A227349 -id:A227349 - cp's OEIS frontend

A284559 a(n) = LCM of run lengths in binary representation of n.

1, 1, 1, 2, 2, 1, 2, 3, 3, 2, 1, 2, 2, 2, 3, 4, 4, 3, 2, 2, 2, 1, 2, 3, 6, 2, 2, 2, 6, 3, 4, 5, 5, 4, 3, 6, 2, 2, 2, 6, 3, 2, 1, 2, 2, 2, 3, 4, 4, 6, 2, 2, 2, 2, 2, 6, 3, 6, 3, 6, 4, 4, 5, 6, 6, 5, 4, 4, 6, 3, 6, 3, 6, 2, 2, 2, 2, 2, 6, 4, 4, 3, 2, 2, 2, 1, 2, 3, 6, 2, 2, 2, 6, 3, 4, 5, 10, 4, 6, 6, 2, 2, 2, 6, 6, 2, 2, 2, 2, 2, 6, 4, 12, 3, 6, 6, 6, 3, 6, 3

Offset: 0

Antti Karttunen, Apr 14 2017

			For n=12, A007088(12) = "1100" in binary, the run lengths are [2,2], thus a(12) = lcm(2,2) = 2.

Cf. A000975 (positions of ones).

Cf. A007088, A167489, A227349, A284558, A284569, A284579.

from math import lcm
from itertools import groupby
def a(n): return lcm(*(len(list(g)) for k, g in groupby(bin(n)[2:])))
print([a(n) for n in range(87)]) # Michael S. Branicky, Oct 15 2022

(define (A284559 n) (apply lcm (binexp->runcount1list n)))
;; Or:
(define (A284559 n) (reduce lcm 1 (binexp->runcount1list n))) ;; For binexp->runcount1list, see the Program section of A227349.

a(n) = A167489(n) / A284558(n).

A284569 a(n) = LCM of the lengths of runs of 1-bits in binary representation of n.

Original entry on oeis.org

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

Offset: 0

Antti Karttunen, Apr 14 2017

			For n = 27, in binary A007088(27) = "11011", the lengths of runs of 1-bits are [2,2], thus a(27) = lcm(2,2) = 2.
For n = 55, in binary A007088(55) = "110111", the lengths of runs of 1-bits are [2,3], thus a(55) = lcm(2,3) = 6.

Antti Karttunen, Table of n, a(n) for n = 0..10922
Index entries for sequences related to binary expansion of n

Cf. A005940, A072411, A227349, A284562, A284559.
Cf. A003714 (positions of ones).
Differs from A227349 for the first time at n=27, where a(27)=2, while A227349(27)= 4.
Differs from A038374 for the first time at n=55, where a(55) = 6, while A038374(55) = 3.

(define (A284569 n) (apply lcm (bisect (reverse (binexp->runcount1list n)) (- 1 (modulo n 2))))) ;; For bisect and binexp->runcount1list, see the Program section of A227349.
(define (A284569 n) (A072411 (A005940 (+ 1 n))))

a(n) = A072411(A005940(1+n)).

a(n) = A227349(n) / A284562(n).

A247282 Run Length Transform of A001317.

Original entry on oeis.org

1, 1, 1, 3, 1, 1, 3, 5, 1, 1, 1, 3, 3, 3, 5, 15, 1, 1, 1, 3, 1, 1, 3, 5, 3, 3, 3, 9, 5, 5, 15, 17, 1, 1, 1, 3, 1, 1, 3, 5, 1, 1, 1, 3, 3, 3, 5, 15, 3, 3, 3, 9, 3, 3, 9, 15, 5, 5, 5, 15, 15, 15, 17, 51, 1, 1, 1, 3, 1, 1, 3, 5, 1, 1, 1, 3, 3, 3, 5, 15, 1, 1, 1, 3, 1, 1, 3, 5, 3, 3, 3, 9, 5, 5, 15, 17, 3, 3, 3, 9, 3, 3, 9, 15, 3, 3, 3, 9, 9, 9, 15, 45

Offset: 0

Antti Karttunen, Sep 22 2014

nonn

The Run Length Transform of a sequence {S(n), n>=0} is defined to be the sequence {T(n), n>=0} given by T(n) = Product_i S(i), where i runs through the lengths of runs of 1's in the binary expansion of n. E.g. 19 is 10011 in binary, which has two runs of 1's, of lengths 1 and 2. So T(19) = S(1)*S(2). T(0)=1 (the empty product).

This sequence is obtained by applying Run Length Transform to the right-shifted version of the sequence A001317: 1, 3, 5, 15, 17, 51, 85, 255, 257, ...

			115 is '1110011' in binary. The run lengths of 1-runs are 2 and 3, thus a(115) = A001317(2-1) * A001317(3-1) = 3*5 = 15.
From _Omar E. Pol_, Feb 15 2015: (Start)
Written as an irregular triangle in which row lengths are the terms of A011782:
1;
1;
1,3;
1,1,3,5;
1,1,1,3,3,3,5,15;
1,1,1,3,1,1,3,5,3,3,3,9,5,5,15,17;
1,1,1,3,1,1,3,5,1,1,1,3,3,3,5,15,3,3,3,9,3,3,9,15,5,5,5,15,15,15,17,51;
...
Right border gives 1 together with A001317.
(End)

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

Cf. A003714 (gives the positions of ones).
Cf. A001317, A051179.
A001316 is obtained when the same transformation is applied to A000079, the powers of two.
Run Length Transforms of other sequences: A071053, A227349, A246588, A246595, A246596, A246660, A246661, A246674, A246685.

a1317[n_] := FromDigits[ Table[ Mod[Binomial[n-1, k], 2], {k, 0, n-1}], 2];
Table[ Times @@ (a1317[Length[#]]&) /@ Select[Split[IntegerDigits[n, 2]], #[[1]] == 1&], {n, 0, 100}] (* Jean-François Alcover, Jul 11 2017 *)

# uses RLT function from A278159
def A247282(n): return RLT(n,lambda m: int(''.join(str(int(not(~(m-1)&k))) for k in range(m)),2)) # Chai Wah Wu, Feb 04 2022

For all n >= 0, a(A051179(n)) = A246674(A051179(n)) = A051179(n).

A286575 Run-length transform of A001316.

Original entry on oeis.org

1, 2, 2, 2, 2, 4, 2, 4, 2, 4, 4, 4, 2, 4, 4, 2, 2, 4, 4, 4, 4, 8, 4, 8, 2, 4, 4, 4, 4, 8, 2, 4, 2, 4, 4, 4, 4, 8, 4, 8, 4, 8, 8, 8, 4, 8, 8, 4, 2, 4, 4, 4, 4, 8, 4, 8, 4, 8, 8, 8, 2, 4, 4, 4, 2, 4, 4, 4, 4, 8, 4, 8, 4, 8, 8, 8, 4, 8, 8, 4, 4, 8, 8, 8, 8, 16, 8, 16, 4, 8, 8, 8, 8, 16, 4, 8, 2, 4, 4, 4, 4, 8, 4, 8, 4, 8, 8, 8, 4

Offset: 0

Antti Karttunen, May 28 2017

			For n = 0, there are no 1-runs, and thus a(0) = 1 as an empty product.
For n = 29, "11101" in binary, there are two 1-runs, of lengths 1 and 3, thus a(29) = A001316(1) * A001316(3) = 2*4 = 8.

Cf. A000079, A001316, A005940, A037445, A227349, A247282, A286574.

Table[Times @@ Map[Sum[Mod[#, 2] &@ Binomial[#, k], {k, 0, #}] &@ Length@ # &, DeleteCases[Split@ IntegerDigits[n, 2], ?(First@ # == 0 &)]], {n, 0, 108}] (* _Michael De Vlieger, May 29 2017 *)

from sympy import factorint, prime, log
import math
def wt(n): return bin(n).count("1")
def a037445(n):
    f=factorint(n)
    return 2**sum([wt(f[i]) for i in f])
def A(n): return n - 2**int(math.floor(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 a(n): return a037445(b(n)) # Indranil Ghosh, May 30 2017

# use RLT function from A278159
def A286575(n): return RLT(n,lambda m: 2**(bin(m).count('1'))) # Chai Wah Wu, Feb 04 2022

(define (A286575 n) (fold-left (lambda (a r) (* a (A001316 r))) 1 (bisect (reverse (binexp->runcount1list n)) (- 1 (modulo n 2)))))
(define (bisect lista parity) (let loop ((lista lista) (i 0) (z (list))) (cond ((null? lista) (reverse! z)) ((eq? i parity) (loop (cdr lista) (modulo (1+ i) 2) (cons (car lista) z))) (else (loop (cdr lista) (modulo (1+ i) 2) z)))))
(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 (A001316 n) (let loop ((n n) (z 1)) (cond ((zero? n) z) ((even? n) (loop (/ n 2) z)) (else (loop (/ (- n 1) 2) (* z 2))))))

a(n) = A037445(A005940(1+n)).

a(n) = A000079(A286574(n)).

A278161 Run length transform of A008619 (floor(n/2)+1).

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 1, 2, 2, 2, 2, 3, 1, 1, 1, 2, 1, 1, 2, 2, 2, 2, 2, 4, 2, 2, 3, 3, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 1, 2, 2, 2, 2, 3, 2, 2, 2, 4, 2, 2, 4, 4, 2, 2, 2, 4, 3, 3, 3, 4, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 1, 2, 2, 2, 2, 3, 1, 1, 1, 2, 1, 1, 2, 2, 2, 2, 2, 4, 2, 2, 3, 3, 2, 2, 2, 4, 2, 2, 4, 4, 2, 2, 2, 4, 4, 4, 4, 6, 2, 2, 2, 4, 2, 2, 4, 4, 3

Offset: 0

Antti Karttunen, Nov 14 2016

			n=111 is "1101111" in binary, which has two runs of 1-bits: the other has length 2, and the other has length 4, thus we take the product A008619(2)*A008619(4) = (floor(2/2)+1) * (floor(4/2)+1) = 2*3, which is the result, so a(111) = 6.

Antti Karttunen, Table of n, a(n) for n = 0..16383
Chai Wah Wu, Sums of products of binomial coefficients mod 2 and run length transforms of sequences, arXiv:1610.06166 [math.CO], 2016.
Index entries for sequences related to binary expansion of n

Cf. A005940, A008619, A046951, A156552.

Cf. A106737, A227349 for other run length transforms, and also A278222.

f[n_] := Floor[n/2] + 1; Table[Times @@ (f[Length[#]]&) /@ Select[ Split[ IntegerDigits[n, 2]], #[[1]] == 1&], {n, 0, 120}] (* Jean-François Alcover, Jul 11 2017 *)

def A278161(n): return sum(int(not (~(n+3*k) & 6*k) | (~n & k)) for k in range(n+1)) # Chai Wah Wu, Sep 28 2021

(define (A278161 n) (fold-left (lambda (a r) (* a (A008619 r))) 1 (bisect (reverse (binexp->runcount1list n)) (- 1 (modulo n 2)))))
(define (A008619 n) (+ 1 (/ (- n (modulo n 2)) 2)))
;; See A227349 for the required other functions.

a(n) = A046951(A005940(1+n)), a(A156552(n)) = A046951(n).

a(n) = Sum_{k=0..n} ((binomial(n+3k,6k)*binomial(n,k)) mod 2). - Chai Wah Wu, Nov 19 2019

A227193 Difference of (product of runlengths of 1-bits) and (product of runlengths of 0-bits) in binary representation of n.

Original entry on oeis.org

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

Offset: 0

Antti Karttunen, Jul 08 2013

sign

The sequence seems to consist of palindromic subsequences centered around each (2^k)-1 and 2^k (with end points near the terms of A000975), which is easily explained by symmetric pairing of binary expansion of n and its complement.

Antti Karttunen, Table of n, a(n) for n = 0..10922

Cf. A145037, A227190.

a:= proc(n) local i, j, m, r, s; m, r, s:= n, 1, 1;
      while m>0 do
        for i from 0 while irem(m, 2, 'h')=0 do m:=h od;
        for j from 0 while irem(m, 2, 'h')=1 do m:=h od;
        r, s:= r*j, s*max(i, 1)
      od; r-s
    end:
seq(a(n), n=0..100);  # Alois P. Heinz, Jul 11 2013

a[n_] := With[{s = Split @ IntegerDigits[n, 2]}, Times @@ Length /@ Select[ s, First[#]==1&] - Times @@ Length /@ Select[s , First[#]==0&]]; Table[ a[n], {n, 0, 100}] (* Jean-François Alcover, Feb 28 2016 *)

(define (A227193 n) (- (A227349 n) (A227350 n)))

a(n) = A227349(n) - A227350(n).

A246685 Run Length Transform of sequence 1, 3, 5, 17, 257, 65537, ... (1 followed by Fermat numbers).

Original entry on oeis.org

1, 1, 1, 3, 1, 1, 3, 5, 1, 1, 1, 3, 3, 3, 5, 17, 1, 1, 1, 3, 1, 1, 3, 5, 3, 3, 3, 9, 5, 5, 17, 257, 1, 1, 1, 3, 1, 1, 3, 5, 1, 1, 1, 3, 3, 3, 5, 17, 3, 3, 3, 9, 3, 3, 9, 15, 5, 5, 5, 15, 17, 17, 257, 65537, 1, 1, 1, 3, 1, 1, 3, 5, 1, 1, 1, 3, 3, 3, 5, 17, 1, 1, 1, 3, 1, 1, 3, 5, 3, 3, 3, 9, 5, 5, 17, 257

Offset: 0

Antti Karttunen, Sep 22 2014

nonn

The Run Length Transform of a sequence {S(n), n>=0} is defined to be the sequence {T(n), n>=0} given by T(n) = Product_i S(i), where i runs through the lengths of runs of 1's in the binary expansion of n. E.g. 19 is 10011 in binary, which has two runs of 1's, of lengths 1 and 2. So T(19) = S(1)*S(2). T(0)=1 (the empty product).

This sequence is obtained by applying Run Length Transform to sequence b = 1, 3, 5, 17, 257, 65537, ... (1 followed by Fermat numbers, with b(1) = 1, b(2) = 3, b(3) = 5, ..., b(n) = 2^(2^(n-2)) + 1 for n >= 2).

			115 is '1110011' in binary. The run lengths of 1-runs are 2 and 3, thus we multiply the second and the third elements of the sequence 1, 3, 5, 17, 257, 65537, ... to get a(115) = 3*5 = 15.

Antti Karttunen, Table of n, a(n) for n = 0..1024

Cf. A003714 (gives the positions of ones).
Cf. A000215.
A001316 is obtained when the same transformation is applied to A000079, the powers of two. Cf. also A001317.
Run Length Transforms of other sequences: A071053, A227349, A246588, A246595, A246596, A246660, A246661, A246674, A247282.

f[n_] := Switch[n, 0|1, 1, , 2^(2^(n-2))+1]; Table[Times @@ (f[Length[#]] &) /@ Select[s = Split[IntegerDigits[n, 2]], #[[1]] == 1&], {n, 0, 95}] (* _Jean-François Alcover, Jul 11 2017 *)

# use RLT function from A278159
def A246685(n): return RLT(n,lambda m: 1 if m <= 1 else 2**(2**(m-2))+1) # Chai Wah Wu, Feb 04 2022

A284580 Carryless base-2 product (A048720) of lengths of runs of 1-bits in binary representation of n.

Original entry on oeis.org

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

Offset: 0

Antti Karttunen, Apr 14 2017

			a(119) = 5, as 119 is "1110111" in binary, and A048720(3,3) = 5.

Antti Karttunen, Table of n, a(n) for n = 0..10922
Index entries for sequences related to binary expansion of n

Cf. A003714 (positions of ones).
Cf. A048720, A284579.
Differs from similar A227349 for the first time at n=119, where a(119) = 5, while A227349(119) = 9.

(define (A284580 n) (reduce A048720bi 1 (bisect (reverse (binexp->runcount1list n)) (- 1 (modulo n 2))))) ;; Where A048720bi is a two-argument function implementing carryless binary product, A048720. For bisect and binexp->runcount1list, see under A227349.

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A284559 a(n) = LCM of run lengths in binary representation of n.

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A284569 a(n) = LCM of the lengths of runs of 1-bits in binary representation of n.

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A247282 Run Length Transform of A001317.

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A286575 Run-length transform of A001316.

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A278161 Run length transform of A008619 (floor(n/2)+1).

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A227193 Difference of (product of runlengths of 1-bits) and (product of runlengths of 0-bits) in binary representation of n.

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A246685 Run Length Transform of sequence 1, 3, 5, 17, 257, 65537, ... (1 followed by Fermat numbers).

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