A000523 -id:A000523 - cp's OEIS frontend

A173419 Length of shortest computation yielding n using addition, subtraction and multiplication (starting from 1).

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

Offset: 1

Charles R Greathouse IV, Feb 17 2010, Apr 22 2010

Let x_0 = 1 and x_m = n, with each x_k = x_i + x_j, x_k = x_i * x_j, or x_k = x_i - x_j for some 0 <= i,j < k. a(n) is the least such m.
Shub & Smale ask if there is a c such that a(n!) <= (log n)^c for all n.
If for any sequence of nonzero integers (m_i) there is no constant c such that a(n! * m_n) <= (log n)^c, then "the Hilbert Nullstellensatz is intractable, and consequently the algebraic version of 'NP != P' is true" (Shub & Smale).
Conjecture: if n is prime then a(n) >= a(n-1). The conjecture is true for n < 1800. - Dmitry Kamenetsky, Dec 26 2019

			For n = 9, one sequence is (1, 1 + 1 = 2, 1 + 2 = 3, 3 * 3 = 9). Since no shorter sequence is possible, a(9) = 3.
For n = 96, one sequence is (1, 1 + 1 = 2, 2 + 2 = 4, 2 + 4 = 6, 4*4 = 16, 6*16 = 96); no shorter is possible so a(96) = 5.

R. K. Guy, Unsolved Problems Number Theory, Sect. F26.

Alois P. Heinz, Table of n, a(n) for n = 1..1800
Peter Borwein and Joe Hobart, The extraordinary power of division in straight line programs, American Mathematical Monthly 119:7 (2012), pp. 584-592.
F. Cucker, M. Shub, and S. Smale, Separation of Complexity classes in Koiran's weak model, Theoretical Computer Science 133:1 (1994), pp. 3-14.
W. DeMelo and B. F. Svaiter, The cost of computing integers, Proc. Amer. Math. Soc. 124 (1996), pp. 1377-1378.
P. Koiran, Valiant's model and the cost of computing integers, Comput. Complex. 13 (2004), pp. 131-146.
Carlos Gustavo T. de A. Moreira, On asymptotic estimates for arithmetic cost functions, Proceedings of the American Mathematical Society 125:2 (1997), pp. 347-353.
Richard J. Mathar, Extended list of chain examples
Michael Shub and Steve Smale, On the intractability of Hilbert's Nullstellensatz and an algebraic version of "NP = P", Duke Mathematical Journal 81:1 (1995), pp. 47-54.
Al Zimmermann's Programming Contests, Factorials
Index to sequences related to the complexity of n

Records are essentially A141414.

Cf. A003313 (shortest chain using just addition), A005245 (number of 1s using just addition and multiplication), A217032(n):=A173419(n!).

g:= f->seq(f union {t}, t={seq(seq({i+j, i-j, i*j}[], j=f), i=f)} minus f):
F:= proc(n) F(n):= map(g, F(n-1)) end: F(0):= {{1}}:
S:= proc(n) S(n):= map(x->x[], F(n)) end:
a:= proc(n) local k; for k from 0 while not(n in S(k)) do od; k end:
seq(a(n), n=1..110);  # Alois P. Heinz, Sep 24 2012

a(n) <= 2 log_2(n).
a(n) >= log_2(log_2(n)) + 1.
a(n) >= log_2(n)/log_2(log_2(n)) for almost all n, as proved by Moreira (improving DeMelo & Svaiter).
a(n) <= A005245(n) <= A003313(n) <= A014701(n) <= 2*A000523(n). - Charles R Greathouse IV, Feb 07 2022

A272011 Irregular triangle read by rows: strictly decreasing sequences of nonnegative numbers given in lexicographic order.

Original entry on oeis.org

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

Offset: 0

Peter Kagey, Apr 17 2016

Length of n-th row given by A000120(n);
Maximum of n-th row given by A000523(n);
Minimum of n-th row given by A007814(n);
GCD of n-th row given by A064894(n);
Sum of n-th row given by A073642(n + 1).
n-th row begins at index A000788(n - 1) for n > 0.
The first appearance of n is at A001787(n).
a(A001787(n) + 1) = a(A001787(n)) for all n > 0.
a(A001787(n) + 2) = 0 for all n > 0.
a(A001787(n) + 3) = a(A001787(n)) for all n > 1.
a(A001787(n) + 4) = 1 for all n > 1.
a(A001787(n) + 5) = a(A001787(n)) for all n > 1.
Row n < 1024 lists the digits of A262557(n). - M. F. Hasler, Dec 11 2019

			Row n is given by the exponents in the binary expansion of n. For example, row 5 = [2, 0] because 5 = 2^2 + 2^0.
Row 0: []
Row 1: [0]
Row 2: [1]
Row 3: [1, 0]
Row 4: [2]
Row 5: [2, 0]
Row 6: [2, 1]
Row 7: [2, 1, 0]

Peter Kagey, Table of n, a(n) for n = 0..10000

Cf. A000120, A000523, A001787, A007814, A064894, A073642.
Cf. A133457 gives the rows in reverse order.
Cf. A272020, A262557.

Map[Length[#] - Flatten[Position[#, 1]] &, IntegerDigits[Range[50], 2]] (* Paolo Xausa, Feb 13 2024 *)

apply( A272011_row(n)=Vecrev(vecextract([0..exponent(n+!n)],n)), [0..39]) \\ For n < 2^10: row(n)=digits(A262557[n]). There are 2^k rows starting with k, they start at row 2^k. - M. F. Hasler, Dec 11 2019

A306393 Number T(n,k) of defective (binary) heaps on n elements where k ancestor-successor pairs do not have the correct order; triangle T(n,k), n >= 0, 0 <= k <= A061168(n), read by rows.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 2, 3, 6, 6, 6, 3, 8, 16, 24, 24, 24, 16, 8, 20, 60, 100, 120, 120, 120, 100, 60, 20, 80, 240, 480, 640, 720, 720, 720, 640, 480, 240, 80, 210, 840, 1890, 3150, 4200, 4830, 5040, 5040, 4830, 4200, 3150, 1890, 840, 210

Offset: 0

Alois P. Heinz, Feb 12 2019

T(n,k) is the number of permutations p of [n] having exactly k pairs (i,j) in {1,...,n} X {1,...,floor(log_2(i))} such that p(i) > p(floor(i/2^j)).

T(n,0) counts perfect (binary) heaps on n elements (A056971).

			T(4,0) = 3: 4231, 4312, 4321.
T(4,1) = 6: 3241, 3412, 3421, 4123, 4132, 4213.
T(4,2) = 6: 2341, 2413, 2431, 3124, 3142, 3214.
T(4,3) = 6: 1342, 1423, 1432, 2134, 2143, 2314.
T(4,4) = 3: 1234, 1243, 1324.
T(5,1) = 16: 43512, 43521, 45123, 45132, 45213, 45231, 45312, 45321, 52314, 52341, 52413, 52431, 53124, 53142, 53214, 53241.
(The examples use max-heaps.)
Triangle T(n,k) begins:
   1;
   1;
   1,   1;
   2,   2,   2;
   3,   6,   6,   6,   3;
   8,  16,  24,  24,  24,  16,   8;
  20,  60, 100, 120, 120, 120, 100,  60,  20;
  80, 240, 480, 640, 720, 720, 720, 640, 480, 240, 80;
  ...

Alois P. Heinz, Rows n = 0..100, flattened
Marko Riedel, math.stackexchange.com, Average number of inversions in a random binary heap on 2^n-1 elements.
Marko Riedel, Average number of inversions in a random binary heap on 2^n-1 elements (PDF).
Eric Weisstein's World of Mathematics, Heap,
Wikipedia, Binary heap.
Wikipedia, Permutation.

Columns k=0-10 give: A056971, A324062, A324063, A324064, A324065, A324066, A324067, A324068, A324069, A324070, A324071.
Row sums give A000142.
Central terms (also maxima) of rows give A324075.
Cf. A000523, A008302, A061168, A120385, A306343, A324074, A372640.
Average number of inversions of a full binary heap on 2^n-1 elements is A000337.

b:= proc(u, o) option remember; local n, g, l; n:= u+o;
      if n=0 then 1
    else g:= 2^ilog2(n); l:= min(g-1, n-g/2); expand(
         add(x^(n-j)*add(binomial(j-1, i)*binomial(n-j, l-i)*
         b(i, l-i)*b(j-1-i, n-l-j+i), i=0..min(j-1, l)), j=1..u)+
         add(x^(j-1)*add(binomial(j-1, i)*binomial(n-j, l-i)*
         b(l-i, i)*b(n-l-j+i, j-1-i), i=0..min(j-1, l)), j=1..o))
      fi
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n, 0)):
seq(T(n), n=0..10);

b[u_, o_] := b[u, o] = Module[{n, g, l}, n = u + o;
     If[n == 0, 1, g = 2^Floor@Log[2, n]; l = Min[g - 1, n - g/2]; Expand[
     Sum[x^(n-j)*Sum[Binomial[j - 1, i]*Binomial[n - j, l - i]*
     b[i, l-i]*b[j-1-i, n-l-j+i], {i, 0, Min[j - 1, l]}], {j, 1, u}] +
     Sum[x^(j-1)*Sum[Binomial[j - 1, i]*Binomial[n - j, l - i]*
     b[l-i, i]*b[n-l-j+i, j-1-i], {i, 0, Min[j-1, l]}], {j, 1, o}]]]];
T[n_] := CoefficientList[b[n, 0], x];
T /@ Range[0, 10] // Flatten (* Jean-François Alcover, Feb 15 2021, after Alois P. Heinz *)

T(n,k) = T(n,A061168(n)-k) for n > 0.

Sum_{k=0..A061168(n)} k * T(n,k) = A324074(n).

A010096 log2*(n) (version 1): number of times floor(log_2(x)) is used in floor(log_2(floor(log_2(...(floor(log_2(n)))...)))) = 0.

Original entry on oeis.org

1, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4

Offset: 1

Leonid Broukhis

From Hieronymus Fischer, Apr 08 2012: (Start)
A possibly simpler definition could be: "Number of iterations log_2(log_2(log_2(...(n)...))) such that the result is < 1".
Changing "< 1" to "<= 1" produces version 3, A230864.
With the only difference in the termination criterion, the definition is essentially the same as version 2, A001069. If we change the definition to "floor(log_2(... = 1" we get A001069. Therefore we get A001069 when subtracting 1 from each term. (End)

			Becomes 5 at 65536, 6 at 2^65536, etc.

Cf. A063510, A000523, A001069 (version 2), A230864 (version 3).

a010096 = length . takeWhile (/= 0) . iterate a000523
-- Reinhard Zumkeller, Mar 16 2012

f[n_] := Length@ NestWhileList[ Log[2, #] &, n, # >= 1 &] - 1; Array[f, 105] (* Robert G. Wilson v, Apr 19 2012 *)

a(n)=if(n<1,0,1+a(log(n)\log(2))) \\ Charles R Greathouse IV, Apr 17 2012

a(n)=if(n<1,0,1+a(logint(n,2))) \\ Charles R Greathouse IV, Oct 23 2015

From Hieronymus Fischer, Apr 08 2012: (Start)
a(n) = A001069(n) + 1.
With the exponentiation definition E_{i=1..n} c(i) := c(1)^(c(2)^(c(3)^(...(c(n-1)^(c(n)))...))); E_{i=1..0} := 1; example: E_{i=1..4} 2 = 2^(2^(2^2)) = 2^16, we get:
a(E_{i=1..n} 2) = a(E_{i=1..n-1} 2) +1, for n >= 1.
G.f.: g(x) = 1/(1-x)*Sum_{k>=0} x^(E_{i=1..k} 2).
The explicit first terms of this g.f. are
g(x) = (x + x^2 + x^4 + x^16 + x^65536 + ...)/(1-x). (End)

Edited by Hieronymus Fischer, Apr 08 2012

Edited by N. J. A. Sloane, Nov 03 2013

A256292 Numbers which have only digits 6 and 7 in base 10.

Original entry on oeis.org

6, 7, 66, 67, 76, 77, 666, 667, 676, 677, 766, 767, 776, 777, 6666, 6667, 6676, 6677, 6766, 6767, 6776, 6777, 7666, 7667, 7676, 7677, 7766, 7767, 7776, 7777, 66666, 66667, 66676, 66677, 66766, 66767, 66776, 66777, 67666, 67667

Offset: 1

M. F. Hasler, Mar 27 2015

Vincenzo Librandi, Table of n, a(n) for n = 1..8190
Index entries for 10-automatic sequences.

Cf. A007088 (digits 0 & 1), A007931 (digits 1 & 2), A032810 (digits 2 & 3), A032834 (digits 3 & 4), A256290 (digits 4 & 5), A256291 (digits 5 & 6), A256340 (digits 7 & 8), A256341 (digits 8 & 9).

[n: n in [1..35000] | Set(IntegerToSequence(n, 10)) subset {7, 6}];

[n: n in [1..100000] | Set(Intseq(n)) subset {6,7}]; // Vincenzo Librandi, Aug 19 2016

Flatten[Table[FromDigits[#,10]&/@Tuples[{6,7},n],{n,5}]]

A256292(n)=vector(#n=binary(n+1)[2..-1],i,10^(#n-i))*n~+10^#n\9*6

a(n) = A007931(n) + A002279(A000523(n+1)) = A256291(n) + A256077(n) etc.

A286357 One more than the exponent of the highest power of 2 dividing sigma(n): a(n) = A001511(A000203(n)).

Original entry on oeis.org

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

Offset: 1

Antti Karttunen, May 10 2017

nonn

Antti Karttunen, Table of n, a(n) for n = 1..10000
Index entries for sequences related to sigma(n)

Cf. A000203, A000523, A001511, A053866, A070939, A082903, A161942, A286358.

Table[IntegerExponent[DivisorSigma[1,n],2]+1,{n,120}] (* Harvey P. Dale, Sep 04 2023 *)

A001511(n) = (1+valuation(n,2));
A286357(n) = A001511(sigma(n));
for(n=1, 10000, write("b286357.txt", n, " ", A286357(n)));

from sympy import divisor_sigma as D
def a001511(n): return bin(n)[2:][::-1].index("1") + 1
def a(n): return a001511(D(n)) # Indranil Ghosh, May 12 2017

from sympy import divisor_sigma
def A286357(n): return ((m:=int(divisor_sigma(n)))&-m).bit_length() # Chai Wah Wu, Jul 10 2022

(define (A286357 n) (A001511 (A000203 n)))
(define (A286357 n) (A070939 (/ (A000203 n) (A161942 n))))

a(n) = A001511(A000203(n)).
a(n) = 1 + A000523(A000203(n)/A161942(n)). [See also A082903.]
a(n) = 1 iff A053866(n) = 1.

A295989 Irregular triangle T(n, k), read by rows, n >= 0 and 0 <= k < A001316(n): T(n, k) is the (k+1)-th nonnegative number m such that n AND m = m (where AND denotes the bitwise AND operator).

Original entry on oeis.org

0, 0, 1, 0, 2, 0, 1, 2, 3, 0, 4, 0, 1, 4, 5, 0, 2, 4, 6, 0, 1, 2, 3, 4, 5, 6, 7, 0, 8, 0, 1, 8, 9, 0, 2, 8, 10, 0, 1, 2, 3, 8, 9, 10, 11, 0, 4, 8, 12, 0, 1, 4, 5, 8, 9, 12, 13, 0, 2, 4, 6, 8, 10, 12, 14, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 0

Offset: 0

Rémy Sigrist, Dec 02 2017

The (n+1)-th row has A001316(n) terms and sums to n * A001316(n) / 2.
For any n >= 0 and k such that 0 <= k < A001316(n):
- if A000120(n) > 0 then T(n, 1) = A006519(n),
- if A000120(n) > 1 then T(n, 2) = 2^A285099(n),
- if A000120(n) > 0 then T(n, A001316(n)/2 - 1) = A053645(n),
- if A000120(n) > 0 then T(n, A001316(n)/2) = 2^A000523(n),
- if A000120(n) > 0 then T(n, A001316(n) - 2) = A129760(n),
- T(n, A001316(n) - 1) = n,
- the six previous relations correspond respectively (when applicable) to the second term, the third term, the pair of central terms, the penultimate term and the last term of a row,
- T(n, k) AND T(n, A001316(n) - k - 1) = 0,
- T(n, k) + T(n, A001316(n) - k - 1) = n,
- T(n, k) = k for any k < A006519(n+1),
- A000120(T(n, k)) = A000120(k).
If we plot (n, T(n,k)) then we obtain a skewed Sierpinski triangle (see Links section).
If interpreted as a flat sequence a(n) for n >= 0:
- a(n) = 0 iff n = A006046(k) for some k >= 0,
- a(n) = 1 iff n = A006046(2*k + 1) + 1 for some k >= 0,
- a(A006046(k) - 1) = k - 1 for any k > 0.

			Triangle begins:
  0:   [0]
  1:   [0, 1]
  2:   [0, 2]
  3:   [0, 1, 2, 3]
  4:   [0, 4]
  5:   [0, 1, 4, 5]
  6:   [0, 2, 4, 6]
  7:   [0, 1, 2, 3, 4, 5, 6, 7]
  8:   [0, 8]
  9:   [0, 1, 8, 9]
  10:  [0, 2, 8, 10]
  11:  [0, 1, 2, 3, 8, 9, 10, 11]
  12:  [0, 4, 8, 12]
  13:  [0, 1, 4, 5, 8, 9, 12, 13]
  14:  [0, 2, 4, 6, 8, 10, 12, 14]
  15:  [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15]

Rémy Sigrist, Rows n = 0..256, flattened
Rémy Sigrist, Scatterplot of (n, T(n, k)) for n = 0..1023 and k = 0..A001316(n)-1

Cf. A000120, A000523, A001316 (row lengths), A006046, A006519, A053645, A129760, A285099.

First column of array in A352909.

A295989row[n_] := Select[Range[0, n], BitAnd[#, n-#] == 0 &];
Array[A295989row, 25, 0] (* Paolo Xausa, Feb 24 2024 *)

T(n,k) = if (k==0, 0, n%2==0, 2*T(n\2,k), k%2==0, 2*T(n\2, k\2), 2*T(n\2, k\2)+1)

For any n >= 0 and k such that 0 <= k < A001316(n):
- T(n, 0) = 0,
- T(2*n, k) = 2*T(n, k),
- T(2*n+1, 2*k) = 2*T(n, k),
- T(2*n+1, 2*k+1) = 2*T(n, k) + 1.

A053398 Nim-values from game of Kopper's Nim.

Original entry on oeis.org

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

Offset: 1

David W. Wilson

Rows/columns 1-10 are A007814, A050603, A053399, A053384-A053890.

Comment from R. K. Guy: David Singmaster (zingmast(AT)sbu.ac.uk) sent me, about 5 years ago, a game he'd received from Bodo Koppers. It is played with two heaps of beans. The move is to remove one heap and split the other into two nonempty heaps. I'm not sure if Koppers invented it, or got it from elsewhere. I do not think that he analyzed it, but Singmaster did.

Reinhard Zumkeller, Rows n = 1..125 of triangle, flattened

Cf. A007814, A050603, A053399, A053384-A053890.

Cf. A003986, A007814 (both edges & central terms & minima per row), A000523 (max per row), A245836 (row sums), A003987, A051775.

a053398 :: Int -> Int -> Int
a053398 n k = a007814 $ a003986 (n - 1) (k - 1) + 1
a053398_row n = map (a053398 n) [1..n]
a053398_tabl = map a053398_row [1..]
-- Reinhard Zumkeller, Aug 04 2014

a(x, y) = place of last zero bit of (x-1) OR (y-1).

T(n,k) = A007814(A003986(n-1,k-1)+1). - Reinhard Zumkeller, Aug 04 2014

A079946 Numbers k whose binary expansion begins with two or more 1's and ends with at least one 0.

Original entry on oeis.org

6, 12, 14, 24, 26, 28, 30, 48, 50, 52, 54, 56, 58, 60, 62, 96, 98, 100, 102, 104, 106, 108, 110, 112, 114, 116, 118, 120, 122, 124, 126, 192, 194, 196, 198, 200, 202, 204, 206, 208, 210, 212, 214, 216, 218, 220, 222, 224, 226, 228, 230, 232, 234, 236, 238, 240, 242, 244, 246

Offset: 1

N. J. A. Sloane, Feb 21 2003

a(n) = b(n+1), with b(2n) = 2b(n), b(2n+1) = 2b(n)+2+4[n==0]. - Ralf Stephan, Oct 11 2003

Yifan Xie, Table of n, a(n) for n = 1..10001 (first 1000 terms from Harvey P. Dale)
Benoit Cloitre, N. J. A. Sloane and M. J. Vandermast, Numerical analogues of Aronson's sequence, J. Integer Seqs., Vol. 6 (2003), #03.2.2.
Benoit Cloitre, N. J. A. Sloane and M. J. Vandermast, Numerical analogues of Aronson's sequence, arXiv:math/0305308 [math.NT], 2003.
Ralf Stephan, Some divide-and-conquer sequences ...
Ralf Stephan, Table of generating functions

A004755 = union of this and A080565. A057547(n) = a(A014486(n)) for n >= 1.

Maple
```
A079946 := n -> 2*(2^(1+A000523(n))+n);
```

Table[Union[FromDigits[Join[{1,1},#,{0}],2]&/@Tuples[{1,0},n]],{n,0,5}]//Flatten (* Harvey P. Dale, Jan 16 2018 *)

for(n=0,6, for(k=2^(n-1),2^n-1,print1((2^n+k)*2,",")))

for(n=1,59,print1((2^(floor(log(n)/log(2))+1)+n)*2,","))

a(n) = n*2 + 4<Ruud H.G. van Tol, May 10 2024

def A079946(n): return n+(1<Chai Wah Wu, Jul 13 2022

a(n) = 2^floor(log_2(4*n))+2*n. - Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Feb 22 2003
a(n) = (2^(floor(log_2(n))+1)+n)*2. - Klaus Brockhaus, Feb 23 2003
a(2n) = 2a(n), a(2n+1) = 2a(n) + 2 + 4[n==0]. Twice A004755. - Ralf Stephan, Oct 12 2003

Definition clarified by N. J. A. Sloane, May 10 2024

A093659 First column of lower triangular matrix A093658; factorial of the number of 1's in binary expansion of n.

Original entry on oeis.org

1, 1, 1, 2, 1, 2, 2, 6, 1, 2, 2, 6, 2, 6, 6, 24, 1, 2, 2, 6, 2, 6, 6, 24, 2, 6, 6, 24, 6, 24, 24, 120, 1, 2, 2, 6, 2, 6, 6, 24, 2, 6, 6, 24, 6, 24, 24, 120, 2, 6, 6, 24, 6, 24, 24, 120, 6, 24, 24, 120, 24, 120, 120, 720, 1, 2, 2, 6, 2, 6, 6, 24, 2, 6, 6, 24, 6

Offset: 0

Paul D. Hanna, Apr 08 2004

a(n) is the number of compositions of n into distinct powers of 2. - Vladimir Shevelev, Jan 15 2014

Antti Karttunen, Table of n, a(n) for n = 0..8192
Index entries for sequences related to binary expansion of n

Cf. A000079, A000523, A023359, A093658, A093677, A000120, A000142, A139329.

a:= n-> add(i,i=Bits[Split](n))!:
seq(a(n), n=0..80);  # Alois P. Heinz, Nov 02 2024

Table[DigitCount[n,2,1]!,{n,0,70}] (* Harvey P. Dale, Jul 09 2019 *)

from math import factorial
def a(n): return factorial(n.bit_count()) # Michael S. Branicky, Nov 02 2024

a(2^n) = n! for n>=0. a(2^n+2^m) = a(2^(m+1)) for n>m>=0.

a(n) = A000120(n)! = A000142(A000120(n)).

cp's OEIS Frontend

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A306393 Number T(n,k) of defective (binary) heaps on n elements where k ancestor-successor pairs do not have the correct order; triangle T(n,k), n >= 0, 0 <= k <= A061168(n), read by rows.

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A010096 log2*(n) (version 1): number of times floor(log_2(x)) is used in floor(log_2(floor(log_2(...(floor(log_2(n)))...)))) = 0.

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A256292 Numbers which have only digits 6 and 7 in base 10.

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A286357 One more than the exponent of the highest power of 2 dividing sigma(n): a(n) = A001511(A000203(n)).

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A295989 Irregular triangle T(n, k), read by rows, n >= 0 and 0 <= k < A001316(n): T(n, k) is the (k+1)-th nonnegative number m such that n AND m = m (where AND denotes the bitwise AND operator).

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