A358170 -id:A358170 - cp's OEIS frontend

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.

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

N. J. A. Sloane and J. H. Conway

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

			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)

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

Antti Karttunen, Table of n, a(n) for n = 1..8192 (terms 1..1024 from Reinhard Zumkeller)
Michael De Vlieger, 6 row Doudna Tree diagram as mentioned in Comments.
Michael De Vlieger, Annotated fan-style binary tree showing 10 levels, with a color function where 2^m is shown in medium blue in row m, k < 2^m is darker blue, and k > 2^m is brighter green, with records in each row shown in red.
Ronald E. Kutz, Two unusual sequences, Two-Year College Mathematics Journal, Vol. 12, No. 5 (1981), pp. 316-319.
Index entries for sequences that are permutations of the natural numbers

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. also A000142, A001511, A002450, A112798, A252463, A252464, A252745, A252750, A324054, A324106, A323505, A323508.
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).
Cf. A001222, A029837, A059893, A241916, A242628, A253566, A359042.

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

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

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 *)

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

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

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

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

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

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

Marc LeBrun

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

			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)

Reinhard Zumkeller, Table of n, a(n) for n = 0..8191

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. A007088, A030308, A000040, A013929, A005117, A103785, A103786, A110765, A064273, A246353, A283475, A283477, A285319, A285331, A285332, A288569, A293442.
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.
Cf. A000120, A000720, A005940, A066099, A358137, A358170.

a019565 n = product $ zipWith (^) a000040_list (a030308_row n)
-- Reinhard Zumkeller, Apr 27 2013

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

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 *)

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

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

(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

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)

Definition corrected by Klaus-R. Löffler, Aug 20 2014

New name from Peter Munn, Jun 14 2020

A253565 Permutation of natural numbers: a(0) = 1, a(1) = 2; after which, a(2n) = A253550(a(n)), a(2n+1) = A253560(a(n)).

Original entry on oeis.org

1, 2, 3, 4, 5, 9, 6, 8, 7, 25, 15, 27, 10, 18, 12, 16, 11, 49, 35, 125, 21, 75, 45, 81, 14, 50, 30, 54, 20, 36, 24, 32, 13, 121, 77, 343, 55, 245, 175, 625, 33, 147, 105, 375, 63, 225, 135, 243, 22, 98, 70, 250, 42, 150, 90, 162, 28, 100, 60, 108, 40, 72, 48, 64, 17, 169, 143, 1331, 91, 847, 539, 2401, 65, 605, 385, 1715, 275, 1225, 875, 3125, 39

Offset: 0

Antti Karttunen, Jan 03 2015

This sequence can be represented as a binary tree. Each child to the left is obtained by applying A253550 to the parent, and each child to the right is obtained by applying A253560 to the parent:
                                    1
                                    |
                 ...................2...................
                3                                       4
      5......../ \........9                   6......../ \........8
     / \                 / \                 / \                 / \
    /   \               /   \               /   \               /   \
   /     \             /     \             /     \             /     \
  7       25         15       27         10       18         12       16
11 49   35  125    21  75   45  81     14  50   30  54     20  36   24  32
etc.
Sequence A253563 is the mirror image of the same tree. Also in binary trees A005940 and A163511 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. Of these four trees, this is the one where the left child is always smaller than the right child.
Note that the indexing of sequence starts from 0, although its range starts from one.
The term a(n) is the Heinz number of the adjusted partial sums of the n-th composition in standard order, where (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, (2) the Heinz number of a partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k), and (3) we define the adjusted partial sums of a composition to be obtained by subtracting one from all parts, taking partial sums, and adding one back to all parts. See formula for a simplification. A triangular form is A242628. The inverse is A253566. The non-adjusted version is A358170. - Gus Wiseman, Dec 17 2022

			From _Gus Wiseman_, Dec 23 2022: (Start)
This represents the following bijection between compositions and partitions. The n-th composition in standard order together with the reversed prime indices of a(n) are:
   0:        () -> ()
   1:       (1) -> (1)
   2:       (2) -> (2)
   3:     (1,1) -> (1,1)
   4:       (3) -> (3)
   5:     (2,1) -> (2,2)
   6:     (1,2) -> (2,1)
   7:   (1,1,1) -> (1,1,1)
   8:       (4) -> (4)
   9:     (3,1) -> (3,3)
  10:     (2,2) -> (3,2)
  11:   (2,1,1) -> (2,2,2)
  12:     (1,3) -> (3,1)
  13:   (1,2,1) -> (2,2,1)
  14:   (1,1,2) -> (2,1,1)
  15: (1,1,1,1) -> (1,1,1,1)
(End)

Antti Karttunen, Table of n, a(n) for n = 0..8191
Index entries for sequences that are permutations of the natural numbers

Inverse: A253566.
Cf. A252737 (row sums), A252738 (row products).
Cf. A122111, A163511, A253550, A253560, A252464, A253563, A054429.
Applying A001222 gives A000120.
A reverse version is A005940.
These are the Heinz numbers of the rows of A242628.
Sum of prime indices of a(n) is A359043, reverse A161511.
A048793 gives partial sums of reversed standard comps, Heinz number A019565.
A066099 lists standard compositions.
A112798 list prime indices, sum A056239.
A358134 gives partial sums of standard compositions, Heinz number A358170.
Cf. A029837, A059893, A070939, A241916, A358135, A358137, A358195, A359042.

stc[n_]:=Differences[Prepend[Join @@ Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
Times@@Prime/@#&/@Table[Accumulate[stc[n]-1]+1,{n,0,60}] (* Gus Wiseman, Dec 17 2022 *)

a(0) = 1, a(1) = 2; after which, a(2n) = A253550(a(n)), a(2n+1) = A253560(a(n)).
As a composition of related permutations:
a(n) = A122111(A163511(n)).
a(n) = A253563(A054429(n)).
Other identities and observations. For all n >= 0:
a(2n+1) - a(2n) > 0. [See the comment above.]
If n = 2^(x_1)+...+2^(x_k) then a(n) = Product_{i=1..k} prime(x_k-x_{i-1}-k+i) where x_0 = 0. - Gus Wiseman, Dec 23 2022

A359043 Sum of adjusted partial sums of the n-th composition in standard order (A066099). Row sums of A242628.

Original entry on oeis.org

0, 1, 2, 2, 3, 4, 3, 3, 4, 6, 5, 6, 4, 5, 4, 4, 5, 8, 7, 9, 6, 8, 7, 8, 5, 7, 6, 7, 5, 6, 5, 5, 6, 10, 9, 12, 8, 11, 10, 12, 7, 10, 9, 11, 8, 10, 9, 10, 6, 9, 8, 10, 7, 9, 8, 9, 6, 8, 7, 8, 6, 7, 6, 6, 7, 12, 11, 15, 10, 14, 13, 16, 9, 13, 12, 15, 11, 14, 13

Offset: 0

Gus Wiseman, Dec 21 2022

nonn

We define the adjusted partial sums of a composition to be obtained by subtracting one from all parts, taking partial sums, and adding one back to all parts.

The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.

			The 29th composition in standard order is (1,1,2,1), with adjusted partial sums (1,1,2,2), with sum 6, so a(29) = 6.

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

See link for sequences related to standard compositions.
The unadjusted reverse version is A029931, row sums of A048793.
The reverse version is A161511, row sums of A125106.
Row sums of A242628, ranked by A253565.
The unadjusted version is A359042, row sums of A358134.
A011782 counts compositions.
A066099 lists standard compositions.
A358135 gives last minus first of standard compositions.
A358194 counts partitions by sum and weighted sum.
Cf. A000120, A005940, A019565, A029837, A059893, A253566, A358133, A358170.

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

A253566 Permutation of natural numbers: a(n) = A243071(A122111(n)).

Original entry on oeis.org

0, 1, 2, 3, 4, 6, 8, 7, 5, 12, 16, 14, 32, 24, 10, 15, 64, 13, 128, 28, 20, 48, 256, 30, 9, 96, 11, 56, 512, 26, 1024, 31, 40, 192, 18, 29, 2048, 384, 80, 60, 4096, 52, 8192, 112, 22, 768, 16384, 62, 17, 25, 160, 224, 32768, 27, 36, 120, 320, 1536, 65536, 58, 131072, 3072, 44, 63, 72, 104, 262144, 448, 640, 50, 524288, 61, 1048576, 6144, 21

Offset: 1

Antti Karttunen, Jan 03 2015

nonn

Note the indexing: domain starts from one, while the range includes also zero. See also comments in A253564.

The a(n)-th composition in standard order (graded reverse-lexicographic, A066099) is one plus the first differences of the weakly increasing sequence of prime indices of n with 1 prepended. See formula for a simplification. The triangular form is A358169. The inverse is A253565. Not prepending 1 gives A358171. For Heinz numbers instead of standard compositions we have A325351 (without prepending A325352). - Gus Wiseman, Dec 23 2022

			From _Gus Wiseman_, Dec 23 2022: (Start)
This represents the following bijection between partitions and compositions. The reversed prime indices of n together with the a(n)-th composition in standard order are:
   1:        () -> ()
   2:       (1) -> (1)
   3:       (2) -> (2)
   4:     (1,1) -> (1,1)
   5:       (3) -> (3)
   6:     (2,1) -> (1,2)
   7:       (4) -> (4)
   8:   (1,1,1) -> (1,1,1)
   9:     (2,2) -> (2,1)
  10:     (3,1) -> (1,3)
  11:       (5) -> (5)
  12:   (2,1,1) -> (1,1,2)
  13:       (6) -> (6)
  14:     (4,1) -> (1,4)
  15:     (3,2) -> (2,2)
  16: (1,1,1,1) -> (1,1,1,1)
(End)

Antti Karttunen, Table of n, a(n) for n = 1..1024
Index entries for sequences that are permutations of the natural numbers

Inverse: A253565.
Cf. A122111, A243071, A253564, A054429, A336120, A336125.
Applying A000120 gives A001222.
A reverse version is A156552, inverse essentially A005940.
The inverse is A253565, triangular form A242628.
The triangular form is A358169.
A048793 gives partial sums of reversed standard comps, Heinz number A019565.
A066099 lists standard compositions, lengths A000120, sums A070939.
A112798 list prime indices, sum A056239.
A358134 gives partial sums of standard compositions, Heinz number A358170.
Cf. A029837, A241916, A243055, A287352, A325390, A355534, A358171, A359042.

primeMS[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
stcinv[q_]:=Total[2^(Accumulate[Reverse[q]])]/2;
stcinv/@Table[Differences[Prepend[primeMS[n],1]]+1,{n,100}] (* Gus Wiseman, Dec 23 2022 *)

(define (A253566 n) (A243071 (A122111 n)))

a(n) = A243071(A122111(n)).
As a composition of other permutations:
a(n) = A054429(A253564(n)).
a(n) = A336120(n) + A336125(n). - Antti Karttunen, Jul 18 2020
If 2n = Product_{i=1..k} prime(x_i) then a(n) = Sum_{i=1..k-1} 2^(x_k-x_{k-i}+i-1). - Gus Wiseman, Dec 23 2022

cp's OEIS Frontend

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.

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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.

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A253565 Permutation of natural numbers: a(0) = 1, a(1) = 2; after which, a(2n) = A253550(a(n)), a(2n+1) = A253560(a(n)).

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A359043 Sum of adjusted partial sums of the n-th composition in standard order (A066099). Row sums of A242628.

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A253566 Permutation of natural numbers: a(n) = A243071(A122111(n)).

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A358195 Heinz number of the partial sums plus one of the reversed first differences of the prime indices of n.

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