A004198 -id:A004198 - cp's OEIS frontend

A059895 Table a(i,j) = product prime[k]^(Ei[k] AND Ej[k]) where Ei and Ej are the vectors of exponents in the prime factorizations of i and j; AND is the bitwise operation on binary representation of the exponents.

1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 4, 1, 2, 1, 1, 1, 3, 1, 1, 3, 1, 1, 1, 2, 1, 1, 5, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 4, 1, 6, 1, 4, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 7, 2, 1, 1, 1, 1, 1, 1, 1, 3, 1, 5, 1, 1, 1, 1, 5, 1, 3, 1, 1, 1, 2, 1, 4, 1, 2, 1, 8, 1, 2, 1, 4, 1, 2, 1

Offset: 1

Marc LeBrun, Feb 06 2001

Analogous to GCD, with AND replacing MIN.

			The top left 18 X 18 corner of the array:
1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  1
1,  2,  1,  1,  1,  2,  1,  2,  1,  2,  1,  1,  1,  2,  1,  1,  1,  2
1,  1,  3,  1,  1,  3,  1,  1,  1,  1,  1,  3,  1,  1,  3,  1,  1,  1
1,  1,  1,  4,  1,  1,  1,  4,  1,  1,  1,  4,  1,  1,  1,  1,  1,  1
1,  1,  1,  1,  5,  1,  1,  1,  1,  5,  1,  1,  1,  1,  5,  1,  1,  1
1,  2,  3,  1,  1,  6,  1,  2,  1,  2,  1,  3,  1,  2,  3,  1,  1,  2
1,  1,  1,  1,  1,  1,  7,  1,  1,  1,  1,  1,  1,  7,  1,  1,  1,  1
1,  2,  1,  4,  1,  2,  1,  8,  1,  2,  1,  4,  1,  2,  1,  1,  1,  2
1,  1,  1,  1,  1,  1,  1,  1,  9,  1,  1,  1,  1,  1,  1,  1,  1,  9
1,  2,  1,  1,  5,  2,  1,  2,  1, 10,  1,  1,  1,  2,  5,  1,  1,  2
1,  1,  1,  1,  1,  1,  1,  1,  1,  1, 11,  1,  1,  1,  1,  1,  1,  1
1,  1,  3,  4,  1,  3,  1,  4,  1,  1,  1, 12,  1,  1,  3,  1,  1,  1
1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  1, 13,  1,  1,  1,  1,  1
1,  2,  1,  1,  1,  2,  7,  2,  1,  2,  1,  1,  1, 14,  1,  1,  1,  2
1,  1,  3,  1,  5,  3,  1,  1,  1,  5,  1,  3,  1,  1, 15,  1,  1,  1
1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  1, 16,  1,  1
1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  1, 17,  1
1,  2,  1,  1,  1,  2,  1,  2,  9,  2,  1,  1,  1,  2,  1,  1,  1, 18
A(864,1944) = A(2^5*3^3,2^3*3^5) = 2^(5 AND 3)* 3^(3 AND 5) = 2^1*3^1 = 6.

Antti Karttunen, Table of n, a(n) for n = 1..10440; the first 144 antidiagonals of the array

Cf. A003985 (A004198), A003989, A028234, A059896, A059897, A067029, A267115, A284578.

a[i_, i_] := i;
a[i_, j_] := Module[{f1 = FactorInteger[i], f2 = FactorInteger[j], e1, e2}, Scan[(e1[#[[1]]] = #[[2]])&, f1]; Scan[(e2[#[[1]]] = #[[2]])&, f2]; Times @@ (#^BitAnd[e1[#], e2[#]]& /@ Intersection[f1[[All, 1]], f2[[All, 1]]]) ];
Table[a[i - j + 1, j], {i, 1, 15}, {j, 1, i}] // Flatten (* Jean-François Alcover, Jun 19 2018 *)

(define (A059895 n) (A059895bi (A002260 n) (A004736 n)))
(define (A059895bi a b) (let loop ((a a) (b b) (m 1)) (cond ((= 1 a) m) ((= 1 b) m) ((equal? (A020639 a) (A020639 b)) (loop (A028234 a) (A028234 b) (* m (expt (A020639 a) (A004198bi (A067029 a) (A067029 b)))))) ((< (A020639 a) (A020639 b)) (loop (A028234 a) b m)) (else (loop a (A028234 b) m)))))
;; Antti Karttunen, Apr 11 2017

From Antti Karttunen, Apr 11 2017: (Start)
A(x,y) = A059896(x,y) / A059897(x,y).
A(x,y) * A059896(x,y) = A(x,y)^2 * A059897(x,y) = x*y.
(End)

Data section extended to 120 terms by Antti Karttunen, Apr 11 2017

A283475 a(n) = A019565(A005187(n)).

Original entry on oeis.org

1, 2, 6, 5, 30, 7, 21, 42, 210, 11, 33, 66, 165, 330, 154, 231, 2310, 13, 39, 78, 195, 390, 182, 273, 1365, 2730, 286, 429, 1430, 2145, 1001, 2002, 30030, 17, 51, 102, 255, 510, 238, 357, 1785, 3570, 374, 561, 1870, 2805, 1309, 2618, 19635, 39270, 442, 663, 2210, 3315, 1547, 3094, 15470, 23205, 2431, 4862, 12155

Offset: 0

Antti Karttunen, Mar 15 2017

Antti Karttunen, Table of n, a(n) for n = 0..8191

Cf. A000040, A000051, A001221, A001222, A004198, A005117, A005187, A019565, A046523, A097248, A108951, A280700, A280705, A283477, A283478.

Cf. A283476 (same sequence sorted into ascending order).

Map[Times @@ Prime@ Flatten@ Position[#, 1] &@ Reverse@ IntegerDigits[#, 2] &, Table[2 n - DigitCount[2 n, 2, 1], {n, 0, 60}]] (* Michael De Vlieger, Mar 16 2017 *)

(define (A283475 n) (A019565 (A005187 n)))

a(n) = A019565(A005187(n)).
a(n) = A097248(A283477(n)) = A097248(A108951(A019565(n))) = A283478(A019565(n)).
Other identities:
If A004198(x,y) = 0, then a(x+y) = A097248(a(x)*a(y)).
For all n >= 1, a(A000051(n)) = A000040(n+2).
For all n >= 0, A001221(a(n)) = A001222(a(n)) = A280700(n).
For all n >= 0, A046523(a(n)) = A280705(n).

A269174 Formula for Wolfram's Rule 124 cellular automaton: a(n) = (n OR 2n) AND ((n XOR 2n) OR (n XOR 4n)).

Original entry on oeis.org

0, 3, 6, 7, 12, 15, 14, 11, 24, 27, 30, 31, 28, 31, 22, 19, 48, 51, 54, 55, 60, 63, 62, 59, 56, 59, 62, 63, 44, 47, 38, 35, 96, 99, 102, 103, 108, 111, 110, 107, 120, 123, 126, 127, 124, 127, 118, 115, 112, 115, 118, 119, 124, 127, 126, 123, 88, 91, 94, 95, 76, 79, 70, 67, 192, 195, 198, 199, 204, 207, 206, 203, 216

Offset: 0

Antti Karttunen, Feb 22 2016

nonn

Antti Karttunen, Table of n, a(n) for n = 0..32767
Eric Weisstein's World of Mathematics, Elementary Cellular Automaton
Stephen Wolfram, A New Kind of Science
Index entries for sequences related to cellular automata
Index to Elementary Cellular Automata

Cf. A003986, A003987, A004198, A048724, A048725, A057889, A269173, A161903.
Cf. A269175.
Cf. A269176 (numbers not present in this sequence).
Cf. A269177 (same sequence sorted into ascending order, duplicates removed).
Cf. A269178 (numbers that occur only once).
Cf. A267357 (iterates from 1 onward).

package main
import "fmt"
func main() {
    for n:=0; n<=100; n++{
    fmt.Println((n|2*n)&((n^(2*n))|(n^(4*n))))}
} // Indranil Ghosh, Apr 19 2017

a[n_] := BitAnd[BitOr[n, 2n], BitOr[BitXor[n, 2n], BitXor[n, 4n]]];
a /@ Range[0, 100] (* Jean-François Alcover, Feb 23 2020 *)

def a(n): return (n|2*n)&((n^(2*n))|(n^(4*n))) # Indranil Ghosh, Apr 19 2017

(define (A269174 n) (A004198bi (A163617 n) (A003986bi (A048724 n) (A048725 n))))

a(n) = A163617(n) AND A269173(n).
a(n) = A163617(n) AND (A048724(n) OR A048725(n)).
a(n) = (n OR 2n) AND ((n XOR 2n) OR (n XOR 4n)).
Other identities. For all n >= 0:
a(2*n) = 2*a(n).
a(n) = A057889(A161903(A057889(n))). [Rule 124 is the mirror image of rule 110.]
G.f.: (-3*x^3 - 2*x^2 - 3*x)/(x^4 - 1) + Sum_{k>=1}((2^(k + 1)*x^(2^k) - 2^(k + 1)*x^(14*2^(k - 2)))/((x^(2^(k + 2)) - 1)*(x - 1))). - Miles Wilson, Jan 25 2025

A292274 a(n) = A292383(A163511(n)).

Original entry on oeis.org

0, 0, 0, 1, 0, 0, 2, 2, 0, 1, 0, 0, 4, 5, 4, 5, 0, 0, 2, 2, 0, 1, 0, 0, 8, 8, 10, 11, 8, 8, 10, 11, 0, 1, 0, 0, 4, 5, 4, 5, 0, 0, 2, 2, 0, 1, 0, 0, 16, 17, 16, 17, 20, 20, 22, 22, 16, 17, 16, 17, 20, 20, 22, 22, 0, 0, 2, 2, 0, 1, 0, 0, 8, 8, 10, 11, 8, 8, 10, 11, 0, 1, 0, 0, 4, 5, 4, 5, 0, 0, 2, 2, 0, 1, 0, 0, 32, 32, 34, 35, 32, 32, 34, 35, 40, 41, 40, 40, 44

Offset: 0

Antti Karttunen, Sep 16 2017

Because A292383(n) = a(A243071(n)), the sequence works as a "masking function" where the 1-bits in a(n) (always a subset of the 1-bits in binary expansion of n) indicate which numbers are of the form 4k+3 in binary tree A005940 on that trajectory which leads from the root of the tree to the node containing A163511(n). This works because A243071(n) = A054429(A156552(n)), a bit-flipped variant of Leonid Broukhis's unary-binary encoded compressed factorization of natural numbers, A156552(n) being an inverse of Doudna map f(n) = A005940(1+n).

			A163511(18) = 54, that is, at "node address" 18 in binary tree A163511 (which is the mirror image of A005940) sits number 54. 18 in binary is "10010", which when read from left to right (after the most significant bit which is always 1) gives the directions to follow in either tree when starting from the root, so that we land in number 54. (E.g. in A005940-tree, turn right from 2, turn right from 4, turn left from 8 and then turn right from 27 and one lands in 54, this corresponds with the four lowermost bits of the code, "0010". In A163511 the sense of direction is just reversed). When one selects the numbers of the form 4k+3 from this path 1 -> 2 -> 4 -> 8 -> 27 -> 54, one sees that only one is 27, which corresponds with the second rightmost bit (which also is the only 1-bit) in the code, which can be masked with 2 (binary "10"), thus a(18) = 2.
A163511(15) = 7, that is, at "node address" 15 in binary tree A163511 sits number 7. 15 in binary is "1111", which tells that 7 can be located in mirror-image tree A005940 by going (after the initial root 1 and 2) three steps towards left from 2: 1 -> 2 -> 3 -> 5 -> 7. Of these numbers, only 3 and 7 are of the form 4k+3, thus the mask with which to obtain the corresponding bits from "1111" is "00101" (5 in binary), thus a(15) = 5.
A163511(31) = 11, that is, at "node address" 31 in binary tree A163511 sits number 11. 31 in binary is "11111", which tells that 11 can be located in mirror-image tree A005940 by going (after the initial root 1 and 2) four steps towards left from 2: 1 -> 2 -> 3 -> 5 -> 7 -> 11. Of these numbers, only 3, 7 and 11 are of the form 4k+3, thus the mask with which to obtain the corresponding bits from "11111" is "001011" (11 in binary), thus a(31) = 11.

Cf. A004198, A163511, A243071, A292271, A292383.

Differs from related A292592 for the first time at n=31, where a(31) = 11, while A292592(31) = 10. Compare also the scatter plots.

f[n_] := Reverse@ Map[Ceiling[(Length@ # - 1)/2] &, DeleteCases[Split@ Join[Riffle[IntegerDigits[n, 2], 0], {0}], {k__} /; k == 1]]; Map[FromDigits[Reverse@ NestWhileList[Function[k, Which[k == 1, 1, EvenQ@ k, k/2, True, Times @@ Power[Which[# == 1, 1, # == 2, 1, True, NextPrime[#, -1]] & /@ First@ #, Last@ #] &@ Transpose@ FactorInteger@ k]], #, # > 1 &] /. k_ /; IntegerQ@ k :> If[Mod[k, 4] == 3, 1, 0], 2] &, {1}~Join~Table[Function[t, Prime[t] Product[Prime[m]^(f[n][[m]]), {m, t}]][DigitCount[n, 2, 1]], {n, 120}]] (* Michael De Vlieger, Sep 22 2017 *)

a(n) = A292383(A163511(n)).
a(n) + A292271(n) = n, a(n) AND A292271(n) = 0.
a(n) AND n = a(n), where AND is bitwise-AND (A004198).

Comments and examples from Antti Karttunen, Sep 22 2017

A317934 Multiplicative with a(p^n) = 2^A011371(n); denominators for certain "Dirichlet Square Roots" sequences.

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 2, 1, 1, 1, 8, 1, 2, 1, 2, 1, 1, 1, 2, 2, 1, 2, 2, 1, 1, 1, 8, 1, 1, 1, 4, 1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 8, 2, 2, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 2, 16, 1, 1, 1, 2, 1, 1, 1, 4, 1, 1, 2, 2, 1, 1, 1, 8, 8, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 8, 1, 2, 2, 4, 1, 1, 1, 2, 1

Offset: 1

Antti Karttunen, Aug 12 2018

a(n) is the denominator of certain rational valued sequences f(n), that have been defined as f(n) = (1/2) * (b(n) - Sum_{d|n, d>1, dA034444 and A037445.
Many of the same observations as given in A046644 apply also here. Note that A011371 shares with A005187 the property that A011371(x+y) <= A011371(x) + A011371(y), with equivalence attained only when A004198(x,y) = 0, and also the property that A011371(2^(k+1)) = 1 + 2*A011371(2^k).
The following list gives such pairs num(n), b(n) for which b(n) is Dirichlet convolution of num(n)/a(n).
  Numerators   Dirichlet convolution of numerator(n)/a(n) yields
  -------      -----------
  A317933      A034444
  A317941      A037445
  A317940      A046644
Expansion of Dirichlet g.f. Product_{prime} 1/(1 - 2/p^s)^(1/2) is A046643/A317934. - Vaclav Kotesovec, May 08 2025

Antti Karttunen, Table of n, a(n) for n = 1..65537
Vaclav Kotesovec, Graph - the asymptotic ratio (1000000 terms)
Wikipedia, Dirichlet convolution

Cf. A011371, A034444, A317940, A317941, A317946.
Cf. A317933, A317940, A317941 (numerator-sequences).
Cf. also A046644, A299150, A299152, A317832, A317932, A317926 (for denominator sequences of other similar constructions).
Cf. A046643, A167864, A347195.

A011371(n) = (n - hammingweight(n));
A317934(n) = factorback(apply(e -> 2^A011371(e),factor(n)[,2]));

for(n=1, 100, print1(denominator(direuler(p=2, n, 1/(1-2*X)^(1/2))[n]), ", ")) \\ Vaclav Kotesovec, May 07 2025

for(n=1, 100, print1(denominator(direuler(p=2, n, ((1+X)/(1-X))^(1/2))[n]), ", ")) \\ Vaclav Kotesovec, May 09 2025

a(n) = 2^A317946(n).
a(n) = denominator of f(n), where f(1) = 1, f(n) = (1/2) * (b(n) - Sum_{d|n, d>1, d 1, where b is A034444, A037445 or A046644 for example.
Sum_{k=1..n} A046643(k)/a(k) ~ n * sqrt(A167864*log(n)/(Pi*log(2))) * (1 + (4*(gamma - 1) + 5*log(2) - 4*A347195)/(8*log(n))), where gamma is the Euler-Mascheroni constant A001620. - Vaclav Kotesovec, May 08 2025

A080099 Triangle T(n,k) = n AND k, 0<=k<=n, bitwise logical AND, read by rows.

Original entry on oeis.org

0, 0, 1, 0, 0, 2, 0, 1, 2, 3, 0, 0, 0, 0, 4, 0, 1, 0, 1, 4, 5, 0, 0, 2, 2, 4, 4, 6, 0, 1, 2, 3, 4, 5, 6, 7, 0, 0, 0, 0, 0, 0, 0, 0, 8, 0, 1, 0, 1, 0, 1, 0, 1, 8, 9, 0, 0, 2, 2, 0, 0, 2, 2, 8, 8, 10, 0, 1, 2, 3, 0, 1, 2, 3, 8, 9, 10, 11, 0, 0, 0, 0, 4, 4, 4, 4, 8, 8, 8, 8, 12, 0, 1, 0, 1, 4, 5, 4, 5, 8, 9, 8, 9

Offset: 0

Reinhard Zumkeller, Jan 28 2003

A080100(n) = number of numbers k such that n AND k = 0 in n-th row of the triangular array.

			Triangle starts:
0
0 1
0 0 2
0 1 2 3
0 0 0 0 4
0 1 0 1 4 5
0 0 2 2 4 4 6
0 1 2 3 4 5 6 7
...

Rick L. Shepherd, Rows n=0..500 of triangle, flattened
Eric Weisstein's World of Mathematics, AND.

Cf. A080100, A222423 (row sums), A004198 (array).

Other triangles: A080098 (OR), A051933 (XOR), A265705 (IMPL), A102037 (CNIMPL).

import Data.Bits ((.&.))
a080099 n k = n .&. k :: Int
a080099_row n = map (a080099 n) [0..n]
a080099_tabl = map a080099_row [0..]
-- Reinhard Zumkeller, Aug 03 2014, Jul 05 2012

Column[Table[BitAnd[n, k], {n, 0, 15}, {k, 0, n}], Center] (* Alonso del Arte, Jun 19 2012 *)

T(n,k)=bitand(n,k) \\ Charles R Greathouse IV, Jan 26 2013

def T(n, k): return n & k
print([T(n, k) for n in range(14) for k in range(n+1)]) # Michael S. Branicky, Dec 16 2021

A218542 Number of times when an even number is encountered, when going from 2^(n+1)-1 to (2^n)-1 using the iterative process described in A071542.

Original entry on oeis.org

1, 0, 1, 1, 2, 3, 8, 12, 23, 44, 86, 163, 308, 576, 1074, 1991, 3680, 6800, 12626, 23644, 44751, 85567, 164941, 319694, 621671, 1211197, 2362808, 4614173, 9018299, 17635055, 34486330, 67408501, 131642673, 256795173, 500346954, 973913365, 1894371802, 3683559071

Offset: 0

Antti Karttunen, Nov 02 2012

nonn

Ratio a(n)/A213709(n) develops as: 1, 0, 0.5, 0.333..., 0.4, 0.333..., 0.471..., 0.400..., 0.426..., 0.449..., 0.480..., 0.494..., 0.502..., 0.501..., 0.497..., 0.489..., 0.479..., 0.469..., 0.461..., 0.455..., 0.453..., 0.454..., 0.458..., 0.464..., 0.469..., 0.475..., 0.480..., 0.484..., 0.488..., 0.492..., 0.496..., 0.499..., 0.502..., 0.503..., 0.505..., 0.505..., 0.505..., 0.505..., 0.505..., 0.504..., 0.504..., 0.503..., 0.503..., 0.502..., 0.502..., 0.502..., 0.503..., 0.503... (See further comments at A218543).

			(2^0)-1 (0) is reached from (2^1)-1 (1) with one step by subtracting A000120(1) from 1. Zero is an even number, so a(0)=1.
(2^1)-1 (1) is reached from (2^2)-1 (3) with one step by subtracting A000120(3) from 3. One is not an even number, so a(1)=0.
(2^2)-1 (3) is reached from (2^3)-1 (7) with two steps by first subtracting A000120(7) from 7 -> 4, and then subtracting A000120(4) from 4 -> 3. Four is an even number, but three is not, so a(2)=1.

Antti Karttunen, Table of n, a(n) for n = 0..47

Cf. A219662 (analogous sequence for factorial number system).

a(n) = Sum_{i=A218600(n) .. (A218600(n+1)-1)} A213728(i).

a(n) = A213709(n) - A218543(n).

More terms from Antti Karttunen, Jun 05 2013

A218543 Number of times when an odd number is encountered, when going from 2^(n+1)-1 to (2^n)-1 using the iterative process described in A071542.

Original entry on oeis.org

0, 1, 1, 2, 3, 6, 9, 18, 31, 54, 93, 167, 306, 574, 1088, 2081, 3998, 7696, 14792, 28335, 54049, 102742, 194948, 369955, 703335, 1340834, 2563781, 4915378, 9444799, 18180238, 35047841, 67660623, 130806130, 253252243, 491034479, 953404380, 1853513715, 3607440034

Offset: 0

Antti Karttunen, Nov 02 2012

nonn

Ratio a(n)/A213709(n) develops as: 0, 1, 0.5, 0.666..., 0.6, 0.666..., 0.529..., 0.6, 0.574..., 0.551..., 0.520..., 0.506..., 0.498..., 0.499..., 0.503..., 0.511..., 0.521..., 0.531..., 0.539..., 0.545..., 0.547..., 0.546..., 0.542..., 0.536..., 0.531..., 0.525..., 0.520..., 0.516..., 0.512..., 0.508..., 0.504..., 0.501..., 0.498..., 0.497..., 0.495..., 0.495..., 0.495..., 0.495..., 0.495..., 0.496..., 0.496..., 0.497..., 0.497..., 0.498..., 0.498..., 0.498..., 0.497..., 0.497...
Ratio a(n)/A218542(n) develops as follows from n>=2 onward:
1, 2, 1.5, 2, 1.125, 1.5, 1.348..., 1.227..., 1.081..., 1.025..., 0.994..., 0.997..., 1.013..., 1.045..., 1.086..., 1.132..., 1.172..., 1.198..., 1.208..., 1.201..., 1.182..., 1.157..., 1.131..., 1.107..., 1.085..., 1.065..., 1.047..., 1.031..., 1.016..., 1.004..., 0.994..., 0.986..., 0.981..., 0.979..., 0.978..., 0.979..., 0.981..., 0.983..., 0.986..., 0.988..., 0.989..., 0.990..., 0.991..., 0.991..., 0.989..., 0.987...
Observation: A179016 seems to alternatively slightly favor the odd numbers and then again the even numbers, at least for the terms computed so far.
Please plot this sequence against A218542 in the "ratio mode" (given as a link) to see how smoothly (almost "continuously") the ratios given above develop.
What is the reason for that smoothness? (Conjecture: The distribution of "tendrils", i.e. finite subtrees in the beanstalk and its almost fractal nature? Cf: A218787.)

			(2^0)-1 (0) is reached from (2^1)-1 (1) with one step by subtracting A000120(1) from 1. Zero is not an odd number, so a(0)=0.
(2^1)-1 (1) is reached from (2^2)-1 (3) with one step by subtracting A000120(3) from 3. One is an odd number, so a(1)=1.
(2^2)-1 (3) is reached from (2^3)-1 (7) with two steps by first subtracting A000120(7) from 7 -> 4, and then subtracting A000120(4) from 4 -> 3. Four is not an odd number, but three is, so a(2)=1.

Antti Karttunen, Table of n, a(n) for n = 0..47
OEIS Server, Sequence plotted together with A218542 showing how their ratio develops.

a(n) = A213709(n)-A218542(n). Cf. A213733, A218787, A218789.

Analogous sequence for factorial number system: A219663.

a(n) = Sum_{i=A218600(n) .. (A218600(n+1)-1)} A213729(i)

A218616 The infinite trunk of beanstalk (A179016) with reversed subsections.

Original entry on oeis.org

0, 1, 3, 7, 4, 15, 11, 8, 31, 26, 23, 19, 16, 63, 57, 53, 49, 46, 42, 39, 35, 32, 127, 120, 116, 112, 109, 104, 101, 97, 94, 89, 85, 81, 78, 74, 71, 67, 64, 255, 247, 240, 236, 231, 225, 221, 215, 209, 205, 200, 197, 193, 190, 184, 180, 176, 173, 168, 165, 161

Offset: 0

Antti Karttunen, Nov 10 2012

This can be viewed as an irregular table: after the initial zero on row 0, start each row n with (2^n)-1 and subtract repeatedly the number of 1-bits to get successive terms, until the number that has already been listed (which is always (2^(n-1))-1) is encountered, which is not listed second time, but instead, the current row is finished and the next row starts with (2^(n+1))-1, with the same process repeated.
This contains the terms in the infinite trunk of beanstalk (A179016) listed in partially reversed manner: after the initial zero each subsequence lists A213709(n) successive terms  from A179016, descending from (2^n)-1 downwards, usually down to 2^(n-1) (conjectured to indeed be a power of 2 in each case, apart from 2 itself missing from the beginning of the sequence).
Currently A179016 and many of the derived sequences are much easier and somewhat faster to compute with the help of this sequence, especially if the program computes any other required values incrementally in the same loop.

			After zero, we start with (2^1)-1 = 1, subtract A000120(1)=1 from it, resulting 1-1=0 (which is of the form (2^0)-1, thus not listed second time), instead, start the next row with (2^2)-1 = 3, subtract A000120(3)=2 from it, resulting 3-2=1, which has been already encountered, thus start the next row with (2^3)-1 = 7, subtract A000120(7)=3 from it, resulting 7-3=4, which is listed after 7, then 4-A000120(4)=4-1=3, which is of the form (2^k)-1 and already encountered, thus start the next row with (2^4)-1 = 15, etc. This results an irregular table which begins as:
0; 1; 3; 7, 4; 15, 11, 8; 31, 26, 23, 19, 16; 63, 57, ...
After zero, each row n is A213709(n-1) elements long.

Antti Karttunen, Rows 0..16, flattened

a(n) = A179016(A218602(n)). The rows are the initial portions of every (2^n)-1:th row in A218254.

a(0)=0, a(1)=1, and for n > 1, if n = A213710(A213711(n)-1) then a(n) = (2^A213711(n)) - 1, and in other cases, a(n) = A011371(a(n-1)).

Alternatively: For n < 4, a(n) = (2^n)-1, and for n >= 4, a(n) = A004755(A004755(A011371(a(n-1)))) if A011371(a(n-1))+1 is power of 2, otherwise just A011371(a(n-1)).

A267115 Bitwise-AND of the exponents of primes in the prime factorization of n, a(1) = 0.

Original entry on oeis.org

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

Offset: 1

Antti Karttunen, Feb 03 2016

nonn

The sums of the first 10^k terms, for k = 1, 2, ..., are 13, 105, 826, 7440, 71558, 707625, 7053959, 70473172, 704531711, 7044701307, 70445097231, ... . Apparently, the asymptotic mean of this sequence is limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 0.7044... . - Amiram Eldar, Sep 09 2022

			For n = 24 = 2^3 * 3^1, bitwise-and of 3 and 1 ("11" and "01" in binary) gives 1, thus a(24) = 1.
For n = 210 = 2^1 * 3^1 * 5^1 * 7^1, bitwise-and of 1, 1, 1 and 1 gives 1, thus a(210) = 1.
For n = 720 = 2^4 * 3^2 * 5^1, bitwise-and of 4, 2 and 1 ("100", "10" and "1" in binary) gives zero, thus a(720) = 0.

Antti Karttunen, Table of n, a(n) for n = 1..10000
Index entries for sequences computed from exponents in factorization of n

Cf. A000961, A004198, A028234, A067029, A267114, A267116, A268387.
Cf. A002035 (indices of odd numbers), A072587 (indices of even numbers that occur after a(1)).
Cf. A267117 (indices of zeros).

{0}~Join~Table[BitAnd @@ Map[Last, FactorInteger@ n], {n, 2, 120}] (* Michael De Vlieger, Feb 07 2016 *)

a(n)=my(f = factor(n)[,2]); if (#f == 0, return (0)); my(b = f[1]); for (k=2, #f, b = bitand(b, f[k]);); b; \\ Michel Marcus, Feb 07 2016

a(n)=if(n>1, fold(bitand, factor(n)[,2]), 0) \\ Charles R Greathouse IV, Aug 04 2016

from functools import reduce
from operator import and_
from sympy import factorint
def A267115(n): return reduce(and_,factorint(n).values()) if n > 1 else 0 # Chai Wah Wu, Aug 31 2022

If A028234(n) = 1 [when n is a power of prime, in A000961], a(n) = A067029(n), otherwise a(n) = A067029(n) AND a(A028234(n)). [Here AND stands for bitwise-and, A004198.]

cp's OEIS Frontend

A059895 Table a(i,j) = product prime[k]^(Ei[k] AND Ej[k]) where Ei and Ej are the vectors of exponents in the prime factorizations of i and j; AND is the bitwise operation on binary representation of the exponents.

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A283475 a(n) = A019565(A005187(n)).

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A269174 Formula for Wolfram's Rule 124 cellular automaton: a(n) = (n OR 2n) AND ((n XOR 2n) OR (n XOR 4n)).

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A292274 a(n) = A292383(A163511(n)).

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A317934 Multiplicative with a(p^n) = 2^A011371(n); denominators for certain "Dirichlet Square Roots" sequences.

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A080099 Triangle T(n,k) = n AND k, 0<=k<=n, bitwise logical AND, read by rows.

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A218542 Number of times when an even number is encountered, when going from 2^(n+1)-1 to (2^n)-1 using the iterative process described in A071542.

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