A299960 -id:A299960 - cp's OEIS frontend

A100706 Bisection of A002275.

1, 111, 11111, 1111111, 111111111, 11111111111, 1111111111111, 111111111111111, 11111111111111111, 1111111111111111111, 111111111111111111111, 11111111111111111111111

Offset: 0

N. J. A. Sloane, Nov 19 2004

Also the binary representation of the n-th iteration of the elementary cellular automaton starting with a single ON (black) cell for Rules 151, 159, 183, 191, 215, 222, 223, 247, 254 and 255. - Robert Price, Feb 21 2016

The aerated sequence 1, 0, 111, 0, 11111, 0, 1111111, ... is a linear divisibility sequence of order 4. It is the case P1 = 0, P2 = -9^2, Q = -10 of the 3-parameter family of 4th-order linear divisibility sequences found by Williams and Guy. Cf. A007583, A095372 and A299960. - Peter Bala, Aug 28 2019

S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 55.

Eric Weisstein's World of Mathematics, Elementary Cellular Automaton
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences Integers, Volume 12A (2012) The John Selfridge Memorial Volume.
S. Wolfram, A New Kind of Science
Index to Elementary Cellular Automata
Index entries for sequences related to cellular automata
Index entries for linear recurrences with constant coefficients, signature (101,-100).

Cf. A002275, A099814 (other bisection), A007583, A095372, A299960.

seq((10^(2*n+1) - 1)/9,n=0..15); # C. Ronaldo (aga_new_ac(AT)hotmail.com), Jan 19 2005

Table[(10^(2*n + 1) - 1)/9, {n, 0, 100}] (* Robert Price, Feb 21 2016 *)

a(n) = (10^(2*n + 1) - 1)/9; \\ Michel Marcus, Mar 12 2023

def A100706(n): return (10**((n<<1)+1)-1)//9 # Chai Wah Wu, Nov 04 2022

Numbers composed entirely of 2*n+1 concatenated 1's for n >= 0.
O.g.f.: (1+10*x)/((-1+x)*(-1+100*x)). - R. J. Mathar, Apr 03 2008
From Klaus Purath, Sep 23 2020: (Start)
a(n) = Sum_{i = 0..2*n} 10^i.
a(n) = 101*a(n-1) - 100*a(n-2).
a(n) = 110*10^(2*n-2) + a(n-1).
a(n) = 100*a(n-1) + 11.
a(n) = (a(n-1)^2 - 1210*10^(2*n-4))/a(n-2). (End)

More terms from C. Ronaldo (aga_new_ac(AT)hotmail.com), Jan 19 2005

A207262 a(n) = 2^(4n - 2) + 1.

Original entry on oeis.org

5, 65, 1025, 16385, 262145, 4194305, 67108865, 1073741825, 17179869185, 274877906945, 4398046511105, 70368744177665, 1125899906842625, 18014398509481985, 288230376151711745, 4611686018427387905, 73786976294838206465, 1180591620717411303425, 18889465931478580854785, 302231454903657293676545

Offset: 1

Alonso del Arte, Feb 16 2012

With the exception of the first term, all these numbers are composite, and in fact are all multiples of 5. The other factors can be considerably larger than 5, as is the case with say, 2^158 + 1. These numbers can be factored as (2^(2n - 1) + 2^n + 1)(2^(2n - 1) - 2^n + 1). For example, 2^6 + 1 = 65 = (2^3 + 2^2 + 1)(2^3 - 2^2 + 1) = 13 * 5.
This formula was discovered by Leon-Francois-Antoine Aurifeuille in 1873. Wells (2005) remarks that knowledge of this formula would have saved Fortune Landry years of work he spent factoring 2^58 + 1.
Aurifeuille actually rediscovered a very special case of the identity 4x^4+1 = (2x^2-2x+1)(2x^2+2x+1), which Euler communicated to Goldbach in 1742. (The Fuss reference is in my book Seminumerical Algorithms, 3rd ed., p. 392; I had cited Aurifeuille in the 1st and 2nd editions.) - Don Knuth, Feb 09 2013
An Engel expansion of 4 to the base 16 as defined in A181565, with the associated series expansion 4 = 16/5 + 16^2/(5*65) + 16^3/(5*65*1025) + 16^4/(5*65*1025*16385) + .... Cf. A087289 and A199561. - Peter Bala, Oct 29 2013
Conjecture: Let m = 4n - 2. a(n) equals the sum of the m-th powers of the divisors of m divided by the sum of the m-th powers of the odd divisors of m. - Ivan N. Ianakiev, Jan 29 2020

David Wells, Prime Numbers: The Most Mysterious Figures in Math. Hoboken, New Jersey: John Wiley & Sons (2005) p. 15

Vincenzo Librandi, Table of n, a(n) for n = 1..200
FactorDB, Factorizations of 2^(4*n-2)+1
P. H. Fuss, Correspondance math. et physique, 1 (1843) p. 145.
Primenumbers Yahoo Group, Aurifeuille and factoring, search results.
Eric Weisstein's World of Mathematics, Aurifeuillean Factorization.
Yahoo Groups, Aurifeuille and factoring
Index entries for linear recurrences with constant coefficients, signature (17,-16).

Cf. A000051, A052539 (supersets). A016825. A087289, A199561.

2^(4*Range[20] - 2) + 1
LinearRecurrence[{17, -16}, {5, 65}, 50] (* Vincenzo Librandi, Mar 03 2012 *)

a(n)=4^(2*n-1)+1 \\ Charles R Greathouse IV, Oct 07 2015

a(n) = 4^(2n - 1) + 1.
G.f.: 5*x*(1-4*x)/((1-x)*(1-16*x)). - Bruno Berselli, Feb 17 2012
a(1) = 5, a(n) = 16*(a(n-1) - 1) + 1. - Arkadiusz Wesolowski, Feb 17 2012
a(n) = 5*A299960(n-1). - R. J. Mathar, Feb 28 2018
E.g.f.: exp(x) + (exp(16*x) - 5)/4. - Stefano Spezia, Jan 30 2020

A229747 Largest prime factor of 4^(2*n+1)+1.

Original entry on oeis.org

5, 13, 41, 113, 109, 2113, 1613, 1321, 26317, 525313, 14449, 30269, 268501, 279073, 536903681, 384773, 4327489, 47392381, 231769777, 21841, 43249589, 1759217765581, 29247661, 140737471578113, 4981857697937, 1326700741, 1801439824104653, 3630105520141

Offset: 0

Colin Barker, Sep 28 2013

nonn

4^(2*n+1)+1 = 2^(2*(2*n+1))+1 = (2^(2*n+1)-2^(n+1)+1) * (2^(2*n+1)+2^(n+1)+1).
For all n, the smallest prime factor of 4^(2*n+1)+1 is 5.
Therefore, the present sequence also gives the largest prime factor of (4^(2*n+1)+1)/5 = A299960(n), for all n > 0. See A299959 for the smallest prime factor of this. - M. F. Hasler, Feb 27 2018

			For n=7, 4^(2*n+1)+1 = 1073741825 = 5*5*13*41*61*1321. So a(7)=1321.

Daniel Suteu, Table of n, a(n) for n = 0..547

Cf. A207262. Bisection of A274903.

Cf. A299960, A299959, A052539.

Table[FactorInteger[4^(2n+1)+1][[-1,1]],{n,0,30}] (* Harvey P. Dale, Mar 10 2018 *)

a(n) = {
  f=factor(2^(2*n+1)-2^(n+1)+1);
  g=factor(2^(2*n+1)+2^(n+1)+1);
  max(f[matsize(f)[1],1], g[matsize(g)[1],1])
}

a(n) = A006530(A052539(2n+1)) = A006530(A207262(n+1)), and for n > 1, a(n) = A006530(A299960(n)) = A006530(A052539(2n+1)/5). \\ M. F. Hasler, Feb 27 2018

a(n) = max(A229767(n), A229768(n)), for n >= 1. - Daniel Suteu, Jun 08 2022

A299959 Least prime factor of (4^(2n+1)+1)/5, a(0) = 1.

Original entry on oeis.org

1, 13, 5, 29, 13, 397, 53, 5, 137, 229, 13, 277, 5, 13, 107367629, 5581, 13, 5, 149, 13, 10169, 173, 5, 3761, 29, 13, 15358129, 5, 13, 1181, 733, 13, 5, 269, 13, 569, 293, 5, 29, 317, 13, 997, 5, 13, 1069, 29, 13, 5, 389, 13, 809, 41201, 5, 857, 5669, 13, 58309, 5, 13, 29, 397, 13, 5, 509, 13

Offset: 0

M. F. Hasler, Feb 22 2018

nonn

The range of this sequence with a(0) = 1 omitted, {5, 13, 29, 53, 137, ...}, appears to be a subset of A261580 (and of the Pythagorean primes A002144). Is there a smaller superset sequence in OEIS? - M. F. Hasler, Jan 07 2025

			For n = 0, A299960(0) = (4^1+1)/5 = 5/5 = 1, therefore we let a(0) = 1.
For n = 1, A299960(1) = (4^3+1)/5 = 65/5 = 13 is prime, therefore a(1) = 13.
For n = 2, A299960(2) = (4^5+1)/5 = 1025/5 = 205 = 5*41, therefore a(2) = 5.

Chai Wah Wu, Table of n, a(n) for n = 0..715

Cf. A299960, A052539.

a(n)=A020639(4^(2*n+1)\5+1) \\ Using factor(...)[1,1] requires complete factorization and is much less efficient for large n.

a(n) = A020639(A299960(n)) = A020639(A052539(2n+1)/5).
a(n) = 5 iff n = 2 (mod 5); otherwise, a(n) = 13 if n = 1 (mod 3).
Otherwise, a(n) = 29 if n = 3 (mod 7), else a(n) = 53 if n = 6 (mod 13), else a(n) = 137 if n = 8 (mod 17), else a(n) = 149 if n = 18 (mod 27), else a(n) = 173 if n = 21 (mod 43), etc... - M. F. Hasler, Jan 07 2025

A340441 Square array, read by ascending antidiagonals, where row n gives all odd solutions k > 1 and n > 0 to A000120(2n+1) = A000120((2n+1)*k), A000120 is the Hamming weight.

Original entry on oeis.org

3, 13, 11, 3, 205, 43, 57, 5, 3277, 171, 35, 3641, 7, 52429, 683, 21, 47, 233017, 19, 838861, 2731, 3, 79, 99, 14913081, 23, 13421773, 10923, 241, 5, 197, 187, 954437177, 37, 214748365, 43691, 7, 61681, 7, 325, 419, 61083979321, 39, 3435973837, 174763

Offset: 1

Thomas Scheuerle, Jan 07 2021

Solutions to related equation A000120(k) = A000120(k*n) are A340351.

			Five initial terms of rows 1-5 are listed below:
   1:  3,   11,     43,       171,       683, ...
   2: 13,  205,   3277,     52429,    838861, ...
   3:  3,    5,      7,        19,        23, ...
   4: 57, 3641, 233017,  14913081, 954437177, ...
   5: 35,   47,      99,      187,       419, ...
T(3,4) = 19 because: (3*2+1) in binary is 111 and (3*2+1)*19 = 133 in binary is 10000101, both have 3 bits set to 1.

Pontus von Brömssen, Antidiagonals n = 1..100, flattened

Cf. A000120, A340351, A340069.

Cf. A263132 (superset of 1st row), A007583 (1st row), A299960 (2nd row).

If 2*n = 2^j, then T(n, m) = (1+2^(j+2*j*m))/(2*n+1) for m > 0. In particular:
  T(1, m) = (1+2^(1+2*m))/3 = A007583(m),
  T(2, m) = (1+2^(2+4*m))/5 = A299960(m),
  T(4, m) = (1+2^(3+6*m))/9.
The third row consists of all numbers of the form (1+2^(1+b*3)+2^(2+c*3))/7, where b and c are natural numbers >= 0 and b+c > 0.
The seventh row consists of all numbers of the form (1+2^(1+b*2)+2^(2+c*2)+2^(3+d*2))/15 where b, c, and d are natural numbers >= 0 and b+c+d > 1.

More terms from Pontus von Brömssen, Jan 08 2021

A362783 Square array A(n,k) = (n^(2*k + 1) + 1)/(n + 1), n >= 0, k >= 0, read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 11, 7, 1, 1, 1, 43, 61, 13, 1, 1, 1, 171, 547, 205, 21, 1, 1, 1, 683, 4921, 3277, 521, 31, 1, 1, 1, 2731, 44287, 52429, 13021, 1111, 43, 1, 1, 1, 10923, 398581, 838861, 325521, 39991, 2101, 57, 1, 1, 1, 43691, 3587227, 13421773, 8138021, 1439671

Offset: 0

Juri-Stepan Gerasimov, May 03 2023

			Array begins:
=====================================================================
n/k |  0    1      2       3         4          5            6   ...
----+----------------------------------------------------------------
0   |  1    1      1       1         1          1            1   ...
1   |  1    1      1       1         1          1            1   ...
2   |  1    3     11      43       171        683         2731   ...
3   |  1    7     61     547      4921      44287       398581   ...
4   |  1   13    205    3277     52429     838861     13421773   ...
5   |  1   21    521   13021    325521    8138021    203450521   ...
6   |  1   31   1111   39991   1439671   51828151   1865813431   ...
   ...

Andrew Howroyd, Table of n, a(n) for n = 0..1325 (rows 0..50)

Columns k=0..3 are A000012, A002061, A060884, A060888.
Rows n=2..4 are A007583, A066443, A299960.
Main diagonal is A179897.

/* as array */ [[&+[(-n)^j: j in [0..2*k]]: k in [0..6]]: n in [0..6]]; // Juri-Stepan Gerasimov, May 06 2023

A(n,k) = (n^(2*k + 1) + 1)/(n + 1) \\ Andrew Howroyd, May 03 2023

A(n,k) = Sum_{j=0..2*k} (-n)^j.

a(49) corrected by Andrew Howroyd, Jan 20 2024

A318236 a(n) = (32^(4n+3) + 1)/5.

Original entry on oeis.org

5, 77, 1229, 19661, 314573, 5033165, 80530637, 1288490189, 20615843021, 329853488333, 5277655813325, 84442493013197, 1351079888211149, 21617278211378381, 345876451382054093, 5534023222112865485, 88544371553805847757, 1416709944860893564109, 22667359117774297025741

Offset: 0

Jianing Song, Aug 21 2018

a(n) is the smallest positive multiplicative inverse of 5 modulo 2^(4*n+3).

In binary, a(n) is written as 10011001...1001101 where "1001" appears n times. When n approaches infinity we get the 2-adic expansion of 1/5: ...10011001101. Similarly, the 2-adic expansion of 1/3 is ...101010101011.

			The smallest solution to 5*x == 1 (mod 8) is x = (3*2^3 + 1)/5 = 5.
The smallest solution to 5*x == 1 (mod 128) is x = (3*2^7 + 1)/5 = 77.

Jianing Song, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (17,-16).

A007583 gives the smallest positive multiplicative inverse of 3 modulo 2^(2*n) and 2^(2*n+1), A299960 gives the smallest positive multiplicative inverse of 5 modulo 2^(4*n), 2^(4*n+1) and 2^(4*n+2).

[(3*2^(4*n + 3) + 1)/5: n in [0..20]]; // Vincenzo Librandi, Aug 24 2018

Table[(3 2^(4 n + 3) + 1) / 5, {n, 0, 20}] (* Vincenzo Librandi, Aug 24 2018 *)
LinearRecurrence[{17,-16},{5,77},20] (* Harvey P. Dale, Sep 25 2020 *)

PARI
```
a(n) = (3*2^(4*n + 3) + 1)/5
```

def A318236(n): return (3*(1<<(n<<2)+3)+1)//5 # Chai Wah Wu, Jul 29 2022

O.g.f.: (5 - 8*x)/((1 - x)*(1 - 16*x)).
E.g.f.: (24*exp(16*x) + exp(x))/5.
a(0) = 5, a(1) = 77; for n>1, a(n) = 17*a(n-1) - 16*a(n-2).

A340547 Square array, read by ascending antidiagonals, where row n gives all solutions n > 0 to A000120(n+1) = A000120((n+1)*k), A000120 is the Hamming weight.

Original entry on oeis.org

1, 1, 2, 1, 2, 4, 1, 2, 3, 8, 1, 2, 4, 4, 16, 1, 2, 4, 8, 6, 32, 1, 2, 3, 8, 16, 8, 64, 1, 2, 3, 4, 13, 32, 11, 128, 1, 2, 4, 4, 6, 16, 64, 12, 256, 1, 2, 2, 8, 5, 8, 26, 128, 16, 512, 1, 2, 4, 8, 16, 6, 11, 32, 256, 22, 1024

Offset: 1

Thomas Scheuerle, Jan 11 2021

Solutions to related equation A000120(k) = A000120(k*n) are A340351.

The same sequence without leading ones and only odd solutions is A340441.

			Eight initial terms of rows 1-8 are listed below:
   1: 1, 2, 4, 8, 16, 32, 64, 128, ...
   2: 1, 2, 3, 4,  6,  8, 11,  12, ...
   3: 1, 2, 4, 8, 16, 32, 64, 128, ...
   4: 1, 2, 4, 8, 13, 16, 26,  32, ...
   5: 1, 2, 3, 4,  6,  8, 11,  12, ...
   6: 1, 2, 3, 4,  5,  6,  7,   8, ...
   7: 1, 2, 4, 8, 16, 32, 64, 128, ...
   8: 1, 2, 4, 8, 16, 32, 57,  64, ...
T(3,4) = 8 because: (3+1) in binary is 100 and (3*1)*8 = 32 in binary is 100000, both have 1 bit set to 1.

Cf. A000120, A340351, A340069, A340441.

Cf. A263132 (superset of 1st row), A007583 (1st row), A299960 (2nd row).

T(2n, ...)   = 2^{0,1,2,...}, 2^{0,1,2,...} * row n of A340441.
T(4n+1, ...) = 2^{0,1,2,...}, 2^{0,1,2,...} * row n of A340441.
T(2^n, ...)  = 2^{0,1,2,...}.

A360967 Array T(n,m) = (2^(m(2n+1))+1)/(2^m+1) read by antidiagonals.

Original entry on oeis.org

3, 13, 11, 57, 205, 43, 241, 3641, 3277, 171, 993, 61681, 233017, 52429, 683, 4033, 1016801, 15790321, 14913081, 838861, 2731, 16257, 16519105, 1041204193, 4042322161, 954437177, 13421773, 10923, 65281, 266354561, 67662254017, 1066193093601, 1034834473201, 61083979321, 214748365

Offset: 1

R. J. Mathar, Feb 27 2023

			The array starts in row n=1 and column n=1 as
      3      13      57     241
     11     205    3641   61681
     43    3277  233017 15790321
    171   52429 14913081 4042322161

Quynh Nguyen, Jean Pedersen, and Hien T. Vu, New Integer Sequences Arising From 3-Period Folding Numbers, JIS vol 19 (2016) #16.3.1 table 3

Cf. A007583 (col 1), A299960 (col 2).

cp's OEIS Frontend

A100706 Bisection of A002275.

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A207262 a(n) = 2^(4n - 2) + 1.

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A229747 Largest prime factor of 4^(2*n+1)+1.

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A299959 Least prime factor of (4^(2n+1)+1)/5, a(0) = 1.

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A340441 Square array, read by ascending antidiagonals, where row n gives all odd solutions k > 1 and n > 0 to A000120(2*n+1) = A000120((2*n+1)*k), A000120 is the Hamming weight.

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A362783 Square array A(n,k) = (n^(2*k + 1) + 1)/(n + 1), n >= 0, k >= 0, read by antidiagonals.

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A318236 a(n) = (3*2^(4*n+3) + 1)/5.

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A340441 Square array, read by ascending antidiagonals, where row n gives all odd solutions k > 1 and n > 0 to A000120(2n+1) = A000120((2n+1)*k), A000120 is the Hamming weight.

A318236 a(n) = (32^(4n+3) + 1)/5.

A360967 Array T(n,m) = (2^(m(2n+1))+1)/(2^m+1) read by antidiagonals.