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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A005940 The Doudna sequence: write n-1 in binary; power of prime(k) in a(n) is # of 1's that are followed by k-1 0's.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 9, 8, 7, 10, 15, 12, 25, 18, 27, 16, 11, 14, 21, 20, 35, 30, 45, 24, 49, 50, 75, 36, 125, 54, 81, 32, 13, 22, 33, 28, 55, 42, 63, 40, 77, 70, 105, 60, 175, 90, 135, 48, 121, 98, 147, 100, 245, 150, 225, 72, 343, 250, 375, 108, 625, 162, 243, 64, 17, 26, 39
Offset: 1

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Author

Keywords

Comments

A permutation of the natural numbers. - Robert G. Wilson v, Feb 22 2005
Fixed points: A029747. - Reinhard Zumkeller, Aug 23 2006
The even bisection, when halved, gives the sequence back. - Antti Karttunen, Jun 28 2014
From Antti Karttunen, Dec 21 2014: (Start)
This irregular table can be represented as a binary tree. Each child to the left is obtained by applying A003961 to the parent, and each child to the right is obtained by doubling the parent:
1
|
...................2...................
3 4
5......../ \........6 9......../ \........8
/ \ / \ / \ / \
/ \ / \ / \ / \
/ \ / \ / \ / \
7 10 15 12 25 18 27 16
11 14 21 20 35 30 45 24 49 50 75 36 125 54 81 32
etc.
Sequence A163511 is obtained by scanning the same tree level by level, from right to left. Also in binary trees A253563 and A253565 the terms on level of the tree are some permutation of the terms present on the level n of this tree. A252464(n) gives the distance of n from 1 in all these trees.
A252737(n) gives the sum and A252738(n) the product of terms on row n (where 1 is on row 0, 2 on row 1, 3 and 4 on row 2, etc.). A252745(n) gives the number of nodes on level n whose left child is larger than the right child, A252750 the difference between left and right child for each node from node 2 onward.
(End)
-A008836(a(1+n)) gives the corresponding numerator for A323505(n). - Antti Karttunen, Jan 19 2019
(a(2n+1)-1)/2 [= A244154(n)-1, for n >= 0] is a permutation of the natural numbers. - George Beck and Antti Karttunen, Dec 08 2019
From Peter Munn, Oct 04 2020: (Start)
Each term has the same even part (equivalently, the same 2-adic valuation) as its index.
Using the tree depicted in Antti Karttunen's 2014 comment:
Numbers are on the right branch (4 and descendants) if and only if divisible by the square of their largest prime factor (cf. A070003).
Numbers on the left branch, together with 2, are listed in A102750.
(End)
According to Kutz (1981), he learned of this sequence from American mathematician Byron Leon McAllister (1929-2017) who attributed the invention of the sequence to a graduate student by the name of Doudna (first name Paul?) in the mid-1950's at the University of Wisconsin. - Amiram Eldar, Jun 17 2021
From David James Sycamore, Sep 23 2022: (Start)
Alternative (recursive) definition: If n is a power of 2 then a(n)=n. Otherwise, if 2^j is the greatest power of 2 not exceeding n, and if k = n - 2^j, then a(n) is the least m*a(k) that has not occurred previously, where m is an odd prime.
Example: Use recursion with n = 77 = 2^6 + 13. a(13) = 25 and since 11 is the smallest odd prime m such that m*a(13) has not already occurred (see a(27), a(29),a(45)), then a(77) = 11*25 = 275. (End)
The odd bisection, when transformed by replacing all prime(k)^e in a(2*n - 1) with prime(k-1)^e, returns a(n), and thus gives the sequence back. - David James Sycamore, Sep 28 2022

Examples

			From _N. J. A. Sloane_, Aug 22 2022: (Start)
Let c_i = number of 1's in binary expansion of n-1 that have i 0's to their right, and let p(j) = j-th prime.  Then a(n) = Product_i p(i+1)^c_i.
If n=9, n-1 is 1000, c_3 = 1, a(9) = p(4)^1 = 7.
If n=10, n-1 = 1001, c_0 = 1, c_2 = 1, a(10) = p(1)*p(3) = 2*5 = 10.
If n=11, n-1 = 1010, c_1 = 1, c_2 = 1, a(11) = p(2)*p(3) = 15. (End)
		

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A103969. Inverse is A005941 (A156552).
Cf. A125106. [From Franklin T. Adams-Watters, Mar 06 2010]
Cf. A252737 (gives row sums), A252738 (row products), A332979 (largest on row).
Related permutations of positive integers: A163511 (via A054429), A243353 (via A006068), A244154, A253563 (via A122111), A253565, A332977, A334866 (via A225546).
A000120, A003602, A003961, A006519, A053645, A070939, A246278, A250246, A252753, A253552 are used in a formula defining this sequence.
Formulas for f(a(n)) are given for f = A000265, A003963, A007949, A055396, A056239.
Numbers that occur at notable sets of positions in the binary tree representation of the sequence: A000040, A000079, A002110, A070003, A070826, A102750.
Cf. A106737, A290077, A323915, A324052, A324054, A324055, A324056, A324057, A324058, A324114, A324335, A324340, A324348, A324349 for various number-theoretical sequences applied to (i.e., permuted by) this sequence.
k-adic valuation: A007814 (k=2), A337821 (k=3).
Positions of multiples of 3: A091067.
Primorial deflation: A337376 / A337377.
Sum of prime indices of a(n) is A161511, reverse version A359043.
A048793 lists binary indices, ranked by A019565.
A066099 lists standard comps, partial sums A358134 (ranked by A358170).

Programs

  • Haskell
    a005940 n = f (n - 1) 1 1 where
       f 0 y _          = y
       f x y i | m == 0 = f x' y (i + 1)
               | m == 1 = f x' (y * a000040 i) i
               where (x',m) = divMod x 2
    -- Reinhard Zumkeller, Oct 03 2012
    (Scheme, with memoization-macro definec from Antti Karttunen's IntSeq-library)
    (define (A005940 n) (A005940off0 (- n 1))) ;; The off=1 version, utilizing any one of three different offset-0 implementations:
    (definec (A005940off0 n) (cond ((< n 2) (+ 1 n)) (else (* (A000040 (- (A070939 n) (- (A000120 n) 1))) (A005940off0 (A053645 n))))))
    (definec (A005940off0 n) (cond ((<= n 2) (+ 1 n)) ((even? n) (A003961 (A005940off0 (/ n 2)))) (else (* 2 (A005940off0 (/ (- n 1) 2))))))
    (define (A005940off0 n) (let loop ((n n) (i 1) (x 1)) (cond ((zero? n) x) ((even? n) (loop (/ n 2) (+ i 1) x)) (else (loop (/ (- n 1) 2) i (* x (A000040 i)))))))
    ;; Antti Karttunen, Jun 26 2014
    
  • Maple
    f := proc(n,i,x) option remember ; if n = 0 then x; elif type(n,'even') then procname(n/2,i+1,x) ; else procname((n-1)/2,i,x*ithprime(i)) ; end if; end proc:
    A005940 := proc(n) f(n-1,1,1) ; end proc: # R. J. Mathar, Mar 06 2010
  • Mathematica
    f[n_] := Block[{p = Partition[ Split[ Join[ IntegerDigits[n - 1, 2], {2}]], 2]}, Times @@ Flatten[ Table[q = Take[p, -i]; Prime[ Count[ Flatten[q], 0] + 1]^q[[1, 1]], {i, Length[p]}] ]]; Table[ f[n], {n, 67}] (* Robert G. Wilson v, Feb 22 2005 *)
    Table[Times@@Prime/@(Join@@Position[Reverse[IntegerDigits[n,2]],1]-Range[DigitCount[n,2,1]]+1),{n,0,100}] (* Gus Wiseman, Dec 28 2022 *)
  • PARI
    A005940(n) = { my(p=2, t=1); n--; until(!n\=2, n%2 && (t*=p) || p=nextprime(p+1)); t } \\ M. F. Hasler, Mar 07 2010; update Aug 29 2014
    
  • PARI
    a(n)=my(p=2, t=1); for(i=0,exponent(n), if(bittest(n,i), t*=p, p=nextprime(p+1))); t \\ Charles R Greathouse IV, Nov 11 2021
    
  • Python
    from sympy import prime
    import math
    def A(n): return n - 2**int(math.floor(math.log(n, 2)))
    def b(n): return n + 1 if n<2 else prime(1 + (len(bin(n)[2:]) - bin(n)[2:].count("1"))) * b(A(n))
    print([b(n - 1) for n in range(1, 101)]) # Indranil Ghosh, Apr 10 2017
    
  • Python
    from math import prod
    from itertools import accumulate
    from collections import Counter
    from sympy import prime
    def A005940(n): return prod(prime(len(a)+1)**b for a, b in Counter(accumulate(bin(n-1)[2:].split('1')[:0:-1])).items()) # Chai Wah Wu, Mar 10 2023

Formula

From Reinhard Zumkeller, Aug 23 2006, R. J. Mathar, Mar 06 2010: (Start)
a(n) = f(n-1, 1, 1)
where f(n, i, x) = x if n = 0,
= f(n/2, i+1, x) if n > 0 is even
= f((n-1)/2, i, x*prime(i)) otherwise. (End)
From Antti Karttunen, Jun 26 2014: (Start)
Define a starting-offset 0 version of this sequence as:
b(0)=1, b(1)=2, [base cases]
and then compute the rest either with recurrence:
b(n) = A000040(1+(A070939(n)-A000120(n))) * b(A053645(n)).
or
b(2n) = A003961(b(n)), b(2n+1) = 2 * b(n). [Compare this to the similar recurrence given for A163511.]
Then define a(n) = b(n-1), where a(n) gives this sequence A005940 with the starting offset 1.
Can be also defined as a composition of related permutations:
a(n+1) = A243353(A006068(n)).
a(n+1) = A163511(A054429(n)). [Compare the scatter plots of this sequence and A163511 to each other.]
This permutation also maps between the partitions as enumerated in the lists A125106 and A112798, providing identities between:
A161511(n) = A056239(a(n+1)). [The corresponding sums ...]
A243499(n) = A003963(a(n+1)). [... and the products of parts of those partitions.]
(End)
From Antti Karttunen, Dec 21 2014 - Jan 04 2015: (Start)
A002110(n) = a(1+A002450(n)). [Primorials occur at (4^n - 1)/3 in the offset-0 version of the sequence.]
a(n) = A250246(A252753(n-1)).
a(n) = A122111(A253563(n-1)).
For n >= 1, A055396(a(n+1)) = A001511(n).
For n >= 2, a(n) = A246278(1+A253552(n)).
(End)
From Peter Munn, Oct 04 2020: (Start)
A000265(a(n)) = a(A000265(n)) = A003961(a(A003602(n))).
A006519(a(n)) = a(A006519(n)) = A006519(n).
a(n) = A003961(a(A003602(n))) * A006519(n).
A007814(a(n)) = A007814(n).
A007949(a(n)) = A337821(n) = A007814(A003602(n)).
a(n) = A225546(A334866(n-1)).
(End)
a(2n) = 2*a(n), or generally a(2^k*n) = 2^k*a(n). - Amiram Eldar, Oct 03 2022
If n-1 = Sum_{i} 2^(q_i-1), then a(n) = Product_{i} prime(q_i-i+1). These are the Heinz numbers of the rows of A125106. If the offset is changed to 0, the inverse is A156552. - Gus Wiseman, Dec 28 2022

Extensions

More terms from Robert G. Wilson v, Feb 22 2005
Sign in a formula switched and Maple program added by R. J. Mathar, Mar 06 2010
Binary tree illustration and keyword tabf added by Antti Karttunen, Dec 21 2014

A019565 The squarefree numbers ordered lexicographically by their prime factorization (with factors written in decreasing order). a(n) = Product_{k in I} prime(k+1), where I is the set of indices of nonzero binary digits in n = Sum_{k in I} 2^k.

Original entry on oeis.org

1, 2, 3, 6, 5, 10, 15, 30, 7, 14, 21, 42, 35, 70, 105, 210, 11, 22, 33, 66, 55, 110, 165, 330, 77, 154, 231, 462, 385, 770, 1155, 2310, 13, 26, 39, 78, 65, 130, 195, 390, 91, 182, 273, 546, 455, 910, 1365, 2730, 143, 286, 429, 858, 715, 1430, 2145, 4290
Offset: 0

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Author

Keywords

Comments

A permutation of the squarefree numbers A005117. The missing positive numbers are in A013929. - Alois P. Heinz, Sep 06 2014
From Antti Karttunen, Apr 18 & 19 2017: (Start)
Because a(n) toggles the parity of n there are neither fixed points nor any cycles of odd length.
Conjecture: there are no finite cycles of any length. My grounds for this conjecture: any finite cycle in this sequence, if such cycles exist at all, must have at least one member that occurs somewhere in A285319, the terms that seem already to be quite rare. Moreover, any such a number n should satisfy in addition to A019565(n) < n also that A048675^{k}(n) is squarefree, not just for k=0, 1 but for all k >= 0. As there is on average a probability of only 6/(Pi^2) = 0.6079... that any further term encountered on the trajectory of A048675 is squarefree, the total chance that all of them would be squarefree (which is required from the elements of A019565-cycles) is soon minuscule, especially as A048675 is not very tightly bounded (many trajectories seem to skyrocket, at least initially). I am also assuming that usually there is no significant correlation between the binary expansions of n and A048675(n) (apart from their least significant bits), or, for that matter, between their prime factorizations.
See also the slightly stronger conjecture in A285320, which implies that there would neither be any two-way infinite cycles.
If either of the conjectures is false (there are cycles), then certainly neither sequence A285332 nor its inverse A285331 can be a permutation of natural numbers. (End)
The conjecture made in A087207 (see also A288569) implies the two conjectures mentioned above. A further constraint for cycles is that in any A019565-trajectory which starts from a squarefree number (A005117), every other term is of the form 4k+2, while every other term is of the form 6k+3. - Antti Karttunen, Jun 18 2017
The sequence satisfies the exponential function identity, a(x + y) = a(x) * a(y), whenever x and y do not have a 1-bit in the same position, i.e., when A004198(x,y) = 0. See also A283475. - Antti Karttunen, Oct 31 2019
The above identity becomes unconditional if binary exclusive OR, A003987(.,.), is substituted for addition, and A059897(.,.), a multiplicative equivalent of A003987, is substituted for multiplication. This gives us a(A003987(x,y)) = A059897(a(x), a(y)). - Peter Munn, Nov 18 2019
Also the Heinz number of the binary indices of n, where the Heinz number of a sequence (y_1,...,y_k) is prime(y_1)*...*prime(y_k), and a number's binary indices (A048793) are the positions of 1's in its reversed binary expansion. - Gus Wiseman, Dec 28 2022

Examples

			5 = 2^2+2^0, e_1 = 2, e_2 = 0, prime(2+1) = prime(3) = 5, prime(0+1) = prime(1) = 2, so a(5) = 5*2 = 10.
From _Philippe Deléham_, Jun 03 2015: (Start)
This sequence regarded as a triangle withs rows of lengths 1, 1, 2, 4, 8, 16, ...:
   1;
   2;
   3,  6;
   5, 10, 15, 30;
   7, 14, 21, 42, 35,  70, 105, 210;
  11, 22, 33, 66, 55, 110, 165, 330, 77, 154, 231, 462, 385, 770, 1155, 2310;
  ...
(End)
From _Peter Munn_, Jun 14 2020: (Start)
The initial terms are shown below, equated with the product of their prime factors to exhibit the lexicographic order. We start with 1, since 1 is factored as the empty product and the empty list is first in lexicographic order.
   n     a(n)
   0     1 = .
   1     2 = 2.
   2     3 = 3.
   3     6 = 3*2.
   4     5 = 5.
   5    10 = 5*2.
   6    15 = 5*3.
   7    30 = 5*3*2.
   8     7 = 7.
   9    14 = 7*2.
  10    21 = 7*3.
  11    42 = 7*3*2.
  12    35 = 7*5.
(End)
		

Crossrefs

Row 1 of A285321.
Equivalent sequences for k-th-power-free numbers: A101278 (k=3), A101942 (k=4), A101943 (k=5), A054842 (k=10).
Cf. A109162 (iterates).
Cf. also A048675 (a left inverse), A087207, A097248, A260443, A054841.
Cf. A285315 (numbers for which a(n) < n), A285316 (for which a(n) > n).
Cf. A276076, A276086 (analogous sequences for factorial and primorial bases), A334110 (terms squared).
For partial sums see A288570.
A003961, A003987, A004198, A059897, A089913, A331590, A334747 are used to express relationships between sequence terms.
Column 1 of A329332.
Even bisection (which contains the odd terms): A332382.
A160102 composed with A052330, and subsequence of the latter.
Related to A000079 via A225546, to A057335 via A122111, to A008578 via A336322.
Least prime index of a(n) is A001511.
Greatest prime index of a(n) is A029837 or A070939.
Taking prime indices gives A048793, reverse A272020, row sums A029931.
A112798 lists prime indices, length A001222, sum A056239.

Programs

  • Haskell
    a019565 n = product $ zipWith (^) a000040_list (a030308_row n)
    -- Reinhard Zumkeller, Apr 27 2013
    
  • Maple
    a:= proc(n) local i, m, r; m:=n; r:=1;
          for i while m>0 do if irem(m,2,'m')=1
            then r:=r*ithprime(i) fi od; r
        end:
    seq(a(n), n=0..60);  # Alois P. Heinz, Sep 06 2014
  • Mathematica
    Do[m=1;o=1;k1=k;While[ k1>0, k2=Mod[k1, 2];If[k2\[Equal]1, m=m*Prime[o]];k1=(k1-k2)/ 2;o=o+1];Print[m], {k, 0, 55}] (* Lei Zhou, Feb 15 2005 *)
    Table[Times @@ Prime@ Flatten@ Position[#, 1] &@ Reverse@ IntegerDigits[n, 2], {n, 0, 55}]  (* Michael De Vlieger, Aug 27 2016 *)
    b[0] := {1}; b[n_] := Flatten[{ b[n - 1], b[n - 1] * Prime[n] }];
      a = b[6] (* Fred Daniel Kline, Jun 26 2017 *)
  • PARI
    a(n)=factorback(vecextract(primes(logint(n+!n,2)+1),n))  \\ M. F. Hasler, Mar 26 2011, updated Aug 22 2014, updated Mar 01 2018
    
  • Python
    from operator import mul
    from functools import reduce
    from sympy import prime
    def A019565(n):
        return reduce(mul,(prime(i+1) for i,v in enumerate(bin(n)[:1:-1]) if v == '1')) if n > 0 else 1
    # Chai Wah Wu, Dec 25 2014
    
  • Scheme
    (define (A019565 n) (let loop ((n n) (i 1) (p 1)) (cond ((zero? n) p) ((odd? n) (loop (/ (- n 1) 2) (+ 1 i) (* p (A000040 i)))) (else (loop (/ n 2) (+ 1 i) p))))) ;; (Requires only the implementation of A000040 for prime numbers.) - Antti Karttunen, Apr 20 2017

Formula

G.f.: Product_{k>=0} (1 + prime(k+1)*x^2^k), where prime(k)=A000040(k). - Ralf Stephan, Jun 20 2003
a(n) = f(n, 1, 1) with f(x, y, z) = if x > 0 then f(floor(x/2), y*prime(z)^(x mod 2), z+1) else y. - Reinhard Zumkeller, Mar 13 2010
For all n >= 0: A048675(a(n)) = n; A013928(a(n)) = A064273(n). - Antti Karttunen, Jul 29 2015
a(n) = a(2^x)*a(2^y)*a(2^z)*... = prime(x+1)*prime(y+1)*prime(z+1)*..., where n = 2^x + 2^y + 2^z + ... - Benedict W. J. Irwin, Jul 24 2016
From Antti Karttunen, Apr 18 2017 and Jun 18 2017: (Start)
a(n) = A097248(A260443(n)), a(A005187(n)) = A283475(n), A108951(a(n)) = A283477(n).
A055396(a(n)) = A001511(n), a(A087207(n)) = A007947(n). (End)
a(2^n - 1) = A002110(n). - Michael De Vlieger, Jul 05 2017
a(n) = A225546(A000079(n)). - Peter Munn, Oct 31 2019
From Peter Munn, Mar 04 2022: (Start)
a(2n) = A003961(a(n)); a(2n+1) = 2*a(2n).
a(x XOR y) = A059897(a(x), a(y)) = A089913(a(x), a(y)), where XOR denotes bitwise exclusive OR (A003987).
a(n+1) = A334747(a(n)).
a(x+y) = A331590(a(x), a(y)).
a(n) = A336322(A008578(n+1)).
(End)

Extensions

Definition corrected by Klaus-R. Löffler, Aug 20 2014
New name from Peter Munn, Jun 14 2020

A029931 If 2n = Sum 2^e_i, a(n) = Sum e_i.

Original entry on oeis.org

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

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Author

Keywords

Comments

Write n in base 2, n = sum b(i)*2^(i-1), then a(n) = sum b(i)*i. - Benoit Cloitre, Jun 09 2002
May be regarded as a triangular array read by rows, giving weighted sum of compositions in standard order. The standard order of compositions is given by A066099. - Franklin T. Adams-Watters, Nov 06 2006
Sum of all positive integer roots m_i of polynomial {m,k} - see link [Shevelev]; see also A264613. - Vladimir Shevelev, Dec 13 2015
Also the sum of binary indices of n, where a binary index of n (A048793) is any position of a 1 in its reversed binary expansion. For example, the binary indices of 11 are {1,2,4}, so a(11) = 7. - Gus Wiseman, May 22 2024

Examples

			14 = 8+4+2 so a(7) = 3+2+1 = 6.
Composition number 11 is 2,1,1; 1*2+2*1+3*1 = 7, so a(11) = 7.
The triangle starts:
  0
  1
  2 3
  3 4 5 6
The reversed binary expansion of 18 is (0,1,0,0,1) with 1's at positions {2,5}, so a(18) = 2 + 5 = 7. - _Gus Wiseman_, Jul 22 2019
		

Crossrefs

Other sequences that are built by replacing 2^k in the binary representation with other numbers: A022290 (Fibonacci), A059590 (factorials), A073642, A089625 (primes), A116549, A326031.
Cf. A001793 (row sums), A011782 (row lengths), A059867, A066099, A124757.
Row sums of A048793 and A272020.
Contains exactly A000009(n) copies of n.
For length instead of sum we have A000120, complement A023416.
For minimum instead of sum we have A001511, opposite A000012.
For maximum instead of sum we have A029837 or A070939, opposite A070940.
For product instead of sum we have A096111.
The reverse version is A230877, row sums of A371572.
The reverse complement is A359359, row sums of A371571.
The complement is A359400, row sums of A368494.
Numbers k such that a(k) is prime are A372689.
A014499 lists binary indices of prime numbers.
A019565 gives Heinz number of binary indices, inverse A048675.
A372471 lists binary indices of primes, row-sums A372429.

Programs

  • Haskell
    a029931 = sum . zipWith (*) [1..] . a030308_row
    -- Reinhard Zumkeller, Feb 28 2014
    
  • Maple
    HammingWeight := n -> add(i, i = convert(n, base, 2)):
    a := proc(n) option remember; `if`(n = 0, 0,
    ifelse(n::even, a(n/2) + HammingWeight(n/2), a(n-1) + 1)) end:
    seq(a(n), n = 0..78); # Peter Luschny, Oct 30 2021
  • Mathematica
    a[n_] := (b = IntegerDigits[n, 2]).Reverse @ Range[Length @ b]; Array[a,78,0] (* Jean-François Alcover, Apr 28 2011, after B. Cloitre *)
  • PARI
    for(n=0,100,l=length(binary(n)); print1(sum(i=1,l, component(binary(n),i)*(l-i+1)),","))
    
  • PARI
    a(n) = my(b=binary(n)); b*-[-#b..-1]~; \\ Ruud H.G. van Tol, Oct 17 2023
    
  • Python
    def A029931(n): return sum(i if j == '1' else 0 for i, j in enumerate(bin(n)[:1:-1],1)) # Chai Wah Wu, Dec 20 2022
    (C#)
    ulong A029931(ulong n) {
        ulong result = 0, counter = 1;
        while(n > 0) {
            if (n % 2 == 1)
              result += counter;
            counter++;
            n /= 2;
        }
        return result;
    } // Frank Hollstein, Jan 07 2023

Formula

a(n) = a(n - 2^L(n)) + L(n) + 1 [where L(n) = floor(log_2(n)) = A000523(n)] = sum of digits of A048794 [at least for n < 512]. - Henry Bottomley, Mar 09 2001
a(0) = 0, a(2n) = a(n) + e1(n), a(2n+1) = a(2n) + 1, where e1(n) = A000120(n). a(n) = log_2(A029930(n)). - Ralf Stephan, Jun 19 2003
G.f.: (1/(1-x)) * Sum_{k>=0} (k+1)*x^2^k/(1+x^2^k). - Ralf Stephan, Jun 23 2003
a(n) = Sum_{k>=0} A030308(n,k)*A000027(k+1). - Philippe Deléham, Oct 15 2011
a(n) = sum of n-th row of the triangle in A213629. - Reinhard Zumkeller, Jun 17 2012
From Reinhard Zumkeller, Feb 28 2014: (Start)
a(A089633(n)) = n and a(m) != n for m < A089633(n).
a(n) = Sum_{k=1..A070939(n)} k*A030308(n,k-1). (End)
a(n) = A073642(n) + A000120(n). - Peter Kagey, Apr 04 2016

Extensions

More terms from Erich Friedman

A333489 Numbers k such that the k-th composition in standard order is an anti-run (no adjacent equal parts).

Original entry on oeis.org

0, 1, 2, 4, 5, 6, 8, 9, 12, 13, 16, 17, 18, 20, 22, 24, 25, 32, 33, 34, 37, 38, 40, 41, 44, 45, 48, 49, 50, 52, 54, 64, 65, 66, 68, 69, 70, 72, 76, 77, 80, 81, 82, 88, 89, 96, 97, 98, 101, 102, 104, 105, 108, 109, 128, 129, 130, 132, 133, 134, 137, 140, 141
Offset: 1

Views

Author

Gus Wiseman, Mar 28 2020

Keywords

Comments

A composition of n is a finite sequence of positive integers summing to n. The k-th composition in standard order (row k of 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.

Examples

			The sequence together with the corresponding compositions begins:
    0: ()          33: (5,1)         70: (4,1,2)
    1: (1)         34: (4,2)         72: (3,4)
    2: (2)         37: (3,2,1)       76: (3,1,3)
    4: (3)         38: (3,1,2)       77: (3,1,2,1)
    5: (2,1)       40: (2,4)         80: (2,5)
    6: (1,2)       41: (2,3,1)       81: (2,4,1)
    8: (4)         44: (2,1,3)       82: (2,3,2)
    9: (3,1)       45: (2,1,2,1)     88: (2,1,4)
   12: (1,3)       48: (1,5)         89: (2,1,3,1)
   13: (1,2,1)     49: (1,4,1)       96: (1,6)
   16: (5)         50: (1,3,2)       97: (1,5,1)
   17: (4,1)       52: (1,2,3)       98: (1,4,2)
   18: (3,2)       54: (1,2,1,2)    101: (1,3,2,1)
   20: (2,3)       64: (7)          102: (1,3,1,2)
   22: (2,1,2)     65: (6,1)        104: (1,2,4)
   24: (1,4)       66: (5,2)        105: (1,2,3,1)
   25: (1,3,1)     68: (4,3)        108: (1,2,1,3)
   32: (6)         69: (4,2,1)      109: (1,2,1,2,1)
		

Crossrefs

Anti-runs summing to n are counted by A003242(n).
A triangle counting maximal anti-runs of compositions is A106356.
A triangle counting maximal runs of compositions is A238279 or A238130.
Partitions whose first differences are an anti-run are A238424.
All of the following pertain to compositions in standard order (A066099):
- Adjacent equal pairs are counted by A124762.
- Weakly decreasing runs are counted by A124765.
- Weakly increasing runs are counted by A124766.
- Equal runs are counted by A124767.
- Strictly increasing runs are counted by A124768.
- Strictly decreasing runs are counted by A124769.
- Strict compositions are ranked by A233564.
- Constant compositions are ranked by A272919.
- Normal compositions are ranked by A333217.
- Anti-runs are counted by A333381.
- Adjacent unequal pairs are counted by A333382.

Programs

  • Mathematica
    stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    Select[Range[0,100],!MatchQ[stc[#],{_,x_,x_,_}]&]

A106356 Triangle T(n,k) 0<=k

Original entry on oeis.org

1, 1, 1, 3, 0, 1, 4, 3, 0, 1, 7, 6, 2, 0, 1, 14, 7, 8, 2, 0, 1, 23, 20, 10, 8, 2, 0, 1, 39, 42, 22, 13, 9, 2, 0, 1, 71, 72, 58, 28, 14, 10, 2, 0, 1, 124, 141, 112, 72, 33, 16, 11, 2, 0, 1, 214, 280, 219, 150, 92, 36, 18, 12, 2, 0, 1, 378, 516, 466, 311, 189, 112, 40, 20, 13, 2, 0, 1
Offset: 1

Views

Author

Christian G. Bower, Apr 29 2005

Keywords

Comments

For n > 0, also the number of compositions of n with k + 1 maximal anti-runs (sequences without adjacent equal terms). - Gus Wiseman, Mar 23 2020

Examples

			T(4,1) = 3 because the compositions of 4 with 1 adjacent equal part are 1+1+2, 2+1+1, 2+2.
Triangle begins:
   1;
   1,  1;
   3,  0,  1;
   4,  3,  0, 1;
   7,  6,  2, 0, 1;
  14,  7,  8, 2, 0, 1;
  23, 20, 10, 8, 2, 0, 1;
  ...
From _Gus Wiseman_, Mar 23 2020 (Start)
Row n = 6 counts the following compositions (empty column shown by dot):
  (6)     (33)    (222)    (11112)  .  (111111)
  (15)    (114)   (1113)   (21111)
  (24)    (411)   (1122)
  (42)    (1131)  (2211)
  (51)    (1221)  (3111)
  (123)   (1311)  (11121)
  (132)   (2112)  (11211)
  (141)           (12111)
  (213)
  (231)
  (312)
  (321)
  (1212)
  (2121)
(End)
		

Crossrefs

Row sums: 2^(n-1)=A000079(n-1). Columns 0-4: A003242, A106357-A106360.
The version counting adjacent unequal parts is A238279.
The k-th composition in standard-order has A124762(k) adjacent equal parts and A333382(k) adjacent unequal parts.
The k-th composition in standard-order has A124767(k) maximal runs and A333381(k) maximal anti-runs.
The version for ascents/descents is A238343.
The version for weak ascents/descents is A333213.

Programs

  • Maple
    b:= proc(n, h, t) option remember;
          if n=0 then `if`(t=0, 1, 0)
        elif t<0 then 0
        else add(b(n-j, j, `if`(j=h, t-1, t)), j=1..n)
          fi
        end:
    T:= (n, k)-> b(n, -1, k):
    seq(seq(T(n, k), k=0..n-1), n=1..15); # Alois P. Heinz, Oct 23 2011
  • Mathematica
    b[n_, h_, t_] := b[n, h, t] = If[n == 0, If[t == 0, 1, 0], If[t<0, 0, Sum[b[n-j, j, If [j == h, t-1, t]], {j, 1, n}]]]; T[n_, k_] := b[n, -1, k]; Table[Table[T[n, k], {k, 0, n-1}], {n, 1, 15}] // Flatten (* Jean-François Alcover, Feb 20 2015, after Alois P. Heinz *)
    Table[Length[Select[Join@@Permutations/@IntegerPartitions[n],n==0||Length[Split[#,#1!=#2&]]==k+1&]],{n,0,12},{k,0,n}] (* Gus Wiseman, Mar 23 2020 *)

A272919 Numbers of the form 2^(n-1)*(2^(n*m)-1)/(2^n-1), n >= 1, m >= 1.

Original entry on oeis.org

1, 2, 3, 4, 7, 8, 10, 15, 16, 31, 32, 36, 42, 63, 64, 127, 128, 136, 170, 255, 256, 292, 511, 512, 528, 682, 1023, 1024, 2047, 2048, 2080, 2184, 2340, 2730, 4095, 4096, 8191, 8192, 8256, 10922, 16383, 16384, 16912, 18724, 32767, 32768, 32896, 34952, 43690, 65535, 65536, 131071
Offset: 1

Views

Author

Ivan Neretin, May 10 2016

Keywords

Comments

In other words, numbers whose binary representation consists of one or more repeating blocks with only one 1 in each block.
Also, fixed points of the permutations A139706 and A139708.
Each a(n) is a term of A064896 multiplied by some power of 2. As such, this sequence must also be a subsequence of A125121.
Also the numbers that uniquely index a Haar graph (i.e., 5 and 6 are not in the sequence since H(5) is isomorphic to H(6)). - Eric W. Weisstein, Aug 19 2017
From Gus Wiseman, Apr 04 2020: (Start)
The k-th composition in standard order (row k of 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. This sequence lists all positive integers k such that the k-th composition in standard order is constant. For example, the sequence together with the corresponding constant compositions begins:
0: () 136: (4,4)
1: (1) 170: (2,2,2,2)
2: (2) 255: (1,1,1,1,1,1,1,1)
3: (1,1) 256: (9)
4: (3) 292: (3,3,3)
7: (1,1,1) 511: (1,1,1,1,1,1,1,1,1)
8: (4) 512: (10)
10: (2,2) 528: (5,5)
15: (1,1,1,1) 682: (2,2,2,2,2)
16: (5) 1023: (1,1,1,1,1,1,1,1,1,1)
31: (1,1,1,1,1) 1024: (11)
32: (6) 2047: (1,1,1,1,1,1,1,1,1,1,1)
36: (3,3) 2048: (12)
42: (2,2,2) 2080: (6,6)
63: (1,1,1,1,1,1) 2184: (4,4,4)
64: (7) 2340: (3,3,3,3)
127: (1,1,1,1,1,1,1) 2730: (2,2,2,2,2,2)
128: (8) 4095: (1,1,1,1,1,1,1,1,1,1,1,1)
(End)

Crossrefs

Cf. A137706 (smallest number indexing a new Haar graph).
Compositions in standard order are A066099.
Strict compositions are ranked by A233564.

Programs

  • Maple
    N:= 10^6: # to get all terms <= N
    R:= select(`<=`,{seq(seq(2^(n-1)*(2^(n*m)-1)/(2^n-1), m = 1 .. ilog2(2*N)/n), n = 1..ilog2(2*N))},N):
    sort(convert(R,list)); # Robert Israel, May 10 2016
  • Mathematica
    Flatten@Table[d = Reverse@Divisors[n]; 2^(d - 1)*(2^n - 1)/(2^d - 1), {n, 17}]

Formula

From Gus Wiseman, Apr 04 2020: (Start)
A333381(a(n)) = A027750(n).
For n > 0, A124767(a(n)) = 1.
If n is a power of two, A333628(a(n)) = 0, otherwise = 1.
A333627(a(n)) is a power of 2.
(End)

A124767 Number of level runs for compositions in standard order.

Original entry on oeis.org

0, 1, 1, 1, 1, 2, 2, 1, 1, 2, 1, 2, 2, 3, 2, 1, 1, 2, 2, 2, 2, 2, 3, 2, 2, 3, 2, 3, 2, 3, 2, 1, 1, 2, 2, 2, 1, 3, 3, 2, 2, 3, 1, 2, 3, 4, 3, 2, 2, 3, 3, 3, 3, 3, 4, 3, 2, 3, 2, 3, 2, 3, 2, 1, 1, 2, 2, 2, 2, 3, 3, 2, 2, 2, 2, 3, 3, 4, 3, 2, 2, 3, 3, 3, 2, 2, 3, 2, 3, 4, 3, 4, 3, 4, 3, 2, 2, 3, 3, 3, 2, 4, 4, 3, 3
Offset: 0

Views

Author

Keywords

Comments

The standard order of compositions is given by A066099.
For n > 0, a(n) is one more than the number of adjacent unequal terms in the n-th composition in standard order. Also the number of runs in the same composition. - Gus Wiseman, Apr 08 2020

Examples

			Composition number 11 is 2,1,1; the level runs are 2; 1,1; so a(11) = 2.
The table starts:
  0
  1
  1 1
  1 2 2 1
  1 2 1 2 2 3 2 1
  1 2 2 2 2 2 3 2 2 3 2 3 2 3 2 1
  1 2 2 2 1 3 3 2 2 3 1 2 3 4 3 2 2 3 3 3 3 3 4 3 2 3 2 3 2 3 2 1
The 1234567th composition in standard order is (3,2,1,2,2,1,2,5,1,1,1) with runs ((3),(2),(1),(2,2),(1),(2),(5),(1,1,1)), so a(1234567) = 8. - _Gus Wiseman_, Apr 08 2020
		

Crossrefs

Row-lengths are A011782.
Compositions counted by number of runs are A238279 or A333755.
All of the following pertain to compositions in standard order (A066099):
- Length is A000120.
- Sum is A070939.
- Weakly decreasing compositions are A114994.
- Adjacent equal pairs are counted by A124762.
- Weakly decreasing runs are counted by A124765.
- Weakly increasing runs are counted by A124766.
- Equal runs are counted by A124767 (this sequence).
- Weakly increasing compositions are A225620.
- Strict compositions A233564.
- Constant compositions are A272919.
- Anti-runs are counted by A333381.
- Adjacent unequal pairs are counted by A333382.
- Anti-run compositions are A333489.
- Runs-resistance is A333628.
- Run-lengths are A333769 (triangle).

Programs

  • Mathematica
    stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    Table[Length[Split[stc[n]]],{n,0,100}] (* Gus Wiseman, Apr 17 2020 *)

Formula

a(0) = 0, a(n) = 1 + Sum_{1<=i=1 0.
For n > 0, a(n) = A333382(n) + 1. - Gus Wiseman, Apr 08 2020

A333755 Triangle read by rows where T(n,k) is the number of compositions of n with k runs, n >= 0, 0 <= k <= n.

Original entry on oeis.org

1, 0, 1, 0, 2, 0, 0, 2, 2, 0, 0, 3, 4, 1, 0, 0, 2, 10, 4, 0, 0, 0, 4, 12, 14, 2, 0, 0, 0, 2, 22, 29, 10, 1, 0, 0, 0, 4, 26, 56, 36, 6, 0, 0, 0, 0, 3, 34, 100, 86, 31, 2, 0, 0, 0, 0, 4, 44, 148, 200, 99, 16, 1, 0, 0, 0, 0, 2, 54, 230, 374, 278, 78, 8, 0, 0, 0, 0
Offset: 0

Views

Author

Gus Wiseman, Apr 10 2020

Keywords

Comments

Except for a(1) = 0, the data is identical to A238130 shifted right once. However, in A238130, each row after the first ends with a zero, while here each row after the first starts with a zero.

Examples

			Triangle begins:
   1
   0   1
   0   2   0
   0   2   2   0
   0   3   4   1   0
   0   2  10   4   0   0
   0   4  12  14   2   0   0
   0   2  22  29  10   1   0   0
   0   4  26  56  36   6   0   0   0
   0   3  34 100  86  31   2   0   0   0
   0   4  44 148 200  99  16   1   0   0   0
   0   2  54 230 374 278  78   8   0   0   0   0
Row n = 6 counts the following compositions (empty column indicated by dot):
  .  (6)       (15)     (123)    (1212)
     (33)      (24)     (132)    (2121)
     (222)     (42)     (141)
     (111111)  (51)     (213)
               (114)    (231)
               (411)    (312)
               (1113)   (321)
               (1122)   (1131)
               (2211)   (1221)
               (3111)   (1311)
               (11112)  (2112)
               (21111)  (11121)
                        (11211)
                        (12111)
		

Crossrefs

Removing all zeros gives A238279.
The version for anti-runs is A106356.
The k-th composition in standard-order has A124767(k) runs.
The version counting descents is A238343.
The version counting weak ascents is A333213.

Programs

  • Mathematica
    Table[Length[Select[Join@@Permutations/@IntegerPartitions[n],Length[Split[#]]==k&]],{n,0,10},{k,0,n}]

A233564 c-squarefree numbers: positive integers which in binary are concatenation of distinct parts of the form 10...0 with nonnegative number of zeros.

Original entry on oeis.org

0, 1, 2, 4, 5, 6, 8, 9, 12, 16, 17, 18, 20, 24, 32, 33, 34, 37, 38, 40, 41, 44, 48, 50, 52, 64, 65, 66, 68, 69, 70, 72, 80, 81, 88, 96, 98, 104, 128, 129, 130, 132, 133, 134, 137, 140, 144, 145, 152, 160, 161, 176, 192, 194, 196, 200, 208, 256, 257, 258, 260, 261
Offset: 1

Views

Author

Vladimir Shevelev, Dec 13 2013

Keywords

Comments

Number of terms in interval [2^(n-1), 2^n) is the number of compositions of n with distinct parts (cf. A032020). For example, if n=6, then interval [2^5, 2^6) contains 11 terms {32,...,52}. This corresponds to 11 compositions with distinct parts of 6: 6, 5+1, 1+5, 4+2, 2+4, 3+2+1, 3+1+2, 2+3+1, 2+1+3, 1+3+2, 1+2+3.
From Gus Wiseman, Apr 06 2020: (Start)
The k-th composition in standard order (row k of 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. This sequence lists all numbers k such that the k-th composition in standard order is strict. For example, the sequence together with the corresponding strict compositions begins:
0: () 38: (3,1,2) 98: (1,4,2)
1: (1) 40: (2,4) 104: (1,2,4)
2: (2) 41: (2,3,1) 128: (8)
4: (3) 44: (2,1,3) 129: (7,1)
5: (2,1) 48: (1,5) 130: (6,2)
6: (1,2) 50: (1,3,2) 132: (5,3)
8: (4) 52: (1,2,3) 133: (5,2,1)
9: (3,1) 64: (7) 134: (5,1,2)
12: (1,3) 65: (6,1) 137: (4,3,1)
16: (5) 66: (5,2) 140: (4,1,3)
17: (4,1) 68: (4,3) 144: (3,5)
18: (3,2) 69: (4,2,1) 145: (3,4,1)
20: (2,3) 70: (4,1,2) 152: (3,1,4)
24: (1,4) 72: (3,4) 160: (2,6)
32: (6) 80: (2,5) 161: (2,5,1)
33: (5,1) 81: (2,4,1) 176: (2,1,5)
34: (4,2) 88: (2,1,4) 192: (1,7)
37: (3,2,1) 96: (1,6) 194: (1,5,2)
(End)

Examples

			49 in binary has the following parts of the form 10...0 with nonnegative number of  zeros: (1),(1000),(1). Two of them are the same. So it is not in the sequence. On the other hand, 50 has distinct parts (1)(100)(10), thus it is a term.
		

Crossrefs

A subset of A333489 and superset of A333218.
All of the following pertain to compositions in standard order (A066099):
- Length is A000120.
- Weighted sum is A029931.
- Partial sums from the right are A048793.
- Sum is A070939.
- Runs are counted by A124767.
- Reversed initial intervals A164894.
- Initial intervals are A246534.
- Constant compositions are A272919.
- Strictly decreasing compositions are A333255.
- Strictly increasing compositions are A333256.
- Anti-runs are counted by A333381.
- Anti-runs are A333489.

Programs

  • Mathematica
    bitPatt[n_]:=bitPatt[n]=Split[IntegerDigits[n,2],#1>#2||#2==0&];
    Select[Range[0,300],bitPatt[#]==DeleteDuplicates[bitPatt[#]]&] (* Peter J. C. Moses, Dec 13 2013 *)
    stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    Select[Range[0,100],UnsameQ@@stc[#]&] (* Gus Wiseman, Apr 04 2020 *)

Extensions

More terms from Peter J. C. Moses, Dec 13 2013
0 prepended by Gus Wiseman, Apr 04 2020

A333217 Numbers k such that the k-th composition in standard order covers an initial interval of positive integers.

Original entry on oeis.org

0, 1, 3, 5, 6, 7, 11, 13, 14, 15, 21, 22, 23, 26, 27, 29, 30, 31, 37, 38, 41, 43, 44, 45, 46, 47, 50, 52, 53, 54, 55, 58, 59, 61, 62, 63, 75, 77, 78, 83, 85, 86, 87, 89, 90, 91, 92, 93, 94, 95, 101, 102, 105, 106, 107, 108, 109, 110, 111, 114, 116, 117, 118
Offset: 1

Views

Author

Gus Wiseman, Mar 15 2020

Keywords

Comments

The k-th composition in standard order (row k of 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.

Examples

			The sequence of terms together with the corresponding compositions begins:
    0: ()              37: (3,2,1)           75: (3,2,1,1)
    1: (1)             38: (3,1,2)           77: (3,1,2,1)
    3: (1,1)           41: (2,3,1)           78: (3,1,1,2)
    5: (2,1)           43: (2,2,1,1)         83: (2,3,1,1)
    6: (1,2)           44: (2,1,3)           85: (2,2,2,1)
    7: (1,1,1)         45: (2,1,2,1)         86: (2,2,1,2)
   11: (2,1,1)         46: (2,1,1,2)         87: (2,2,1,1,1)
   13: (1,2,1)         47: (2,1,1,1,1)       89: (2,1,3,1)
   14: (1,1,2)         50: (1,3,2)           90: (2,1,2,2)
   15: (1,1,1,1)       52: (1,2,3)           91: (2,1,2,1,1)
   21: (2,2,1)         53: (1,2,2,1)         92: (2,1,1,3)
   22: (2,1,2)         54: (1,2,1,2)         93: (2,1,1,2,1)
   23: (2,1,1,1)       55: (1,2,1,1,1)       94: (2,1,1,1,2)
   26: (1,2,2)         58: (1,1,2,2)         95: (2,1,1,1,1,1)
   27: (1,2,1,1)       59: (1,1,2,1,1)      101: (1,3,2,1)
   29: (1,1,2,1)       61: (1,1,1,2,1)      102: (1,3,1,2)
   30: (1,1,1,2)       62: (1,1,1,1,2)      105: (1,2,3,1)
   31: (1,1,1,1,1)     63: (1,1,1,1,1,1)    106: (1,2,2,2)
		

Crossrefs

Sequences covering an initial interval are counted by A000670.
Composition in standard order are A066099.
The case of strictly increasing initial intervals is A164894.
The case of strictly decreasing initial intervals is A246534.
The case of permutations is A333218.
The weakly increasing version is A333379.
The weakly decreasing version is A333380.

Programs

  • Mathematica
    normQ[m_]:=Or[m=={},Union[m]==Range[Max[m]]];
    stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    Select[Range[0,100],normQ[stc[#]]&]
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