A075159 -id:A075159 - cp's OEIS frontend

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

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");

A278217 Filter-sequence related to base-2 run-length encoding: a(n) = A046523(A075159(1+n)) = A046523(1+A075157(n)).

Original entry on oeis.org

1, 2, 2, 4, 6, 2, 4, 8, 12, 6, 2, 6, 12, 4, 8, 16, 24, 12, 6, 30, 6, 2, 6, 12, 36, 12, 4, 12, 24, 8, 16, 32, 48, 24, 12, 60, 30, 6, 30, 60, 12, 6, 2, 6, 30, 6, 12, 24, 72, 36, 12, 60, 12, 4, 12, 36, 72, 24, 8, 24, 48, 16, 32, 64, 96, 48, 24, 120, 60, 12, 60, 180, 60, 30, 6, 30, 210, 30, 60, 120, 24, 12, 6, 30, 6, 2, 6, 12, 60, 30, 6, 30, 60, 12, 24, 48, 144, 72

Offset: 0

Antti Karttunen, Nov 16 2016

nonn

Antti Karttunen, Table of n, a(n) for n = 0..16384

Cf. A046523, A075157, A075159, A286617 (rgs-version of this filter).
Other base-2 related filter sequences: A278219, A278222.
Sequences that partition N into same or coarser equivalence classes are at least these: A092339, A227185.

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)); \\ From Antti Karttunen, May 17 2017

(define (A278217 n) (A046523 (+ 1 (A075157 n))))
(define (A278217 n) (A046523 (A075159 (+ 1 n))))

a(n) = A046523(1+A075157(n)) = A046523(A075159(1+n)).

A075157 Run lengths in the binary expansion of n gives the vector of exponents in prime factorization of a(n)+1, with the least significant run corresponding to the exponent of the least prime, 2; with one subtracted from each run length, except for the most significant run of 1's.

Original entry on oeis.org

0, 1, 2, 3, 5, 4, 8, 7, 11, 14, 6, 9, 17, 24, 26, 15, 23, 44, 34, 29, 13, 10, 20, 19, 35, 74, 48, 49, 53, 124, 80, 31, 47, 134, 174, 89, 69, 76, 104, 59, 27, 32, 12, 21, 41, 54, 62, 39, 71, 224, 244, 149, 97, 120, 146, 99, 107, 374, 342, 249, 161, 624, 242, 63, 95, 404

Offset: 0

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(11) = 9, as 11 = 1011 and 9+1 = 10 = 5^1 * 3^(1-1) * 2^(2-1).

Paul Tek (terms 0..10000) & Antti Karttunen, Table of n, a(n) for n = 0..16384
Index entries for sequences that are permutations of the natural numbers

Inverse of A075158.
Cf. A008578, A056539, A059900, A075162.
Cf. A000040, A000975, A006093, A030308, A007088, A075159, A278217, A286617.

import Data.List (group)
a075157 0 = 0
a075157 n = product (zipWith (^) a000040_list rs') - 1 where
   rs' = reverse $ r : map (subtract 1) rs
   (r:rs) = reverse $ map length $ group $ a030308_row n
-- Reinhard Zumkeller, Aug 04 2014

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);

(define (A075157 n) (if (zero? n) n (+ -1 (* (A000040 (A005811 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, using the code of A286468:
(define (A075157 n) (if (zero? n) n (- (* (A000040 (A005811 n)) (A286468 n)) 1)))

a(n) = A075159(n+1) - 1.
a(0) = 0; for n >= 1, a(n) = (A000040(A005811(n)) * A286468(n)) - 1.
Other identities. For all n >= 1:
a(A000975(n)) = A006093(n) = A000040(n)-1.

Entry revised, PARI-program added and the old incorrect Scheme-program replaced with a new one by Antti Karttunen, May 17 2017

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).

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).

A075160 Prime factorization of n encoded with the run lengths of binary expansion + 1.

Original entry on oeis.org

1, 2, 3, 4, 6, 5, 11, 8, 7, 12, 22, 9, 43, 21, 10, 16, 86, 13, 171, 24, 23, 44, 342, 17, 14, 85, 15, 41, 683, 20, 1366, 32, 42, 172, 19, 25, 2731, 341, 87, 48, 5462, 45, 10923, 88, 18, 684, 21846, 33, 27, 28, 170, 169, 43691, 29, 46, 81, 343, 1365, 87382, 40, 174763

Offset: 1

Antti Karttunen, Sep 13 2002

nonn

See the comment at A075158.

			a(9) = 7 = 1+6 (1 + 110) as 9 = 3^2 * 2^(1-1). (The run lengths of 6, 110 in binary are 2 and 1).

Index entries for sequences that are permutations of the natural numbers

Inverse of A075159. a(n) = A075158(n-1)+1.

cp's OEIS Frontend

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

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A278217 Filter-sequence related to base-2 run-length encoding: a(n) = A046523(A075159(1+n)) = A046523(1+A075157(n)).

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A075157 Run lengths in the binary expansion of n gives the vector of exponents in prime factorization of a(n)+1, with the least significant run corresponding to the exponent of the least prime, 2; with one subtracted from each run length, except for the most significant run of 1's.

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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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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.

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A075160 Prime factorization of n encoded with the run lengths of binary expansion + 1.

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