A284562
Product / LCM of the lengths of 1-runs in binary representation of n: a(n) = A227349(n) / A284569(n).
Original entry on oeis.org
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 2, 2, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 3, 1
Offset: 0
A156552
Unary-encoded compressed factorization of natural numbers.
Original entry on oeis.org
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
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.
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
A072411
LCM of exponents in prime factorization of n, a(1) = 1.
Original entry on oeis.org
1, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 2, 1, 1, 1, 4, 1, 2, 1, 2, 1, 1, 1, 3, 2, 1, 3, 2, 1, 1, 1, 5, 1, 1, 1, 2, 1, 1, 1, 3, 1, 1, 1, 2, 2, 1, 1, 4, 2, 2, 1, 2, 1, 3, 1, 3, 1, 1, 1, 2, 1, 1, 2, 6, 1, 1, 1, 2, 1, 1, 1, 6, 1, 1, 2, 2, 1, 1, 1, 4, 4, 1, 1, 2, 1, 1, 1, 3, 1, 2, 1, 2, 1, 1, 1, 5, 1, 2, 2, 2, 1, 1, 1, 3, 1, 1, 1, 6, 1, 1, 1, 4, 1, 1, 1, 2, 2, 1, 1, 3
Offset: 1
n = 288 = 2*2*2*2*2*3*3; lcm(5,2) = 10; Product(5,2) = 10, max(5,2) = 5;
n = 180 = 2*2*3*3*5; lcm(2,2,1) = 2; Product(2,2,1) = 4; max(2,2,1) = 2; it deviates both from maximum of exponents (A051903, for the first time at n=72), and product of exponents (A005361, for the first time at n=36).
For n = 36 = 2*2*3*3 = 2^2 * 3^2 we have a(36) = lcm(2,2) = 2.
For n = 72 = 2*2*2*3*3 = 2^3 * 3^2 we have a(72) = lcm(2,3) = 6.
For n = 144 = 2^4 * 3^2 we have a(144) = lcm(2,4) = 4.
For n = 360 = 2^3 * 3^2 * 5^1 we have a(360) = lcm(1,2,3) = 6.
Differs from
A290107 for the first time at n=144.
After the initial term, differs from
A157754 for the first time at n=360.
-
Table[LCM @@ Last /@ FactorInteger[n], {n, 2, 100}] (* Ray Chandler, Jan 24 2006 *)
-
a(n) = lcm(factor(n)[,2]); \\ Michel Marcus, Mar 25 2017
-
from sympy import lcm, factorint
def a(n):
l=[]
f=factorint(n)
for i in f: l+=[f[i],]
return lcm(l)
print([a(n) for n in range(1, 151)]) # Indranil Ghosh, Mar 25 2017
a(1) = 1 prepended and the data section filled up to 120 terms by
Antti Karttunen, Aug 09 2016
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
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)
Cf.
A000120 (sum of lengths of runs of 1-bits),
A167489,
A227350,
A227193,
A278222,
A245562,
A284562,
A284569,
A283972,
A284582,
A284583.
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)))))))
A284559
a(n) = LCM of run lengths in binary representation of n.
Original entry on oeis.org
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
For n=12, A007088(12) = "1100" in binary, the run lengths are [2,2], thus a(12) = lcm(2,2) = 2.
-
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.
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