A070939 -id:A070939 - cp's OEIS frontend

A265917 a(n) = floor(A070939(n)/A000120(n)) where A070939(n) is the binary length of n and A000120(n) is the binary weight of n.

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

Offset: 1

Alex Ratushnyak, Dec 18 2015

1/a(n) gives a very rough approximation of the density of 1-bits in the binary representation (A007088) of n. This is 1 if more than half of the bits of n are 1. - Antti Karttunen, Dec 19 2015

Michael De Vlieger, Table of n, a(n) for n = 1..10000

Cf. A000120, A007088, A070939, A135941, A199238, A265918.

Table[Floor[IntegerLength[n, 2]/Total@ IntegerDigits[n, 2]], {n, 120}] (* Michael De Vlieger, Dec 21 2015 *)

a(n) = #binary(n)\hammingweight(n); \\ Michel Marcus, Dec 19 2015

for n in range(1, 88):
    print(str((len(bin(n))-2) // bin(n).count('1')), end=',')

A320673 Positive integers m with binary expansion (b_1, ..., b_k) (where k = A070939(m)) such that b_i = [m == 0 (mod i)] for i = 1..k (where [] is an Iverson bracket).

Original entry on oeis.org

1, 50, 52, 104, 114, 3460, 12298, 29442, 31368, 856592, 1713184, 54822416, 109578256, 109644832, 219156512, 219289664, 438313024, 438579328, 876626048, 877158656, 1034367516, 1753252096, 1754317312, 112208117792, 113290248736, 224416235584, 226580497472

Offset: 1

Rémy Sigrist, Oct 19 2018

In other words, the binary representation of a term of this sequence encodes the first divisors and nondivisors of this term respectively as ones and zeros.
Is this sequence infinite?
See A320674 and A320675 for similar sequences.

			The first terms, alongside their binary representation and the divisors encoded therein, are:
  n  a(n)   bin(a(n))        First divisors
  -  -----  ---------------  --------------------
  1      1  1                1
  2     50  110010           1, 2, 5
  3     52  110100           1, 2, 4
  4    104  1101000          1, 2, 4
  5    114  1110010          1, 2, 3, 6
  6   3460  110110000100     1, 2, 4, 5, 10
  7  12298  11000000001010   1, 2, 11, 13
  8  29442  111001100000010  1, 2, 3, 6, 7, 14
  9  31368  111101010001000  1, 2, 3, 4, 6, 8, 12

Cf. A070939, A320674, A320675.

is(n) = my (b=binary(n)); b==vector(#b, k, n%k==0)

from itertools import count, islice
def A320673_gen(startvalue=0): # generator of terms >= startvalue
    return filter(lambda n:not any(int(b)==bool(n%i) for i,b in enumerate(bin(n)[2:],1)),count(max(startvalue,0)))
A320673_list = list(islice(A320673_gen(),10)) # Chai Wah Wu, Dec 12 2022

A379537 Frugal numbers in base 2: numbers k such that A377369(k) < A070939(k).

Original entry on oeis.org

1, 27, 32, 49, 64, 81, 121, 125, 128, 135, 147, 162, 169, 189, 192, 243, 250, 256, 289, 297, 320, 338, 343, 351, 361, 363, 375, 384, 405, 448, 486, 507, 512, 513, 529, 539, 567, 576, 578, 605, 621, 625, 637, 640, 648, 675, 686, 704, 722, 729, 750, 768, 783, 832

Offset: 1

Paolo Xausa, Dec 25 2024

A frugal number in base 2 is a number with more bits than the total number of bits of its prime factorization (including exponents > 1).
Following the definition by Pinch (1998), 1 is considered a frugal number.
Some authors call these numbers "economical numbers", as in A046759 which, according to the definition provided here, lists frugal numbers in base 10 (additionally, A046759 does not include 1).

			32 is a term because 32 = 2^5 = 10_2^101_2; the total number of bits of (10_2, 101_2) = 5 < the number of bits of 32 = 100000_2 (6).
135 is a term because 135 = 3^3*5 = 11_2^11_2*101_2; the total number of bits of (11_2, 11_2, 101_2) = 7 < the number of bits of 135 = 10000111_2 (8).

Paolo Xausa, Table of n, a(n) for n = 1..10000
Richard G. E. Pinch, Economical numbers, arXiv:math/9802046 [math.NT], 1998.
Wikipedia, Frugal number.

Row n = 2 of A379538.

Cf. A046759, A070939, A377369, A379373.

A379537Q[k_] := Total[BitLength[Select[Flatten[FactorInteger[k]], # > 1 &]]] < BitLength[k];
Select[Range[1000], A379537Q]

A232085 Primes that can be written in binary representation as concatenation of two primes. That is, primes representable as p * 2^L + q, where p and q are primes, and L is the length of binary representation of q: L = A070939(q).

Original entry on oeis.org

11, 23, 29, 31, 43, 47, 59, 61, 71, 79, 83, 109, 113, 127, 151, 157, 167, 179, 181, 191, 229, 233, 239, 241, 251, 271, 283, 317, 349, 353, 359, 367, 373, 379, 383, 431, 433, 439, 457, 463, 467, 479, 487, 491, 509, 541, 563, 599, 607, 631, 643, 661, 691, 701, 709, 719

Offset: 1

Alex Ratushnyak, Nov 17 2013

A subsequence of A090423.

Cf. A000040, A070939, A090423.

A232238 Primes p such that p+2 and q are primes, where q is concatenation of binary representations of p and p+2: q = p * 2^L + p+2, where L is the length of binary representation of p+2: L=A070939(p+2).

Original entry on oeis.org

3, 5, 17, 71, 269, 1049, 1151, 1721, 5099, 5279, 5657, 6299, 6569, 6779, 7307, 7589, 16451, 16649, 16691, 19079, 19139, 19211, 19841, 19961, 20771, 20981, 21011, 21059, 21599, 22619, 22961, 23201, 23369, 23741, 23909, 24419, 26729, 26951, 27689, 28109, 28409, 28751, 29129

Offset: 1

Alex Ratushnyak, Nov 20 2013

			3 is 11 in binary, 5 is 101. Because 11101 = 29d is a prime, 3 is in the sequence.
5 is 101 in binary, 7 is 111, and because 101111 = 47d is a prime, 5 is in the sequence.

Cf. A000040, A070939, A232236.

A232236(n) = a(n) * 2^A070939(a(n)+2) + a(n)+2.

A265370 a(0) = 0; for n >= 1, a(n) = A070939(A078510(n)).

Original entry on oeis.org

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

Offset: 0

Antti Karttunen, Dec 16 2015

nonn

Number of significant bits in the binary representation of n-th Spironacci number, A078510(n).

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

Cf. A070939, A078510.

Cf. also A020909.

a(0) = 0; for n >= 1, a(n) = A070939(A078510(n)).

A265918 a(n) = A070939(n) mod A000120(n), where A070939(n) is the binary length of n and A000120(n) is the binary weight of n.

Original entry on oeis.org

0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 2, 1, 2, 2, 1, 1, 2, 2, 1, 2, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 2, 0, 2, 2, 1, 0, 0, 0, 2, 0, 2, 2, 1, 0, 2, 2, 1, 2, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 3, 1, 3, 3, 2, 1, 1, 1, 3, 1, 3, 3

Offset: 1

Alex Ratushnyak, Dec 18 2015

Michael De Vlieger, Table of n, a(n) for n = 1..10000

Cf. A000120, A070939, A135941, A199238, A265917.

Table[Mod[IntegerLength[n, 2], Total@ IntegerDigits[n, 2]], {n, 120}] (* Michael De Vlieger, Dec 21 2015 *)

a(n) = #binary(n) % hammingweight(n); \\ Michel Marcus, Dec 19 2015

for n in range(1, 88): print((len(bin(n))-2) % bin(n).count('1'), end=', ')

A283982 a(0) = 0, and for n > 0, a(n) = A070939(n) - A280700(n).

Original entry on oeis.org

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

Offset: 0

Antti Karttunen, Mar 19 2017

nonn

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

Cf. A005187, A070939, A124757, A280700, A283475, A283477, A283981.

(define (A283982 n) (if (zero? n) n (- (A070939 n) (A280700 n))))

a(0) = 0, for n > 0, a(n) = A070939(n) - A280700(n).

a(n) = A283981(n) - A124757(n).

A293198 a(n) is the least positive k such that f(k) = f(k + n) where f(k) = A000120(k) / A070939(k).

Original entry on oeis.org

1, 5, 1, 9, 3, 21, 1, 2, 2, 19, 2, 38, 3, 37, 1, 33, 15, 35, 38, 37, 84, 35, 76, 12, 7, 10, 9, 10, 3, 4, 1, 10, 4, 2, 5, 2, 2, 6, 5, 2, 2, 5, 2, 9, 4, 6, 5, 2, 2, 5, 2, 6, 5, 5, 2, 5, 7, 137, 138, 134, 3, 133, 1, 129, 63, 131, 134, 133, 140, 131, 138, 137, 152, 139, 134, 133, 148, 131, 146, 56, 336, 135, 150, 52

Offset: 0

Altug Alkan, Oct 05 2017

Numbers m such that a(2^m*(2^(m + 1) - 1) + 1) = 2^m are 0, 1, 2, 3, 4, 5, 6, 8, 9, 11, 14, 15, 18, 20, 21, ...
Numbers t such that a(t) = 2 are 7, 8, 10, 33, 35, 36, 39, 40, 42, 47, 48, 50, ...
Numbers t such that a(t) > t are 0, 1, 3, 5, 9, 11, 13, 15, 17, 18, 19, 20, 21, ...

			a(5) = 21 because 21 = 2^4 + 2^2 + 2^0, 21 + 5 = 2^4 + 2^3 + 2^1; A000120(21) / A070939(21) = A000120(21 + 5) / A070939(21 + 5) and 21 is the least number with this property.

Rémy Sigrist, Table of n, a(n) for n = 0..8192
Altug Alkan, Line graph of a(n+1)-a(n) for n <= 2^13
Altug Alkan, A logarithmic scatterplot of a(n)

Cf. A000120, A070939, A099245, A099246, A292895.

a(n) = {my(k=1); while (hammingweight(k+n)/#binary(k+n) != hammingweight(k) /#binary(k), k++); k;}

a(n) <> n for all n >= 0.
a(n) <= 5*n for all n >= 1.
a(2^m - 1) = 1 for all m >= 1.
a(2^m - 2^2) = 2^2 - 1 for all m >= 3.
a(2^m - 2^3) = 2^3 - 1 for all m >= 5.
a(2^m - 2^4) = 2^4 - 1 for all m >= 7.
a(2^m - 2^5) = 2^5 - 1 for all m >= 10.
a(2^m - 2^6) = 2^6 - 1 for all m >= 13.
a(2^m - 2^7) = 2^7 - 1 for all m >= 17.
a(2^m - 2^8) = 2^8 - 1 for all m >= 21.
a(2^m - 2^9) = 2^9 - 1 for all m >= 26.
a(2^(p - 1)) = 2^(p - 1) - 1 and a(2^(p - 1) - 1) = 2^p + 1 for all primes p.
a(2^(p - 1) + 1) = 2^p + 3 for all primes p >= 5.

A309576 Table read by rows: T(n, k) is the last k bits of n interpreted as a base-2 representation of a number and converted to decimal, 0 <= k <= A070939(n).

Original entry on oeis.org

0, 1, 0, 0, 2, 0, 1, 3, 0, 0, 0, 4, 0, 1, 1, 5, 0, 0, 2, 6, 0, 1, 3, 7, 0, 0, 0, 0, 8, 0, 1, 1, 1, 9, 0, 0, 2, 2, 10, 0, 1, 3, 3, 11, 0, 0, 0, 4, 12, 0, 1, 1, 5, 13, 0, 0, 2, 6, 14, 0, 1, 3, 7, 15, 0, 0, 0, 0, 0, 16, 0, 1, 1, 1, 1, 17, 0, 0, 2, 2, 2, 18, 0, 1

Offset: 1

Peter Kagey, Aug 08 2019

			For n = 26 and k = 3, T(26, 3) = 2 because 26 = 11010_2, and looking at only the last three bits gives 010_2 = 2.
Table begins:
  n\k| 0 1 2 3 4
  ---+-----------
   1 | 0 1
   2 | 0 0 2
   3 | 0 1 3
   4 | 0 0 0 4
   5 | 0 1 1 5
   6 | 0 0 2 6
   7 | 0 1 3 7
   8 | 0 0 0 0 8
   9 | 0 1 1 1 9

Peter Kagey, Table of n, a(n) for n = 1..9987 (first 1000 rows)

Cf. A070939, A309577.

T[n_, k_] := BitAnd[n, 2^k-1]; Table[T[n, k], {n, 1, 20}, {k, 0, BitLength[n]}] // Flatten (* Amiram Eldar, Aug 09 2019 *)

Ruby
```
def t(n,k); n & (1 << k) - 1 end
```

T(n,0) = 0 and T(n, A070939(n)) = n.

cp's OEIS Frontend

A265917 a(n) = floor(A070939(n)/A000120(n)) where A070939(n) is the binary length of n and A000120(n) is the binary weight of n.

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A320673 Positive integers m with binary expansion (b_1, ..., b_k) (where k = A070939(m)) such that b_i = [m == 0 (mod i)] for i = 1..k (where [] is an Iverson bracket).

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A379537 Frugal numbers in base 2: numbers k such that A377369(k) < A070939(k).

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A232085 Primes that can be written in binary representation as concatenation of two primes. That is, primes representable as p * 2^L + q, where p and q are primes, and L is the length of binary representation of q: L = A070939(q).

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A232238 Primes p such that p+2 and q are primes, where q is concatenation of binary representations of p and p+2: q = p * 2^L + p+2, where L is the length of binary representation of p+2: L=A070939(p+2).

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A265370 a(0) = 0; for n >= 1, a(n) = A070939(A078510(n)).

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A265918 a(n) = A070939(n) mod A000120(n), where A070939(n) is the binary length of n and A000120(n) is the binary weight of n.

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A283982 a(0) = 0, and for n > 0, a(n) = A070939(n) - A280700(n).

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A293198 a(n) is the least positive k such that f(k) = f(k + n) where f(k) = A000120(k) / A070939(k).

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