A070939 -id:A070939 - cp's OEIS frontend

A309577 Table read by rows: T(n, k) is n with the first k bits removed from its binary expansion and then converted back to decimal, for 0 <= k <= A070939(n).

1, 0, 2, 0, 0, 3, 1, 0, 4, 0, 0, 0, 5, 1, 1, 0, 6, 2, 0, 0, 7, 3, 1, 0, 8, 0, 0, 0, 0, 9, 1, 1, 1, 0, 10, 2, 2, 0, 0, 11, 3, 3, 1, 0, 12, 4, 0, 0, 0, 13, 5, 1, 1, 0, 14, 6, 2, 0, 0, 15, 7, 3, 1, 0, 16, 0, 0, 0, 0, 0, 17, 1, 1, 1, 1, 0, 18, 2, 2, 2, 0, 0, 19

Offset: 1

Peter Kagey, Aug 08 2019

			For n = 26 and k = 2, T(26, 2) = 2 because 26 = 11010_2, and removing the first two bits leaves 010_2 = 2.
Table begins:
  n\k| 0 1 2 3 4
  ---+-----------
   1 | 1 0
   2 | 2 0 0
   3 | 3 1 0
   4 | 4 0 0 0
   5 | 5 1 1 0
   6 | 6 2 0 0
   7 | 7 3 1 0
   8 | 8 0 0 0 0
   9 | 9 1 1 1 0

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

Cf. A070939, A309576.

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

def t(n,k); n & (1 << n.bit_length - k) - 1 end

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

A317441 a(n) = number of k with 1 <= k <= n-1 such that a(k) and a(n-k) have the same binary length (A070939).

Original entry on oeis.org

0, 1, 2, 1, 2, 3, 0, 5, 4, 1, 6, 3, 0, 5, 4, 1, 10, 5, 2, 7, 0, 7, 10, 5, 6, 7, 2, 5, 8, 5, 8, 7, 10, 9, 6, 7, 8, 7, 10, 11, 6, 7, 8, 11, 6, 17, 8, 9, 8, 19, 8, 13, 10, 15, 6, 19, 10, 9, 6, 19, 8, 13, 14, 13, 10, 17, 14, 15, 10, 19, 14, 9, 16, 15, 10, 11, 20

Offset: 1

Rémy Sigrist, Jul 28 2018

See A317420 for similar sequences.

			For n = 4:
- A070939(a(1)) = 1 <> 2 = A070939(a(3)),
- A070939(a(2)) = 1 =  1 = A070939(a(2)),
- A070939(a(3)) = 2 <> 1 = A070939(a(1)),
- hence a(4) = 1.

Rémy Sigrist, Table of n, a(n) for n = 1..10000
Rémy Sigrist, Scatterplot of the first 500000 terms

Cf. A070939, A317420.

l = vector(77); for (n=1, #l, l[n] = #binary(max(1, v=sum(k=1, n-1, l[k]==l[n-k]))); print1 (v ", "))

A320674 Positive integers m with binary expansion (b_1, ..., b_k) (where k = A070939(m)) such that b_i = [m == 0 (mod prime(i))] for i = 1..k (where prime(i) denotes the i-th prime number and [] is an Iverson bracket).

Original entry on oeis.org

2, 4, 6, 8, 10, 12, 16, 20, 24, 32, 40, 48, 64, 80, 96, 128, 160, 192, 256, 320, 384, 512, 640, 768, 1024, 1280, 1536, 2048, 2560, 3072, 4096, 5120, 6144, 8192, 10240, 12288, 16384, 20480, 24576, 32768, 40960, 49152, 65536, 81920, 98304, 131072, 163840, 196608

Offset: 1

Rémy Sigrist, Oct 19 2018

In other words, the 1's in the binary representation of a term of this sequence encode the first prime divisors of this term.
All terms are even.
All even terms in A029747 belong to this sequence.
The term a(71) = 33554434 is the first one that does not belong to A029747.
See A320673 for similar sequences.

			The initial terms, alongside their binary representation and the prime divisors encoded therein, are:
  n   a(n)      bin(a(n))                   First prime divisors
  --  --------  --------------------------  --------------------
   1         2  10                          2
   2         4  100                         2
   3         6  110                         2, 3
   4         8  1000                        2
   5        10  1010                        2, 5
   6        12  1100                        2, 3
   7        16  10000                       2
   8        20  10100                       2, 5
   9        24  11000                       2, 3
  ...
  71  33554434  10000000000000000000000010  2, 97
  ...
33554434 is in the sequence because its binary expansion 10000000000000000000000010 of length 26 has a 1 in the 1st place and in the 25th place from the left and 0 elsewhere. As it is divisible by the 1st and 25th prime and by no other prime with index <= 26, 33554434 in the sequence. - _David A. Corneth_, Oct 20 2018

Cf. A029747, A070939, A320673.

selQ[n_] := With[{bb = IntegerDigits[n, 2]}, (Prime /@ Flatten[Position[bb, 1]]) == FactorInteger[n][[All, 1]]];
Select[Range[2, 200000], selQ] (* Jean-François Alcover, Nov 01 2018 *)

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

A338433 Values of n for which A070939(n^3) differs from A004221(n).

Original entry on oeis.org

1, 20, 40, 80, 101, 126, 127, 159, 160, 161, 200, 201, 202, 203, 252, 253, 254, 255, 317, 318, 319, 320, 321, 322, 399, 400, 401, 402, 403, 404, 405, 406, 502, 503, 504, 505, 506, 507, 508, 509, 510, 511, 631, 632, 633, 634, 635, 636, 637, 638, 639, 640, 641, 642, 643, 644, 645

Offset: 1

Jeremy Gardiner, Oct 27 2020

nonn

Sequence gives the values of n for which the length of the binary representation of n^3 differs from ceiling(10*log_10(n)) rounded up.

The largest number not in the sequence is 158489319246111348520210137339 = floor(10^29.2). - Robert Israel, Oct 27 2020

Jeremy Gardiner, Table of n, a(n) for n = 1..1083
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A004221, A070939, A111393, A329193.

filter:= n -> evalb(ilog2(n^3)+1 <> ceil(10*log[10](n))):
select(filter, [$1..1000]); # Robert Israel, Oct 27 2020

Select[Range[1000], IntegerLength[#^3, 2] != Ceiling[10*Log10[#]] &] (* Amiram Eldar, Oct 27 2020 *)

A344851 a(n) = (n^2) mod (2^A070939(n)).

Original entry on oeis.org

0, 1, 0, 1, 0, 1, 4, 1, 0, 1, 4, 9, 0, 9, 4, 1, 0, 1, 4, 9, 16, 25, 4, 17, 0, 17, 4, 25, 16, 9, 4, 1, 0, 1, 4, 9, 16, 25, 36, 49, 0, 17, 36, 57, 16, 41, 4, 33, 0, 33, 4, 41, 16, 57, 36, 17, 0, 49, 36, 25, 16, 9, 4, 1, 0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100

Offset: 0

Rémy Sigrist, May 30 2021

Informally, if n has w binary digits, a(n) is obtained by keeping the w final binary digits of n^2.
For n > 0, a(n) is the final digit of n^2 in base A062383(n).
This sequence has interesting graphical features (see illustration in Links section).

			For n = 42:
- A070939(42) = 6,
- a(42) = (42^2) mod (2^6) = 1764 mod 64 = 36.

Rémy Sigrist, Table of n, a(n) for n = 0..8192
John S. McCaskill and Peter R.Wills, On permutations derived from integer powers x^n, arXiv:1907.01890 [math.NT], 2019.
Rémy Sigrist, Scatterplot of the sequence for n = 2^18..2^19

Cf. A000290, A048152, A062383, A070939, A086341, A116882, A316347 (decimal analog).

{0}~Join~Table[Mod[n^2,2^(1+Floor@Log2@n)],{n,100}] (* Giorgos Kalogeropoulos, Jun 02 2021 *)

PARI
```
a(n) = (n^2) % 2^#binary(n)
```

def a(n): return (n**2) % (2**n.bit_length())
print([a(n) for n in range(75)]) # Michael S. Branicky, May 30 2021

a(n) = 0 iff n = 0 or n > 1 and n belongs to A116882.
a(n) = 1 iff n belongs to A086341.
a(2^k + m) = a(2^(k+1)-m) for any k > 0 and m = 0..2^k.

A352961 a(0) = 0, a(1) = 1, and for any n > 1, a(n) = a(n-2^e) + a(n-2^(e+1)) with e as large as possible (e = A070939(n) - 2).

Original entry on oeis.org

0, 1, 1, 2, 1, 3, 2, 5, 1, 4, 3, 7, 2, 7, 5, 12, 1, 5, 4, 9, 3, 10, 7, 17, 2, 9, 7, 16, 5, 17, 12, 29, 1, 6, 5, 11, 4, 13, 9, 22, 3, 13, 10, 23, 7, 24, 17, 41, 2, 11, 9, 20, 7, 23, 16, 39, 5, 22, 17, 39, 12, 41, 29, 70, 1, 7, 6, 13, 5, 16, 11, 27, 4, 17, 13

Offset: 0

Rémy Sigrist, Apr 11 2022

nonn

This sequence is a variant of the Fibonacci sequence (A000045) with variable steps.

			a(0) = 0 by definition.
a(1) = 1 by definition.
a(2) = a(2-2^0) + a(2-2^1) = a(1) + a(0) = 1 + 0 = 1.
a(3) = a(3-2^0) + a(3-2^1) = a(2) + a(1) = 1 + 1 = 2.
a(4) = a(4-2^1) + a(4-2^2) = a(2) + a(0) = 1 + 0 = 1.
a(5) = a(5-2^1) + a(5-2^2) = a(3) + a(1) = 2 + 1 = 3.
a(6) = a(6-2^1) + a(6-2^2) = a(4) + a(2) = 1 + 1 = 2.
a(7) = a(7-2^1) + a(7-2^2) = a(5) + a(3) = 3 + 2 = 5.
a(8) = a(8-2^2) + a(8-2^3) = a(4) + a(0) = 1 + 0 = 1.

Rémy Sigrist, Table of n, a(n) for n = 0..8191
Rémy Sigrist, Logarithmic scatterplot of the first 2^17 terms

See A352964 for a similar sequence.

Cf. A000045, A070939.

{ for (n=1, #a=vector(75), print1 (a[n]=if (n==1, 0, n==2, 1, e=#binary(n-1)-2; a[n-2^e]+a[n-2^(e+1)]),", ")) }

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

A359450 a(1) = 1, a(2) = 2; thereafter a(n) = n * a(A070939(n)).

Original entry on oeis.org

1, 2, 6, 24, 30, 36, 42, 192, 216, 240, 264, 288, 312, 336, 360, 480, 510, 540, 570, 600, 630, 660, 690, 720, 750, 780, 810, 840, 870, 900, 930, 1152, 1188, 1224, 1260, 1296, 1332, 1368, 1404, 1440, 1476, 1512, 1548, 1584, 1620, 1656, 1692, 1728, 1764, 1800, 1836

Offset: 1

Amiram Eldar, Jan 02 2023

Problem A6 of the 63rd Putnam Competition (2002) asked to prove that when this sequence is generalized to base-b digits, the sum of reciprocals converges only for b = 2.

Problem 2 in Appendix D of Bornemann et al. (2004) asked to calculate the sum of the reciprocals of this sequence.

Daniel D. Bonar and Michael J. Khoury, Jr., Real infinite Series, The Mathematical Association of America, 2006, pp. 159, 190-191.
Hongwei Chen, Classical Analysis: An Approach through Problems, CRC Press, 2022, p. 118, exercise 34.
Kiran S. Kedlaya, Daniel M. Kane, Jonathan M. Kane, and Evan M. O'Dorney, The William Lowell Putnam Mathematical Competition 2001-2016: Problems, Solutions, and Commentary, American Mathematical Society, 2020, pp. 86-87.

Amiram Eldar, Table of n, a(n) for n = 1..10000
Folkmar Bornemann, Dirk Laurie, Stan Wagon, and Jörg Waldvogel, The SIAM 100-Digit Challenge, A Study in High-Accuracy Numerical Computing, SIAM, Philadelphia, 2004. See Appendix D, Problem 2, p. 281.
Kiran Kedlaya and Lenny Ng, Solutions to the 63rd William Lowell Putnam Mathematical Competition, Saturday, December 7, 2002. See Problem A-6, p. 2.
Leonard F. Klosinski, Gerald L. Alexanderson, and Loren C. Larson, The Sixty-Third William Lowell Putnam Mathematical Competition, The American Mathematical Monthly, Vol. 110, No. 8 (2003), pp. 718-726.
John Scholes, Problem A6, The 63rd Putnam Competition, 2002.
David Smith, On a slowly converging sum, Solution to Problem 2, 2003.

Cf. A070939, A359451.

a[1] = 1; a[2] = 2; a[n_] := a[n] = n * a[BitLength[n]]; Array[a, 100]

PARI
```
a(n) = if(n < 3, n, n * a(#binary(n)));
```

Sum_{n>=1} 1/a(n) = A359451.

A368877 a(n) = f^k(n) where f(n) = A014682(n), the Collatz map, and k = A070939(n), the length of n in base 2.

Original entry on oeis.org

2, 2, 8, 2, 2, 8, 26, 2, 17, 2, 20, 8, 8, 26, 80, 2, 5, 17, 17, 2, 2, 20, 20, 8, 22, 8, 71, 26, 26, 80, 242, 2, 44, 5, 5, 17, 17, 17, 152, 2, 161, 2, 56, 20, 20, 20, 182, 8, 7, 22, 22, 8, 8, 71, 71, 26, 74, 26, 76, 80, 80, 242, 728, 2, 14, 44, 44, 5, 5, 5, 137, 17, 47, 17, 16

Offset: 1

Michel Marcus, Jan 08 2024

nonn

This is the jump function jp in the paper of Eliahou et al.

Paolo Xausa, Table of n, a(n) for n = 1..10000
Shalom Eliahou, Jean Fromentin, and Rénald Simonetto, Is the Syracuse falling time bounded by 12?, hal-03294829, 2021.

Cf. A014682, A070939, A368878.

A368877[n_] := Nest[If[OddQ[#], (3*#+1)/2, #/2]&, n, BitLength[n]];
Array[A368877, 100] (* Paolo Xausa, Jan 08 2024 *)

T(n) = if (n%2, (3*n+1)/2, n/2); \\ A014682
a(n) = my(N=1+logint(n, 2)); for (i=1, N, n = T(n)); n;

A380230 a(n) = n*A070939(n).

Original entry on oeis.org

0, 1, 4, 6, 12, 15, 18, 21, 32, 36, 40, 44, 48, 52, 56, 60, 80, 85, 90, 95, 100, 105, 110, 115, 120, 125, 130, 135, 140, 145, 150, 155, 192, 198, 204, 210, 216, 222, 228, 234, 240, 246, 252, 258, 264, 270, 276, 282, 288, 294, 300, 306, 312, 318, 324, 330, 336, 342

Offset: 0

Paolo Xausa, Jan 17 2025

Paolo Xausa, Table of n, a(n) for n = 0..10000

Cf. A070939, A110803, A168160.

Mathematica
```
Table[n*BitLength[n], {n, 0, 100}]
```

A226836 Squares s such that first m and last m digits of the binary representation are perfect positive squares written in binary, and m = floor(binaryLength(s)/2), where binaryLength(s) = A070939(s) is the binary length of s.

Original entry on oeis.org

36, 289, 4624, 10404, 115600, 248004, 1083681, 1281424, 2232036, 2509056, 21307456, 23892544, 31494544, 40144896, 66357316, 271359729, 340919296, 479785216, 512026384, 597215844, 767068416, 4831918144, 5454708736, 8126661904, 8522982400, 12273094656, 16705045504

Offset: 1

Alex Ratushnyak, Jun 19 2013

The sequence of roots of a(n) begins: 6, 17, 68, 102, 340, 498, 1041, 1132, 1494, 1584, 4616, 4888, 5612, 6336, 8146, 16473, 18464, 21904, 22628, 24438, 27696, 69512, 73856, 90148, 92320, ...

Cf. A070939, A226736.

#include 
#include 
typedef unsigned long long U64;
U64 isSquare(U64 a) {
    U64 s = sqrt(a);
    return (s*s==a);
}
int main() {
  U64 i, j, n, sq, s, S;
  for (n = 1; n < (1ULL<<20); ++n) {
    for (i = 64, j = sq = n*n; j < (1ULL<<63); j += j)
      --i;  // binary length of sq
    j = i >> 1;  //  Sbs or Ss, binary length of s is j
    s = sq & ((1ULL<> (j+(i&1));
    if (isSquare(S) && s && isSquare(s)) printf("%llu, ", sq);
  }
  return 0;
}

cp's OEIS Frontend

A309577 Table read by rows: T(n, k) is n with the first k bits removed from its binary expansion and then converted back to decimal, for 0 <= k <= A070939(n).

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Formula

A317441 a(n) = number of k with 1 <= k <= n-1 such that a(k) and a(n-k) have the same binary length (A070939).

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PARI

A320674 Positive integers m with binary expansion (b_1, ..., b_k) (where k = A070939(m)) such that b_i = [m == 0 (mod prime(i))] for i = 1..k (where prime(i) denotes the i-th prime number and [] is an Iverson bracket).

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Author

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Comments

Examples

Crossrefs

Programs

Mathematica

PARI

A338433 Values of n for which A070939(n^3) differs from A004221(n).

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Maple

Mathematica

A344851 a(n) = (n^2) mod (2^A070939(n)).

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Mathematica

PARI

Python

Formula

A352961 a(0) = 0, a(1) = 1, and for any n > 1, a(n) = a(n-2^e) + a(n-2^(e+1)) with e as large as possible (e = A070939(n) - 2).

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Keywords

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Examples

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Programs

PARI

Formula

A359450 a(1) = 1, a(2) = 2; thereafter a(n) = n * a(A070939(n)).

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Programs

Mathematica

PARI

Formula

A368877 a(n) = f^k(n) where f(n) = A014682(n), the Collatz map, and k = A070939(n), the length of n in base 2.

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