A278161 -id:A278161 - 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

A046951 a(n) is the number of squares dividing n.

Original entry on oeis.org

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

Offset: 1

Simon Colton (simonco(AT)cs.york.ac.uk)

Rediscovered by the HR automatic theory formation program.
a(n) depends only on prime signature of n (cf. A025487, A046523). So a(24) = a(375) since 24 = 2^3*3 and 375 = 3*5^3 both have prime signature (3, 1).
First differences of A013936. Average value tends towards Pi^2/6 = 1.644934... (A013661, A013679). - Henry Bottomley, Aug 16 2001
We have a(n) = A159631(n) for all n < 125, but a(125) = 2 < 3 = A159631(125). - Steven Finch, Apr 22 2009
Number of 2-generated Abelian groups of order n, if n > 1. - Álvar Ibeas, Dec 22 2014 [In other words, number of order-n abelian groups with rank <= 2. Proof: let b(n) be such number. A finite abelian group is the inner direct product of all Sylow-p subgroups, so {b(n)} is multiplicative. Obviously b(p^e) = floor(e/2)+1 (corresponding to the groups C_(p^r) X C_(p^(e-r)) for 0 <= r <= floor(e/2)), hence b(n) = a(n) for all n. - Jianing Song, Nov 05 2022]
Number of ways of writing n = r*s such that r|s. - Eric M. Schmidt, Jan 08 2015
The number of divisors of the square root of the largest square dividing n. - Amiram Eldar, Jul 07 2020
The number of unordered factorizations of n into cubefree powers of primes (1, primes and squares of primes, A166684). - Amiram Eldar, Jun 12 2025

			a(16) = 3 because the squares 1, 4, and 16 divide 16.
G.f. = x + x^2 + x^3 + 2*x^4 + x^5 + x^6 + x^7 + 2*x^8 + 2*x^9 + x^10 + ...

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Antonio Amariti, Claudius Klare, Domenico Orlando and Susanne Reffert, The M-theory origin of global properties of gauge theories, Nuclear Physics B, Vol. 901 (2015), pp. 318-337, arXiv preprint, arXiv:1507.04743 [hep-th], 2015 (see (A.13)).
Simon Colton, Refactorable Numbers - A Machine Invention, J. Integer Sequences, Vol. 2 (1999), Article 99.1.2.
Simon Colton, HR - Automatic Theory Formation in Pure Mathematics
Ian G. Connell, A number theory problem concerning finite groups and rings, Canad. Math. Bull, 7 (1964), 23-34. See delta(n).
Andrew V. Lelechenko, Average number of squares dividing mn, arXiv preprint arXiv:1407.1222 [math.NT], 2014.
Werner Georg Nowak and László Tóth, On the average number of subgroups of the group Z_m X Z_n, International Journal of Number Theory, Vol. 10, No. 2 (2014), pp. 363-374, arXiv preprint, arXiv:1307.1414 [math.NT], 2013.
N. J. A. Sloane, Transforms.
Index entries for sequences computed from exponents in factorization of n.

Cf. A000005, A000188, A004101, A005117 (positions of 1's), A008619, A008833, A013936 (partial sums), A038538, A046952, A052304, A056595, A159631, A007814, A010052, A027750, A239930, A007862, A046523, A064989, A065704, A130279, A156552, A166684, A278161, A307445 (Dirichlet inverse).
One more than A071325.
Differs from A096309 for the first time at n=32, where a(32) = 3, while A096309(32) = 2 (and also A185102(32) = 2).
Sum of the k-th powers of the square divisors of n for k=0..10: this sequence (k=0), A035316 (k=1), A351307 (k=2), A351308 (k=3), A351309 (k=4), A351310 (k=5), A351311 (k=6), A351313 (k=7), A351314 (k=8), A351315 (k=9), A351315 (k=10).
Sequences of the form n^k * Sum_{d^2|n} 1/d^k for k = 0..10: this sequence (k=0), A340774 (k=1), A351600 (k=2), A351601 (k=3), A351602 (k=4), A351603 (k=5), A351604 (k=6), A351605 (k=7), A351606 (k=8), A351607 (k=9), A351608 (k=10).
Cf. A082293 (a(n)==2), A082294 (a(n)==3).

a046951 = sum . map a010052 . a027750_row
-- Reinhard Zumkeller, Dec 16 2013

[#[d: d in Divisors(n)|IsSquare(d)]:n in [1..120]]; // Marius A. Burtea, Jan 21 2020

A046951 := proc(n)
    local a,s;
    a := 1 ;
    for p in ifactors(n)[2] do
        a := a*(1+floor(op(2,p)/2)) ;
    end do:
    a ;
end proc: # R. J. Mathar, Sep 17 2012
# Alternatively:
isbidivisible := (n, d) -> igcd(n, d) = d and igcd(n/d, d) = d:
a := n -> nops(select(k -> isbidivisible(n, k), [seq(1..n)])): # Peter Luschny, Jun 13 2025

a[n_] := Length[ Select[ Divisors[n], IntegerQ[Sqrt[#]]& ] ]; Table[a[n], {n, 1, 105}] (* Jean-François Alcover, Jun 26 2012 *)
Table[Length[Intersection[Divisors[n], Range[10]^2]], {n, 100}] (* Alonso del Arte, Dec 10 2012 *)
a[ n_] := If[ n < 1, 0, Sum[ Mod[ DivisorSigma[ 0, d], 2], {d, Divisors @ n}]]; (* Michael Somos, Jun 13 2014 *)
a[ n_] := If[ n < 2, Boole[ n == 1], Times @@ (Quotient[ #[[2]], 2] + 1 & /@ FactorInteger @ n)]; (* Michael Somos, Jun 13 2014 *)
a[ n_] := If[ n < 0, 0, SeriesCoefficient[ Sum[ x^k^2 / (1 - x^k^2), {k, Sqrt @ n}], {x, 0, n}]]; (* Michael Somos, Jun 13 2014 *)
f[p_, e_] := 1 + Floor[e/2]; a[1] = 1; a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Sep 15 2020 *)

a(n)=my(f=factor(n));for(i=1,#f[,1],f[i,2]\=2);numdiv(factorback(f)) \\ Charles R Greathouse IV, Dec 11 2012

a(n) = direuler(p=2, n, 1/((1-X^2)*(1-X)))[n]; \\ Michel Marcus, Mar 08 2015

a(n)=factorback(apply(e->e\2+1, factor(n)[,2])) \\ Charles R Greathouse IV, Sep 17 2015

from math import prod
from sympy import factorint
def A046951(n): return prod((e>>1)+1 for e in factorint(n).values()) # Chai Wah Wu, Aug 04 2024

def is_bidivisible(n, d) -> bool: return gcd(n, d) == d and gcd(n//d, d) == d
def aList(n) -> list[int]: return [k for k in range(1, n+1) if is_bidivisible(n, k)]
print([len(aList(n)) for n in range(1, 126)])  # Peter Luschny, Jun 13 2025

(definec (A046951 n) (if (= 1 n) 1 (* (A008619 (A007814 n)) (A046951 (A064989 n)))))
(define (A008619 n) (+ 1 (/ (- n (modulo n 2)) 2)))
;; Antti Karttunen, Nov 14 2016

a(p^k) = A008619(k) = [k/2] + 1. a(A002110(n)) = 1 for all n. (This is true for any squarefree number, A005117). - Original notes clarified by Antti Karttunen, Nov 14 2016
a(n) = |{(i, j) : i*j = n AND i|j}| = |{(i, j) : i*j^2 = n}|. Also tau(A000188(n)), where tau = A000005.
Multiplicative with p^e --> floor(e/2) + 1, p prime. - Reinhard Zumkeller, May 20 2007
a(A130279(n)) = n and a(m) <> n for m < A130279(n); A008966(n)=0^(a(n) - 1). - Reinhard Zumkeller, May 20 2007
Inverse Moebius transform of characteristic function of squares (A010052). Dirichlet g.f.: zeta(s)*zeta(2s).
G.f.: Sum_{k > 0} x^(k^2)/(1 - x^(k^2)). - Vladeta Jovovic, Dec 13 2002
a(n) = Sum_{k=1..A000005(n)} A010052(A027750(n,k)). - Reinhard Zumkeller, Dec 16 2013
a(n) = Sum_{k = 1..n} ( floor(n/k^2) - floor((n-1)/k^2) ). - Peter Bala, Feb 17 2014
From Antti Karttunen, Nov 14 2016: (Start)
a(1) = 1; for n > 1, a(n) = A008619(A007814(n)) * a(A064989(n)).
a(n) = A278161(A156552(n)). (End)
G.f.: Sum_{k>0}(theta(q^k)-1)/2, where theta(q)=1+2q+2q^4+2q^9+2q^16+... - Mamuka Jibladze, Dec 04 2016
From  Antti Karttunen, Nov 12 2017: (Start)
a(n) = A000005(n) - A056595(n).
a(n) = 1 + A071325(n).
a(n) = 1 + A001222(A293515(n)). (End)
L.g.f.: -log(Product_{k>=1} (1 - x^(k^2))^(1/k^2)) = Sum_{n>=1} a(n)*x^n/n. - Ilya Gutkovskiy, Jul 30 2018
a(n) = Sum_{d|n} A000005(d) * A008836(n/d). - Torlach Rush, Jan 21 2020
a(n) = A000005(sqrt(A008833(n))). - Amiram Eldar, Jul 07 2020
a(n) = Sum_{d divides n} mu(core(d)^2), where core(n) = A007913(n). - Peter Bala, Jan 24 2024

Data section filled up to 125 terms and wrong claim deleted from Crossrefs section by Antti Karttunen, Nov 14 2016

A278159 Run length transform of primorials, A002110.

Original entry on oeis.org

1, 2, 2, 6, 2, 4, 6, 30, 2, 4, 4, 12, 6, 12, 30, 210, 2, 4, 4, 12, 4, 8, 12, 60, 6, 12, 12, 36, 30, 60, 210, 2310, 2, 4, 4, 12, 4, 8, 12, 60, 4, 8, 8, 24, 12, 24, 60, 420, 6, 12, 12, 36, 12, 24, 36, 180, 30, 60, 60, 180, 210, 420, 2310, 30030, 2, 4, 4, 12, 4, 8, 12, 60, 4, 8, 8, 24, 12, 24, 60, 420, 4, 8, 8, 24, 8, 16, 24, 120, 12, 24, 24, 72, 60, 120, 420

Offset: 0

Antti Karttunen, Nov 16 2016

Like every run length transform this sequence satisfies for all i, j: A278222(i) = A278222(j) => a(i) = a(j).

			For n=7, "111" in binary, there is a run of 1-bits of length 3, thus a(7) = product of A002110(3), = A002110(3) = 30.
For n=39, "10111" in binary, there are two runs, of lengths 1 and 3, thus a(39) = A002110(1) * A002110(3) = 2*30 = 60.

Cf. A002110, A005940, A124859, A227349, A246660, A278161, A278222.

f[n_] := Product[Prime[k], {k, 1, n}]; Table[Times @@ (f[Length[#]]&) /@ Select[Split[IntegerDigits[n, 2]], #[[1]] == 1&], {n, 0, 94}] (* Jean-François Alcover, Jul 11 2017 *)

from math import prod
from re import split
from sympy import primorial
def RLT(n,f):
    """ run length transform of a function f """
    return prod(f(len(d)) for d in split('0+', bin(n)[2:]) if d != '') if n > 0 else 1
def A278159(n): return RLT(n,primorial) # Chai Wah Wu, Feb 04 2022

(define (A278159 n) (fold-left (lambda (a r) (* a (A002110 r))) 1 (bisect (reverse (binexp->runcount1list n)) (- 1 (modulo n 2)))))
;; See A227349 for the required other functions.

a(n) = A124859(A005940(1+n)).

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A156552 Unary-encoded compressed factorization of natural numbers.

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A046951 a(n) is the number of squares dividing n.

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A278159 Run length transform of primorials, A002110.

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