A048896 -id:A048896 - cp's OEIS frontend

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.

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

Offset: 0

Gus Wiseman, Sep 28 2022

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.

			Composition 92 in standard order is (2,1,1,3), with compositions ((2),(1),(1),(1,1)) so a(92) = 2 + 1 + 1 + 1 + 1 = 6.

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

See link for sequences related to standard compositions.
This is the sum of A029837 over the n-th composition in standard order.
Vertex degrees are A133494.
The version for Heinz numbers of partitions is A325033.
Row sums of A357135.
First differences are A357187.
Cf. A000120, A001511, A003963, A029931, A048896, A058891, A070939, A096111, A329395, A333766, A335404, A357137.

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

a(n) = A029837(A357134(n)).

A101692 A modular binomial sum transform of 2^n.

Original entry on oeis.org

1, 1, 5, 1, 5, 17, 85, 1, 5, 17, 85, 257, 1285, 4369, 21845, 1, 5, 17, 85, 257, 1285, 4369, 21845, 65537, 327685, 1114129, 5570645, 16843009, 84215045, 286331153, 1431655765, 1, 5, 17, 85, 257, 1285, 4369, 21845, 65537, 327685, 1114129, 5570645, 16843009, 84215045

Offset: 0

Paul Barry, Dec 11 2004

a(2^n) is 1, 5, 5, 5, 5, ...
a(2^n+1) is 5, 1, 17, 17, 17, ...
a(2*(2^n+1)) is 5, 85, 85, 85, ...
a(2^n)*a(2^n+1) is 5, 5, 85, 85, 85, ...
Also, decimal representation of the x-axis, from the origin to the right edge, of the n-th stage of growth of the two-dimensional cellular automaton defined by "Rule 4", based on the 5-celled von Neumann neighborhood. Initialized with a single black (ON) cell at stage zero. - Robert Price, Nov 03 2016

S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 170.

Stefano Spezia, Table of n, a(n) for n = 0..1500
Robert Price, Diagrams of first 20 stages of the cellular automaton
N. J. A. Sloane, On the Number of ON Cells in Cellular Automata, arXiv:1503.01168 [math.CO], 2015
Eric Weisstein's World of Mathematics, Elementary Cellular Automaton
S. Wolfram, A New Kind of Science
Index entries for sequences related to cellular automata
Index to 2D 5-Neighbor Cellular Automata
Index to Elementary Cellular Automata

Cf. A001045, A048896.

Cf. A277916, A277917, A277918.

a[n_]:= Sum[Mod[Binomial[2*n+2, k],2]*2^k,{k,0,n}]; Array[a,43,0] (* Stefano Spezia, Aug 04 2025 *)

a(n) = Sum_{k=0..n} (binomial(2*n+2, k) mod 2)*2^k.

a(n) = Sum_{k=0..n} A128937(n, k)*2^(n-k). - Philippe Deléham, Oct 09 2007

A128937 Triangle formed by reading A039598 mod 2.

Original entry on oeis.org

1, 0, 1, 1, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1

Offset: 0

Philippe Deléham, Apr 27 2007, May 02 2007

Also triangle formed by reading triangles A052179, A053121, A124575, A126075, A126093.

Also triangle formed by reading A065600 mod 2. - Philippe Deléham, Oct 15 2007

			Triangle begins:
  1;
  0, 1;
  1, 0, 1;
  0, 0, 0, 1;
  0, 0, 1, 0, 1;
  0, 1, 0, 0, 0, 1;
  1, 0, 1, 0, 1, 0, 1;
  0, 0, 0, 0, 0, 0, 0, 1;
  0, 0, 0, 0, 0, 0, 1, 0, 1;
  0, 0, 0, 0, 0, 1, 0, 0, 0, 1;
  0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1;
  0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1;
  0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1;
  0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1;
  1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1;
  0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1; ...

Cf. A048896 (row sums).

A039598 := proc(n,k)
        binomial(2*n,n-k)-binomial(2*n,n-k-2) ;
end proc:
A128937 := proc(n,k)
        modp(A039598(n,k),2) ;
end proc: # R. J. Mathar, Apr 21 2021

Sum_{k=0..n} T(n,k) = A048896(n).
Sum_{k=0..n} T(n,k)*2^(n-k) = A101692(n). - Philippe Deléham, Oct 09 2007
Sum_{k=0..n} T(n,k)*2^k = A062878(n+1)/3. - Philippe Deléham, Aug 31 2009

A133179 A modular binomial sum transform of 2^n .

Original entry on oeis.org

1, 1, 1, 3, 1, 3, 5, 15, 1, 3, 5, 15, 17, 51, 85, 255, 1, 3, 5, 15, 17, 51, 85, 255, 257, 771, 1285, 3855, 4369, 13107, 21845, 65535, 1, 3, 5, 15, 17, 51, 85, 255, 257, 771, 1285, 3855, 4369, 13107, 21845, 65535

Offset: 0

Philippe Deléham, Oct 10 2007

			A034868 is:
1;
1;
1, 2;
1, 3;
1, 4, 6;
1, 5, 10 ;...
A034868 modulo 2:
1;
1;
1, 0;
1, 1;
1, 0, 0;
1, 1, 0 ;...
a(0)=1*2^0 = 1;
a(1)=1*2^0 = 1;
a(2)=1*2^0+0*2^1 = 1;
a(3)=1*2^0+1*2^1 = 3;
a(4)=1*2^0+0*2^1+0*2^2 = 1
a(5)=1*2^0+1*2^1+0*2^2 = 3

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened

Cf. A034868, A048896, A101692, A130047.

A133179[n_] := Sum[2^k*Mod[Binomial[n, k], 2], {k, 0, Floor[n/2]}]; Table[A133179[n], {n,0,50}] (* G. C. Greubel, Aug 11 2017 *)

a(n) = Sum_{k=0..floor(n/2)} mod(binomial(n,k),2) * 2^k.

A347928 Triangle read by rows, T(n, k) are the coefficients of the scaled Mandelbrot-Larsen polynomials P(n, x) = 2^(2n-1)M(n, x), where M(n, x) are the Mandelbrot-Larsen polynomials; for 0 <= k <= n.

Original entry on oeis.org

0, 0, 1, 0, 2, 1, 0, 0, 4, 2, 0, 16, 12, 12, 5, 0, 0, 32, 40, 40, 14, 0, 0, 192, 208, 168, 140, 42, 0, 0, 0, 640, 800, 720, 504, 132, 0, 2048, 1792, 2688, 3920, 3584, 3080, 1848, 429, 0, 0, 4096, 4608, 11520, 17760, 16512, 13104, 6864, 1430

Offset: 0

Peter Luschny, Oct 27 2021

To avoid confusion: the polynomials which are called 'Mandelbrot polynomials' by some authors are listed in A137560. The name 'Mandelbrot-Larsen' polynomials was introduced in Calkin, Chan, & Corless to distinguish them from the Mandelbrot polynomials.

			Triangle starts:
[0]  0;
[1]  0,    1;
[2]  0,    2,    1;
[3]  0,    0,    4,    2;
[4]  0,   16,   12,   12,     5;
[5]  0,    0,   32,   40,    40,    14;
[6]  0,    0,  192,  208,   168,   140,    42;
[7]  0,    0,    0,  640,   800,   720,   504,   132;
[8]  0, 2048, 1792, 2688,  3920,  3584,  3080,  1848,  429;
[9]  0,    0, 4096, 4608, 11520, 17760, 16512, 13104, 6864, 1430.

Neil J. Calkin, Eunice Y. S. Chan, and Robert M. Corless, Some Facts and Conjectures about Mandelbrot Polynomials, Maple Trans., Vol. 1, No. 1, Article 14037 (July 2021).
V. P. Johnson, Enumeration Results on Leaf Labeled Trees, Ph. D. Dissertation, Univ. Southern Calif., 2012.
Michael Larsen, Multiplicative series, modular forms, and Mandelbrot polynomials, in: Mathematics of Computation 90.327 (Sept. 2020), pp. 345-377. Preprint: arXiv:1908.09974 [math.NT], 2019.

Cf. A137560, A000108, A000984, A001190, A083563, A088674, A048896, A107087, A220816, A220817.

Cf. A319539, A348678, A348679, A348686.

M := proc(n, x) local k; option remember;
if n = 0 then 0 elif n = 1 then x else add(M(k, x)*M(n-k, x), k = 1..n-1) +
ifelse(n::even, M(n/2, x), 0) fi; expand(%/2) end:
P := n -> 2^(2*n - 1)*M(n, x):
row := n -> seq(coeff(P(n), x, k), k = 0..n): seq(row(n), n = 0..9);

M[n_, x_] := M[n, x] = Module[{k, w}, w = Which[n == 0, 0, n == 1, x, True, Sum[M[k, x]*M[n-k, x], {k, 1, n-1}] + If[EvenQ[n], M[n/2, x], 0]]; Expand[w/2]];
P[n_] := 2^(2*n - 1)*M[n, x];
row [n_] := If[n == 0, {0}, CoefficientList[P[n], x]];
Table[row[n], {n, 0, 9}] // Flatten (* Jean-François Alcover, Jul 07 2022, after Maple code *)

The Mandelbrot-Larsen polynomials are defined: M(0, x) = 0; M(1, x) = x/2;
M(n, x) = (1/2)*(even(n)*M(n/2, x) + Sum_{k=1..n-1} M(k, x)*M(n-k, x)) for n > 1. Here even(n) = 1 if n is even, otherwise 0.
P(n, x) = 2^(2*n-1)*M(n, x) (scaled Mandelbrot-Larsen polynomials).
T(n, k) = [x^k] P(n, x).
[x^k] M(n,k) = A348679(n, k) / A348678(n, k).
M(n, 2*k) = P(n, 2*k) / 2^(2*n-1) = A319539(n, k).
P(n, k) = A348686(n, k).
T(n, n) = A000108(n-1) for n >= 1, Catalan numbers.
T(n+2, n+1) / 2 = A000984(n) for n >= 0, central binomials.
P(n, 1) = A088674(n-1) for n >= 1, also row sums.
M(n, 2) = A001190(n) for n >= 0.
M(n, 4) = A083563(n) for n >= 0.
M(n,-4) = -A107087(n) for n >= 1.
M(n, 6) = A220816(n) for n >= 1.
M(n, 8) = A220817(n) for n >= 1.
Conjecture (Calkin, Chan, & Corless): content(P(n)) = gcd(row(n)) = A048896(n-1) for n >= 1.

A130047 Left half of Pascal's triangle (A034868) modulo 2.

Original entry on oeis.org

1, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0

Offset: 0

Philippe Deléham, Oct 10 2007

Row sums yield: 1, 1, 1, 2, 1, 2, 2, 4, 1, 2, 2, 4, 2, 4, 4, 8, 1, 2, 2, 4, 2, 4, 4, 8, ...(see A048896).

			Triangle begins:
1,
1,
1, 0,
1, 1,
1, 0, 0,
1, 1, 0,
1, 0, 1, 0,
1, 1, 1, 1,
1, 0, 0, 0, 0,
1, 1, 0, 0, 0,
1, 0, 1, 0, 0, 0,
1, 1, 1, 1, 0, 0,
1, 0, 0, 0, 1, 0, 0,
1, 1, 0, 0, 1, 1, 0,
1, 0, 1, 0, 1, 0, 1, 0,
1, 1, 1, 1, 1, 1, 1, 1,
1, 0, 0, 0, 0, 0, 0, 0, 0,
...
Triangle (right aligned) begins:
                                  1,
                                1,
                              1,  0,
                            1,  1,
                          1,  0,  0,
                        1,  1,  0,
                      1,  0,  1,  0,
                    1,  1,  1,  1,
                  1,  0,  0,  0,  0,
                1,  1,  0,  0,  0,
              1,  0,  1,  0,  0,  0,
            1,  1,  1,  1,  0,  0,
          1,  0,  0,  0,  1,  0,  0,
        1,  1,  0,  0,  1,  1,  0,
      1,  0,  1,  0,  1,  0,  1,  0,
    1,  1,  1,  1,  1,  1,  1,  1,
  1,  0,  0,  0,  0,  0,  0,  0,  0,
1,  1,  0,  0,  0,  0,  0,  0,  0,
...

G. C. Greubel, Table of n, a(n) for the first 100 rows, flattened

Cf. A007318, A034868, A048896, A133179.

# From N. J. A. Sloane, Mar 22 2015:
for n from 0 to 20 do
lprint(seq(binomial(n,k) mod 2, k=0..floor(n/2))); od:
# For row sums:
f:=n->add(binomial(n,k) mod 2, k=0..floor(n/2));
[seq(f(n),n=0..60)];

Table[Mod[Binomial[n, k], 2], {n, 0, 10}, {k, 0, Floor[n/2]}] (* G. C. Greubel, Aug 12 2017 *)

T(n,k) = mod(binomial(n, k), 2), 0 <= k <= floor(n/2). - G. C. Greubel, Aug 12 2017

Corrected by N. J. A. Sloane, Mar 22 2015 at the suggestion of Kevin Ryde

A151784 a(n) = 6^(wt(n) - 1) where wt(n) = A000120(n).

Original entry on oeis.org

1, 1, 6, 1, 6, 6, 36, 1, 6, 6, 36, 6, 36, 36, 216, 1, 6, 6, 36, 6, 36, 36, 216, 6, 36, 36, 216, 36, 216, 216, 1296, 1, 6, 6, 36, 6, 36, 36, 216, 6, 36, 36, 216, 36, 216, 216, 1296, 6, 36, 36, 216, 36, 216, 216, 1296, 36, 216, 216, 1296, 216, 1296, 1296, 7776, 1, 6, 6, 36, 6, 36, 36

Offset: 1

N. J. A. Sloane, Jun 25 2009

nonn

			From _Omar E. Pol_, Jul 21 2009: (Start)
If written as a triangle:
  1;
  1,6;
  1,6,6,36;
  1,6,6,36,6,36,36,216;
  1,6,6,36,6,36,36,216,6,36,36,216,36,216,216,1296;
  1,6,6,36,6,36,36,216,6,36,36,216,36,216,216,1296,6,36,36,216,36,216,216,...
(End)

Cf. A000120, A048881, A048896, A147610, A151780, A151783, A000120.

Cf. A000400. - Omar E. Pol, Jul 21 2009

a(n) = 6^(hammingweight(n)-1); \\ Michel Marcus, Nov 15 2022

A167364 Triangle read by rows, A047999 * A010060 (diagonalized); as infinite lower triangular matrices.

Original entry on oeis.org

1, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0

Offset: 1

Gary W. Adamson & Mats Granvik, Nov 01 2009

Row sums = A048896: (1, 1, 2, 1, 2, 2, 4, 1, 2, 2, 4,...). Right border = Thue-Morse sequence A010060, starting with offset 1.

			First few rows of the triangle =
1;
0, 1;
1, 1, 0;
0, 0, 0, 1;
1, 0, 0, 1, 0;
0, 1, 0, 1, 0, 0;
1, 1, 0, 1, 0, 0, 1;
0, 0, 0, 0, 0, 0, 0, 1;
1, 0, 0, 0, 0, 0, 0, 1, 0;
0, 1, 0, 0, 0, 0, 0, 1, 0, 0;
1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1;
0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0
1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1;
0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1;
1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0;
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1
1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0;
0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0;
1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1;
...

Cf. A047999, A010060.

Let S = Sierpinski's gasket, A047999. Let Q = a diagonalized version of the Thue-Morse sequence, A010060: [0; 0,1; 0,0,1; 0,0,0,0; 0,0,0,0,1;...], (i.e. A010060 as the rightmost diagonal and the rest zeros). A167364 = S * Q, as infinite lower triangular matrices. Delete leftmost column of zeros.

A243036 Number of entries of length n in A240602.

Original entry on oeis.org

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

Offset: 1

Lior Manor, May 29 2014

For n>0, a(2^n) = 1.
For any base b, a(1) = b, a(2) = b-1, a(3) = b*(b-1), for m>1, a(2m) = a(m), a(2m+1) = b*a(m).
For any base b, a(n) = (b-1)*b^(A000120(n)-1).
Essentially the same as A048896. - R. J. Mathar, Jun 27 2014

Cf. A240602, A243035, A000120.

a(1) = 2, a(2) = 1, a(3) = 2, For m>1, a(2m) = a(m), a(2m+1) = 2*a(m).

a(n) = 2^(A000120(n)-1).

A256257 6 times numbers of Gould's sequence A001316.

Original entry on oeis.org

6, 12, 12, 24, 12, 24, 24, 48, 12, 24, 24, 48, 24, 48, 48, 96, 12, 24, 24, 48, 24, 48, 48, 96, 24, 48, 48, 96, 48, 96, 96, 192, 12, 24, 24, 48, 24, 48, 48, 96, 24, 48, 48, 96, 48, 96, 96, 192, 24, 48, 48, 96, 48, 96, 96, 192, 48, 96, 96, 192, 96, 192, 192, 384, 12, 24, 24, 48, 24, 48, 48, 96, 24, 48, 48, 96, 48, 96, 96, 192

Offset: 0

Omar E. Pol, Mar 20 2015

Also, number of triangular cells turned ON at (n+1)-st stage in the structure of A256256.

First differences of A256256.

			Written as an irregular triangle in which the row lengths are the terms of A011782, the sequence begins:
6;
12;
12, 24;
12, 24, 24, 48;
12, 24, 24, 48, 24, 48, 48, 96;
12, 24, 24, 48, 24, 48, 48, 96, 24, 48, 48, 96, 48, 96, 96, 192;
...

Cf. A001316, A011782, A047999, A048896, A117973, A151787, A160713, A160721, A256256.

a(n) = 6*A001316(n) = 3*A117973(n) = 2*A160713(n).

a(n) = 12*A048896(n-1), n >= 1.

cp's OEIS Frontend

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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A101692 A modular binomial sum transform of 2^n.

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A128937 Triangle formed by reading A039598 mod 2.

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A133179 A modular binomial sum transform of 2^n .

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A347928 Triangle read by rows, T(n, k) are the coefficients of the scaled Mandelbrot-Larsen polynomials P(n, x) = 2^(2*n-1)*M(n, x), where M(n, x) are the Mandelbrot-Larsen polynomials; for 0 <= k <= n.

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A130047 Left half of Pascal's triangle (A034868) modulo 2.

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A151784 a(n) = 6^(wt(n) - 1) where wt(n) = A000120(n).

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A167364 Triangle read by rows, A047999 * A010060 (diagonalized); as infinite lower triangular matrices.

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A347928 Triangle read by rows, T(n, k) are the coefficients of the scaled Mandelbrot-Larsen polynomials P(n, x) = 2^(2n-1)M(n, x), where M(n, x) are the Mandelbrot-Larsen polynomials; for 0 <= k <= n.