A077241 -id:A077241 - cp's OEIS frontend

A144797 Numbers k such that 2*k^2 + 17 is a square.

2, 4, 16, 26, 94, 152, 548, 886, 3194, 5164, 18616, 30098, 108502, 175424, 632396, 1022446, 3685874, 5959252, 21482848, 34733066, 125211214, 202439144, 729784436, 1179901798, 4253495402, 6876971644, 24791187976, 40081928066, 144493632454, 233614596752

Offset: 1

Richard Choulet, Sep 21 2008

			a(1)=2 because 2*4 + 17 = 25 = 5^2.

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

Cf. A133301.

Select[Range[6000000],IntegerQ[Sqrt[2#^2+17]]&] (* Harvey P. Dale, Aug 18 2012 *)
LinearRecurrence[{0, 6, 0, -1}, 2{1, 2, 8, 13}, 30] (* Robert G. Wilson v, Dec 02 2014 *)

Vec(2*x*(1+x)*(1+x+x^2) / ((x^2+2*x-1)*(x^2-2*x-1)) + O(x^50)) \\ Colin Barker, Oct 20 2014

G.f.: 2*x*(1+x)*(1+x+x^2) / ( (x^2+2*x-1)*(x^2-2*x-1) ). - R. J. Mathar, Nov 27 2011
a(n) = 2*A077241(n-1). - R. J. Mathar, Nov 27 2011
a(n) = 6*a(n-2) - a(n-4). - Colin Barker, Oct 20 2014

Corrected by R. J. Mathar, Nov 27 2011

Editing and more terms from Colin Barker, Oct 20 2014

A305292 Numbers k such that k-1 is a square and k+1 is a triangular number.

Original entry on oeis.org

2, 5, 65, 170, 2210, 5777, 75077, 196250, 2550410, 6666725, 86638865, 226472402, 2943171002, 7693394945, 99981175205, 261348955730, 3396416785970, 8878171099877, 115378189547777, 301596468440090, 3919462027838450, 10245401755863185, 133146330756959525

Offset: 1

Bruno Berselli, Jun 11 2018

It is easy to prove that there are no numbers k such that k-1 is a triangular number and k+1 is a square.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,34,-34,-1,1).

Cf. A000045, A000290, A077241, A214838.

LinearRecurrence[{1, 34, -34, -1, 1}, {2, 5, 65, 170, 2210}, 25]

Vec(x*(2 + 3*x - 8*x^2 + 3*x^3 + 2*x^4)/((1 - x)*(1 - 6*x + x^2)*(1 + 6*x + x^2)) + O(x^30)) \\ Colin Barker, Jun 14 2018

G.f.: x*(2 + 3*x - 8*x^2 + 3*x^3 + 2*x^4)/((1 - x)*(1 - 6*x + x^2)*(1 + 6*x + x^2)).
a(n) = a(-n-1) = a(n-1) + 34*a(n-2) - 34*a(n-3) - a(n-4) + a(n-5).
a(n) = 35*a(n-2) - 35*a(n-4) + a(n-6) = 34*a(n-2) - a(n-4) + 2.
a(n) = (-2 + (13*sqrt(2) + 7*(-1)^n)*(1 + sqrt(2))^(2*n+1) - (13*sqrt(2) - 7* (-1)^n)*(1 - sqrt(2))^(2*n+1))/32.
a(n) = A214838(n) - 1.
a(n) = A077241(n-1)^2 + 1.

A165356 Primes p such that p + (p^2 - 1)/8 is a perfect square.

Original entry on oeis.org

3, 19, 211, 1249, 4513, 1445953, 30381331, 286292179, 2959257735801707821729

Offset: 1

Vincenzo Librandi, Sep 16 2009

The primes p = A000040(j) at j= 2, 8, 47, 204, 612, 110340 etc. generating the squares 2^2, 8^2, 76^2, 443^2 etc.

From the ansatz p + (p^2 - 1)/8 = s^2 we conclude p = -4 + sqrt(17 + 8*s^2), so all s are members of A077241.

			For p=3, p + (p^2-1)/8 = 4 = 2^2. For p=19, p + (p^2-1)/8 = 64 = 8^2. For p=211, p + (p^2-1)/8 = 5776 = 76^2.

Cf. A165352, A165353, A165354.

A077241 := proc(n) if n <= 3 then op(n+1,[1,2,8,13]) ; else 6*procname(n-2)-procname(n-4) ; fi; end:
for n from 0 do s := A077241(n) ; p := sqrt(17+8*s^2)-4 ; if isprime(p) then printf("%d,\n",p) ; fi; od: # R. J. Mathar, Sep 21 2009
a := proc (n) if isprime(n) = true and type(sqrt(n+(1/8)*n^2-1/8), integer) = true then n else end if end proc; seq(a(n), n = 1 .. 10000000); # Emeric Deutsch, Sep 21 2009

p = 2; lst = {}; While[p < 10^12, If[ IntegerQ@ Sqrt[p + (p^2 - 1)/8], AppendTo[lst, p]; Print@p]; p = NextPrime@p] (* Robert G. Wilson v, Sep 30 2009 *)

6 more terms from R. J. Mathar, Sep 21 2009

cp's OEIS Frontend

A144797 Numbers k such that 2*k^2 + 17 is a square.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

Extensions

A305292 Numbers k such that k-1 is a square and k+1 is a triangular number.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A165356 Primes p such that p + (p^2 - 1)/8 is a perfect square.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Maple

Mathematica

Extensions