A028871 -id:A028871 - cp's OEIS frontend

A008865 a(n) = n^2 - 2.

-1, 2, 7, 14, 23, 34, 47, 62, 79, 98, 119, 142, 167, 194, 223, 254, 287, 322, 359, 398, 439, 482, 527, 574, 623, 674, 727, 782, 839, 898, 959, 1022, 1087, 1154, 1223, 1294, 1367, 1442, 1519, 1598, 1679, 1762, 1847, 1934, 2023, 2114, 2207, 2302, 2399, 2498

Offset: 1

N. J. A. Sloane, R. K. Guy

For n >= 2, least m >= 1 such that f(m, n) = 0 where f(m,n) = Sum_{i=0..m} Sum_{k= 0..i} (-1)^k*(floor(i/n^k) - n*floor(i/n^(k+1))). - Benoit Cloitre, May 02 2004
For n >= 3, the a(n)-th row of Pascal's triangle always contains a triple forming an arithmetic progression. - Lekraj Beedassy, Jun 03 2004
Let C = 1 + sqrt(2) = 2.414213...; and 1/C = 0.414213... Then a(n) = (n + 1 + 1/C) * (n + 1 - C). Example: a(6) = 34 = (7 + 0.414...) * (7 - 2.414...). - Gary W. Adamson, Jul 29 2009
The sequence (n-4)^2-2, n = 7, 8, ... enumerates the number of non-isomorphic sequences of length n, with entries from {1, 2, 3} and no two adjacent entries the same, that minimally contain each of the thirteen rankings of three players (111, 121, 112, 211, 122, 212, 221, 123, 132, 213, 231, 312, 321) as embedded order isomorphic subsequences. By "minimally", we mean that the n-th symbol is necessary for complete inclusion of all thirteen words. See the arXiv paper below for proof. If n = 7, these sequences are 1213121, 1213212, 1231213, 1231231, 1231321, 1232123, and 1232132, and for each case, there are 3! = 6 isomorphs. - Anant Godbole, Feb 20 2013
a(n), n >= 0, with a(0) = -2, gives the values for a*c of indefinite binary quadratic forms [a, b, c] of discriminant D = 8 for b = 2*n. In general D = b^2 - 4*a*c > 0 and the form [a, b, c] is a*x^2 + b*x*y + c*y^2. - Wolfdieter Lang, Aug 15 2013
With a different offset, this is 2*n^2 - (n + 1)^2, which arises in one explanation of why Bertrand's postulate does not automatically prove Legendre's conjecture: as n gets larger, so does the range of numbers that can have primes that satisfy Bertrand's postulate yet do nothing for Legendre's conjecture. - Alonso del Arte, Nov 06 2013
x*(x + r*y)^2 + y*(y + r*x)^2 can be written as (x + y)*(x^2 + s*x*y + y^2). For r >= 0, the sequence gives the values of s: in fact, s = (r + 1)^2 - 2. - Bruno Berselli, Feb 20 2019
For n >= 2, the continued fraction expansion of sqrt(a(n)) is [n-1; {1, n-2, 1, 2n-2}]. For n=2, this collapses to [1; {2}]. - Magus K. Chu, Sep 06 2022

			G.f. = -x + 2*x^2 + 7*x^3 + 14*x^4 + 23*x^5 + 34*x^6 + 47*x^7 + 62*x^8 + 79*x^9 + ...

T. D. Noe, Table of n, a(n) for n = 1..1000
Anant Godbole and Martha Liendo, Waiting time distribution for the emergence of superpatterns, arxiv 1302.4668 [math.PR], 2013.
Eric Weisstein's World of Mathematics, Near-Square Prime.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A145067 (Zero followed by partial sums of A008865).
Cf. A000027, A013648.
Cf. A028871 (primes).
Cf. A263766 (partial products).
Cf. A270109. [Bruno Berselli, Mar 17 2016]

a008865 = (subtract 2) . (^ 2) :: Integral t => t -> t
a008865_list = scanl (+) (-1) [3, 5 ..]
-- Reinhard Zumkeller, May 06 2013

[n^2 - 2: n in [1..60]]; // Vincenzo Librandi, May 01 2014

Range[50]^2 - 2 (* Harvey P. Dale, Mar 14 2011 *)

{for(n=1, 47, print1(n^2-2, ","))} \\ Klaus Brockhaus, Oct 17 2008

For n > 1: a(n) = A143053(A000290(n)), A143054(a(n)) = A000290(n). - Reinhard Zumkeller, Jul 20 2008
G.f.: (x-5*x^2+2*x^3)/(-1+3*x-3*x^2+x^3). - Klaus Brockhaus, Oct 17 2008
E.g.f.: (x^2 + x -2)*exp(x) + 2. - G. C. Greubel, Aug 19 2017
a(n+1) = A101986(n) - A101986(n-1) = A160805(n) - A160805(n-1). - Reinhard Zumkeller, May 26 2009
For n > 1, a(n) = floor(n^5/(n^3 + n + 1)). - Gary Detlefs, Feb 10 2010
a(n) = a(n-1) + 2*n - 1 for n > 1, a(1) = -1. - Vincenzo Librandi, Nov 18 2010
Right edge of the triangle in A195437: a(n) = A195437(n-2, n-2). - Reinhard Zumkeller, Nov 23 2011
a(n)*a(n-1) + 2 = (a(n) - n)^2 = A028552(n-2)^2. - Bruno Berselli, Dec 07 2011
a(n+1) = A000096(n) + A000096(n-1) for all n in Z. - Michael Somos, Nov 11 2015
From Amiram Eldar, Jul 13 2020: (Start)
Sum_{n>=1} 1/a(n) = (1 - sqrt(2)*Pi*cot(sqrt(2)*Pi))/4.
Sum_{n>=1} (-1)^n/a(n) = (1 - sqrt(2)*Pi*cosec(sqrt(2)*Pi))/4. (End)
Assume offset 0. Then a(n) = 2*LaguerreL(2, 1 - n). - Peter Luschny, May 09 2021
From Amiram Eldar, Feb 05 2024: (Start)
Product_{n>=1} (1 - 1/a(n)) = sqrt(2/3)*sin(sqrt(3)*Pi)/sin(sqrt(2)*Pi).
Product_{n>=2} (1 + 1/a(n)) = -Pi/(sqrt(2)*sin(sqrt(2)*Pi)). (End)

A146326 Length of the period of the continued fraction of (1+sqrt(n))/2.

Original entry on oeis.org

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

Offset: 1

Artur Jasinski, Oct 30 2008

nonn

First occurrence of n in this sequence see A146343.
Records see A146344.
Indices where records occurred see A146345.
a(n) =0 for n = k^2 (A000290).
a(n) =1 for n = 4 k^2 + 4 k + 5 (A078370). For primes see A005473.
a(n) =2 for n in A146327. For primes see A056899.
a(n) =3 for n in A146328. For primes see A146348.
a(n) =4 for n in A146329. For primes see A028871 - {2}.
a(n) =5 for n in A146330. For primes see A146350.
a(n) =6 for n in A146331. For primes see A146351.
a(n) =7 for n in A146332. For primes see A146352.
a(n) =8 for n in A146333. For primes see A146353.
a(n) =9 for n in A143577. For primes see A146354.
a(n)=10 for n in A146334. For primes see A146355.
a(n)=11 for n in A146335. For primes see A146356.
a(n)=12 for n in A146336. For primes see A146357.
a(n)=13 for n in A333640. For primes see A146358.
a(n)=14 for n in A146337. For primes see A146359.
a(n)=15 for n in A146338. For primes see A146360.
a(n)=16 for n in A146339. For primes see A146361.
a(n)=17 for n in A146340. For primes see A146362.

			a(2) = 2 because continued fraction of (1+sqrt(2))/2 = 1, 4, 1, 4, 1, 4, 1, ... has period (1,4) length 2.

R. J. Mathar, Table of n, a(n) for n = 1..20000.

Cf. A003285, A146343, A146344, A146345.

Cf. also A000290, A078370, A146327-A146342; A005473, A056899, A146348-A146362.

A146326 := proc(n) if not issqr(n) then numtheory[cfrac]( (1+sqrt(n))/2, 'periodic','quotients') ; nops(%[2]) ; else 0 ; fi; end: seq(A146326(n),n=1..100) ; # R. J. Mathar, Sep 06 2009

Table[cf = ContinuedFraction[(1 + Sqrt[n])/2]; If[Head[cf[[-1]]] === List, Length[cf[[-1]]], 0], {n, 100}]
f[n_] := Length@ ContinuedFraction[(1 + Sqrt[n])/2][[-1]]; Array[f, 100] (* Robert G. Wilson v, Apr 11 2017 *)

a(n) = 0 iff n is a square (A000290). - Robert G. Wilson v, Apr 11 2017

a(39) and a(68) corrected by R. J. Mathar, Sep 06 2009

A028870 Numbers k such that k^2 - 2 is prime.

Original entry on oeis.org

2, 3, 5, 7, 9, 13, 15, 19, 21, 27, 29, 33, 35, 37, 43, 47, 49, 55, 61, 63, 69, 71, 75, 77, 89, 93, 103, 107, 117, 119, 121, 127, 131, 135, 139, 145, 155, 161, 169, 173, 177, 183, 191, 205, 211, 217, 223, 231, 233, 237, 239, 247, 253, 257, 259, 265, 267, 273, 279, 285

Offset: 1

Patrick De Geest

It is conjectured that this sequence is infinite.

Primes 2,3,5,7,13,... are in A062326. - Zak Seidov, Oct 05 2014

			5^2 - 2 = 23 is prime, so 5 is in the sequence.

D. Shanks, Solved and Unsolved Problems in Number Theory, 2nd. ed., Chelsea, 1978, p. 31.

Nathaniel Johnston, Table of n, a(n) for n = 1..10000
P. De Geest, Palindromic Quasipronics of the form n(n+x)
Eric Weisstein's World of Mathematics, Near-Square Prime

Cf. A028871.

[n: n in [1..1000] |IsPrime( n^2 - 2)]; // Vincenzo Librandi, Nov 18 2010

select(k->isprime(k^2-2),[$1..300]); # Muniru A Asiru, Jul 15 2018

a[n_]:=n^x-y;lst={};x=2;y=2;Do[If[PrimeQ[a[n]],AppendTo[lst,n]],{n,0,6!}];lst (* Vladimir Joseph Stephan Orlovsky, Jan 03 2009 *)
Select[Range[300],PrimeQ[#^2-2]&] (* Harvey P. Dale, Mar 21 2013 *)

is(n)=isprime(n^2-2) \\ Charles R Greathouse IV, Jul 01 2013

a(n) = sqrt(2 + A028871(n)). - Zak Seidov, Oct 05 2014

A054269 Length of period of continued fraction for sqrt(prime(n)).

Original entry on oeis.org

1, 2, 1, 4, 2, 5, 1, 6, 4, 5, 8, 1, 3, 10, 4, 5, 6, 11, 10, 8, 7, 4, 2, 5, 11, 1, 12, 6, 15, 9, 12, 6, 9, 18, 9, 20, 17, 18, 4, 5, 14, 21, 16, 13, 1, 20, 26, 4, 2, 5, 11, 12, 17, 14, 1, 12, 3, 24, 21, 13, 18, 5, 14, 16, 17, 11, 34, 19, 14, 7, 15, 4, 20, 5, 30, 8, 9, 21, 1, 21, 18, 37, 16

Offset: 1

N. J. A. Sloane, May 05 2000

The following sequences (allowing offset of first term) all appear to have the same parity: A034953, triangular numbers with prime indices; A054269, length of period of continued fraction for sqrt(p), p prime; A082749, difference between the sum of next prime(n) natural numbers and the sum of next n primes; A006254, numbers n such that 2n-1 is prime; A067076, 2n+3 is a prime. - Jeremy Gardiner, Sep 10 2004
Note that primes of the form n^2+1 (A002496) have a continued fraction whose period length is 1; odd primes of the form n^2+2 (A056899) have length 2; odd primes of the form n^2-2 (A028871) have length 4. - T. D. Noe, Nov 03 2006
For an odd prime p, the length of the period is odd if p=1 (mod 4) or even if p=3 (mod 4). - T. D. Noe, May 22 2007

T. D. Noe, Table of n, a(n) for n = 1..10000
A. I. Gliga, On continued fractions of the square root of prime numbers

Cf. A003285, A130272 (primes at which the period length sets a new record).

with(numtheory): for i from 1 to 150 do cfr := cfrac(ithprime(i)^(1/2), 'periodic','quotients'); printf(`%d,`, nops(cfr[2])) od:

Table[p=Prime[n]; Length[Last[ContinuedFraction[Sqrt[p]]]],{n,100}] (* T. D. Noe, May 22 2007 *)
Length[ContinuedFraction[Sqrt[#]][[2]]]&/@Prime[Range[100]] (* Harvey P. Dale, Sep 28 2024 *)

More terms from James Sellers, May 05 2000

A118337 Nonnegative values x of solutions (x, y) to the Diophantine equation x^2 + (x + 23)^2 = y^2.

Original entry on oeis.org

0, 12, 33, 69, 133, 252, 460, 832, 1525, 2737, 4905, 8944, 16008, 28644, 52185, 93357, 167005, 304212, 544180, 973432, 1773133, 3171769, 5673633, 10334632, 18486480, 33068412, 60234705, 107747157, 192736885, 351073644, 627996508, 1123352944, 2046207205

Offset: 1

Mohamed Bouhamida, May 14 2006

Also values x of Pythagorean triples (x, x+23, y).
Corresponding values y of solutions (x, y) are in A156567.
For the generic case x^2 + (x + p)^2 = y^2 with p = m^2 - 2 a (prime) number in A028871, m>=5, the x values are given by the sequence defined by a(n) = 6*a(n-3) - a(n-6) + 2p with a(1)=0, a(2) = 2m + 2, a(3) = 3*m^2 - 10m + 8, a(4)=3p, a(5) = 3*m^2 + 10m + 8, a(6) = 20*m^2 - 58m + 42. Pairs (p, m) are (23, 5), (47, 7), (79, 9), (167, 13), (223, 15), (359, 19), (439, 21), (727, 27), (839, 29), ...
Limit_{n -> oo} a(n)/a(n-3) = 3 + 2*sqrt(2).
Limit_{n -> oo} a(n)/a(n-1) = (27 + 10*sqrt(2))/23 for n mod 3 = {1, 2}.
Limit_{n -> oo} a(n)/a(n-1) = (3 + 2*sqrt(2))/((27 + 10*sqrt(2))/23)^2 for n mod 3 = 0.
For the generic case x^2 + (x + p)^2=y^2 with p = m^2 - 2 a prime number in A028871, m>=5, Y values are given by the sequence defined by b(n) = 6*b(n-3) - b(n-6) with b(1) = p, b(2) = m^2 + 2m + 2, b(3) = 5m^2 - 14m + 10, b(4) = 5p, b(5) = 5m^2 + 14m + 10, b(6) = 29m^2 - 82m + 58. - Mohamed Bouhamida, Sep 09 2009
For the generic case x^2 + (x + p)^2 = y^2 with p = m^2 - 2 a prime number, m>=5, the first three consecutive solutions are: (0;p), (2m+2; m^2+2m+2), (3*m^2-10m+8; 5*m^2-14m+10) and the other solutions are defined by: (X(n); Y(n))= (3*X(n-3)+2*Y(n-3)+p; 4*X(n-3)+3*Y(n-3)+2p). - Mohamed Bouhamida, Aug 19 2019
X(n) = 6*X(n-3) - X(n-6) + 2*p, and Y(n) = 6*Y(n-3) - Y(n-6) (can be easily proved using X(n) = 3*X(n-3) + 2*Y(n-3) + p, and Y(n) = 4*X(n-3) + 3*Y(n-3) + 2*p). - Mohamed Bouhamida, Aug 20 2019

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,6,-6,0,-1,1).

Cf. A156567, A028871 (primes of form n^2 - 2), A156035 (decimal expansion of 3 + 2*sqrt(2)), A156571 (decimal expansion of (27 + 10*sqrt(2))/23).

Cf. A118675 (p=47), A118676 (p=79), A130608 (p=167), A130609 (p=223), A130610 (p=359), A130645 (p=439), A130646 (p=727), A130647 (p=839).

I:=[0,12,33,69,133,252,460]; [n le 7 select I[n] else Self(n-1) +6*Self(n-3) -6*Self(n-4) -Self(n-6) +Self(n-7): n in [1..30]]; // G. C. Greubel, May 04 2018

Select[Range[0,100000],IntegerQ[Sqrt[#^2+(#+23)^2]]&] (* or *) LinearRecurrence[{1,0,6,-6,0,-1,1},{0,12,33,69,133,252,460},50] (* Vladimir Joseph Stephan Orlovsky, Feb 02 2012 *)

forstep(n=0, 1124000000, [1, 3], if(issquare(2*n*(n+23)+529), print1(n, ",")))

x='x+O('x^30); concat([0], Vec(x*(12+21*x+36*x^2-8*x^3-7*x^4-8*x^5)/((1-x)*(1-6*x^3+x^6)))) \\ G. C. Greubel, May 04 2018

a(n) = 6*a(n-3) - a(n-6) + 46 for n > 6; a(1)=0, a(2)=12, a(3)=33, a(4)=69, a(5)=133, a(6)=252.

G.f.: x*(12 + 21*x + 36*x^2 - 8*x^3 - 7*x^4 - 8*x^5)/((1-x)*(1 - 6*x^3 + x^6)).

Edited by Klaus Brockhaus, Feb 10 2009

A073577 a(n) = 4n^2 + 4n - 1.

Original entry on oeis.org

7, 23, 47, 79, 119, 167, 223, 287, 359, 439, 527, 623, 727, 839, 959, 1087, 1223, 1367, 1519, 1679, 1847, 2023, 2207, 2399, 2599, 2807, 3023, 3247, 3479, 3719, 3967, 4223, 4487, 4759, 5039, 5327, 5623, 5927, 6239, 6559, 6887, 7223, 7567, 7919, 8279, 8647

Offset: 1

M. N. Deshpande (dpratap(AT)nagpur.dot.net.in), Aug 27 2002

The sum of the squares of two consecutive terms multiplied (or divided) by 2 is always a perfect square. In general, numbers represented by the quadratic form a(n) = (2*i*n + j)^2 - 2*i^2 for any i and j have 2*(a(n)^2 + a(n+1)^2) and (a(n)^2 + a(n+1)^2)/2 as perfect squares: in this case, i=j=1.
The terms of this sequence may be seen to be 2 less than the odd squares. As such they run parallel to those in the square spiral as well as the Ulam square spiral. - Stuart M. Ellerstein (ellerstein(AT)aol.com), Oct 01 2002
Primes in the sequence are in A028871. - Russ Cox, Aug 26 2019
The continued fraction expansion of sqrt(4*a(n)) is [4n+1; {1, n-1, 2, 2n, 2, n-1, 1, 8n+2}]. For n=1, this collapses to [5; {3, 2, 3, 10}]. - Magus K. Chu, Sep 12 2022

			a(2) = 8*2 + 7 = 23;
a(3) = 8*3 + 23 = 47;
a(4) = 8*4 + 47 = 79. - _Vincenzo Librandi_, Aug 08 2010

G. C. Greubel, Table of n, a(n) for n = 1..5000
Soren Laing Aletheia-Zomlefer, Lenny Fukshansky, and Stephan Ramon Garcia, The Bateman-Horn Conjecture: Heuristics, History, and Applications, Expositiones Mathematicae, Vol. 38, No. 4 (2020), pp. 430-479; arXiv preprint, arXiv:1807.08899 [math.NT], 2018-2019. See 6.6.7, p. 36 (p. 35 in the preprint).
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A008590, A028871, A214345.

List([1..50],n->4*n^2+4*n-1); # Muniru A Asiru, Nov 01 2018

[4*n^2 + 4*n - 1: n in [1..50]]; // Wesley Ivan Hurt, Apr 18 2016

seq(4*n^2+4*n-1,n=1..100); # Robert Israel, Jan 13 2015

Table[4*n^2+4*n-1,{n,60}] (* Vladimir Joseph Stephan Orlovsky, Nov 18 2009 *)
LinearRecurrence[{3,-3,1},{7,23,47},50] (* Harvey P. Dale, Dec 04 2018 *)

A073577(n):=4*n^2+4*n-1$
makelist(A073577(n),n,1,30); /* Martin Ettl, Nov 01 2012 */

vector(50, n, 4*n^2 + 4*n - 1) \\ Michel Marcus, Jan 14 2015

for n in range(1,50): print(4*n**2+4*n-1, end=', ') # Stefano Spezia, Nov 01 2018

a(n) = FrobeniusNumber(2*n+1, 2*n+3). - Darrell Minor, Jul 29 2008
a(n) = 8*n + a(n-1) (with a(1)=7). - Vincenzo Librandi, Aug 08 2010
G.f.: x*(7+2*x-x^2)/(1-x)^3. - Robert Israel, Jan 13 2015
E.g.f.: 1 - (1-8*x-4*x^2)*exp(x). - Robert Israel, Jan 13 2015
a(n+1) = a(n) + A008590(n+1), a(1) = 7. - Altug Alkan, Sep 28 2015
a(n) = (2*n+1)+(2*n-1) + (2*n+1)*(2*n-1). - J. M. Bergot, Apr 17 2016
a(n) = (2*n+1)^2 - 2. - Zhandos Mambetaliyev, Jun 13 2017
From Stefano Spezia, Nov 04 2018: (Start)
L.g.f.: 4*x*(2+x)/(1+x)^2-log(1+x).
L.h.g.f.: -4*(-2+x)*x/(-1+x)^2+log(1-x).
(End)
Sum_{n>=1} 1/a(n) = 1 + sqrt(2)*Pi*tan(Pi/sqrt(2))/8. - Amiram Eldar, Jan 03 2021

Edited and extended by Henry Bottomley, Oct 10 2002

A118676 Nonnegative values x of solutions (x, y) to the Diophantine equation x^2+(x+79)^2 = y^2.

Original entry on oeis.org

0, 20, 161, 237, 341, 1140, 1580, 2184, 6837, 9401, 12921, 40040, 54984, 75500, 233561, 320661, 440237, 1361484, 1869140, 2566080, 7935501, 10894337, 14956401, 46251680, 63497040, 87172484, 269574737, 370088061, 508078661, 1571196900, 2157031484, 2961299640

Offset: 1

Mohamed Bouhamida, May 19 2006

Also values x of Pythagorean triples (x, x+79, y).
Corresponding values y of solutions (x, y) are in A159758.
For the generic case x^2+(x+p)^2 = y^2 with p = m^2-2 a (prime) number > 7 in A028871, see A118337.
lim_{n -> infinity} a(n)/a(n-3) = 3+2*sqrt(2).
lim_{n -> infinity} a(n)/a(n-1) = (83+18*sqrt(2))/79 for n mod 3 = {1, 2}.
lim_{n -> infinity} a(n)/a(n-1) = (10659+6110*sqrt(2))/79^2 for n mod 3 = 0.

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,6,-6,0,-1,1).

Cf. A159758, A028871, A118337, A001652, A156035 (decimal expansion of 3+2*sqrt(2)), A159759 (decimal expansion of (83+18*sqrt(2))/79), A159760 (decimal expansion of (10659+6110*sqrt(2))/79^2).

m:=25; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(x*(20+141*x+76*x^2-16*x^3-47*x^4-16*x^5)/((1-x)*(1- 6*x^3+x^6)))); // G. C. Greubel, May 07 2018

LinearRecurrence[{1,0,6,-6,0,-1,1},{0,20,161,237,341,1140,1580},75] (* Vladimir Joseph Stephan Orlovsky, Feb 07 2012 *)

forstep(n=0, 100000000, [1, 3], if(issquare(2*n^2+158*n+6241), print1(n, ",")))

a(n) = 6*a(n-3) -a(n-6) +158 for n > 6; a(1)=0, a(2)=20, a(3)=161, a(4)=237, a(5)=341, a(6)=1140.
G.f.: x*(20+141*x+76*x^2-16*x^3-47*x^4-16*x^5)/((1-x)*(1-6*x^3+x^6)).
a(3*k+1) = 79*A001652(k) for k >= 0.

Edited by Klaus Brockhaus, Apr 30 2009

A078142 a(n) is the sum of the differences of the distinct prime factors p of n and the next square larger than p.

Original entry on oeis.org

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

Offset: 1

Jason Earls, Nov 20 2002

From Bernard Schott, Dec 05 2019: (Start)
a(n) = 1 iff n > 1 is a power of 3.
a(n) = 2 iff n > 1 is a power of p prime with p in A028871. (End)

			a(6)=3 because 6=2*3 and (4-2)+(4-3)=3.
a(20)=6 because 20=2^2*5 and (4-2)+(9-5)=6.

Amiram Eldar, Table of n, a(n) for n = 1..10000

f[n_]:=Module[{difs=Transpose[FactorInteger[n]][[1]]}, Total[Ceiling[ Sqrt[difs]]^2-difs]]; Array[f,120] (* Harvey P. Dale, Apr 18 2011 *)

A118675 Nonnegative values x of solutions (x, y) to the Diophantine equation x^2+(x+47)^2 = y^2.

Original entry on oeis.org

0, 16, 85, 141, 225, 616, 940, 1428, 3705, 5593, 8437, 21708, 32712, 49288, 126637, 190773, 287385, 738208, 1112020, 1675116, 4302705, 6481441, 9763405, 25078116, 37776720, 56905408, 146166085, 220178973, 331669137, 851918488, 1283297212, 1933109508

Offset: 1

Mohamed Bouhamida, May 19 2006

nonn

Also values x of Pythagorean triples (x, x+47, y).
Corresponding values y of solutions (x, y) are in A159750.
For the generic case x^2+(x+p)^2 = y^2 with p = m^2-2 a (prime) number > 7 in A028871, see A118337.
lim_{n -> infinity} a(n)/a(n-3) = 3+2*sqrt(2).
lim_{n -> infinity} a(n)/a(n-1) = (51+14*sqrt(2))/47 for n mod 3 = {1, 2}.
lim_{n -> infinity} a(n)/a(n-1) = (3267+1702*sqrt(2))/47^2 for n mod 3 = 0.

Index entries for linear recurrences with constant coefficients, signature (1,0,6,-6,0,-1,1).

Cf. A159750, A028871, A118337, A001652, A156035 (decimal expansion of 3+2*sqrt(2)), A159751 (decimal expansion of (51+14*sqrt(2))/47), A159752 (decimal expansion of (3267+1702*sqrt(2))/47^2).

m:=25; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(x*(16+69*x+56*x^2-12*x^3-23*x^4-12*x^5)/((1-x)*(1-6*x^3 +x^6)))); // G. C. Greubel, May 07 2018

Select[Range[0,100000],IntegerQ[Sqrt[#^2+(#+47)^2]]&] (* or *) LinearRecurrence[{1,0,6,-6,0,-1,1},{0,16,85,141,225,616,940},50] (* Vladimir Joseph Stephan Orlovsky, Feb 02 2012 *)

{forstep(n=0, 100000000, [1, 3], if(issquare(2*n^2+94+2209), print1(n, ",")))}

a(n) = 6*a(n-3) -a(n-6) +94 for n > 6; a(1)=0, a(2)=16, a(3)=85, a(4)=141, a(5)=225, a(6)=616.
G.f.: x*(16+69*x+56*x^2-12*x^3-23*x^4-12*x^5)/((1-x)*(1-6*x^3+x^6)).
a(3*k+1) = 47*A001652(k) for k >= 0.

Edited by Klaus Brockhaus, Apr 30 2009

A156567 Positive numbers y such that y^2 is of the form x^2+(x+23)^2 with integer x.

Original entry on oeis.org

17, 23, 37, 65, 115, 205, 373, 667, 1193, 2173, 3887, 6953, 12665, 22655, 40525, 73817, 132043, 236197, 430237, 769603, 1376657, 2507605, 4485575, 8023745, 14615393, 26143847, 46765813, 85184753, 152377507, 272571133, 496493125

Offset: 1

Klaus Brockhaus, Feb 11 2009 , Feb 16 2009

nonn

(-8, a(1)) and(A118337(n), a(n+1)) are solutions (x, y) to the Diophantine equation x^2+(x+23)^2 = y^2.
lim_{n -> infinity} a(n)/a(n-3) = 3+2*sqrt(2).
lim_{n -> infinity} a(n)/a(n-1) = (27+10*sqrt(2))/23 for n mod 3 = {0, 2}.
lim_{n -> infinity} a(n)/a(n-1) = (627+238*sqrt(2))/23^2 for n mod 3 = 1.
For the generic case x^2+(x+p)^2=y^2 with p=m^2-2 a prime number in A028871, m>=2, the x values are given by the sequence defined by: a(n)=6*a(n-3)-a(n-6)+2p with a(1)=0, a(2)=2m+2, a(3)=3m^2-10m+8, a(4)=3p, a(5)=3m^2+10m+8, a(6)=20m^2-58m+42.Y values are given by the sequence defined by: b(n)=6*b(n-3)-b(n-6) with b(1)=p, b(2)=m^2+2m+2, b(3)=5m^2-14m+10, b(4)=5p, b(5)=5m^2+14m+10, b(6)=29m^2-82m+58. [From Mohamed Bouhamida, Sep 09 2009]

			(-8, a(1)) = (-8, 17) is a solution: (-8)^2+(-8+23)^2 = 64+225 = 289 = 17^2.
(A118337(1), a(2)) = (0, 23) is a solution: 0^2+(0+23)^2 = 529 = 23^2.
(A118337(3), a(4)) = (33, 65) is a solution: 33^2+(33+23)^2 = 1089+3136 = 4225 = 65^2.

Cf. A118337, A156035 (decimal expansion of 3+2*sqrt(2)), A156571 (decimal expansion of (27+10*sqrt(2))/23), A157472 (decimal expansion of (627+238*sqrt(2))/23^2).

A156570 (first trisection), A156568 (second trisection), A156569 (third trisection).

{forstep(n=-8, 360000000, [1,3], if(issquare(2*n*(n+23)+529, &k), print1(k, ",")))}

a(n) = 6*a(n-3)-a(n-6) for n > 6; a(1)=17, a(2)=23, a(3)=37, a(4)=65, a(5)=115, a(6)=205.

G.f.: x*(1-x)*(17+40*x+77*x^2+40*x^3+17*x^4)/(1-6*x^3+x^6).

G.f. corrected, third and fourth comment edited, cross-reference added by Klaus Brockhaus, Sep 18 2009

cp's OEIS Frontend

A008865 a(n) = n^2 - 2.

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A146326 Length of the period of the continued fraction of (1+sqrt(n))/2.

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A028870 Numbers k such that k^2 - 2 is prime.

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A054269 Length of period of continued fraction for sqrt(prime(n)).

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A118337 Nonnegative values x of solutions (x, y) to the Diophantine equation x^2 + (x + 23)^2 = y^2.

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A073577 a(n) = 4*n^2 + 4*n - 1.

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A073577 a(n) = 4n^2 + 4n - 1.