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A239678 Least numbers k such that k*2^n+1 is a square. 2
3, 4, 2, 1, 3, 7, 15, 31, 63, 127, 255, 511, 1023, 2047, 4095, 8191, 16383, 32767, 65535, 131071, 262143, 524287, 1048575, 2097151, 4194303, 8388607, 16777215, 33554431, 67108863, 134217727, 268435455, 536870911, 1073741823, 2147483647, 4294967295 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,1

COMMENTS

Excluding a(1) and a(2), these numbers equal 2^n-1 (See A000225).

The sequence A088041 is the square roots of the squares produced.

LINKS

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (3,-2).

FORMULA

a(n+2) == A000225(n) for n >= 1.

a(n) = 3*a(n-1)-2*a(n-2) for n>4. G.f.: (4*x^4+3*x^3-4*x^2-5*x+3) / ((x-1)*(2*x-1)). - Colin Barker, Mar 24 2014

EXAMPLE

1*2^1+1 = 3 is not a square. 2*2^1+1 = 5 is not a square. 3*2^1+1 = 7 is not a square. 4*2^1+1 = 9 is a square. Thus, a(1) = 4.

MATHEMATICA

CoefficientList[Series[(4 x^4 + 3 x^3 - 4 x^2 - 5 x + 3)/((x - 1) (2 x - 1)), {x, 0, 50}], x] (* Vincenzo Librandi, Mar 24 2014 *)

PROG

(Python)

import sympy

from sympy import isprime

def TwoSq(n):

..for k in range(1, 10**10):

....for i in range(10**4):

......if k*(2**n)+1 == i**2:

........return k

......if k*(2**n)+1 < i**2:

........break

n = 1

while n < 100:

..print(TwoSq(n))

..n += 1

(PARI)

Vec((4*x^4+3*x^3-4*x^2-5*x+3)/((x-1)*(2*x-1)) + O(x^100)) \\ Colin Barker, Mar 24 2014

CROSSREFS

Cf. A088041, A000225.

Sequence in context: A201909 A070352 A227216 * A136374 A303869 A081246

Adjacent sequences:  A239675 A239676 A239677 * A239679 A239680 A239681

KEYWORD

nonn,easy

AUTHOR

Derek Orr, Mar 23 2014

STATUS

approved

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Last modified October 2 23:51 EDT 2022. Contains 357230 sequences. (Running on oeis4.)