A001511 -id:A001511 - cp's OEIS frontend

A286460 Compound filter (2-adic valuation & sum of the divisors): a(n) = P(A001511(n), A000203(n)), where P(n,k) is sequence A000027 used as a pairing function.

1, 8, 7, 39, 16, 80, 29, 157, 79, 173, 67, 438, 92, 302, 277, 600, 154, 782, 191, 949, 497, 668, 277, 1957, 466, 905, 781, 1656, 436, 2630, 497, 2284, 1129, 1487, 1129, 4281, 704, 1832, 1541, 4282, 862, 4658, 947, 3658, 3004, 2630, 1129, 8133, 1597, 4373, 2557, 4953, 1432, 7262, 2557, 7507, 3161, 4097, 1771, 14368, 1892, 4658, 5357, 8785, 3487, 10442, 2279

Offset: 1

Antti Karttunen, May 10 2017

nonn

Antti Karttunen, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, Pairing Function
Index entries for sequences related to sigma(n)

Cf. A000027, A286360.

Cf. A000593, A146076 (sequences matching to this filter), also A000203, A161942, A286260, A286357.

A000203(n) = sigma(n);
A001511(n) = (1+valuation(n,2));
A286460(n) = (1/2)*(2 + ((A001511(n)+A000203(n))^2) - A001511(n) - 3*A000203(n));
for(n=1, 10000, write("b286460.txt", n, " ", A286460(n)));

from sympy import divisor_sigma as D
def a001511(n): return bin(n)[2:][::-1].index("1") + 1
def T(n, m): return ((n + m)**2 - n - 3*m + 2)//2
def a(n): return T(a001511(n), D(n))
print([a(n) for n in range(1, 101)]) # Indranil Ghosh, May 12 2017

(define (A286460 n) (* (/ 1 2) (+ (expt (+ (A001511 n) (A000203 n)) 2) (- (A001511 n)) (- (* 3 (A000203 n))) 2)))

a(n) = (1/2)*(2 + ((A001511(n)+A000203(n))^2) - A001511(n) - 3*A000203(n)).

A246678 Permutation of natural numbers: a(1) = 1, a(2n+1) = 1 + 2*a(n), a(2n) = A242378(A001511(n), (1+A000265(n))) - 1.

Original entry on oeis.org

1, 2, 3, 4, 5, 8, 7, 6, 9, 14, 11, 24, 17, 26, 15, 10, 13, 20, 19, 34, 29, 44, 23, 48, 49, 32, 35, 124, 53, 80, 31, 12, 21, 74, 27, 54, 41, 62, 39, 76, 69, 38, 59, 174, 89, 134, 47, 120, 97, 50, 99, 64, 65, 98, 71, 342, 249, 104, 107, 624, 161, 242, 63, 16, 25, 56, 43, 244, 149, 224, 55, 90, 109, 68, 83

Offset: 1

Antti Karttunen, Sep 01 2014

nonn

See the comments in A246676. This is a similar permutation, except for odd numbers, which are here recursively permuted by the emerging permutation itself. The even bisection halved gives A246680, the odd bisection from a(3) onward with one subtracted and then halved gives this sequence back.

Antti Karttunen, Table of n, a(n) for n = 1..8192
Index entries for sequences that are permutations of the natural numbers

Inverse: A246677.
Variants: A246676, A246684.
Even bisection halved: A246680.
Cf. A000265, A001511, A003961, A242378.

a(1) = 1, a(2n+1) = 1 + 2*a(n), a(2n) = A242378(A001511(n), (1+A000265(n))) - 1. [Where the bivariate function A242378(k,n) changes each prime p(i) in the prime factorization of n to p(i+k), i.e., it's the result of A003961 iterated k times starting from n].

A060571 Tower of Hanoi: the optimal way to move an even number of disks from peg 0 to peg 2 or an odd number from peg 0 to peg 1 is on move n to move disk A001511 from peg A060571 (here) to peg A060572.

Original entry on oeis.org

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

Offset: 1

Henry Bottomley, Apr 03 2001

a(n) is equal to a(2n) with the 1's and 2s reversed, thus a(n) = a(4n). - Donald Sampson (marsquo(AT)hotmail.com), Dec 01 2003

			Start by moving first disk from peg 0 (to peg 1), second disk from peg 0 (to peg 2), first disk form peg 1 (to peg 2), etc. so sequence starts 0,0,1,...

J.-P. Allouche, D. Astoorian, J. Randall, and J. Shallit, Morphisms, squarefree strings, and the Tower of Hanoi puzzle, Amer. Math. Monthly 101 (1994), 651-658.
Index entries for sequences related to Towers of Hanoi

Cf. A001511, A055662, A060572, A060573, A060574, A060575.

If n>2^A001511(n) then a(n)=a(n-2^A001511(n))-(-1)^A001511(n) mod 3 =A060572(n-2^A001511(n)), otherwise a(k)=0. Also A001511(n)-th digit from right of A055662(n-1).

A129628 Inverse Moebius transform of A001511.

Original entry on oeis.org

1, 3, 2, 6, 2, 6, 2, 10, 3, 6, 2, 12, 2, 6, 4, 15, 2, 9, 2, 12, 4, 6, 2, 20, 3, 6, 4, 12, 2, 12, 2, 21, 4, 6, 4, 18, 2, 6, 4, 20, 2, 12, 2, 12, 6, 6, 2, 30, 3, 9, 4, 12, 2, 12, 4, 20, 4, 6, 2, 24, 2, 6, 6, 28, 4, 12, 2, 12, 4, 12, 2, 30, 2, 6, 6, 12, 4, 12, 2, 30, 5, 6, 2, 24, 4, 6, 4, 20

Offset: 1

Ralf Stephan, May 31 2007

Dirichlet convolution of A000005 with A209229. - Ridouane Oudra, Jul 25 2025

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

Row sums of A129353.
Cf. A000005, A001511, A001620, A129628.
Cf. A007814, A099777, A115364, A001227, A209229.

seq(add(padic[ordp](2*d, 2), d in numtheory[divisors](n)), n=1..100); # Ridouane Oudra, Sep 30 2024

f[p_, e_] := If[p==2, (e+1)*(e+2)/2, e+1]; a[1] = 1; a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Sep 30 2020 *)

a(n)={sumdiv(n, d, 1 + valuation(d, 2))} \\ Andrew Howroyd, Aug 04 2018

a(2n) = a(n) + A000005(2n), a(2n+1) = A000005(2n+1).
Dirichlet g.f.: zeta(s)^2 * 2^s/(2^s-1). - Ralf Stephan, Jun 17 2007, corrected by Vaclav Kotesovec, Feb 02 2019
a(n) = Sum_{d|n} A001511(d). - Andrew Howroyd, Aug 04 2018
Sum_{k=1..n} a(k) ~ 2*n * (2*gamma - 1 + log(n/2)), where gamma is the Euler-Mascheroni constant A001620. - Vaclav Kotesovec, Feb 02 2019
Multiplicative with a(2^e) = (e+1)*(e+2)/2, and a(p^e) = e+1 for p > 2. - Amiram Eldar, Sep 30 2020
From Ridouane Oudra, Sep 30 2024: (Start)
a(n) = Sum_{i=0..A007814(n)} tau(n/2^i).
a(n) = Sum_{d|2*n} A007814(d).
a(n) = (1/2)*A001511(n)*A099777(n).
a(n) = (1/2)*(A001511(n) + 1)*A000005(n).
a(n) = A115364(n)*A001227(n). (End)

A244316 a(0) = 0, after which, if A176137(n) = 1, a(n) = A001511(A244230(n)), otherwise a(n) = a(n-A197433(A244230(n)-1)).

Original entry on oeis.org

0, 1, 2, 1, 1, 3, 1, 2, 1, 1, 2, 1, 1, 3, 4, 1, 2, 1, 1, 3, 1, 2, 1, 1, 2, 1, 1, 3, 1, 2, 1, 1, 2, 1, 1, 3, 4, 1, 2, 1, 1, 3, 5, 1, 2, 1, 1, 3, 1, 2, 1, 1, 2, 1, 1, 3, 4, 1, 2, 1, 1, 3, 1, 2, 1, 1, 2, 1, 1, 3, 1, 2, 1, 1, 2, 1, 1, 3, 4, 1, 2, 1, 1, 3, 1, 2, 1, 1, 2, 1, 1, 3, 1, 2, 1, 1, 2, 1, 1, 3, 4, 1, 2, 1, 1, 3, 5, 1, 2, 1, 1, 3, 1, 2, 1, 1, 2, 1, 1, 3, 4, 1, 2, 1, 1, 3, 1, 2, 1, 1, 2, 1, 6

Offset: 0

Antti Karttunen, Jun 25 2014

nonn

For n >= 1, a(n) tells the one-based position of the digit (from the right) where the iteration stopped at, when constructing a Semigreedy Catalan representation of n as described in A244159.
Algorithm for constructing the sequence: Find the largest Catalan number which is less than or equal to n (this is A081290(n) = A000108(k), where k = A244160(n), that is, the corresponding index of that Catalan number), and subtract that from n. Then check whether the previous Catalan number, C(m) = A000108(m), where m = k-1, exceeds the remaining n, and if it does not, then subtract that also from n, and keep on doing the same for lesser and lesser Catalan numbers, comparing and also subtracting them (whenever it is possible without going less than zero) from n, until either n becomes zero, or after subtracting C(1) = 1 from n, it still has not reached zero. In the latter case, find again the largest Catalan number which is less than or equal to remaining n, and start the process again. However, when at some point n finally reaches zero, then the index k of the last Catalan number, A000108(k) which was subtracted from n before it reached zero, is our result, a(n) = k. [Here n = the original value of n, from which we started subtracting initially from].
If n is one of the terms of A197433, meaning that if it can be represented as a sum of distinct Catalan numbers as n = C(i) + C(j) + ... + C(k) (which representation then necessarily is unique), then a(n) = min(i,j,...,k).

Antti Karttunen, Table of n, a(n) for n = 0..16796

One more than A244315.

Cf. A000108, A001511, A176137, A197433, A244230, A244159, A244160, A081290.

a(0) = 0, and for n >= 1, if A176137(n) = 1, a(n) = A001511(A244230(n)), otherwise a(n) = a(n-A197433(A244230(n)-1)).
For n >= 1, a(n) = A244315(n)+1.
For n >= 1, a(A000108(n)) = n and a(A014138(n)) = a(A014143(n)) = 1.

A269374 Permutation of natural numbers: a(1) = 1, a(n) = A255551(A001511(n), a(A003602(n))) - 1.

Original entry on oeis.org

1, 2, 3, 6, 5, 4, 11, 8, 9, 10, 7, 18, 21, 28, 15, 12, 17, 22, 19, 38, 13, 16, 35, 26, 41, 58, 55, 102, 29, 40, 23, 14, 33, 46, 43, 80, 37, 52, 75, 56, 25, 34, 31, 60, 69, 100, 51, 44, 81, 118, 115, 206, 109, 160, 203, 152, 57, 82, 79, 144, 45, 64, 27, 20, 65, 94, 91, 164, 85, 124, 159, 120, 73, 106, 103, 186, 149, 220, 111, 96, 49

Offset: 1

Antti Karttunen, Mar 01 2016

Permutation obtained from the Lucky sieve.
This sequence can be represented as a binary tree. For n > 2, each left hand child is obtained by doubling the contents of the parent node and subtracting one, and each right hand child is obtained by applying A269372(n), when the parent node contains n:
                                    1
                                    |
                 ...................2...................
                3                                       6
      5......../ \........4                  11......../ \........8
     / \                 / \                 / \                 / \
    /   \               /   \               /   \               /   \
   /     \             /     \             /     \             /     \
  9       10          7       18         21       28         15       12
17 22   19  38      13 16   35  26     41  58   55  102    29  40   23  14
etc.

Antti Karttunen, Table of n, a(n) for n = 1..4096
Index entries for sequences that are permutations of the natural numbers

Inverse: A269373.
Cf. A000035, A001511, A003602, A255551, A269372.
Cf. also A269375, A269377 and also A249814, A269384.

a(1) = 1, a(n) = A255551(A001511(n), a(A003602(n))) - 1.
a(1) = 1, a(2n) = A269372(a(n)), a(2n+1) = (2*a(n+1))-1.
Other identities. For all n >= 0:
A000035(a(n)) = A000035(n). [This permutation preserves the parity of n.]

A269384 Permutation of natural numbers: a(1) = 1, a(n) = A255127(A001511(n), a(A003602(n))) - 1.

Original entry on oeis.org

1, 2, 3, 4, 5, 8, 7, 6, 9, 14, 15, 18, 13, 20, 11, 10, 17, 26, 27, 34, 29, 44, 35, 30, 25, 38, 39, 48, 21, 32, 19, 12, 33, 50, 51, 64, 53, 80, 67, 58, 57, 86, 87, 108, 69, 104, 59, 54, 49, 74, 75, 94, 77, 116, 95, 84, 41, 62, 63, 78, 37, 56, 23, 16, 65, 98, 99, 124, 101, 152, 127, 112, 105, 158, 159, 198, 133, 200

Offset: 1

Antti Karttunen, Mar 01 2016

Permutation obtained from the Ludic sieve.
This sequence can be represented as a binary tree. For n > 2, each left hand child is obtained by doubling the contents of the parent node and subtracting one, and each right hand child is obtained by applying A269382(n), when the parent node contains n:
                                    1
                                    |
                 ...................2...................
                3                                       4
      5......../ \........8                   7......../ \........6
     / \                 / \                 / \                 / \
    /   \               /   \               /   \               /   \
   /     \             /     \             /     \             /     \
  9       14         15       18         13       20         11       10
17 26   27  34     29  44   35  30     25  38   39  48     21  32   19  12
etc.

Antti Karttunen, Table of n, a(n) for n = 1..8192
Index entries for sequences that are permutations of the natural numbers

Inverse: A269383.
Cf. A000035, A001511, A003602, A255127, A269382.
Cf. also A269385, A269387 and also A249814, A269374.

a(1) = 1, a(n) = A255127(A001511(n), a(A003602(n))) - 1.
a(1) = 1, a(2n) = A269382(a(n)), a(2n+1) = (2*a(n+1))-1.
Other identities. For all n >= 0:
A000035(a(n)) = A000035(n). [This permutation preserves the parity of n.]

A286252 Compound filter: a(n) = P(A001511(1+n), A278222(n)), where P(n,k) is sequence A000027 used as a pairing function.

Original entry on oeis.org

1, 5, 2, 18, 2, 23, 7, 59, 2, 23, 16, 94, 7, 80, 29, 195, 2, 23, 16, 94, 16, 467, 67, 355, 7, 80, 67, 706, 29, 302, 121, 672, 2, 23, 16, 94, 16, 467, 67, 355, 16, 467, 436, 1894, 67, 1832, 277, 1331, 7, 80, 67, 706, 67, 1832, 631, 2779, 29, 302, 277, 2704, 121, 1178, 497, 2422, 2, 23, 16, 94, 16, 467, 67, 355, 16, 467, 436, 1894, 67, 1832, 277, 1331, 16, 467, 436

Offset: 0

Antti Karttunen, May 07 2017

nonn

Antti Karttunen, Table of n, a(n) for n = 0..16384
MathWorld, Pairing Function

Cf. A000027, A001511, A278222, A286162, A286251, A286253, A286254.

A001511(n) = (1+valuation(n,2));
A005940(n) = { my(p=2, t=1); n--; until(!n\=2, if((n%2), (t*=p), p=nextprime(p+1))); t }; \\ Modified from code of M. F. Hasler
A046523(n) = { my(f=vecsort(factor(n)[, 2], , 4), p); prod(i=1, #f, (p=nextprime(p+1))^f[i]); };  \\ This function from Charles R Greathouse IV, Aug 17 2011
A278222(n) = A046523(A005940(1+n));
A286252(n) = (2 + ((A001511(1+n)+A278222(n))^2) - A001511(1+n) - 3*A278222(n))/2;
for(n=0, 16384, write("b286252.txt", n, " ", A286252(n)));

from sympy import prime, factorint
import math
def T(n, m): return ((n + m)**2 - n - 3*m + 2)/2
def A(n): return n - 2**int(math.floor(math.log(n, 2)))
def b(n): return n + 1 if n<2 else prime(1 + (len(bin(n)[2:]) - bin(n)[2:].count("1"))) * b(A(n))
def a005940(n): return b(n - 1)
def P(n):
    f = factorint(n)
    return sorted([f[i] for i in f])
def a046523(n):
    x=1
    while True:
        if P(n) == P(x): return x
        else: x+=1
def a278222(n): return a046523(a005940(n + 1))
def a001511(n): return 2 + bin(n - 1)[2:].count("1") - bin(n)[2:].count("1")
def a(n): return T(a001511(n + 1), a278222(n)) # Indranil Ghosh, May 07 2017

(define (A286252 n) (* (/ 1 2) (+ (expt (+ (A001511 (+ 1 n)) (A278222 n)) 2) (- (A001511 (+ 1 n))) (- (* 3 (A278222 n))) 2)))

a(n) = (1/2)*(2 + ((A001511(1+n)+A278222(n))^2) - A001511(1+n) - 3*A278222(n)).

A302035 a(1) = 0, for n > 1, a(n) = A001511(A260739(n)); Number of instances of (the smallest) Ludic factor A272565(n) in n.

Original entry on oeis.org

0, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 2, 1, 1, 1, 4, 1, 1, 2, 2, 3, 1, 1, 3, 1, 1, 1, 2, 1, 1, 2, 5, 2, 1, 1, 2, 1, 1, 1, 3, 1, 1, 1, 2, 4, 1, 1, 4, 3, 1, 1, 2, 1, 1, 2, 3, 2, 1, 1, 2, 1, 1, 1, 6, 1, 1, 1, 2, 3, 1, 1, 3, 2, 1, 1, 2, 1, 1, 2, 4, 2, 1, 1, 2, 3, 1, 1, 3, 1, 1, 1, 2, 5, 1, 1, 5, 1, 1, 1, 2, 2, 1, 1, 3, 2

Offset: 1

Antti Karttunen, Apr 01 2018

nonn

An A067029 analog for "Ludic factorization": iterating the map n -> A302034(n) until 1 is reached, and taking the Ludic factor (A272565) of each term gives a sequence of distinct Ludic numbers (A003309) in ascending order, while applying this function (A302035) to those terms gives the corresponding "exponents" of those Ludic factors, that is, the count of consecutive occurrences of each when iterating the map n -> A302032(n), which gives the same factors with repetitions. Permutation pair A302025/A302026 maps between the Ludic factorization and the ordinary prime factorization of n. See also comments and examples in A302032.

Antti Karttunen, Table of n, a(n) for n = 1..10105
Index entries for sequences generated by sieves

Cf. A001511, A003309, A255127, A260739, A302025, A302026, A302032, A302034.

Cf. also A067029, A302045.

a(1) = 0; for n > 1, a(n) = A001511(A260739(n)).

For n > 1, a(n) = A302025(A067029(A302026(n))).

A324724 a(n) = A001511(A324712(n)), with a(n) = 0 if A324712(n) = 0.

Original entry on oeis.org

0, 1, 1, 1, 1, 3, 1, 3, 1, 1, 1, 1, 1, 3, 2, 5, 1, 1, 1, 3, 1, 2, 1, 4, 1, 2, 3, 1, 1, 3, 1, 4, 2, 2, 2, 3, 1, 3, 3, 6, 1, 4, 1, 1, 1, 2, 1, 4, 1, 1, 8, 1, 1, 3, 1, 3, 3, 4, 1, 2, 1, 2, 3, 4, 2, 1, 1, 1, 2, 2, 1, 3, 1, 2, 1, 1, 2, 1, 1, 4, 5, 2, 1, 1, 4, 2, 3, 4, 1, 1, 1, 1, 2, 2, 3, 5, 1, 4, 1, 2, 1, 1, 1, 8, 3

Offset: 1

Antti Karttunen, Mar 17 2019

nonn

Antti Karttunen, Table of n, a(n) for n = 1..10000 (based on Hans Havermann's factorization of A156552)
Index entries for sequences related to binary expansion of n
Index entries for sequences computed from indices in prime factorization
Index entries for sequences related to sigma(n)

Cf. A001511, A323243, A324543, A324712, A324725, A324810, A324828, A324829.

A324712(n) = { my(v=0); fordiv(n, d, if(issquarefree(n/d), v=bitxor(v,
A323243(d)))); (v); }; \\ Needs also code from A323243.
A001511ext(n) = if(!n,n,sign(n)*(1+valuation(n,2))); \\ Like A001511 but gives 0 for 0 and -A001511(-n) for negative numbers.
A324724(n) = A001511ext(A324712(n));

If A324712(n) = 0, then a(n) = 0, otherwise a(n) = A001511(A324712(n)).
a(p) = 1 for all primes p.
A324828(n) = [1 == a(n)], where [ ] is the Iverson bracket.

cp's OEIS Frontend

A286460 Compound filter (2-adic valuation & sum of the divisors): a(n) = P(A001511(n), A000203(n)), where P(n,k) is sequence A000027 used as a pairing function.

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A246678 Permutation of natural numbers: a(1) = 1, a(2n+1) = 1 + 2*a(n), a(2n) = A242378(A001511(n), (1+A000265(n))) - 1.

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A060571 Tower of Hanoi: the optimal way to move an even number of disks from peg 0 to peg 2 or an odd number from peg 0 to peg 1 is on move n to move disk A001511 from peg A060571 (here) to peg A060572.

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A129628 Inverse Moebius transform of A001511.

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A244316 a(0) = 0, after which, if A176137(n) = 1, a(n) = A001511(A244230(n)), otherwise a(n) = a(n-A197433(A244230(n)-1)).

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A269374 Permutation of natural numbers: a(1) = 1, a(n) = A255551(A001511(n), a(A003602(n))) - 1.

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A269384 Permutation of natural numbers: a(1) = 1, a(n) = A255127(A001511(n), a(A003602(n))) - 1.

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A286252 Compound filter: a(n) = P(A001511(1+n), A278222(n)), where P(n,k) is sequence A000027 used as a pairing function.

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A302035 a(1) = 0, for n > 1, a(n) = A001511(A260739(n)); Number of instances of (the smallest) Ludic factor A272565(n) in n.

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