A285099 -id:A285099 - cp's OEIS frontend

A295989 Irregular triangle T(n, k), read by rows, n >= 0 and 0 <= k < A001316(n): T(n, k) is the (k+1)-th nonnegative number m such that n AND m = m (where AND denotes the bitwise AND operator).

0, 0, 1, 0, 2, 0, 1, 2, 3, 0, 4, 0, 1, 4, 5, 0, 2, 4, 6, 0, 1, 2, 3, 4, 5, 6, 7, 0, 8, 0, 1, 8, 9, 0, 2, 8, 10, 0, 1, 2, 3, 8, 9, 10, 11, 0, 4, 8, 12, 0, 1, 4, 5, 8, 9, 12, 13, 0, 2, 4, 6, 8, 10, 12, 14, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 0

Offset: 0

Rémy Sigrist, Dec 02 2017

The (n+1)-th row has A001316(n) terms and sums to n * A001316(n) / 2.
For any n >= 0 and k such that 0 <= k < A001316(n):
- if A000120(n) > 0 then T(n, 1) = A006519(n),
- if A000120(n) > 1 then T(n, 2) = 2^A285099(n),
- if A000120(n) > 0 then T(n, A001316(n)/2 - 1) = A053645(n),
- if A000120(n) > 0 then T(n, A001316(n)/2) = 2^A000523(n),
- if A000120(n) > 0 then T(n, A001316(n) - 2) = A129760(n),
- T(n, A001316(n) - 1) = n,
- the six previous relations correspond respectively (when applicable) to the second term, the third term, the pair of central terms, the penultimate term and the last term of a row,
- T(n, k) AND T(n, A001316(n) - k - 1) = 0,
- T(n, k) + T(n, A001316(n) - k - 1) = n,
- T(n, k) = k for any k < A006519(n+1),
- A000120(T(n, k)) = A000120(k).
If we plot (n, T(n,k)) then we obtain a skewed Sierpinski triangle (see Links section).
If interpreted as a flat sequence a(n) for n >= 0:
- a(n) = 0 iff n = A006046(k) for some k >= 0,
- a(n) = 1 iff n = A006046(2*k + 1) + 1 for some k >= 0,
- a(A006046(k) - 1) = k - 1 for any k > 0.

			Triangle begins:
  0:   [0]
  1:   [0, 1]
  2:   [0, 2]
  3:   [0, 1, 2, 3]
  4:   [0, 4]
  5:   [0, 1, 4, 5]
  6:   [0, 2, 4, 6]
  7:   [0, 1, 2, 3, 4, 5, 6, 7]
  8:   [0, 8]
  9:   [0, 1, 8, 9]
  10:  [0, 2, 8, 10]
  11:  [0, 1, 2, 3, 8, 9, 10, 11]
  12:  [0, 4, 8, 12]
  13:  [0, 1, 4, 5, 8, 9, 12, 13]
  14:  [0, 2, 4, 6, 8, 10, 12, 14]
  15:  [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15]

Rémy Sigrist, Rows n = 0..256, flattened
Rémy Sigrist, Scatterplot of (n, T(n, k)) for n = 0..1023 and k = 0..A001316(n)-1

Cf. A000120, A000523, A001316 (row lengths), A006046, A006519, A053645, A129760, A285099.

First column of array in A352909.

A295989row[n_] := Select[Range[0, n], BitAnd[#, n-#] == 0 &];
Array[A295989row, 25, 0] (* Paolo Xausa, Feb 24 2024 *)

T(n,k) = if (k==0, 0, n%2==0, 2*T(n\2,k), k%2==0, 2*T(n\2, k\2), 2*T(n\2, k\2)+1)

For any n >= 0 and k such that 0 <= k < A001316(n):
- T(n, 0) = 0,
- T(2*n, k) = 2*T(n, k),
- T(2*n+1, 2*k) = 2*T(n, k),
- T(2*n+1, 2*k+1) = 2*T(n, k) + 1.

A285097 a(n) = difference between the positions of two least significant 1-bits in base-2 representation of n, or 0 if there are less than two 1-bits in n (when n is either zero or a power of 2).

Original entry on oeis.org

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

Offset: 0

Antti Karttunen, Apr 20 2017

a(1+n) is the length of the least significant run of 0-bits in n, or 0 if n is one of terms of A000225. - Antti Karttunen, Oct 14 2023

			For n = 3, "11" in binary, the second least significant 1-bit (the second 1-bit from the right) is at position 1 and the rightmost 1-bit is at position 0, thus a(3) = 1-0 = 1.
For n = 4, "100" in binary, there is just one 1-bit present, thus a(4) = 0.
For n = 5, "101" in binary, the second 1-bit from the right is at position 2, and the least significant 1 is at position 0, thus a(5) = 2-0 = 2.
For n = 26, "11010" in binary, the second 1-bit from the right is at position 3, and the least significant 1 is at position 1, thus a(26) = 3-1 = 2.

Antti Karttunen, Table of n, a(n) for n = 0..8192
Index entries for sequences related to binary expansion of n

Cf. A000120, A000225, A000265, A007814, A119387, A129760, A285099, A366370.

a007814[n_]:=IntegerExponent[n, 2]; a285099[n_]:=If[DigitCount[n, 2, 1]<2, 0, a007814[BitAnd[n, n - 1]]]; a[n_]:=If[DigitCount[n, 2, 1]<2, 0,a285099[n] - a007814[n]]; Table[a[n], {n, 0, 150}] (* Indranil Ghosh, Apr 20 2017 *)

A285097(n) = if(!n || !bitand(n,n-1), 0, valuation((n>>valuation(n,2))-1, 2)); \\ Antti Karttunen, Oct 14 2023

import math
def a007814(n): return int(math.log(n - (n & n - 1), 2))
def a285099(n): return 0 if bin(n)[2:].count("1") < 2 else a007814(n & (n - 1))
def a(n): return 0 if bin(n)[2:].count("1")<2 else a285099(n) - a007814(n) # Indranil Ghosh, Apr 20 2017

(define (A285097 n) (if (<= (A000120 n) 1) 0 (- (A285099 n) (A007814 n))))

If A000120(n) < 2, a(n) = 0, otherwise a(n) = A285099(n) - A007814(n) = A007814(A129760(n)) - A007814(n).

a(n) = 0 if n is 0 or of the form 2^k, (k>=0), otherwise a(n) = v_2(A000265(n)-1), where v_2(i) = A007814(i). - Ridouane Oudra, Oct 20 2019

A285110 a(n) = A001222(A285323(n)).

Original entry on oeis.org

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

Offset: 0

Antti Karttunen, Apr 19 2017

nonn

The sequence is completely determined by the positions of two least significant 1-bits of n: After initial zero, if n is a power of two (only one 1-bit present) or if prime(1+A285099(n)) > prime(1+A007814(n))^2, a(n) = 2, otherwise a(n) = 1.

Antti Karttunen, Table of n, a(n) for n = 0..8192

Cf. A000040, A000290, A001222, A007814, A285099, A285323, A285324.

from operator import mul
from sympy import prime, primefactors
from functools import reduce
def a001222(n): return 0 if n<2 else a001222(n//min(primefactors(n))) + 1
def a019565(n): return reduce(mul, (prime(i+1) for i, v in enumerate(bin(n)[:1:-1]) if v == '1')) if n > 0 else 1 # This function from Chai Wah Wu
def a007947(n): return 1 if n<2 else reduce(mul, primefactors(n))
def a065642(n):
    if n==1: return 1
    r=a007947(n)
    n += r
    while a007947(n)!=r:
        n+=r
    return n
def a285323(n): return a065642(a065642(a019565(n)))//a019565(n)
def a(n): return a001222(a285323(n))
print([a(n) for n in range(121)]) # Indranil Ghosh, Apr 20 2017

(define (A285110 n) (A001222 (A285323 n)))
(define (A285110 n) (cond ((zero? n) n) ((or (= 1 (A000120 n)) (> (A000040 (+ 1 (A285099 n))) (A000290 (A000040 (+ 1 (A007814 n)))))) 2) (else 1)))

a(n) = A001222(A285323(n)).

A285323 a(n) = A065642(A065642(A019565(n))) / A019565(n).

Original entry on oeis.org

1, 4, 9, 3, 25, 4, 5, 3, 49, 4, 7, 3, 7, 4, 5, 3, 121, 4, 9, 3, 11, 4, 5, 3, 11, 4, 7, 3, 7, 4, 5, 3, 169, 4, 9, 3, 13, 4, 5, 3, 13, 4, 7, 3, 7, 4, 5, 3, 13, 4, 9, 3, 11, 4, 5, 3, 11, 4, 7, 3, 7, 4, 5, 3, 289, 4, 9, 3, 17, 4, 5, 3, 17, 4, 7, 3, 7, 4, 5, 3, 17, 4, 9, 3, 11, 4, 5, 3, 11, 4, 7, 3, 7, 4, 5, 3, 17, 4, 9, 3, 13, 4, 5, 3, 13, 4, 7, 3, 7, 4, 5, 3

Offset: 0

Antti Karttunen, Apr 19 2017

nonn

After the initial a(0)=1, the third row of array A285321 divided by its first row. After 1, all terms are either primes or squares of primes. See A285110.

Antti Karttunen, Table of n, a(n) for n = 0..10000

Cf. A000040, A007814, A014673, A020639, A019565, A065642, A285099, A285110, A285321, A285327.

A019565(n) = {my(j,v); factorback(Mat(vector(if(n, #n=vecextract(binary(n), "-1..1")), j, [prime(j), n[j]])~))}; \\ This function from M. F. Hasler
A007947(n) = factorback(factorint(n)[, 1]); \\ From Andrew Lelechenko, May 09 2014
A065642(n) = { my(r=A007947(n)); if(1==n,n,n = n+r; while(A007947(n) <> r, n = n+r); n); };
A285323(n) = A065642(A065642(A019565(n))) / A019565(n);

from operator import mul
from sympy import prime, primefactors
from functools import reduce
def a019565(n): return reduce(mul, (prime(i+1) for i, v in enumerate(bin(n)[:1:-1]) if v == '1')) if n > 0 else 1 # This function from Chai Wah Wu
def a007947(n): return 1 if n<2 else reduce(mul, primefactors(n))
def a065642(n):
    if n==1: return 1
    r=a007947(n)
    n += r
    while a007947(n)!=r:
        n+=r
    return n
def a(n): return a065642(a065642(a019565(n)))//a019565(n)
print([a(n) for n in range(101)]) # Indranil Ghosh, Apr 20 2017

(define (A285323 n) (/ (A065642 (A065642 (A019565 n))) (A019565 n)))
(define (A285323 n) (cond ((zero? n) 1) ((or (= 1 (A000120 n)) (> (A000040 (+ 1 (A285099 n))) (A000290 (A000040 (+ 1 (A007814 n)))))) (A000290 (A000040 (+ 1 (A007814 n))))) (else  (A000040 (+ 1 (A285099 n))))))

a(n) = A065642(A065642(A019565(n))) / A019565(n).

cp's OEIS Frontend

A295989 Irregular triangle T(n, k), read by rows, n >= 0 and 0 <= k < A001316(n): T(n, k) is the (k+1)-th nonnegative number m such that n AND m = m (where AND denotes the bitwise AND operator).

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A285097 a(n) = difference between the positions of two least significant 1-bits in base-2 representation of n, or 0 if there are less than two 1-bits in n (when n is either zero or a power of 2).

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Mathematica

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Formula

A285110 a(n) = A001222(A285323(n)).

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A285323 a(n) = A065642(A065642(A019565(n))) / A019565(n).

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