A246596 -id:A246596 - cp's OEIS frontend

A156552 Unary-encoded compressed factorization of natural numbers.

0, 1, 2, 3, 4, 5, 8, 7, 6, 9, 16, 11, 32, 17, 10, 15, 64, 13, 128, 19, 18, 33, 256, 23, 12, 65, 14, 35, 512, 21, 1024, 31, 34, 129, 20, 27, 2048, 257, 66, 39, 4096, 37, 8192, 67, 22, 513, 16384, 47, 24, 25, 130, 131, 32768, 29, 36, 71, 258, 1025, 65536, 43, 131072, 2049, 38, 63, 68, 69, 262144

Offset: 1

Leonid Broukhis, Feb 09 2009

The primes become the powers of 2 (2 -> 1, 3 -> 2, 5 -> 4, 7 -> 8); the composite numbers are formed by taking the values for the factors in the increasing order, multiplying them by the consecutive powers of 2, and summing. See the Example section.
From Antti Karttunen, Jun 27 2014: (Start)
The odd bisection (containing even terms) halved gives A244153.
The even bisection (containing odd terms), when one is subtracted from each and halved, gives this sequence back.
(End)
Question: Are there any other solutions that would satisfy the recurrence r(1) = 0; and for n > 1, r(n) = Sum_{d|n, d>1} 2^A033265(r(d)), apart from simple variants 2^k * A156552(n)? See also A297112, A297113. - Antti Karttunen, Dec 30 2017

			For 84 = 2*2*3*7 -> 1*1 + 1*2 + 2*4 + 8*8 =  75.
For 105 = 3*5*7 -> 2*1 + 4*2 + 8*4 = 42.
For 137 = p_33 -> 2^32 = 4294967296.
For 420 = 2*2*3*5*7 -> 1*1 + 1*2 + 2*4 + 4*8 + 8*16 = 171.
For 147 = 3*7*7 = p_2 * p_4 * p_4 -> 2*1 + 8*2 + 8*4 = 50.

David A. Corneth, Table of n, a(n) for n = 1..10000 (first 1024 terms from Antti Karttunen)
Hans Havermann, Factorization of the first 10000 terms, in format [[primes], [exponents]]
A puzzle by Sergey Orlov (in Russian)
Index entries for sequences related to binary expansion of n
Index entries for sequences that are permutations of the natural numbers
Index entries for sequences computed from indices in prime factorization

One less than A005941.
Inverse permutation: A005940 with starting offset 0 instead of 1.
Cf. A000079, A000120, A001222, A052126, A054429, A061395, A064216, A064989, A003188, A243071, A243065-A243066, A244153, A243354, A112798, A125106, A056239, A161511.
Cf. also A297106, A297112 (Möbius transform), A297113, A153013, A290308, A300827, A323243, A323244, A323247, A324201, A324812 (n for which a(n) is a square), A324813, A324822, A324823, A324398, A324713, A324815, A324819, A324865, A324866, A324867.
Other related permutations: A253551, A253792, A253564, A253791, A277195, A297163, A297164, A297165, A297166, A302023, A305418, A322863, A322864.

Table[Floor@ Total@ Flatten@ MapIndexed[#1 2^(#2 - 1) &, Flatten[ Table[2^(PrimePi@ #1 - 1), {#2}] & @@@ FactorInteger@ n]], {n, 67}] (* Michael De Vlieger, Sep 08 2016 *)

a(n) = {my(f = factor(n), p2 = 1, res = 0); for(i = 1, #f~, p = 1 << (primepi(f[i, 1]) - 1); res += (p * p2 * (2^(f[i, 2]) - 1)); p2 <<= f[i, 2]); res}; \\ David A. Corneth, Mar 08 2019

A064989(n) = {my(f); f = factor(n); if((n>1 && f[1,1]==2), f[1,2] = 0); for (i=1, #f~, f[i,1] = precprime(f[i,1]-1)); factorback(f)};
A156552(n) = if(1==n, 0, if(!(n%2), 1+(2*A156552(n/2)), 2*A156552(A064989(n)))); \\ (based on the given recurrence) - Antti Karttunen, Mar 08 2019

# Program corrected per instructions from Leonid Broukhis. - Antti Karttunen, Jun 26 2014
# However, it gives correct answers only up to n=136, before corruption by a wrap-around effect.
# Note that the correct answer for n=137 is A156552(137) = 4294967296.
$max = $ARGV[0];
$pow = 0;
foreach $i (2..$max) {
@a = split(/ /, `factor $i`);
shift @a;
$shift = 0;
$cur = 0;
while ($n = int shift @a) {
$prime{$n} = 1 << $pow++ if !defined($prime{$n});
$cur |= $prime{$n} << $shift++;
}
print "$cur, ";
}
print "\n";
(Scheme, with memoization-macro definec from Antti Karttunen's IntSeq-library, two different implementations)
(definec (A156552 n) (cond ((= n 1) 0) (else (+ (A000079 (+ -2 (A001222 n) (A061395 n))) (A156552 (A052126 n))))))
(definec (A156552 n) (cond ((= 1 n) (- n 1)) ((even? n) (+ 1 (* 2 (A156552 (/ n 2))))) (else (* 2 (A156552 (A064989 n))))))
;; Antti Karttunen, Jun 26 2014

from sympy import primepi, factorint
def A156552(n): return sum((1<Chai Wah Wu, Mar 10 2023

From Antti Karttunen, Jun 26 2014: (Start)
a(1) = 0, a(n) = A000079(A001222(n)+A061395(n)-2) + a(A052126(n)).
a(1) = 0, a(2n) = 1+2*a(n), a(2n+1) = 2*a(A064989(2n+1)). [Compare to the entanglement recurrence A243071].
For n >= 0, a(2n+1) = 2*A244153(n+1). [Follows from the latter clause of the above formula.]
a(n) = A005941(n) - 1.
As a composition of related permutations:
a(n) = A003188(A243354(n)).
a(n) = A054429(A243071(n)).
For all n >= 1, A005940(1+a(n)) = n and for all n >= 0, a(A005940(n+1)) = n. [The offset-0 version of A005940 works as an inverse for this permutation.]
This permutations also maps between the partition-lists A112798 and A125106:
A056239(n) = A161511(a(n)). [The sums of parts of each partition (the total sizes).]
A003963(n) = A243499(a(n)). [And also the products of those parts.]
(End)
From Antti Karttunen, Oct 09 2016: (Start)
A161511(a(n)) = A056239(n).
A029837(1+a(n)) = A252464(n). [Binary width of terms.]
A080791(a(n)) = A252735(n). [Number of nonleading 0-bits.]
A000120(a(n)) = A001222(n). [Binary weight.]
For all n >= 2, A001511(a(n)) = A055396(n).
For all n >= 2, A000120(a(n))-1 = A252736(n). [Binary weight minus one.]
A252750(a(n)) = A252748(n).
a(A250246(n)) = A252754(n).
a(A005117(n)) = A277010(n). [Maps squarefree numbers to a permutation of A003714, fibbinary numbers.]
A085357(a(n)) = A008966(n). [Ditto for their characteristic functions.]
For all n >= 0:
a(A276076(n)) = A277012(n).
a(A276086(n)) = A277022(n).
a(A260443(n)) = A277020(n).
(End)
From Antti Karttunen, Dec 30 2017: (Start)
For n > 1, a(n) = Sum_{d|n, d>1} 2^A033265(a(d)). [See comments.]
More linking formulas:
A106737(a(n)) = A000005(n).
A290077(a(n)) = A000010(n).
A069010(a(n)) = A001221(n).
A136277(a(n)) = A181591(n).
A132971(a(n)) = A008683(n).
A106400(a(n)) = A008836(n).
A268411(a(n)) = A092248(n).
A037011(a(n)) = A010052(n) [conjectured, depends on the exact definition of A037011].
A278161(a(n)) = A046951(n).
A001316(a(n)) = A061142(n).
A277561(a(n)) = A034444(n).
A286575(a(n)) = A037445(n).
A246029(a(n)) = A181819(n).
A278159(a(n)) = A124859(n).
A246660(a(n)) = A112624(n).
A246596(a(n)) = A069739(n).
A295896(a(n)) = A053866(n).
A295875(a(n)) = A295297(n).
A284569(a(n)) = A072411(n).
A286574(a(n)) = A064547(n).
A048735(a(n)) = A292380(n).
A292272(a(n)) = A292382(n).
A244154(a(n)) = A048673(n), a(A064216(n)) = A244153(n).
A279344(a(n)) = A279339(n), a(A279338(n)) = A279343(n).
a(A277324(n)) = A277189(n).
A037800(a(n)) = A297155(n).
For n > 1, A033265(a(n)) = 1+A297113(n).
(End)
From Antti Karttunen, Mar 08 2019: (Start)
a(n) = A048675(n) + A323905(n).
a(A324201(n)) = A000396(n), provided there are no odd perfect numbers.
The following sequences are derived from or related to the base-2 expansion of a(n):
A000265(a(n)) = A322993(n).
A002487(a(n)) = A323902(n).
A005187(a(n)) = A323247(n).
A324288(a(n)) = A324116(n).
A323505(a(n)) = A323508(n).
A079559(a(n)) = A323512(n).
A085405(a(n)) = A323239(n).
The following sequences are obtained by applying to a(n) a function that depends on the prime factorization of its argument, which goes "against the grain" because a(n) is the binary code of the factorization of n, which in these cases is then factored again:
A000203(a(n)) = A323243(n).
A033879(a(n)) = A323244(n) = 2*a(n) - A323243(n),
A294898(a(n)) = A323248(n).
A000005(a(n)) = A324105(n).
A000010(a(n)) = A324104(n).
A083254(a(n)) = A324103(n).
A001227(a(n)) = A324117(n).
A000593(a(n)) = A324118(n).
A001221(a(n)) = A324119(n).
A009194(a(n)) = A324396(n).
A318458(a(n)) = A324398(n).
A192895(a(n)) = A324100(n).
A106315(a(n)) = A324051(n).
A010052(a(n)) = A324822(n).
A053866(a(n)) = A324823(n).
A001065(a(n)) = A324865(n) = A323243(n) - a(n),
A318456(a(n)) = A324866(n) = A324865(n) OR a(n),
A318457(a(n)) = A324867(n) = A324865(n) XOR a(n),
A318458(a(n)) = A324398(n) = A324865(n) AND a(n),
A318466(a(n)) = A324819(n) = A323243(n) OR 2*a(n),
A318467(a(n)) = A324713(n) = A323243(n) XOR 2*a(n),
A318468(a(n)) = A324815(n) = A323243(n) AND 2*a(n).
(End)

More terms from Antti Karttunen, Jun 28 2014

A227349 Product 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, 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

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).

A246661 Run Length Transform of swinging factorials (A056040).

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 2, 6, 1, 1, 1, 2, 2, 2, 6, 6, 1, 1, 1, 2, 1, 1, 2, 6, 2, 2, 2, 4, 6, 6, 6, 30, 1, 1, 1, 2, 1, 1, 2, 6, 1, 1, 1, 2, 2, 2, 6, 6, 2, 2, 2, 4, 2, 2, 4, 12, 6, 6, 6, 12, 6, 6, 30, 20, 1, 1, 1, 2, 1, 1, 2, 6, 1, 1, 1, 2, 2, 2, 6, 6, 1, 1, 1, 2, 1, 1

Offset: 0

Peter Luschny, Sep 07 2014

For the definition of the Run Length Transform see A246595.

Alois P. Heinz, Table of n, a(n) for n = 0..8191

Cf. A227349, A246588, A246595, A246596, A246660.

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

# use RLT function from A278159
from math import factorial
def A246661(n): return RLT(n,lambda m: factorial(m)//factorial(m//2)**2) # Chai Wah Wu, Feb 04 2022

# uses[RLT from A246660]
A246661_list = lambda len: RLT(lambda n: factorial(n)/factorial(n//2)^2, len)
A246661_list(88)

a(2^n-1) = n$ where n$ is the swinging factorial of n, A056040(n).

A069739 Size of the key space for isomorphism verification of circulant graphs of order n.

Original entry on oeis.org

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

Offset: 1

Valery A. Liskovets, Apr 15 2002

Multiplicative with a(p^m) = Catalan(m) (A000108). Coincides with A066060 up to n=63 except for n=32.

Antti Karttunen, Table of n, a(n) for n = 1..10000
M. Muzychuk, A solution of the isomorphism problem for circulant graphs, Proc. London Math. Soc. (3) 88 (2004), no. 1, 1-41.
Index entries for sequences computed from exponents in factorization of n

Cf. A000108, A066060, A028234, A067029, A156552, A246596.

A000108 := proc(n) binomial(2*n,n)/(n+1) ; end proc:
A069739 := proc(n) local ifa; if n = 1 then 1; else ifa := ifactors(n)[2] ; mul( A000108(op(2,f)), f=ifa) ; end if; end proc:
seq(A069739(n),n=1..90) ; # R. J. Mathar, Feb 08 2011

Table[Times @@ Map[CatalanNumber, FactorInteger[n][[All, -1]]], {n, 90}] (* Michael De Vlieger, May 28 2017 *)

(define (A069739 n) (if (= 1 n) n (* (A000108 (A067029 n)) (A069739 (A028234 n))))) ;; Antti Karttunen, May 28 2017

a(n) = prod_p Catalan(m_p) where n=prod_p p^(m_p), p|n prime.
From Antti Karttunen, May 28-29 2017: (Start)
a(1) = 1; for n > 1, a(n) = A000108(A067029(n)) * a(A028234(n)).
a(n) = A246596(A156552(n)). (End)

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

A156552 Unary-encoded compressed factorization of natural numbers.

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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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A246674 Run Length Transform of A000225.

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