A246661 -id:A246661 - cp's OEIS frontend

A227349 Product of lengths of runs of 1-bits in binary representation of n.

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, 9, 4

Offset: 0

Antti Karttunen, Jul 08 2013

This is the Run Length Transform of S(n) = {0, 1, 2, 3, 4, 5, 6, ...}. 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). - N. J. A. Sloane, Sep 05 2014

Like all run length transforms also this sequence satisfies for all i, j: A278222(i) = A278222(j) => a(i) = a(j). - Antti Karttunen, Apr 14 2017

			a(0) = 1, as zero has no runs of 1's, and an empty product is 1.
a(1) = 1, as 1 is "1" in binary, and the length of that only 1-run is 1.
a(2) = 1, as 2 is "10" in binary, and again there is only one run of 1-bits, of length 1.
a(3) = 2, as 3 is "11" in binary, and there is one run of two 1-bits.
a(55) = 6, as 55 is "110111" in binary, and 2 * 3 = 6.
a(119) = 9, as 119 is "1110111" in binary, and 3 * 3 = 9.
From _Omar E. Pol_, Feb 10 2015: (Start)
Written as an irregular triangle in which row lengths is A011782:
  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;
  ...
Right border gives A028310: 1 together with the positive integers. (End)
From _Omar E. Pol_, Mar 19 2015: (Start)
Also, the sequence can be written as an irregular tetrahedron T(s, r, k) as shown below:
  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;
  ...
Apart from the initial 1, we have that T(s, r, k) = T(s+1, r, k). (End)

Antti Karttunen, Table of n, a(n) for n = 0..8192
N. J. A. Sloane, On the Number of ON Cells in Cellular Automata, arXiv:1503.01168 [math.CO], 2015.
Index entries for sequences computed with run length transform

Cf. A003714 (positions of ones), A005361, A005940.
Cf. A000120 (sum of lengths of runs of 1-bits), A167489, A227350, A227193, A278222, A245562, A284562, A284569, A283972, A284582, A284583.
Run Length Transforms of other sequences: A246588, A246595, A246596, A246660, A246661, A246674.
Differs from similar A284580 for the first time at n=119, where a(119) = 9, while A284580(119) = 5.

a:= proc(n) local i, m, r; m, r:= n, 1;
      while m>0 do
        while irem(m, 2, 'h')=0 do m:=h od;
        for i from 0 while irem(m, 2, 'h')=1 do m:=h od;
        r:= r*i
      od; r
    end:
seq(a(n), n=0..100);  # Alois P. Heinz, Jul 11 2013
ans:=[];
for n from 0 to 100 do lis:=[]; t1:=convert(n, base, 2); L1:=nops(t1); out1:=1; c:=0;
for i from 1 to L1 do
   if out1 = 1 and t1[i] = 1 then out1:=0; c:=c+1;
   elif out1 = 0 and t1[i] = 1 then c:=c+1;
   elif out1 = 1 and t1[i] = 0 then c:=c;
   elif out1 = 0 and t1[i] = 0 then lis:=[c, op(lis)]; out1:=1; c:=0;
   fi;
   if i = L1 and c>0 then lis:=[c, op(lis)]; fi;
                   od:
a:=mul(i, i in lis);
ans:=[op(ans), a];
od:
ans;  # N. J. A. Sloane, Sep 05 2014

onBitRunLenProd[n_] := Times @@ Length /@ Select[Split @ IntegerDigits[n, 2], #[[1]] == 1 & ]; Array[onBitRunLenProd, 100, 0] (* Jean-François Alcover, Mar 02 2016 *)

from operator import mul
from functools import reduce
from re import split
def A227349(n):
    return reduce(mul, (len(d) for d in split('0+',bin(n)[2:]) if d)) if n > 0 else 1 # Chai Wah Wu, Sep 07 2014

# uses[RLT from A246660]
A227349_list = lambda len: RLT(lambda n: n, len)
A227349_list(88) # Peter Luschny, Sep 07 2014

(define (A227349 n) (apply * (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)))))))

A167489(n) = a(n) * A227350(n).
A227193(n) = a(n) - A227350(n).
a(n) = Product_{i in row n of table A245562} i. - N. J. A. Sloane, Aug 10 2014
From Antti Karttunen, Apr 14 2017: (Start)
a(n) = A005361(A005940(1+n)).
a(n) = A284562(n) * A284569(n).
A283972(n) = n - a(n).
(End)
a(4n+1) = a(2n) = a(n). If n is odd, then a(4n+3) = 2*a(2n+1)-a(n). If n is even, then a(4n+3) = 2*a(2n+1) = 2*a(n/2). - Chai Wah Wu, Jul 17 2025

Data section extended up to term a(120) by Antti Karttunen, Apr 14 2017

A246660 Run Length Transform of factorials.

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 2, 6, 1, 1, 1, 2, 2, 2, 6, 24, 1, 1, 1, 2, 1, 1, 2, 6, 2, 2, 2, 4, 6, 6, 24, 120, 1, 1, 1, 2, 1, 1, 2, 6, 1, 1, 1, 2, 2, 2, 6, 24, 2, 2, 2, 4, 2, 2, 4, 12, 6, 6, 6, 12, 24, 24, 120, 720, 1, 1, 1, 2, 1, 1, 2, 6, 1, 1, 1, 2, 2, 2, 6, 24, 1, 1, 1

Offset: 0

Peter Luschny, Sep 07 2014

For the definition of the Run Length Transform see A246595.

Only Jordan-Polya numbers (A001013) are terms of this sequence.

Antti Karttunen, Table of n, a(n) for n = 0..8192
Index entries for sequences computed with run length transform

Cf. A000142, A001013, A007814, A112624, A323505, A323506.
Cf. A003714 (gives the positions of ones).
Run Length Transforms of other sequences: A001316, A071053, A227349, A246588, A246595, A246596, A246661, A246674.

Table[Times @@ (Length[#]!&) /@ Select[Split[IntegerDigits[n, 2]], #[[1]] == 1&], {n, 0, 83}] (* Jean-François Alcover, Jul 11 2017 *)

A246660(n) = { my(i=0, p=1); while(n>0, if(n%2, i++; p = p * i, i = 0); n = n\2); p; };
for(n=0, 8192, write("b246660.txt", n, " ", A246660(n)));
\\ Antti Karttunen, Sep 08 2014

from operator import mul
from functools import reduce
from re import split
from math import factorial
def A246660(n):
    return reduce(mul,(factorial(len(d)) for d in split('0+',bin(n)[2:]) if d)) if n > 0 else 1 # Chai Wah Wu, Sep 09 2014

def RLT(f, size):
    L = lambda n: [a for a in Integer(n).binary().split('0') if a != '']
    return [mul([f(len(d)) for d in L(n)]) for n in range(size)]
A246660_list = lambda len: RLT(factorial, len)
A246660_list(88)

;; A stand-alone loop version, like the Pari-program above:
(define (A246660 n) (let loop ((n n) (i 0) (p 1)) (cond ((zero? n) p) ((odd? n) (loop (/ (- n 1) 2) (+ i 1) (* p (+ 1 i)))) (else (loop (/ n 2) 0 p)))))
;; One based on given recurrence, utilizing memoizing definec-macro from my IntSeq-library:
(definec (A246660 n) (cond ((zero? n) 1) ((even? n) (A246660 (/ n 2))) (else (* (A007814 (+ n 1)) (A246660 (/ (- n 1) 2))))))
;; Yet another implementation, using fold:
(define (A246660 n) (fold-left (lambda (a r) (* a (A000142 r))) 1 (bisect (reverse (binexp->runcount1list n)) (- 1 (modulo n 2)))))
(definec (A000142 n) (if (zero? n) 1 (* n (A000142 (- n 1)))))
;; Other functions are as in A227349 - Antti Karttunen, Sep 08 2014

a(2^n-1) = n!.
a(0) = 1, a(2n) = a(n), a(2n+1) = a(n) * A007814(2n+2). - Antti Karttunen, Sep 08 2014
a(n) = A112624(A005940(1+n)). - Antti Karttunen, May 29 2017
a(n) = A323505(n) / A323506(n). - Antti Karttunen, Jan 17 2019

A246588 Run Length Transform of S(n) = wt(n) = 0,1,1,2,1,2,2,3,1,... (cf. A000120).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 1, 1, 2, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 2, 1, 2, 1, 1, 1, 1, 1

Offset: 0

N. J. A. Sloane, Sep 05 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).

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000

Cf. A000120.
Cf. A167489, A030308.
Run Length Transforms of other sequences: A071053, A227349, A246595, A246596, A246660, A246661, A246674.

import Data.List (group)
a246588 = product . map (a000120 . length) .
          filter ((== 1) . head) . group . a030308_row
-- Reinhard Zumkeller, Feb 13 2015, Sep 05 2014

A000120 := proc(n) local w, m, i; w := 0; m :=n; while m > 0 do i := m mod 2; w := w+i; m := (m-i)/2; od; w; end: wt := A000120;
ans:=[];
for n from 0 to 100 do lis:=[]; t1:=convert(n, base, 2); L1:=nops(t1); out1:=1; c:=0;
for i from 1 to L1 do
   if out1 = 1 and t1[i] = 1 then out1:=0; c:=c+1;
   elif out1 = 0 and t1[i] = 1 then c:=c+1;
   elif out1 = 1 and t1[i] = 0 then c:=c;
   elif out1 = 0 and t1[i] = 0 then lis:=[c, op(lis)]; out1:=1; c:=0;
   fi;
   if i = L1 and c>0 then lis:=[c, op(lis)]; fi;
                   od:
a:=mul(wt(i), i in lis);
ans:=[op(ans), a];
od:
ans;

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

from operator import mul
from functools import reduce
from re import split
def A246588(n):
    return reduce(mul,(bin(len(d)).count('1') for d in split('0+',bin(n)[2:]) if d)) if n > 0 else 1 # Chai Wah Wu, Sep 07 2014

# uses[RLT from A246660]
A246588_list = lambda len: RLT(lambda n: sum(Integer(n).digits(2)), len)
A246588_list(88) # Peter Luschny, Sep 07 2014

A246595 Run Length Transform of squares.

Original entry on oeis.org

1, 1, 1, 4, 1, 1, 4, 9, 1, 1, 1, 4, 4, 4, 9, 16, 1, 1, 1, 4, 1, 1, 4, 9, 4, 4, 4, 16, 9, 9, 16, 25, 1, 1, 1, 4, 1, 1, 4, 9, 1, 1, 1, 4, 4, 4, 9, 16, 4, 4, 4, 16, 4, 4, 16, 36, 9, 9, 9, 36, 16, 16, 25, 36, 1, 1, 1, 4, 1, 1, 4, 9, 1, 1, 1, 4, 4, 4, 9, 16, 1, 1, 1, 4, 1, 1

Offset: 0

N. J. A. Sloane, Sep 06 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).

			From _Omar E. Pol_, Feb 10 2015: (Start)
Written as an irregular triangle in which row lengths is A011782:
1;
1;
1,4;
1,1,4,9;
1,1,1,4,4,4,9,16;
1,1,1,4,1,1,4,9,4,4,4,16,9,9,16,25;
1,1,1,4,1,1,4,9,1,1,1,4,4,4,9,16,4,4,4,16,4,4,16,36,9,9,9,36,16,16,25,36;
...
Right border gives A253909: 1 together with the positive squares.
(End)
From _Omar E. Pol_, Mar 19 2015: (Start)
Also, the sequence can be written as an irregular tetrahedron T(s,r,k) as shown below:
1;
..
1;
..
1;
4;
.......
1,   1;
4;
9;
...............
1,   1,  1,  4;
4,   4;
9;
16;
.............................
1,   1,  1,  4, 1, 1,  4,  9;
4,   4,  4, 16;
9,   9;
16;
25;
......................................................
1,   1,  1,  4, 1, 1,  4,  9, 1, 1, 1, 4, 4, 4, 9, 16;
4,   4,  4, 16, 4, 4, 16, 36;
9,   9,  9, 36;
16, 16;
25;
36;
...
Apart from the initial 1, we have that T(s,r,k) = T(s+1,r,k).
(End)

Chai Wah Wu, Table of n, a(n) for n = 0..8192
N. J. A. Sloane, On the Number of ON Cells in Cellular Automata, arXiv:1503.01168 [math.CO], 2015.

Cf. A000290, A253082.
Cf. A003714 (gives the positions of ones).
Run Length Transforms of other sequences: A071053, A227349, A246588, A246596, A246660, A246661, A246674.

ans:=[];
for n from 0 to 100 do lis:=[]; t1:=convert(n, base, 2); L1:=nops(t1); out1:=1; c:=0;
for i from 1 to L1 do
   if out1 = 1 and t1[i] = 1 then out1:=0; c:=c+1;
   elif out1 = 0 and t1[i] = 1 then c:=c+1;
   elif out1 = 1 and t1[i] = 0 then c:=c;
   elif out1 = 0 and t1[i] = 0 then lis:=[c, op(lis)]; out1:=1; c:=0;
   fi;
   if i = L1 and c>0 then lis:=[c, op(lis)]; fi;
                   od:
a:=mul(i^2, i in lis);
ans:=[op(ans), a];
od:
ans;

Table[Times @@ (Length[#]^2&) /@ Select[Split[IntegerDigits[n, 2]], #[[1]] == 1&], {n, 0, 85}] (* Jean-François Alcover, Jul 11 2017 *)

from operator import mul
from functools import reduce
from re import split
def A246595(n):
    return reduce(mul,(len(d)**2 for d in split('0+',bin(n)[2:]) if d != '')) if n > 0 else 1 # Chai Wah Wu, Sep 07 2014

# uses[RLT from A246660]
A246595_list = lambda len: RLT(lambda n: n^2, len)
A246595_list(86) # Peter Luschny, Sep 07 2014

; using MIT/GNU Scheme
(define (A246595 n) (fold-left (lambda (a r) (* a r r)) 1 (bisect (reverse (binexp->runcount1list n)) (- 1 (modulo n 2)))))
;; Other functions are as in A227349 - Antti Karttunen, Sep 08 2014

a(n) = A227349(n)^2. - Omar E. Pol, Feb 10 2015

A246596 Run Length Transform of Catalan numbers A000108.

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 2, 5, 1, 1, 1, 2, 2, 2, 5, 14, 1, 1, 1, 2, 1, 1, 2, 5, 2, 2, 2, 4, 5, 5, 14, 42, 1, 1, 1, 2, 1, 1, 2, 5, 1, 1, 1, 2, 2, 2, 5, 14, 2, 2, 2, 4, 2, 2, 4, 10, 5, 5, 5, 10, 14, 14, 42, 132, 1, 1, 1, 2, 1, 1, 2, 5, 1, 1, 1, 2, 2, 2, 5, 14, 1, 1, 1, 2, 1, 1, 2, 5

Offset: 0

N. J. A. Sloane, Sep 06 2014

nonn

			From _Omar E. Pol_, Feb 15 2015: (Start)
Written as an irregular triangle in which row lengths are the terms of A011782:
1;
1;
1,2;
1,1,2,5;
1,1,1,2,2,2,5,14;
1,1,1,2,1,1,2,5,2,2,2,4,5,5,14,42;
1,1,1,2,1,1,2,5,1,1,1,2,2,2,5,14,2,2,2,4,2,2,4,10,5,5,5,10,14,14,42,132;
...
Right border gives the Catalan numbers. This is simply a restatement of the theorem that this sequence is the Run Length Transform of A000108.
(End)

Cf. A000108.
Cf. A003714 (gives the positions of ones).
Run Length Transforms of other sequences: A005940, A069739, A071053, A227349, A246588, A246595, A246660, A246661, A246674.

Cat:=n->binomial(2*n,n)/(n+1);
ans:=[];
for n from 0 to 100 do lis:=[]; t1:=convert(n, base, 2); L1:=nops(t1); out1:=1; c:=0;
for i from 1 to L1 do
if out1 = 1 and t1[i] = 1 then out1:=0; c:=c+1;
elif out1 = 0 and t1[i] = 1 then c:=c+1;
elif out1 = 1 and t1[i] = 0 then c:=c;
elif out1 = 0 and t1[i] = 0 then lis:=[c, op(lis)]; out1:=1; c:=0;
fi;
if i = L1 and c>0 then lis:=[c, op(lis)]; fi;
od:
a:=mul(Cat(i), i in lis);
ans:=[op(ans), a];
od:
ans;

f = CatalanNumber; Table[Times @@ (f[Length[#]]&) /@ Select[ Split[ IntegerDigits[n, 2]], #[[1]] == 1&], {n, 0, 87}] (* Jean-François Alcover, Jul 11 2017 *)

from operator import mul
from functools import reduce
from gmpy2 import divexact
from re import split
def A246596(n):
    s, c = bin(n)[2:], [1, 1]
    for m in range(1, len(s)):
        c.append(divexact(c[-1]*(4*m+2),(m+2)))
    return reduce(mul,(c[len(d)] for d in split('0+',s))) if n > 0 else 1
# Chai Wah Wu, Sep 07 2014

# uses[RLT from A246660]
A246596_list = lambda len: RLT(lambda n: binomial(2*n, n)/(n+1), len)
A246596_list(88) # Peter Luschny, Sep 07 2014

; using MIT/GNU Scheme
(define (A246596 n) (fold-left (lambda (a r) (* a (A000108 r))) 1 (bisect (reverse (binexp->runcount1list n)) (- 1 (modulo n 2)))))
(define A000108 (EIGEN-CONVOLUTION 1 *))
;; Note: EIGEN-CONVOLUTION can be found from my IntSeq-library and other functions are as in A227349. - Antti Karttunen, Sep 08 2014

a(n) = A069739(A005940(1+n)). - Antti Karttunen, May 29 2017

A246674 Run Length Transform of A000225.

Original entry on oeis.org

1, 1, 1, 3, 1, 1, 3, 7, 1, 1, 1, 3, 3, 3, 7, 15, 1, 1, 1, 3, 1, 1, 3, 7, 3, 3, 3, 9, 7, 7, 15, 31, 1, 1, 1, 3, 1, 1, 3, 7, 1, 1, 1, 3, 3, 3, 7, 15, 3, 3, 3, 9, 3, 3, 9, 21, 7, 7, 7, 21, 15, 15, 31, 63, 1, 1, 1, 3, 1, 1, 3, 7, 1, 1, 1, 3, 3, 3, 7, 15, 1, 1, 1, 3, 1, 1, 3, 7, 3, 3, 3, 9, 7, 7, 15, 31, 3, 3, 3, 9, 3, 3, 9, 21, 3, 3, 3, 9, 9, 9, 21, 45, 7, 7, 7, 21, 7, 7, 21, 49, 15, 15, 15, 45, 31, 31, 63, 127, 1

Offset: 0

Antti Karttunen, Sep 08 2014

a(n) can be also computed by replacing all consecutive runs of zeros in the binary expansion of n with * (multiplication sign), and then performing that multiplication, still in binary, after which the result is converted into decimal. See the example below.

			115 is '1110011' in binary. The run lengths of 1-runs are 2 and 3, thus a(115) = A000225(2) * A000225(3) = ((2^2)-1) * ((2^3)-1) = 3*7 = 21.
The same result can be also obtained more directly, by realizing that '111' and '11' are the binary representations of 7 and 3, and 7*3 = 21.
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;
1,1,3,7;
1,1,1,3,3,3,7,15;
1,1,1,3,1,1,3,7,3,3,3,9,7,7,15,31;
1,1,1,3,1,1,3,7,1,1,1,3,3,3,7,15,3,3,3,9,3,3,9,21,7,7,7,21,15,15,31,63;
...
Right border gives 1 together with the positive terms of A000225.
(End)

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

Cf. A003714 (gives the positions of ones).
Cf. A000225, 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, A246685, A247282.

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

# uses RLT function in A278159
def A246674(n): return RLT(n,lambda m: 2**m-1) # Chai Wah Wu, Feb 04 2022

For all n >= 0, a(A051179(n)) = A247282(A051179(n)) = A051179(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

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).

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

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

cp's OEIS Frontend

A227349 Product of lengths of runs of 1-bits in binary representation of n.

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A246660 Run Length Transform of factorials.

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A246588 Run Length Transform of S(n) = wt(n) = 0,1,1,2,1,2,2,3,1,... (cf. A000120).

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A246595 Run Length Transform of squares.

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A246596 Run Length Transform of Catalan numbers A000108.

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A246674 Run Length Transform of A000225.

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