A075157 -id:A075157 - cp's OEIS frontend

A243353 Permutation of natural numbers which maps between the partitions as encoded in A227739 (binary based system, zero-based) to A112798 (prime-index based system, one-based).

1, 2, 4, 3, 9, 8, 6, 5, 25, 18, 16, 27, 15, 12, 10, 7, 49, 50, 36, 75, 81, 32, 54, 125, 35, 30, 24, 45, 21, 20, 14, 11, 121, 98, 100, 147, 225, 72, 150, 245, 625, 162, 64, 243, 375, 108, 250, 343, 77, 70, 60, 105, 135, 48, 90, 175, 55, 42, 40, 63, 33, 28, 22, 13, 169, 242, 196, 363, 441, 200, 294, 605, 1225, 450, 144

Offset: 0

Antti Karttunen, Jun 05 2014

nonn

Note the indexing: the domain includes zero, but the range starts from one.

Antti Karttunen, Table of n, a(n) for n = 0..8192
Index entries for sequences that are permutations of the natural numbers

A243354 gives the inverse mapping.

Cf. A227739, A112798, A075157, A241909, A005940, A003188, A226062, A242424.

f[n_, i_, x_] := Which[n == 0, x, EvenQ@ n, f[n/2, i + 1, x], True, f[(n - 1)/2, i, x Prime@ i]]; Table[f[BitXor[n, Floor[n/2]], 1, 1], {n, 0, 74}] (* Michael De Vlieger, May 09 2017 *)

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))
def a005940(n): return b(n - 1)
def a003188(n): return n^int(n/2)
def a243353(n): return a005940(1 + a003188(n)) # Indranil Ghosh, May 07 2017

(define (A243353 n) (A005940 (+ 1 (A003188 n))))

a(n) = A005940(1+A003188(n)).
a(n) = A241909(1+A075157(n)). [With A075157's original starting offset]
For all n >= 0, A243354(a(n)) = n.
A227183(n) = A056239(a(n)). [Maps between the corresponding sums ...]
A227184(n) = A003963(a(n)). [... and products of parts of each partition].
For n >= 0, a(A037481(n)) = A002110(n). [Also "triangular partitions", the fixed points of Bulgarian solitaire, A226062 & A242424].
For n >= 1, a(A227451(n+1)) = 4*A243054(n).

A227184 a(n) = product of parts of the unordered partition encoded with the runlengths of binary expansion of n.

Original entry on oeis.org

1, 1, 1, 2, 4, 1, 2, 3, 9, 4, 1, 8, 6, 2, 3, 4, 16, 9, 4, 18, 16, 1, 8, 27, 12, 6, 2, 12, 8, 3, 4, 5, 25, 16, 9, 32, 36, 4, 18, 48, 81, 16, 1, 32, 54, 8, 27, 64, 20, 12, 6, 24, 24, 2, 12, 36, 15, 8, 3, 16, 10, 4, 5, 6, 36, 25, 16, 50, 64, 9, 32, 75, 144, 36, 4, 72

Offset: 0

Antti Karttunen, Jul 04 2013

a(0) = 1, as 0 is here considered to encode an empty partition {}, and the empty product is one.
Like A129594, this sequence is based on the fact that compositions (i.e., ordered partitions) can be mapped 1-to-1 to partitions by taking the partial sums of the list where one is subtracted from each composant except the first (originally explained by Marc LeBrun in his Jan 11 2006 post on SeqFan mailing list, with an additional twist involving factorization and prime exponents, cf. A129595). The example below show how this works.
Compare the scatterplot of this sequence to those of A002487, A243353, A243499 and A253552.

			8 has binary expansion "1000", whose runs have lengths [3,1] when arranged from the least significant to the most significant end. Taking partial sums of 3 and 0, we get 3 and 3, whose product is 9, thus a(8) = 9.
For 44, in binary "101100", the run lengths are [2,2,1,1] (from the least significant end), and subtracting one from all terms except the first one, we get [2,1,0,0], whose partial sums are [2,3,3,3], and 2*3*3*3 = 54, thus a(44)=54.

Antti Karttunen, Table of n, a(n) for n = 0..8192

For n>=1, a(n) gives the product of nonzero terms on row n of table A227189/A227739.
Cf. A227183 (gives the corresponding sums).
See also A167489 for a similar sequence, which gives the product of parts of the compositions (ordered partitions).
Cf. A243499, A003963, A243504 (other such product sequences) and A003188, A243353, A075157 (associated permutations mapping between these schemes).
Cf. also A002487, A243353, A253552.

Table[Function[b, Times @@ Accumulate@ Prepend[If[Length@ b > 1, Rest[b] - 1, {}], First@ b]]@ Map[Length, Split@ Reverse@ IntegerDigits[n, 2]], {n, 0, 75}] // Flatten (* Michael De Vlieger, May 09 2017 *)

def A227184(n):
  '''Product of parts of the unique unordered partition encoded in the run lengths of the binary expansion of n.'''
  p = 1
  b = n%2
  i = 1
  while (n != 0):
    n >>= 1
    if ((n%2) == b): i += 1
    else:
      b = n%2
      p *= i
  return(p)

(define (A227184 n) (if (zero? n) 1 (apply * (binexp_to_ascpart n))))
(define (binexp_to_ascpart n) (let ((runlist (reverse! (binexp->runcount1list n)))) (PARTSUMS (cons (car runlist) (map -1+ (cdr runlist))))))
(define (binexp->runcount1list n) (if (zero? n) (list) (let loop ((n n) (rc (list)) (count 0) (prev-bit (modulo n 2))) (if (zero? n) (cons count rc) (if (eq? (modulo n 2) prev-bit) (loop (floor->exact (/ n 2)) rc (1+ count) (modulo n 2)) (loop (floor->exact (/ n 2)) (cons count rc) 1 (modulo n 2)))))))
(define (PARTSUMS a) (cdr (reverse! (fold-left (lambda (psums n) (cons (+ n (car psums)) psums)) (list 0) a))))

Can be also obtained by mapping with an appropriate permutation from the products of parts of each partition computed for other enumerations similar to A227739:
a(n) = A243499(A003188(n)).
a(n) = A003963(A243353(n)).
a(n) = A243504(1+A075157(n)).

A243051 Integer sequence induced by Bulgarian solitaire operation on partition list A241918: a(n) = A241909(A242424(A241909(n))).

Original entry on oeis.org

1, 2, 4, 3, 8, 25, 16, 9, 9, 343, 32, 10, 64, 14641, 125, 27, 128, 15, 256, 98, 2401, 371293, 512, 30, 27, 24137569, 6, 2662, 1024, 147, 2048, 81, 161051, 893871739, 625, 50, 4096, 78310985281, 4826809, 28, 8192, 3993, 16384, 57122, 50, 14507145975869, 32768, 90, 81

Offset: 1

Antti Karttunen, May 29 2014

nonn

In "Bulgarian solitaire" a deck of cards or another finite set of objects is divided into one or more piles, and the "Bulgarian operation" is performed by taking one card from each pile, and making a new pile of them, which is added to the remaining set of piles. Essentially, this operation is a function whose domain and range are unordered integer partitions (cf. A000041) and which preserves the total size of a partition (the sum of its parts). This sequence is induced when the operation is implemented on the partitions as ordered by the list A241918.

			For n = 10, we see that as 10 = 2*5 = p_1^1 * p_2^0 * p_3^1, it encodes a partition [2,2,2]. Applying one step of Bulgarian solitaire (subtract one from each part, and add a new part as large as there were parts in the old partition) to this partition results a new partition [1,1,1,3], which is encoded in the prime factorization of p_1^0 * p_2^0 * p_3^0 * p_4^3 = 7^3 = 343. Thus a(10) = 343.
For n = 46, we see that as 46 = 2*23 = p_1 * p_9 = p_1^1 * p_2^0 * p_3^0 * ... * p_9^1, it encodes a partition [2,2,2,2,2,2,2,2,2]. Applying one step of Bulgarian solitaire to this partition results a new partition [1,1,1,1,1,1,1,1,1,9], which is encoded in the prime factorization of p_1^0 * p_2^0 * ... * p_9^0 * p_10^9 = 29^9 = 14507145975869. Thus a(46) = 14507145975869.
For n = 1875, we see that as 1875 = p_1^0 * p_2^1 * p_3^4, it encodes a partition [1,2,5]. Applying Bulgarian Solitaire, we get a new partition [1,3,4]. This in turn is encoded by p_1^0 * p_2^2 * p_3^2 = 3^2 * 5^2 = 225. Thus a(1875)=225.

Martin Gardner, Colossal Book of Mathematics, Chapter 34, Bulgarian Solitaire and Other Seemingly Endless Tasks, pp. 455-467, W. W. Norton & Company, 2001.

Antti Karttunen, Table of n, a(n) for n = 1..256
Ethan Akin and Morton Davis, "Bulgarian solitaire", American Mathematical Monthly 92 (4): 237-250. (1985).
Wikipedia, Bulgarian solitaire

Row 1 of A243060 (table which gives successive "recursive iterates" of this sequence and converges towards A122111).
Fixed points: A243054.
Cf. A243052, A243053, A243503, A242424, A241909, A226062, A075157, A075158.

a(n) = A241909(A242424(A241909(n))).
a(n) = 1 + A075157(A226062(A075158(n-1))).
A243503(a(n)) = A243503(n) for all n. [Because Bulgarian operation doesn't change the total sum of the partition].

A165199 a(n) is obtained by flipping every second bit in the binary representation of n starting at the second-most significant bit and on downwards.

Original entry on oeis.org

0, 1, 3, 2, 6, 7, 4, 5, 13, 12, 15, 14, 9, 8, 11, 10, 26, 27, 24, 25, 30, 31, 28, 29, 18, 19, 16, 17, 22, 23, 20, 21, 53, 52, 55, 54, 49, 48, 51, 50, 61, 60, 63, 62, 57, 56, 59, 58, 37, 36, 39, 38, 33, 32, 35, 34, 45, 44, 47, 46, 41, 40, 43, 42, 106, 107, 104, 105, 110, 111, 108

Offset: 0

Leroy Quet, Sep 07 2009

This is a self-inverse permutation of the positive integers.
Old name was: a(0) = 0, and for n>=1, let b(n,m) be the m-th digit, reading left to right, of binary n. (b(n, 1) is the most significant binary digit, which is 1.) Then a(n) is such that b(a(n),1)=1; and if b(n,m)=b(n,m-1) then b(a(n),m) does not = b(a(n),m-1); and if b(n,m) does not = b(n,m-1) then b(a(n), m) = b(a(n),m-1), for all m where 2 <= m <= number binary digits in n.
From Emeric Deutsch, Oct 06 2020: (Start)
a(n) is the index of the composition that is the conjugate of the composition with index n.
The index of a composition is defined to be the positive integer whose binary form has run-lengths (i.e., runs of 1's, runs of 0's, etc. from left to right) equal to the parts of the composition. Example: the composition 1,1,3,1 has index 46 since the binary form of 46 is 101110.
a(18) = 24. Indeed, since the binary form of 18 is 10010, the composition with index 18 is 1,2,1,1 (the run-lengths of 10010); the conjugate of 1,2,1,1 is 2,3 and so the binary form of a(18) is 11000; consequently, a(18) = 24. (End)

			a(12) = 9 because 12 = 1100_2 and 1100_2 XOR 0101_2 = 1001_2 = 9.

Cf. A000035, A000523, A075157, A075158, A241909, A243353, A245443, A245444.

{A001477, A129594, A165199, A056539} form a 4-group.

a:= n-> Bits[Xor](n, iquo(2^(1+ilog2(n)), 3)):
seq(a(n), n=0..100);  # Alois P. Heinz, Oct 07 2020

for(k=0,67,my(b(n)=vector(#digits(n,2),i,!(i%2)));print1(bitxor(k,fromdigits(b(k),2)),", ")) \\ Hugo Pfoertner, Oct 07 2020

a(n) = if(n, bitxor(n,2<Kevin Ryde, Oct 07 2020

maxrow <- 8 # by choice
a <- 1
for(m in 0: maxrow) for(k in 0:(2^m-1)){
a[2^(m+1) +       k] = a[2^(m+1) - 1 - k] + 2^(m+1)
a[2^(m+1) + 2^m + k] = a[2^(m+1) - 1 - k] + 2^m
}
(a <- c(0, a))
# Yosu Yurramendi, Apr 04 2017

From Antti Karttunen, Jul 22 2014: (Start)
a(0) = 0, and for n >= 1, a(n) = 2*a(floor(n/2)) + A000035(n+A000523(n)).
As a composition of related permutations:
a(n) = A056539(A129594(n)) = A129594(A056539(n)).
a(n) = A245443(A193231(n)) = A193231(A245444(n)).
a(n) = A075158(A243353(n)-1) = A075158((A241909(1+A075157(n))) - 1).
(End)
a(n) = A258746(A054429(n)) = A054429(A258746(n)), n > 0. - Yosu Yurramendi, Mar 29 2017

Extended by Ray Chandler, Sep 10 2009
a(0) = 0 prepended by Antti Karttunen, Jul 22 2014
New name from Kevin Ryde, Oct 07 2020

A075159 Run lengths in the binary expansion of n-1 gives the vector of exponents in prime factorization of a(n), with the least significant run corresponding to the exponent of the least prime, 2.

Original entry on oeis.org

1, 2, 3, 4, 6, 5, 9, 8, 12, 15, 7, 10, 18, 25, 27, 16, 24, 45, 35, 30, 14, 11, 21, 20, 36, 75, 49, 50, 54, 125, 81, 32, 48, 135, 175, 90, 70, 77, 105, 60, 28, 33, 13, 22, 42, 55, 63, 40, 72, 225, 245, 150, 98, 121, 147, 100, 108, 375, 343, 250, 162, 625, 243, 64, 96, 405

Offset: 1

Antti Karttunen, Sep 13 2002

nonn

To make this a permutation of nonnegative integers, we subtract one from each run count except for the most significant run, e.g. a(12) = 10, as 12-1 = 11 = 1011 and 10 = 5^1 * 3^(1-1) * 2^(2-1).

Index entries for sequences that are permutations of the natural numbers

Inverse of A075160. a(n) = A075157(n-1)+1.

A286617 Restricted growth sequence of A278217 (prime-signature of A075159(1+n)).

Original entry on oeis.org

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

Offset: 0

Antti Karttunen, May 17 2017

Antti Karttunen, Table of n, a(n) for n = 0..65537

Cf. A075159, A278217, A286539, A286619, A286622.

rgs_transform(invec) = { my(occurrences = Map(), outvec = vector(length(invec)), u=1); for(i=1, length(invec), if(mapisdefined(occurrences,invec[i]), my(pp = mapget(occurrences, invec[i])); outvec[i] = outvec[pp] , mapput(occurrences,invec[i],i); outvec[i] = u; u++ )); outvec; };
write_to_bfile(start_offset,vec,bfilename) = { for(n=1, length(vec), write(bfilename, (n+start_offset)-1, " ", vec[n])); }
A005811(n) = hammingweight(bitxor(n, n>>1));  \\ This function from Gheorghe Coserea, Sep 03 2015
A286468(n) = { my(p=((n+1)%2), i=0, m=1); while(n>0, if(((n%2)==p), m *= prime(i), p = (n%2); i = i+1); n = n\2); m };
A075157(n) = if(!n,n,(prime(A005811(n))*A286468(n))-1);
A046523(n) = { my(f=vecsort(factor(n)[, 2], , 4), p); prod(i=1, #f, (p=nextprime(p+1))^f[i]); };  \\ This function from Charles R Greathouse IV, Aug 17 2011
A278217(n) = A046523(1+A075157(n));
write_to_bfile(0,rgs_transform(vector(65538,n,A278217(n-1))),"b286617.txt");

A075162 Position of A014486(n) in A075165, minus one.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 8, 7, 15, 6, 9, 14, 11, 23, 24, 17, 26, 31, 63, 80, 511, 255, 65535, 10, 13, 20, 19, 39, 34, 29, 44, 47, 95, 134, 767, 383, 98303, 48, 49, 74, 35, 71, 124, 53, 242, 127, 1023, 728, 32767, 4095, 16777215, 624, 161, 19682, 33554431, 262143, 6560

Offset: 0

Antti Karttunen, Sep 13 2002

nonn

See A075166.

A. Karttunen, Alternative Catalan Orderings (with the complete Scheme source)
Index entries for sequences that are permutations of the natural numbers

Inverse of A075161. a(n) = A075164(n+1)-1. Cf. A075157, A075169.

A286468 Run lengths in the binary expansion of n are one more than the exponents of corresponding primes in the prime factorization of a(n).

Original entry on oeis.org

1, 1, 2, 2, 1, 3, 4, 4, 3, 1, 2, 6, 5, 9, 8, 8, 9, 5, 6, 2, 1, 3, 4, 12, 15, 7, 10, 18, 25, 27, 16, 16, 27, 25, 18, 10, 7, 15, 12, 4, 3, 1, 2, 6, 5, 9, 8, 24, 45, 35, 30, 14, 11, 21, 20, 36, 75, 49, 50, 54, 125, 81, 32, 32, 81, 125, 54, 50, 49, 75, 36, 20, 21, 11, 14, 30, 35, 45, 24, 8, 9, 5, 6, 2, 1, 3, 4, 12, 15, 7, 10, 18, 25, 27, 16, 48, 135, 175, 90, 70

Offset: 1

Antti Karttunen, May 17 2017

nonn

			For n = 25, "11001" in binary, the lengths of runs from the right are 1, 2 and 2, thus we form a product prime(1)^(1-1) * prime(2)^(2-1) * prime(3)^(2-1) = 2^0 * 3^1 * 5^1 = 15, and a(26) = 15.

Antti Karttunen, Table of n, a(n) for n = 1..65536

Cf. A000079, A003961, A007283, A052126, A075157, A075159.

Cf. A000975 (positions of ones).

A286468(n) = { my(p=((n+1)%2), i=0, m=1); while(n>0, if(((n%2)==p), m *= prime(i), p = (n%2); i = i+1); n = n\2); m };

(define (A286468 n) (fold-left (lambda (a r) (* (A003961 a) (A000079 (- r 1)))) 1 (binexp->runcount1list n)))
(define (binexp->runcount1list n) (if (zero? n) (list) (let loop ((n n) (rc (list)) (count 0) (prev-bit (modulo n 2))) (if (zero? n) (cons count rc) (if (eq? (modulo n 2) prev-bit) (loop (floor->exact (/ n 2)) rc (1+ count) (modulo n 2)) (loop (floor->exact (/ n 2)) (cons count rc) 1 (modulo n 2)))))))
;; Or by implementing the given recurrence:
(define (A286468 n) (cond ((= 1 n) n) ((zero? (modulo n 4)) (* 2 (A286468 (/ n 2)))) ((= 1 (modulo n 4)) (A003961 (A286468 (/ (- n 1) 2)))) ((= 2 (modulo n 4)) (A003961 (A286468 (/ n 2)))) ((= 3 (modulo n 4)) (* 2 (A286468 (/ (- n 1) 2))))))

a(1) = 1; and then after, a(4n) = 2*a(2n), a(4n+1) = A003961(a((n-1)/2)), a(4n+2) = A003961(a(n/2)), a(4n+3) = 2*a((n-1)/2).
a(n) = A052126(1+A075157(n)) = A052126(A075159(1+n)).
Other identities. For all n >= 1:
a(A007283(n)) = A007283(n-1).

A129600 Array A(i,j) of binary run-length encoded product of i and j, read by ascending antidiagonals.

Original entry on oeis.org

0, 1, 1, 2, 3, 2, 3, 4, 4, 3, 4, 7, 6, 7, 4, 5, 8, 8, 8, 8, 5, 6, 11, 12, 15, 12, 11, 6, 7, 12, 9, 16, 16, 9, 12, 7, 8, 15, 14, 23, 24, 23, 14, 15, 8, 9, 16, 16, 24, 19, 19, 24, 16, 16, 9, 10, 19, 24, 31, 28, 13, 28, 31, 24, 19, 10, 11, 20, 17, 32, 32, 17, 17, 32, 32, 17, 20, 11, 12

Offset: 0

Antti Karttunen, May 01 2007

Cf. A003991, A129601.

Cf. A129602 (center diagonal), A014601 (row 1 apart from the first term).

A(i,j) = A075158(((A075157(i)+1)*(A075157(j)+1)) - 1).

Name edited by Michel Marcus, Dec 01 2021

A227351 Permutation of nonnegative integers: map each number by lengths of runs of zeros in its Zeckendorf expansion shifted once left to the number which has the same lengths of runs (in the same order, but alternatively of runs of 0's and 1's) in its binary representation.

Original entry on oeis.org

0, 1, 3, 7, 2, 15, 6, 4, 31, 14, 12, 8, 5, 63, 30, 28, 24, 13, 16, 9, 11, 127, 62, 60, 56, 29, 48, 25, 27, 32, 17, 19, 23, 10, 255, 126, 124, 120, 61, 112, 57, 59, 96, 49, 51, 55, 26, 64, 33, 35, 39, 18, 47, 22, 20, 511, 254, 252, 248, 125, 240, 121, 123, 224

Offset: 0

Antti Karttunen, Jul 08 2013

This permutation is based on the fact that by appending one extra zero to the right of Fibonacci number representation of n (aka "Zeckendorf expansion") and then counting the lengths of blocks of consecutive (nonleading) zeros we get bijective correspondence with compositions, and thus also with the binary representation of a unique n. See the chart below:
   n   A014417(n) A014417(A022342(n+1)) Runs of    Binary number   In dec.
                  [shifted once left]   zeros      with same runs  = a(n)
  ......0              ......0    []         .....0           0
  ......1              .....10    [1]        .....1           1
  .....10              ....100    [2]        ....11           3
  ....100              ...1000    [3]        ...111           7
  ....101              ...1010    [1,1]      ....10           2
  ...1000              ..10000    [4]        ..1111          15
  ...1001              ..10010    [2,1]      ...110           6
  ...1010              ..10100    [1,2]      ...100           4
  ..10000              .100000    [5]        .11111          31
  ..10001              .100010    [3,1]      ..1110          14
  ..10010              .100100    [2,2]      ..1100          12
  ..10100              .101000    [1,3]      ..1000           8
  ..10101              .101010    [1,1,1]    ...101           5
  .100000              1000000    [6]        111111          63
Are there any other fixed points after 0, 1, 6, 803, 407483 ?

Antti Karttunen, Table of n, a(n) for n = 0..10946
OEIS Wiki, Run-length encoding
Index entries for sequences that are permutations of the natural numbers

Inverse permutation: A227352. Cf. also A003714, A014417, A006068, A048679.

Could be further composed with A075157 or A075159, also A129594.

(define (A227351 n) (A006068 (A048679 n)))

(define (A227351v2 n) (A006068 (A106151 (* 2 (A003714 n)))))

a(n) = A006068(A048679(n)) = A006068(A106151(2*A003714(n))).
This permutation effects following correspondences:
a(A000045(n)) = A000225(n-1).
a(A027941(n)) = A000975(n).
For n >=3, a(A000204(n)) = A000079(n-2).

cp's OEIS Frontend

A243353 Permutation of natural numbers which maps between the partitions as encoded in A227739 (binary based system, zero-based) to A112798 (prime-index based system, one-based).

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A227184 a(n) = product of parts of the unordered partition encoded with the runlengths of binary expansion of n.

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A243051 Integer sequence induced by Bulgarian solitaire operation on partition list A241918: a(n) = A241909(A242424(A241909(n))).

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A165199 a(n) is obtained by flipping every second bit in the binary representation of n starting at the second-most significant bit and on downwards.

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A075159 Run lengths in the binary expansion of n-1 gives the vector of exponents in prime factorization of a(n), with the least significant run corresponding to the exponent of the least prime, 2.

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A286617 Restricted growth sequence of A278217 (prime-signature of A075159(1+n)).

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A075162 Position of A014486(n) in A075165, minus one.

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A286468 Run lengths in the binary expansion of n are one more than the exponents of corresponding primes in the prime factorization of a(n).

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