A207262 -id:A207262 - cp's OEIS frontend

A087289 a(n) = 2^(2*n+1) + 1.

3, 9, 33, 129, 513, 2049, 8193, 32769, 131073, 524289, 2097153, 8388609, 33554433, 134217729, 536870913, 2147483649, 8589934593, 34359738369, 137438953473, 549755813889, 2199023255553, 8796093022209, 35184372088833, 140737488355329, 562949953421313, 2251799813685249

Offset: 0

W. Edwin Clark, Aug 29 2003

Number of pairs of polynomials (f,g) in GF(2)[x] satisfying deg(f) <= n, deg(g) <= n and gcd(f,g) = 1.
An unpublished result due to Stephen Suen, David desJardins, and W. Edwin Clark. This is the case k = 2, q = 2 of their formula q^((n+1)*k) * (1 - 1/q^(k-1) + (q-1)/q^((n+1)*k)) for the number of ordered k-tuples (f_1, ..., f_k) of polynomials in GF(q)[x] such that deg(f_i) <= n for all i and gcd(f_1, ..., f_k) = 1.
Apparently the same as A084508 shifted left.
Terms in binary are palindromes of the form 1x1 where x is a string of 2*n zeros (A152577). - Brad Clardy, Sep 01 2011
For n > 0, a(n) is the number k such that the number of iterations of the map k -> (3k +1)/8 == 4 (mod 8) until reaching (3k +1)/8 <> 4 (mod 8) equals n. (see the Collatz problem: the start of the parity trajectory of a(n) is n times {100} = 100100100100...100abcd...). - Michel Lagneau, Jan 23 2012
An Engel expansion of 2 to the base 4 as defined in A181565, with the associated series expansion 2 = 4/3 + 4^2/(3*9) + 4^3/(3*9*33) + 4^4/(3*9*33*129) + .... Cf. A199561 and A207262. - Peter Bala, Oct 29 2013
For x = A083420(n), y = A000079(n+1), z = a(n) then x^2 + 2*y^2 = z^2. - Vincenzo Librandi, Jun 09 2014
A254046(n+1) is the 3-adic valuation of a(n). - Fred Daniel Kline, Jan 11 2017

			a(0) = 3 since there are three pairs, (0,1), (1,0) and (1,1) of polynomials (f,g) in GF(2)[x] of degree at most 0 such that gcd(f,g) = 1.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
K. Morrison, Random polynomials over finite fields.
Index entries for linear recurrences with constant coefficients, signature (5,-4).

Cf. A087290, A087291, A087292, A099393.
Equals A004171 + 1.
Cf. also A007583, A181565, A199561, A207262, A254046, A283070.
Cf. A000079, A083420, A084508, A152577.

[2^(2*n+1) + 1: n in [0..30]]; // Vincenzo Librandi, May 16 2011

Table[2^(2 n + 1) + 1, {n, 0, 20}] (* or *) 3 NestList[4 # - 1 &, 1, 20]
(* or *) CoefficientList[Series[(3 - 6 x)/((1 - x) (1 - 4 x)), {x, 0, 20}], x] (* Michael De Vlieger, Mar 03 2017 *)

a(n)=2^(2*n+1)+1 \\ Charles R Greathouse IV, Sep 24 2015

G.f.: (3-6*x)/((1-x)*(1-4*x)).
a(n) = 3*A007583(n).
a(n) = 4*a(n-1) - 3. - Lekraj Beedassy, Apr 29 2005
a(n) = A099393(n+1) - 2*A099393(n). - Brad Clardy, Sep 01 2011
a(n) = 2^(2*n + 1)*a(-1-n) for all n in Z. - Michael Somos, Jan 11 2017
a(n) = A283070(n) - 1. - Peter M. Chema, Mar 02 2017
From Elmo R. Oliveira, Feb 22 2025: (Start)
E.g.f.: exp(x)*(2*exp(3*x) + 1).
a(n) = 5*a(n-1) - 4*a(n-2). (End)

A299960 a(n) = (4^(2*n+1) + 1) / 5.

Original entry on oeis.org

1, 13, 205, 3277, 52429, 838861, 13421773, 214748365, 3435973837, 54975581389, 879609302221, 14073748835533, 225179981368525, 3602879701896397, 57646075230342349, 922337203685477581, 14757395258967641293, 236118324143482260685, 3777893186295716170957

Offset: 0

M. F. Hasler, Feb 22 2018

nonn

It is easily seen that 4^(2n+1)+1 is divisible by 5 for all n, since 4 = -1 (mod 5). For even powers this does not hold.
The aerated sequence 1, 0, 13, 0, 205, 0, 3277, ... is a linear divisibility sequence of order 4. It is the case P1 = 0, P2 = -5^2, Q = 4 of the 3-parameter family of 4th-order linear divisibility sequences found by Williams and Guy. Cf. A007583, A095372 and A100706. - Peter Bala, Aug 28 2019
Let G be a sequence satisfying G(i) = 2*G(i-1) + G(i-2) - 2*G(i-3) for arbitrary integers i and without regard to the initial values of G. Then a(n) = (G(i)*2^(4*n+2) + G(i+8*n+4))/(5*G(i+4*n+2)) as long as G(i+4*n+2) != 0. - Klaus Purath, Feb 02 2021
Ch. Gerbr asks (personal comm.) whether we can prove that 13 is the only prime in this sequence. We can prove divisibility conditions for many residue classes of the index n (cf. formulas), but have not yet found a complete covering set. - M. F. Hasler, Jan 07 2025

			For n = 0, a(0) = (4^1+1)/5 = 5/5 = 1.
For n = 1, a(1) = (4^3+1)/5 = 65/5 = 13.

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.
Index entries for linear recurrences with constant coefficients, signature (17,-16).

Cf. A299959 for the smallest prime factor.

Cf. A052539, A007583, A095372, A100706.

A299960 := n -> (4^(2*n+1)+1)/5: seq(A299960(n), n=0..20);

LinearRecurrence[{17, -16}, {1, 13}, 20] (* Jean-François Alcover, Feb 22 2018 *)

PARI
```
A299960(n)=4^(2*n+1)\5+1
```

def A299960(n): return ((1<<(n<<2)+2)+1)//5 # Chai Wah Wu, Jul 29 2022

a(n) = A052539(2*n+1)/5 = A015521(2*n+1) = A014985(2*n+1) = A007910(4*n+1) = A007909(4*n+1) = A207262(n+1)/5.
O.g.f.: (1 - 4*x)/(1 - 17*x + 16*x^2). - Peter Bala, Aug 28 2019
a(n) = 17*a(n-1) - 16*a(n-2). - Wesley Ivan Hurt, Oct 02 2020
From Klaus Purath, Feb 02 2021: (Start)
a(n) = (2^(4*n+2)+1)/5.
a(n) = (A061654(n) + A001025(n))/2.
a(n) = A091881(n+1) + 7*A131865(n-1) for n > 0.
(End)
E.g.f.: (exp(x) + 4*exp(16*x))/5. - Stefano Spezia, Feb 02 2021
We have d | a(n) for all n in R, for the following pairs (d, R) of divisors d and residue classes R: (13, 1 + 3Z), (5, 2 + 5Z), (29, 3 + 7Z), (397, 5 + 11Z),
  (53, 6 + 13Z), (137, 8 + 17Z), (229, 9 + 19Z), (277, 11 + 23Z),
  (107367629, 14 + 29Z), (5581, 15 + 31Z), (149, 18 + 27Z), (10169, 20 + 41Z),
  (173, 21 + 43Z), (3761, 23 + 47Z), (15358129, 26 + 53Z), (1181, 29 + 59Z),
  (733, 30 + 61Z), (269, 33 + 67Z), (569, 35 + 71Z),(293, 36 + 73Z), (317, 39 + 79Z),
  (997, 41 + 83Z), (1069, 44 + 89Z), (389, 48 + 97Z), (809, 50 + 101Z),
  (41201, 51 + 103Z), (857, 53 + 107Z), (5669, 54 + 109Z), (58309, 56 + 113Z),
  (509, 63 + 127Z), (269665073, 65 + 131Z), (189061, 68 + 137Z), (557, 69 + 139Z),
  (1789, 74 + 149Z), (653, 81 + 163Z), (9413, 90 + 181Z), (3821, 95 + 191Z),
  (773, 96 + 193Z), (4729, 98 + 197Z), (797, 99 + 199Z), ... - M. F. Hasler, Jan 07 2025

A199561 a(n) = 3*9^n + 1.

Original entry on oeis.org

4, 28, 244, 2188, 19684, 177148, 1594324, 14348908, 129140164, 1162261468, 10460353204, 94143178828, 847288609444, 7625597484988, 68630377364884, 617673396283948, 5559060566555524, 50031545098999708, 450283905890997364, 4052555153018976268, 36472996377170786404

Offset: 0

Vincenzo Librandi, Nov 08 2011

An Engel expansion of 3 to the base 9 as defined in A181565, with the associated series expansion 3 = 9/4 + 9^2/(4*28) + 9^3/(4*28*244) + 9^4/(4*28*244*2188) + .... Cf. A087289 and A207262. - Peter Bala, Oct 29 2013

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (10,-9).

Cf. A066443, A087289, A181565, A199560, A207262.

Magma
```
[3*9^n+1: n in [0..30]];
```

3*9^Range[0,20]+1 (* or *) LinearRecurrence[{10,-9},{4,28},20] (* Harvey P. Dale, Jul 30 2019 *)

a(n) = 4*A066443(n).
a(n) = 9*a(n-1) - 8.
a(n) = 10*a(n-1) - 9*a(n-2).
G.f.: 4*(1-3*x)/((1-x)*(1-9*x)).
From Elmo R. Oliveira, Sep 13 2024: (Start)
E.g.f.: exp(x)*(3*exp(8*x) + 1).
a(n) = 2*A199560(n). (End)

A211412 a(n) = 4*n^4 + 1.

Original entry on oeis.org

5, 65, 325, 1025, 2501, 5185, 9605, 16385, 26245, 40001, 58565, 82945, 114245, 153665, 202501, 262145, 334085, 419905, 521285, 640001, 777925, 937025, 1119365, 1327105, 1562501, 1827905, 2125765, 2458625, 2829125, 3240001, 3694085, 4194305, 4743685, 5345345, 6002501, 6718465, 7496645

Offset: 1

Alonso del Arte, Feb 10 2013

Except for the first term, all terms are composite. a(n) is divisible by 5 if n is not.
Long before Aurifeuille, Euler discovered that 4n^4 + 1 = (2n^2 + 2n + 1)*(2n^2 - 2n + 1). For example, 325 = 4 * 3^4 + 1 = (2 * 3^2 + 2 * 3 + 1)*(2 * 3^2 - 2 * 3 + 1) = 25 * 13. Euler shared this discovery with Goldbach in a letter dated August 28, 1742. [Euler identity corrected by Graham Holmes, Jun 02 2023]
The terms of the sequence are the arithmetic mean of eight numbers located on concentric circles (see Avilov link). - Nicolay Avilov, Jan 22 2021

Don Knuth, The Art of Computer Programming: Seminumerical Algorithms, 3rd ed., New York: Addison-Wesley Professional (1997), p. 392.
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..1000
Nicolay Avilov, Drawing with circles
P. H. Fuss, Correspondance mathématique et physique de quelques célèbres géomètres du XVIIIème siècle, Saint-Pétersbourg, 1843, p. 145; alternative link. See in particular Lettre XLVI (Euler to Goldbach), Aug 28 1742
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A207262 (subset).
After the first term, subsequence of A121944.
Cf. A053755.

[4*n^4 + 1: n in [1..50]]; // Vincenzo Librandi, Jun 11 2017

4 Range[44]^4 + 1
Table[4 n^4 + 1, {n, 50}] (* Vincenzo Librandi, Jun 11 2017 *)
LinearRecurrence[{5,-10,10,-5,1},{5,65,325,1025,2501},40] (* Harvey P. Dale, Sep 15 2023 *)

a(n) = 4*n^4+1 \\ Felix Fröhlich, Jun 07 2017

G.f.: -x*(x^4+50*x^2+40*x+5) / (x-1)^5. - Colin Barker, Feb 11 2013
a(n) = A053755(n^2). - Michel Marcus, Sep 18 2015
a(n) = (2*n^2)^2 + 1^2 = (2*n^2-1)^2 + (2*n)^2. - Thomas Ordowski, Sep 18 2015
a(n) = A001844(n) * A001844(n+1) = A141046(n) + 1 = (A000583(n) * 4 ) + 1 = A016742(n) + A173121(n) + 1. - Bruce J. Nicholson, Jun 06 2017
From Amiram Eldar, Jul 26 2022: (Start)
Sum_{n>=1} 1/a(n) = tanh(Pi/2)*Pi/4 - 1/2.
Sum_{n>=1} (-1)^n/a(n) = 1/2 - sech(Pi/2)*Pi/4. (End)

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

A229767 Largest prime factor of 2^(2*n+1)-2^(n+1)+1.

Original entry on oeis.org

5, 5, 113, 37, 397, 1613, 61, 953, 457, 14449, 30269, 8101, 246241, 107367629, 384773, 312709, 47392381, 184481113, 1249, 12112549, 1759217765581, 54001, 140737471578113, 4981857697937, 26317, 1801439824104653, 415878438361, 525313, 174877, 368140581013

Offset: 1

Colin Barker, Sep 29 2013

nonn

2^(2*n+1)-2^(n+1)+1 is a factor of 4^(2*n+1)+1.

			For n=5, 2^(2*n+1)-2^(n+1)+1 = 1985 = 5*397, so a(5)=397.

Daniel Suteu, Table of n, a(n) for n = 1..547

Cf. A092440, A207262, A229747, A229768.

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

A229768 Largest prime factor of 2^(2*n+1)+2^(n+1)+1.

Original entry on oeis.org

13, 41, 29, 109, 2113, 157, 1321, 26317, 525313, 1429, 1657, 268501, 279073, 536903681, 49477, 4327489, 7416361, 231769777, 21841, 43249589, 500177, 29247661, 7484047069, 19707683773, 1326700741, 586477649, 3630105520141, 275415303169, 104399276341

Offset: 1

Colin Barker, Sep 29 2013

nonn

2^(2*n+1)+2^(n+1)+1 is a factor of 4^(2*n+1)+1.

			For n=10, 2^(2*n+1)+2^(n+1)+1 = 2099201 = 13*113*1429, so a(10)=1429.

Daniel Suteu, Table of n, a(n) for n = 1..547

Cf. A085601, A207262, A229747, A229767.

Table[FactorInteger[2^(2n+1)+2^(n+1)+1][[-1,1]],{n,30}] (* Harvey P. Dale, Nov 03 2017 *)

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

cp's OEIS Frontend

A087289 a(n) = 2^(2*n+1) + 1.

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A299960 a(n) = (4^(2*n+1) + 1) / 5.

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A199561 a(n) = 3*9^n + 1.

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A211412 a(n) = 4*n^4 + 1.

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

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A229767 Largest prime factor of 2^(2*n+1)-2^(n+1)+1.

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A229768 Largest prime factor of 2^(2*n+1)+2^(n+1)+1.

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