf. A014081, A245180, A160239, A048896, A245196, A005725, A005940, A011782, A106737, A162510.

# This Maple program applies more generally to a sequence where the recurrence across a block is as follows. The parameters to be set are the sequence G(0), G(1), G(2), ... (the final terms in the blocks), and the multiplier m.
# For n in the range 2^(k-1) <= n < 2^k, write n = 2^k-2^r+j, with 0 <= r <= k-1 and 0 <= j < 2^(r-1), and j=0 if r=0. Then
# (if j=0) a(2^k-2^r) = G(k-r-1),
# (if j>0) a(2^k-2^r+j) = m*G(k-r-1)*a(j).
# Since Maple gives its lists an offset of 1, it is necessary to add 1 to the arguments of G.
# For the present sequence, G(n)=2^n and m=1.
G:=[seq(2^n,n=0..30)];
m:=1;
f:=proc(n) option remember; global m,G; local k,r,j,np;
if n <= 2 then G[0+1] elif n=3 then G[1+1]
elif n=4 then G[0+1] elif n=5 then m*G[0+1] elif n=6 then G[1+1] elif n=7 then G[2+1]
else
   k:=1+floor(log[2](n)); np:=2^k-n;
   if np=1 then r:=0; j:=0; else r:=1+floor(log[2](np-1)); j:=2^r-np; fi;
   if j=0 then G[k-r-1+1]; else m*G[k-r-1+1]*f(j); fi;
fi;
end;
[seq(f(n),n=1..520)]:
# Setting G(n) = A083424(n) and m = 8 gives A245180. Setting G(n) = 2^n and m = 2 gives A048896.
A245195:=n->add(binomial(n,2*k)*binomial(n,k) mod 2, k=0..floor(n/2)): seq(A245195(n), n=0..200); # Wesley Ivan Hurt, Nov 01 2016

Table[Sum[Mod[Binomial[n, 2 k] Binomial[n, k], 2], {k, 0, n}], {n, 0, 85}] (* Michael De Vlieger, Oct 21 2016 *)

a(n) = 2^hammingweight(bitand(n, n>>1)) \\ Charles R Greathouse IV, Jul 16 2016

a(n) = sum(k=0, n, binomial(n, 2*k)*binomial(n,k) % 2); \\ Michel Marcus, Oct 21 2016

from _future_ import division
def A277560(n):
    return sum(int(not (~n & 2*k) | (~n & k)) for k in range(n//2+1))

def A245195(n): return 1<<(n&(n>>1)).bit_count() # Chai Wah Wu, Feb 11 2023

The entries may be arranged into blocks of sizes 1,2,4,8,...:
B_0: 1,
B_1: 1, 2,
B_2: 1, 1, 2, 4,
B_3: 1, 1, 1, 2, 2, 2, 4, 8,
B_4: 1, 1, 1, 2, 1, 1, 2, 4, 2, 2, 2, 4, 4, 4, 8, 16,
B_5: 1, 1, 1, 2, 1, 1, 2, 4, 1, 1, 1, 2, 2, 2, 4, 8, 2, 2, 2, 4, 2, 2, 4, 8, 4, 4, 4, 8, 8, 8, 16, 32,
...
Consider the block B_{k-1} containing terms a(2^(k-1)), a(2^(k-1)+1), ..., a(2^k-1). It is convenient to index the terms working backwards from the next, 2^k-th, term. For n in the range 2^(k-1) <= n < 2^k, write n = 2^k-2^r+j, with 0 <= r <= k-1 and 0 <= j < 2^(r-1), and j=0 if r=0. Then
(if j=0) a(2^k-2^r) = 2^(k-r-1),
(if j>0) a(2^k-2^r+j) = 2^(k-r-1)*a(j).
a(n) = A162510(A005940(1+n)). - Antti Karttunen, Oct 29 2016
From Robert Israel, Nov 02 2016: (Start)
a(2*k)   = a(k).
a(4*k+1) = a(k).
a(4*k+3) = 2*a(2*k+1).
G.f. g(x) satisfies g(x) = x + (2*x+1)*g(x^2) - x*g(x^4). (End)
Also, a(n) = Sum_{k=0..floor(n/2)} ((binomial(n,2k)*binomial(n,k)) mod 2). - Chai Wah Wu, Oct 19 2016 and Robert Israel, Nov 04 2016. For proof, see the article by Chai Wah Wu, Sums of products of binomial coefficients mod 2 and run length transforms of sequences, arXiv:1610.06166, or the Robert Israel link.

Changed offset to 0, merged former entry A277560 from Chai Wah Wu (Oct 19 2016) with this sequence. - N. J. A. Sloane, Nov 05 2016

cp's OEIS Frontend

A358135 Difference of first and last parts of the n-th composition in standard order.

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A160468 Triangle of polynomial coefficients related to the o.g.f.s of the RES1 polynomials.

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Maple

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A261363 Triangle read by rows: partial row sums of Sierpinski's triangle.

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Haskell

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A357187 First differences A357186 = "Take the k-th composition in standard order for each part k of the n-th composition in standard order, then add up everything.".

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A358133 Triangle read by rows whose n-th row lists the first differences of the n-th composition in standard order (row n of A066099).

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Mathematica

A151780 a(n) = 5^(wt(n) - 1) where wt(n) = A000120(n).

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PARI

A368225 Irregular table of nonnegative integers read by rows: the 1's in the binary expansion of n exactly match the nonzero digits in the balanced ternary expansions of the terms in the n-th row.

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A382291 a(n) = A037445(n)/A034444(n).

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