A134538 -id:A134538 - cp's OEIS frontend

A212656 a(n) = 5*n^2 + 1.

1, 6, 21, 46, 81, 126, 181, 246, 321, 406, 501, 606, 721, 846, 981, 1126, 1281, 1446, 1621, 1806, 2001, 2206, 2421, 2646, 2881, 3126, 3381, 3646, 3921, 4206, 4501, 4806, 5121, 5446, 5781, 6126, 6481, 6846, 7221, 7606, 8001, 8406, 8821, 9246, 9681, 10126, 10581, 11046, 11521, 12006, 12501

Offset: 0

Alonso del Arte, May 23 2012

Z[sqrt(-5)] is not a unique factorization domain, and some of the numbers in this sequence have two different factorizations in that domain, e.g., 21 = 3 * 7 = (1 + 2*sqrt(-5))*(1 - 2*sqrt(-5)). And of course some primes in Z are composite in Z[sqrt(-5)], like 181 = (1 + 6*sqrt(-5))*(1 - 6*sqrt(-5)).

These are pentagonal-star numbers. - Mario Cortés, Oct 26 2020

Benjamin Fine & Gerhard Rosenberger, Number Theory: An Introduction via the Distribution of Primes, Boston: Birkhäuser, 2007, page 268.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
F. Javier de Vega, An extension of Furstenberg's theorem of the infinitude of primes, arXiv:2003.13378 [math.NT], 2020.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A033429, A134538, A002522, A053755, A056107, A058331.

Cf. A137530 (primes of the form 1+5*n^2).

[5*n^2 + 1: n in [0..50]]; // Vincenzo Librandi, Jul 10 2012

Table[5n^2 + 1, {n, 0, 49}]
LinearRecurrence[{3,-3,1},{1,6,21},60] (* Harvey P. Dale, Apr 04 2017 *)

a(n)=5*n^2+1 \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 5*n^2 + 1 = (1 + n*sqrt(-5))*(1 - n*sqrt(-5)).
G.f.: (1+3*x+6*x^2)/(1-x)^3. - Bruno Berselli, May 23 2012
a(n) = 3*a(n-1) -3*a(n-2) +a(n-3). - Vincenzo Librandi, Jul 10 2012
From Amiram Eldar, Jul 15 2020: (Start)
Sum_{n>=0} 1/a(n) = (1 + (Pi/sqrt(5))*coth(Pi/sqrt(5)))/2.
Sum_{n>=0} (-1)^n/a(n) = (1 + (Pi/sqrt(5))*csch(Pi/sqrt(5)))/2. (End)
a(n) = A005891(n-1) + 5*A000217(n). - Mario Cortés, Oct 26 2020
From Amiram Eldar, Feb 05 2021: (Start)
Product_{n>=0} (1 + 1/a(n)) = sqrt(2)*csch(Pi/sqrt(5))*sinh(sqrt(2/5)*Pi).
Product_{n>=1} (1 - 1/a(n)) = (Pi/sqrt(5))*csch(Pi/sqrt(5)).(End)
E.g.f.: exp(x)*(1 + 5*x + 5*x^2). - Stefano Spezia, Feb 05 2021

A158491 a(n) = 20*n^2 - 1.

Original entry on oeis.org

19, 79, 179, 319, 499, 719, 979, 1279, 1619, 1999, 2419, 2879, 3379, 3919, 4499, 5119, 5779, 6479, 7219, 7999, 8819, 9679, 10579, 11519, 12499, 13519, 14579, 15679, 16819, 17999, 19219, 20479, 21779, 23119, 24499, 25919, 27379, 28879, 30419, 31999, 33619, 35279

Offset: 1

Vincenzo Librandi, Mar 20 2009

The identity (20*n^2 - 1)^2 - (100*n^2 - 10)*(2*n)^2 = 1 can be written as a(n)^2 - A158490(n)*A005843(n)^2 = 1.

Sequence found by reading the line from 19, in the direction 19, 79, ... in the square spiral whose vertices are the generalized dodecagonal numbers A195162. - Omar E. Pol, Nov 05 2012

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Vincenzo Librandi, X^2-AY^2=1, Math Forum, 2007. [Wayback Machine link]
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A005843, A134538, A158490, A195162.

I:=[19, 79, 179]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..40]];

LinearRecurrence[{3,-3,1},{19,79,179},50]
20*Range[40]^2-1 (* Harvey P. Dale, Aug 24 2021 *)

a(n)=20*n^2-1 \\ Charles R Greathouse IV, Dec 23 2011

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
G.f: x*(-19-22*x+x^2)/(x-1)^3.
From Amiram Eldar, Mar 06 2023: (Start)
Sum_{n>=1} 1/a(n) = (1 - cot(Pi/(2*sqrt(5)))*Pi/(2*sqrt(5)))/2.
Sum_{n>=1} (-1)^(n+1)/a(n) = (cosec(Pi/(2*sqrt(5)))*Pi/(2*sqrt(5)) - 1)/2. (End)
From Elmo R. Oliveira, Jan 25 2025: (Start)
E.g.f.: exp(x)*(20*x^2 + 20*x - 1) + 1.
a(n) = A134538(2*n). (End)

A127561 Array T(n,k) = n^2+5nk+5*k^2 read downwards antidiagonals, n,k>=0.

Original entry on oeis.org

0, 5, 1, 20, 11, 4, 45, 31, 19, 9, 80, 61, 44, 29, 16, 125, 101, 79, 59, 41, 25, 180, 151, 124, 99, 76, 55, 36, 245, 211, 179, 149, 121, 95, 71, 49, 320, 281, 244, 209, 176, 145, 116, 89, 64, 405, 361, 319, 279, 241, 205, 171, 139, 109, 81, 500, 451, 404, 359, 316, 275, 236, 199, 164, 131

Offset: 0

Kenneth J Ramsey, Jan 18 2007, Feb 05 2007, Feb 06 2007

Lattice table of Fibonacci characteristic values from Wechsler's J determinant sequence A022344 uniquely position such that the row and column determine starting a,b values of a Fibonacci sequence having the same characteristic value.
A vector from (0,0) to any prime value P in the array does not pass through any other lattice point. If that vector is extended it passes through lattice points having successively the values 0, P*1^2, P*2^2, P*3^2, P*4^2 ... All primes ending in 1, 5 and 9 or the product thereof appear in the array, no prime ending in 3 or 7 appears in the array except in a square product which may be multiplied by a squarefree product of primes ending in 1, 5 or 9.
The table can be expanded by allowing negative arguments in the formula, but any positive value obtained can be expressed with nonnegative arguments.
The second row is the sequence A062786. The term in every succeeding row is 2* the term immediately above minus the next above term plus 2.
If the table is rearranged by shifting each column down by twice the column number, then the terms in second column would be equal to the row number squared plus the row number minus 1 and every succeeding term to the right would be equal to twice the left-hand term minus the next left-hand term minus 2.
It appears that any prime ending in 1,5, or 9 or any such prime times 5 appears only once in the table and that every power of such a prime or product thereof has one and only one nonnegative row and column position such that the row and column positions are coprime. A method for finding a coprime row and column position of the 2^n th power of any prime ending in 1,5,or 9, or of the product thereof, from the coprime row and column position of that prime or product is suggested by the discussion in the link titled "Wythoff Array, Pythagorean Triples, Primes".
It seems that if you stack the row and column positions of two numbers in the array that the determinant gives a column in which the product appears. Thus since the row and column position of 29 and 41 are 3,1 and 4,1 respectively then the product (41*29) appears in column 1*4 - 3*1 or column 1. The same value appears also in column -1 so 3*1-1*4 is a valid answer also. For our purposes however we choose the order that gives a positive value. Once the column number of the product is known it is easy to find the row number. There may be new determinant based math to find the row directly, but I don't know of any. It may happen that the row is negative, in which case the following transformation works a(r,c) = a(-r,c+r). Applied twice this transformation gives the original starting pair. I have yet to find any case in which one starts out with positive values for the row and column of each factor of a number appearing in the table and using the above determinant math cannot find positive values for the row and column of the product. I posted a few interesting results in the Cut-the-knot forum. Use the link given previously.

			T(0,1) = 5 because (0+2*1)^2 + 1*(0+2*1) - 1^2 = 5 and also because the Fibonacci sequence having the Horadam ID {a,b,1,1} with a = 0+2*1 and b = 1 has the characteristic value a^2 + b*a - b^2.
  0,  5, 20, 45, 80,125,180,245,320,405,500,... A033429
  1, 11, 31, 61,101,151,211,281,361,451,551,..  A062786
  4, 19, 44, 79,124,179,244,319,404,499,604,..  A134538
  9, 29, 59, 99,149,209,279,359,449,549,659,... A143198 (row 9)
 16, 41, 76,121,176,241,316,401,496,601,716,...
 25, 55, 95,145,205,275,355,445,545,655,775,..
 36, 71,116,171,236,311,396,491,596,711,836,...
 49, 89,139,199,269,349,439,539,649,769,899,...
 64,109,164,229,304,389,484,589,704,829,964,...
 81,131,191,261,341,431,531,641,761,891,1031,...
100,155,220,295,380,475,580,695,820,955,1100,...

K. J. Ramsey, Wythoff Array, Pythagorean triples, Primes

Cf. A035513, A022344.

T(a,b) = (a+2b)^2 + b(a+2b) - b^2.

A135064 Numbers n such that the quintic polynomial x^5 - 10nx^2 - 24*n has Galois group A_5 over rationals.

Original entry on oeis.org

1, 11, 29, 76, 199, 521, 1364, 3571, 9349, 24476, 64079, 167761, 439204, 1149851, 3010349, 7881196, 20633239, 54018521, 141422324, 370248451, 969323029, 2537720636, 6643838879, 17393796001, 45537549124, 119218851371, 312119004989, 817138163596, 2139295485799, 5600748293801

Offset: 1

Artur Jasinski, Nov 15 2007

nonn

Sequence appears to agree with the Lucas bisection A002878 for n > 1. - Klaus Brockhaus, Nov 18 2007
A002878(n) is in this sequence for all 1 < n <= 1000, and the sequences agree through a(20) = 370248451. Of course this is not a proof. - Charles R Greathouse IV, Mar 03 2017, updated Mar 20 2017
If this agreement is provable then of course it provides recurrences, generating functions, etc., for this sequence. - N. J. A. Sloane, Nov 24 2007 However, at present this is only a conjecture, and should not be used as the basis for formulas or computer programs. - N. J. A. Sloane, Mar 04 2017
Comparing A135064 with A002878, the number 4 is missing. In this case the Galois group of the quintic polynomial x^5 - 40*x^2 - 96 is dihedral of order 10. - Artur Jasinski, May 27 2010
The relation with A002878 is proved in Wong's article. - Eric M. Schmidt, Nov 25 2017

Siman Wong, Specialization of Galois groups and integral points on elliptic curves, Proceedings of the American Mathematical Society, 145 (2017), 5179-5190.

Cf. A134538, A134547, A002878.

is(n)=my(p=Pol([1,0,0,-10*n,0,-24*n])); polisirreducible(p) && polgalois(p)[1]==60 \\ Charles R Greathouse IV, Mar 03 2017

a(20) corrected by Klaus Brockhaus, Nov 18 2007

Unjustified formulas, programs, and b-file deleted. - N. J. A. Sloane, Mar 04 2017

A158446 a(n) = 25*n^2 - 5.

Original entry on oeis.org

20, 95, 220, 395, 620, 895, 1220, 1595, 2020, 2495, 3020, 3595, 4220, 4895, 5620, 6395, 7220, 8095, 9020, 9995, 11020, 12095, 13220, 14395, 15620, 16895, 18220, 19595, 21020, 22495, 24020, 25595, 27220, 28895, 30620, 32395, 34220, 36095, 38020, 39995, 42020, 44095

Offset: 1

Vincenzo Librandi, Mar 19 2009

The identity (10*n^2 - 1)^2 - (25*n^2 - 5)*(2*n)^2 = 1 can be written as A158447(n)^2 - a(n)*A005843(n)^2 = 1.

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Vincenzo Librandi, X^2-AY^2=1, Math Forum, 2007. [Wayback Machine link]
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A005843, A134538, A158447.

I:=[20, 95, 220]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+Self(n-3): n in [1..50]];

Table[25n^2-5,{n,50}]
LinearRecurrence[{3,-3,1},{20,95,220},40] (* Harvey P. Dale, May 05 2019 *)

PARI
```
a(n) = 25*n^2 - 5.
```

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
G.f: 5*x*(4+7*x-x^2)/(1-x)^3.
From Amiram Eldar, Mar 05 2023: (Start)
Sum_{n>=1} 1/a(n) = (1 - cot(Pi/sqrt(5))*Pi/sqrt(5))/10.
Sum_{n>=1} (-1)^(n+1)/a(n) = (cosec(Pi/sqrt(5))*Pi/sqrt(5) - 1)/10. (End)
From Elmo R. Oliveira, Jan 16 2025: (Start)
E.g.f.: 5*(exp(x)*(5*x^2 + 5*x - 1) + 1).
a(n) = 5*A134538(n). (End)

A134547 a(n) = 5n^2 + 20n + 4.

Original entry on oeis.org

29, 64, 109, 164, 229, 304, 389, 484, 589, 704, 829, 964, 1109, 1264, 1429, 1604, 1789, 1984, 2189, 2404, 2629, 2864, 3109, 3364, 3629, 3904, 4189, 4484, 4789, 5104, 5429, 5764, 6109, 6464, 6829, 7204, 7589, 7984, 8389, 8804, 9229, 9664, 10109, 10564, 11029

Offset: 1

Artur Jasinski, Oct 31 2007, Nov 21 2007

Most quintic polynomials x^5 + 5*x*(5*n^2 + 20*n + 4) + 8*(5*n^2 + 20*n + 4) (with the exception of n=0 or -4 when the polynomial is solvable, or n=-2 when it is reducible) have nonsolvable alternating Galois group A_5 (of order 60) over rational numbers.

Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A134538.

seq(5*n^2+20*n+4,n=1..40); # Robert Israel, May 19 2025

Mathematica
```
Table[5n^2 + 20n + 4, {n, 1, 30}]
```

a(n)=5*n^2+20*n+4 \\ Charles R Greathouse IV, Jun 17 2017

G.f.: -x*(29-23*x+4*x^2)/(-1+x)^3. - R. J. Mathar, Nov 14 2007
From Elmo R. Oliveira, Jun 04 2025: (Start)
E.g.f.: -4 + (4 + 25*x + 5*x^2)*exp(x).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n >= 4. (End)

cp's OEIS Frontend

A212656 a(n) = 5*n^2 + 1.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A158491 a(n) = 20*n^2 - 1.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A127561 Array T(n,k) = n^2+5*n*k+5*k^2 read downwards antidiagonals, n,k>=0.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Formula

A135064 Numbers n such that the quintic polynomial x^5 - 10*n*x^2 - 24*n has Galois group A_5 over rationals.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

PARI

Extensions

A158446 a(n) = 25*n^2 - 5.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A134547 a(n) = 5*n^2 + 20*n + 4.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

A127561 Array T(n,k) = n^2+5nk+5*k^2 read downwards antidiagonals, n,k>=0.

A135064 Numbers n such that the quintic polynomial x^5 - 10nx^2 - 24*n has Galois group A_5 over rationals.

A134547 a(n) = 5n^2 + 20n + 4.