A000120 -id:A000120 - cp's OEIS frontend

A175522 A000120-perfect numbers.

2, 25, 95, 111, 119, 123, 125, 169, 187, 219, 221, 247, 289, 335, 365, 411, 415, 445, 485, 493, 505, 629, 655, 685, 695, 697, 731, 767, 815, 841, 871, 943, 949, 965, 985, 1003, 1139, 1207, 1241, 1261, 1263, 1273, 1343, 1387, 1465, 1469, 1507, 1513, 1529, 1563

Offset: 1

Vladimir Shevelev, Dec 03 2010

nonn

Let A(n), n>=1, be an infinite positive sequence.
We call a number n:
   A-deficient if Sum{d|n, d   A-abundant if Sum{d|n, d A(n),
and
   A-perfect if Sum{d|n, ddepending on the sum over the proper divisors of n.
The definition generalizes the standard nomenclature of deficient (A005100), abundant (A005101) and perfect numbers (A000396), which is recovered by setting A(n) = n = A000027(n).
Conjecture: if there exist infinitely many A-deficient numbers and infinitely many A-abundant numbers, then there exist infinitely many A-perfect numbers.
Note that the sequence contains squares of all Fermat primes larger than 3 (see A019434). [This would also hold for squares of any hypothetical Fermat primes after the fifth one, 65537. Comment clarified by Antti Karttunen, May 14 2015]
A192895(a(n)) = 0. - Reinhard Zumkeller, Jul 12 2011

			Proper divisors of 119 are 1,7,17. Since A000120(1)+A000120(7)+A000120(17)=A000120(119), then 119 is in the sequence.

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Reinhard Zumkeller)

Cf. A175524 (deficient version), A175526 (abundant version), A000120, A000396.
Subsequence of A257691 (non-abundant version).
Positions of zeros in A192895.
Cf. A019434, A253204.

import Data.List (elemIndices)
a175522 n = a175522_list !! (n-1)
a175522_list = map (+ 1) $ elemIndices 0 a192895_list
-- Reinhard Zumkeller, Jul 12 2011

binw[n_] := DigitCount[n, 2, 1]; Select[Range[1500], binw[#] == DivisorSum[#, binw[#1] &]/2 &] (* Amiram Eldar, Dec 14 2020 *)

is(n)=sumdiv(n,d,hammingweight(d))==2*hammingweight(n) \\ Charles R Greathouse IV, Jan 28 2016

from sympy import divisors
def A000120(n): return bin(n).count('1')
def aupto(limit):
  alst = []
  for m in range(1, limit+1):
    if A000120(m) == sum(A000120(d) for d in divisors(m)[:-1]): alst += [m]
  return alst
print(aupto(1563)) # Michael S. Branicky, Feb 25 2021

A000120 = lambda x: x.digits(base=2).count(1)
is_A175522 = lambda x: sum(A000120(d) for d in divisors(x)) == 2*A000120(x)
A175522 = filter(is_A175522, IntegerRange(1, 10**4))
# D. S. McNeil, Dec 04 2010

More terms from Amiram Eldar, Feb 18 2019

A057335 a(0) = 1, and for n > 0, a(n) = A000040(A000120(n)) * a(floor(n/2)); essentially sequence A055932 generated using A000120, hence sorted by number of factors.

Original entry on oeis.org

1, 2, 4, 6, 8, 12, 18, 30, 16, 24, 36, 60, 54, 90, 150, 210, 32, 48, 72, 120, 108, 180, 300, 420, 162, 270, 450, 630, 750, 1050, 1470, 2310, 64, 96, 144, 240, 216, 360, 600, 840, 324, 540, 900, 1260, 1500, 2100, 2940, 4620, 486, 810, 1350, 1890, 2250, 3150, 4410

Offset: 0

Alford Arnold, Aug 27 2000

Note that for n>0 the prime divisors of a(n) are consecutive primes starting with 2. All of the least prime signatures (A025487) are included; with the other values forming A056808.
Using the formula, terms of b(n)= a(n)/A057334(n) are: 1, 1, 2, 2, 4, 4, 6, 6, 8, ..., indeed a(n) repeated. - Michel Marcus, Feb 09 2014
a(n) is the unique normal number whose unsorted prime signature is the k-th composition in standard order (graded reverse-lexicographic). This composition (row k of A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. A number is normal if its prime indices cover an initial interval of positive integers. Unsorted prime signature is the sequence of exponents in a number's prime factorization. - Gus Wiseman, Apr 19 2020

			From _Gus Wiseman_, Apr 19 2020: (Start)
The sequence of terms together with their prime indices begins:
      1: {}
      2: {1}
      4: {1,1}
      6: {1,2}
      8: {1,1,1}
     12: {1,1,2}
     18: {1,2,2}
     30: {1,2,3}
     16: {1,1,1,1}
     24: {1,1,1,2}
     36: {1,1,2,2}
     60: {1,1,2,3}
     54: {1,2,2,2}
     90: {1,2,2,3}
    150: {1,2,3,3}
    210: {1,2,3,4}
     32: {1,1,1,1,1}
     48: {1,1,1,1,2}
For example, the 27th composition in standard order is (1,2,1,1), and the normal number with prime signature (1,2,1,1) is 630 = 2*3*3*5*7, so a(27) = 630.
(End)

Michael De Vlieger, Table of n, a(n) for n = 0..10000

Cf. A000120, A057334, A055932 and A056808.
Cf. A324939.
Unsorted prime signature is A124010.
Numbers whose prime signature is aperiodic are A329139.
The reversed version is A334031.
A partial inverse is A334032.
All of the following pertain to compositions in standard order (A066099):
- Length is A000120.
- Sum is A070939.
- Strict compositions are A233564.
- Constant compositions are A272919.
- Aperiodic compositions are A328594.
- Normal compositions are A333217.
- Permutations are A333218.
- Heinz number is A333219.
Cf. A005867, A029931, A048793, A052409, A056239, A066099, A112798, A124767, A228351, A233249, A333220, A345974.
Related to A019565 via A122111 and to A000079 via A336321.

Table[Times @@ Map[If[# == 0, 1, Prime@ #] &, Accumulate@ IntegerDigits[n, 2]], {n, 0, 54}] (* Michael De Vlieger, May 23 2017 *)

mg(n) = if (n==0, 1, prime(hammingweight(n))); \\ A057334
lista(nn) = {my(v = vector(nn)); v[1] = 1; for (i=2, nn, v[i] = mg(i-1)*v[(i+1)\2];); v;} \\ Michel Marcus, Feb 09 2014

A057335(n) = if(0==n,1,prime(hammingweight(n))*A057335(n\2)); \\ Antti Karttunen, Jul 20 2020

a(n) = A057334(n) * a (repeated).
A334032(a(n)) = n; a(A334032(n)) = A071364(n). - Gus Wiseman, Apr 19 2020
a(n) = A122111(A019565(n)); A019565(n) = A122111(a(n)). - Peter Munn, Jul 18 2020
a(n) = A336321(2^n). - Peter Munn, Mar 04 2022
Sum_{n>=0} 1/a(n) = Sum_{n>=0} 1/A005867(n) = 2.648101... (A345974). - Amiram Eldar, Jun 26 2025

More terms from Antonio G. Astudillo (afg_astudillo(AT)lycos.com), Mar 29 2003

New primary name from Antti Karttunen, Jul 20 2020

A228085 a(n) = number of distinct k which satisfy n = k + wt(k), where wt(k) (A000120) gives the number of 1's in binary representation of a nonnegative integer k.

Original entry on oeis.org

1, 0, 1, 1, 0, 2, 0, 1, 1, 1, 1, 1, 1, 0, 2, 0, 1, 2, 0, 2, 1, 0, 2, 0, 1, 1, 1, 1, 1, 1, 0, 2, 0, 2, 1, 1, 2, 0, 2, 0, 1, 1, 1, 1, 1, 1, 0, 2, 0, 1, 2, 0, 2, 1, 0, 2, 0, 1, 1, 1, 1, 1, 1, 0, 2, 1, 1, 2, 1, 1, 2, 0, 1, 1, 1, 1, 1, 1, 0, 2, 0, 1, 2, 0, 2, 1, 0

Offset: 0

Antti Karttunen, Aug 09 2013

wt(k) is also called bitcount(k).
a(n) = number of times n occurs in A092391.
The first 2 occurs at n = A230303(2) = 5 (as we have two solutions A092391(3) = A092391(4) = 5).
The first 3 occurs at n = A230303(3) = 129 (as we have three solutions A092391(123) = A092391(124) = A092391(128) = 129).
The first 4 occurs at n = A230303(4) = 4102, where we have solutions A092391(4091) = A092391(4092) = A092391(4099) = A092391(4100) = 4102.
For n>=1, a(2^n) = a(n-1) since an integer k = m is a solution to n-1 = m + wt(m) if and only if k = 2^n - 1 - m is a solution to 2^n = k + wt(k). - Max Alekseyev, Feb 23 2021

Antti Karttunen, Table of n, a(n) for n = 0..8191
Max A. Alekseyev and N. J. A. Sloane, On Kaprekar's Junction Numbers, arXiv:2112.14365, 2021; Journal of Combinatorics and Number Theory 12:3 (2022), 115-155.
Index entries for Colombian or self numbers and related sequences

A010061 gives the position of zeros, A228082 the positions of nonzeros, A228088 the positions of ones.

Cf. A000120, A010062, A092391, A228086, A228087, A228091 (positions of 2's), A227643, A230058, A230092 (positions of 3's), A230093, A227915 (positions of 4's), A070939, A230303.

a228085 n = length $ filter ((== n) . a092391) [n - a070939 n .. n]
-- Reinhard Zumkeller, Oct 13 2013

For Maple code see A230091. - N. J. A. Sloane, Oct 10 2013
# Find all inverses of m under x -> x + wt(x) - N. J. A. Sloane, Oct 19 2013
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;
F:=proc(m) local ans,lb,n,i;
lb:=m-ceil(log(m+1)/log(2)); ans:=[];
for n from max(1,lb) to m do if (n+wt(n)) = m then ans:=[op(ans),n]; fi; od:
[seq(ans[i],i=1..nops(ans))];
end;

nmax = 8191; Clear[a]; a[_] = 0;
Scan[Set[a[#[[1]]], #[[2]]]&, Tally[Table[n + DigitCount[n, 2, 1], {n, 0, nmax}]]];
a /@ Range[0, nmax] (* Jean-François Alcover, Oct 29 2019 *)
a[n_] := Module[{k, cnt = 0}, For[k = n - Floor[Log[2, n]] - 1, k < n, k++, If[n == k + DigitCount[k, 2, 1], cnt++]]; cnt];
a /@ Range[0, 100] (* Jean-François Alcover, Nov 28 2020 *)

A284005 a(0) = 1, and for n > 1, a(n) = (1 + A000120(n))*a(floor(n/2)); also a(n) = A000005(A283477(n)).

Original entry on oeis.org

1, 2, 4, 6, 8, 12, 18, 24, 16, 24, 36, 48, 54, 72, 96, 120, 32, 48, 72, 96, 108, 144, 192, 240, 162, 216, 288, 360, 384, 480, 600, 720, 64, 96, 144, 192, 216, 288, 384, 480, 324, 432, 576, 720, 768, 960, 1200, 1440, 486, 648, 864, 1080, 1152, 1440, 1800, 2160, 1536, 1920, 2400, 2880, 3000

Offset: 0

Antti Karttunen, Mar 18 2017

Antti Karttunen, Table of n, a(n) for n = 0..8191
Index entries for sequences related to binary expansion of n

Cf. A000005, A283477, A284001.

Similar recurrences: A124758, A243499, A329369, A341392.

Table[DivisorSigma[0, #] &@ Apply[Times, Map[#1^#2 & @@ # &, FactorInteger[#] /. {p_, e_} /; e == 1 :> {Times @@ Prime@ Range@ PrimePi@ p, e}]] &[Times @@ Prime@ Flatten@ Position[#, 1] &@ Reverse@ IntegerDigits[n, 2]], {n, 0, 71}] (* Michael De Vlieger, Mar 18 2017 *)

A284005(n) = numdiv(A283477(n)); \\ edited by Michel Marcus, May 01 2019, M. F. Hasler, Nov 10 2019

a(n) = my(k=if(n,logint(n,2)),s=1); prod(i=0,k, s+=bittest(n,k-i)); \\ Kevin Ryde, Jan 20 2021

(define (A284005 n) (A000005 (A283477 n)))

a(n) = A000005(A283477(n)).
Conjecture: a(n) = 2*a(f(n)) + Sum_{k=0..floor(log_2(n))-1} a(f(n) + 2^k*(1 - T(n,k))) for n > 1 with a(0) = 1, a(1) = 2, f(n) = A053645(n), T(n,k) = floor(n/2^k) mod 2. - Mikhail Kurkov, Nov 10 2019
From Mikhail Kurkov, Aug 23 2021: (Start)
a(2n+1) = a(n) + a(2n) for n >= 0.
a(2n) = a(n) + a(2n - 2^A007814(n)) for n > 0 with a(0) = 1. (End)
Conjecture: a(n) = Sum_{k=0..n} (binomial(n, k) mod 2)*A329369(k). In other words, this sequence is modulo 2 binomial transform of A329369. - Mikhail Kurkov, Mar 10 2023
Conjecture: a(2^m*(2n+1)) = Sum_{k=0..m+1} binomial(m+1, k)*a(2^k*n) for m >= 0, n >= 0 with a(0) = 1. - Mikhail Kurkov, Apr 24 2023

Made Mikhail Kurkov's Nov 10 2019 formula the new primary name of this sequence - Antti Karttunen, Dec 30 2020

A175526 A000120-abundant numbers.

Original entry on oeis.org

4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 26, 27, 28, 30, 32, 33, 34, 35, 36, 38, 39, 40, 42, 44, 45, 46, 48, 49, 50, 51, 52, 54, 55, 56, 57, 58, 60, 62, 63, 64, 65, 66, 68, 69, 70, 72, 74, 75, 76, 77, 78, 80, 81, 82, 84, 85, 86, 87, 88, 90, 91, 92, 93, 94, 96, 98, 99, 100, 102, 104, 105, 106, 108, 110, 112, 114, 115, 116, 117, 118, 120

Offset: 1

Vladimir Shevelev, Dec 03 2010

nonn

Definition see in A175522. All even numbers > 2 are in the sequence.

A192895(a(n)) > 0. Reinhard Zumkeller, Jul 12 2011

Reinhard Zumkeller (first 1000 terms) & Antti Karttunen, Table of n, a(n) for n = 1..10000

Cf. A175522 (perfect version), A175524 (deficient version), A257691 (complement, non-abundant version).
Cf. A000120, A192895.
Cf. also A005100, A005101.
a(n) differs from A091212(n) and from A205783(n+1) for the first time at n=37, where a(37) = 55, while 55 is missing from both A091212 and A205783.
Differs from A192506 for the first time at n=54, where a(54) = 77, while 77 is missing from A192506.

import Data.List (findIndices)
a175526 n = a175526_list !! (n-1)
a175526_list = map (+ 1) $ findIndices (> 0) a192895_list
-- Reinhard Zumkeller, Jul 12 2011

isA175526 := proc(n) s := 0 ; for d in (numtheory[divisors](n) minus {n}) do s := s+A000120(d) ; end do: evalb(s> A000120(n)) ; end proc:
for n from 1 to 120 do if isA175526(n) then printf("%d,",n); end if; end do: # R. J. Mathar, Jul 11 2011

okQ[n_] := DivisorSum[n, Total[IntegerDigits[#, 2]]*(-1)^Boole[#==n]&]>0; Select[Range[120], okQ] (* Jean-François Alcover, Dec 06 2015 *)

A192895(n) = sumdiv(n, d, hammingweight(d)*(-1)^(d==n)); \\ Charles R Greathouse IV, Feb 07 2013
isA175526(n) = (A192895(n) > 0);
n = 0; i = 0; while(i < 10000, n++; if(isA175526(n), i++; write("b175526.txt", i, " ", n)));
\\ Antti Karttunen, May 11 2015

is(n)=sumdiv(n,d,hammingweight(d))>2*hammingweight(n) \\ Charles R Greathouse IV, Jan 28 2016

is_A175526 = lambda x: sum(A000120(d) for d in divisors(x)) > 2*A000120(x)
A175526 = filter(is_A175526, IntegerRange(1, 10**4))
# D. S. McNeil, Dec 04 2010

A341392 a(n) = A284005(n) / (1 + A000120(n))!.

Original entry on oeis.org

1, 1, 2, 1, 4, 2, 3, 1, 8, 4, 6, 2, 9, 3, 4, 1, 16, 8, 12, 4, 18, 6, 8, 2, 27, 9, 12, 3, 16, 4, 5, 1, 32, 16, 24, 8, 36, 12, 16, 4, 54, 18, 24, 6, 32, 8, 10, 2, 81, 27, 36, 9, 48, 12, 15, 3, 64, 16, 20, 4, 25, 5, 6, 1, 64, 32, 48, 16, 72, 24, 32, 8, 108, 36, 48, 12, 64, 16, 20, 4, 162, 54, 72, 18, 96, 24, 30, 6, 128

Offset: 0

Mikhail Kurkov, Feb 10 2021 [verification needed]

From Antti Karttunen, Feb 10 2021: (Start)
This sequence can be represented as a binary tree. Each child to the left is obtained by multiplying its parent with (1+{binary weight of its breadth-first-wise index in the tree}), while each child to the right is just a clone of its parent:
                                      1
                                      |
                   ...................1...................
                  2                                       1
        4......../ \........2                   3......../ \........1
       / \                 / \                 / \                 / \
      /   \               /   \               /   \               /   \
     /     \             /     \             /     \             /     \
    8       4           6       2           9       3           4       1
  16 8    12 4        18 6     8 2        27 9    12 3        16 4     5 1
etc.
(End)
This sequence and A243499 have the same set of values on intervals from 2^m to 2^(m+1) - 1 for m >= 0. - Mikhail Kurkov, Jun 18 2021 [verification needed]
FindStat provides a sequence of mappings between this sequence and A000110 starting from collection [Set partitions] (see Links section for illustration). - Mikhail Kurkov, May 20 2023 [verification needed]

Alois P. Heinz, Table of n, a(n) for n = 0..16384
Michael De Vlieger, This sequence as a binary tree showing rows 0 <= r <= 5.
Michael De Vlieger, This sequence as a binary tree showing rows 0 <= r <= 8.
FindStat, Illustration of the sequence of mappings

Cf. A000120, A000142, A007814, A036987, A053645, A243499, A284005, A329369 (similar recurrence).

a:= proc(n) option remember; `if`(n=0, 1,
      a(iquo(n, 2, 'd'))*`if`(d=1, 1, add(i, i=Bits[Split](n+1))))
    end:
seq(a(n), n=0..120);  # Alois P. Heinz, Jun 23 2021

Array[DivisorSigma[0, Apply[Times, Map[#1^#2 & @@ # &, FactorInteger[#1] /. {p_, e_} /; e == 1 :> {Times @@ Prime@ Range@ PrimePi@ p, e}]]]/#2 & @@ {Times @@ Prime@ Flatten@ Position[#, 1] &@ Reverse@ #, (1 + Count[#, 1])!} &@ IntegerDigits[#, 2] &, 89, 0] (* Michael De Vlieger, Feb 24 2021 *)

A284005(n) = { my(k=if(n, logint(n, 2)), s=1); prod(i=0, k, s+=bittest(n, k-i)); }; \\ From A284005
A341392(n) = (A284005(n)/((1 + hammingweight(n))!)); \\ Antti Karttunen, Feb 10 2021

A341392(n) = if(!n,1,if(n%2, A341392((n-1)/2), (1+hammingweight(n))*A341392(n/2))); \\ Antti Karttunen, Feb 10 2021

a(n) = A284005(n) / (1 + A000120(n))! = A284005(n) / A000142(1 + A000120(n)).
a(2n+1) = a(n) for n >= 0.
a(2n) = (1 + A000120(n))*a(n) = A243499(2*A059894(n)) = a(n) + a(2n - 2^A007814(n)) for n > 0 with a(0) = 1.
[2*a(n) - 1 = A329369(n)] = A036987(A053645(n)).
From Mikhail Kurkov, Apr 24 2023: (Start)
a(2^m*(2n+1)) = Sum_{k=0..m} binomial(m, k)*a(2^k*n) for m >= 0, n >= 0 with a(0) = 1.
a(n) = a(f(n)) + Sum_{k=0..floor(log_2(n))-1} (1 - T(n, k))*a(f(n) + 2^k*(1 - T(n, k))) for n > 1 with a(0) = 1, a(1) = 1, where f(n) = A053645(n) and where T(n, k) = floor(n/2^k) mod 2. (End) [verification needed]

A192895 A000120-deficiency of n.

Original entry on oeis.org

-1, 0, -1, 1, -1, 2, -2, 2, 1, 2, -2, 5, -2, 2, 1, 3, -1, 6, -2, 5, 3, 2, -3, 8, 0, 2, 1, 6, -3, 10, -4, 4, 4, 2, 3, 11, -2, 2, 2, 8, -2, 12, -3, 6, 7, 2, -4, 11, 1, 6, 1, 6, -3, 10, 1, 10, 2, 2, -4, 19, -4, 2, 5, 5, 4, 12, -2, 5, 4, 12, -3, 16, -2, 2, 8, 6

Offset: 1

Reinhard Zumkeller, Jul 12 2011

sign

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A000120, A033879, A093653, A175522, A175524, A175526, A292257.

Cf. A257691 (positions where a(n) <= 0), A294905 (and its char.fun).

a192895 n =
   sum (map a000120 $ filter ((== 0) . (mod n)) [1..n-1]) - a000120 n
a192895_list = map a192895 [1..]

a[n_] := DivisorSum[n, Total[IntegerDigits[#, 2]]*(-1)^Boole[# == n]&]; Array[a, 80] (* Jean-François Alcover, Dec 05 2015, adapted from PARI *)

a(n)=sumdiv(n,d,hammingweight(d)*(-1)^(d==n)) \\ Charles R Greathouse IV, Feb 07 2013

from sympy import divisors
def A192895(n): return sum((d.bit_count() if dChai Wah Wu, Jul 25 2023

a(n) = Sum(A000120(d): 1 <= d < n and n mod d = 0) - A000120(n); see A175522 for motivation and more information;
a(A175524(n)) < 0; a(A175522(n)) = 0; a(A175526(n)) > 0.
a(n) = A292257(n) - A000120(n). - Antti Karttunen, Nov 10 2017

A213723 a(n) = smallest natural number x such that x=n+A000120(x), otherwise zero.

Original entry on oeis.org

0, 2, 0, 4, 6, 0, 0, 8, 10, 0, 12, 14, 0, 0, 0, 16, 18, 0, 20, 22, 0, 0, 24, 26, 0, 28, 30, 0, 0, 0, 0, 32, 34, 0, 36, 38, 0, 0, 40, 42, 0, 44, 46, 0, 0, 0, 48, 50, 0, 52, 54, 0, 0, 56, 58, 0, 60, 62, 0, 0, 0, 0, 0, 64, 66, 0, 68, 70, 0, 0, 72, 74, 0, 76, 78

Offset: 0

Antti Karttunen, Nov 01 2012

nonn

			a(1) = 2, as 2 is the smallest natural number such that x such that x=1+A000120(x) (as 2=1+A000120(2)=1+1).
a(2) = 0, as there are no solutions for 2, because it belongs to A055938.
a(11) = 14, as 14 is the smallest natural number x such that x=11+A000120(x) (as 14=11+A000120(14)=11+3).

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

a(A055938(n)) = 0. a(A005187(n)) = A005843(n) = 2n.

Cf. A213724. Used for computing A213725-A213727. Cf. A179016.

a213723 = (* 2) . a213714  -- Reinhard Zumkeller, May 01 2015

(define (A213723 n) (A005843 (A213714 n)))

a(n) = 2*A213714(n).

Also, by partitioning into sums of distinct nonzero terms of A000225: if n can be formed as a sum of (2^a)-1 + (2^b)-1 + (2^c)-1, etc. where the exponents a, b, c are distinct and all > 0, then a(n) = 2^a + 2^b + 2^c, etc. If this is not possible, then n is one of the terms of A055938, and a(n)=0.

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cp's OEIS Frontend

A060818 a(n) = 2^(n - HammingWeight(n)) = 2^(n - BitCount(n)) = 2^(n - A000120(n)).

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A130665 a(n) = Sum_{k=0..n} 3^wt(k), where wt() = A000120().

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A175522 A000120-perfect numbers.

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A057335 a(0) = 1, and for n > 0, a(n) = A000040(A000120(n)) * a(floor(n/2)); essentially sequence A055932 generated using A000120, hence sorted by number of factors.

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A228085 a(n) = number of distinct k which satisfy n = k + wt(k), where wt(k) (A000120) gives the number of 1's in binary representation of a nonnegative integer k.

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A284005 a(0) = 1, and for n > 1, a(n) = (1 + A000120(n))*a(floor(n/2)); also a(n) = A000005(A283477(n)).

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