A038198 -id:A038198 - cp's OEIS frontend

A248346 Primes of the form 2^x - y^2, with y^2 < 2^x.

2, 3, 7, 23, 31, 47, 71, 79, 103, 127, 151, 199, 223, 271, 367, 431, 463, 487, 503, 727, 751, 823, 967, 1087, 1303, 1319, 1423, 1439, 1559, 1607, 1759, 1823, 1879, 1951, 1999, 2039, 2143, 3343, 3527, 3623, 3967, 4447, 4943, 5167, 5503, 5591, 5791, 6199, 6343

Offset: 1

Juri-Stepan Gerasimov, Oct 05 2014

nonn

Primes in A051213.

			7 is in this sequence because 7 = 2^3 - 1^2 = 2^4 - 3^2 = 2^5 - 5^2 = 2^7 - 11^2 = 2^15 - 181^2.
1559 is in this sequence because 1559 = 2^19 - 723^2 is prime. - _Sean A. Irvine_, Apr 28 2022

Cf. A038198, A051213, A060728, A248344.

Primes in A056007 form a subset of the numbers in this sequence.

Select[Union[Flatten[Table[2^x - y^2, {x, 16}, {y, 0, Floor[Sqrt[2^x]]}]]], PrimeQ] (* Alonso del Arte, Oct 05 2014 *)

a(24)-a(38) from Alonso del Arte, Oct 05 2014

More terms and missing terms inserted by Sean A. Irvine, Apr 28 2022

A336819 Odd values of D > 0 for which the generalized Ramanujan-Nagell equation x^2 + D = 2^m has two or more solutions in the positive integers.

Original entry on oeis.org

7, 15, 23, 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, 8589934591, 17179869183, 34359738367, 68719476735

Offset: 1

Bernard Schott, Aug 04 2020

nonn

D = 7 corresponds to Ramanujan-Nagell equation x^2 + 7 = 2^m with its 5 solutions (A038198 for x, A060728 for n, Wikipedia link).
If D odd <> 7, R. Apéry proved in 1960 that the equation x^2 + D = 2^m has at most 2 solutions (see links).
If D odd > 0, this equation has 2 solutions iff D = 23 or D = 2^k - 1 for some k >= 4 (link Beukers, theorem 2, p. 395).
For any solution (x,m), m is bounded by m < 435 + 10 * (log(D) / log(2)) [link Beukers, corollary 1, p. 394]. If D < 2^96, then the bound becomes m < 18 + 2 * (log(D) / log(2)) [link Beukers, corollary 2, p. 395].

			For these exceptional cases, the corresponding solutions are:
D = 7,  (x,m) = (1,3), (3,4), (5,5), (11,7), (181,15);
D = 23, (x,m) = (3,5), (45,11);
D = 2^k -1, k >= 4,  (x,m) = (1,k), (2^(k-1) - 1, 2*(k-1)).
For k = 4 and D = 15, then 1^2 + 15 = 2^4 = 16, and 7^2 + 15 = 2^6 = 64.
Remark: for k = 2 and D = 3, the two possible solutions corresponding to 2^k-1 coincide with (1, 2).

Richard K. Guy, Unsolved Problems in Theory of Numbers, Springer-Verlag, 2004, D10.

Roger Apéry, Sur une équation Diophantienne, C. R. Acad. Sci. Paris Sér. A251 (1960), 1263-1264.
Roger Apéry, Sur une équation Diophantienne, C. R. Acad. Sci. Paris Sér. A251 (1960), 1451-1452.
Frits Beukers, On the generalized Ramanujan-Nagell equation, I, Acta arithmetica, XXXVIII, 1980-1981, page 389-410.
Wikipedia, Ramanujan-Nagell equation.
Index entries for linear recurrences with constant coefficients, signature (3,-2).

Cf. A000225, A038198, A060728, A247763, A336881.

From Colin Barker, Aug 05 2020: (Start)
G.f.: x*(7 - 6*x - 8*x^2 - 8*x^3 + 16*x^4) / ((1 - x)*(1 - 2*x)).
a(n) = 3*a(n-1) - 2*a(n-2) for n>5.
a(n) = 2^(1+n)-1 for n>3. (End)
The two formulas with a(n) are true, according to theorem 2 of Beukers' link. - Bernard Schott, Aug 07 2020

A336881 a(n) is the number of solutions (x, m) of the generalized Ramanujan-Nagell equation x^2 + n = 2^m, x > 0, m > 0, n > 0.

Original entry on oeis.org

1, 0, 1, 1, 0, 0, 5, 0, 0, 0, 0, 1, 0, 0, 2, 1, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 5, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 2, 0, 0, 2, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1

Offset: 1

Bernard Schott, Aug 06 2020

nonn

Equivalently, number of representations of n as n = 2^m - x^2, m > 0, x > 0.
a(7) = 5 corresponds to Ramanujan-Nagell equation (A038198 for x, A060728 for m, Wikipedia link).
If n odd <> 7, Apéry proved in 1960 that the equation x^2 + n = 2^m has at most 2 solutions (see link).
If n odd, this equation has 2 solutions iff n = 23 or n = 2^k - 1 for some k >= 4 (link Beukers, theorem 2, p. 395).

			1^2 + 1 = 2^1 hence a(1) = 1.
3^2 + 23 = 2^5 and 45^2 + 23 = 2^11 hence a(23) = 2.
28 = 2^5 - 2^2 = 2^6 - 6^2 = 2^7 - 10^2 = 2^9 - 22^2 = 2^17 - 362^2 hence a(28) = 5.

Roger Apéry, Sur une équation Diophantienne, C. R. Acad. Sci. Paris Sér. A251 (1960), 1263-1264.
Frits Beukers, On the generalized Ramanujan-Nagell equation, I, Acta arithmetica, XXXVIII, 1980-1981, page 389-410.
Wikipedia, Ramanujan-Nagell equation.

Cf. A247763, A336819.

More terms from Jinyuan Wang, Aug 07 2020

A098808 a(n) = 2^(n + 11) - 11.

Original entry on oeis.org

2037, 4085, 8181, 16373, 32757, 65525, 131061, 262133, 524277, 1048565, 2097141, 4194293, 8388597, 16777205, 33554421, 67108853, 134217717, 268435445, 536870901, 1073741813, 2147483637, 4294967285, 8589934581, 17179869173

Offset: 0

Parthasarathy Nambi, Oct 06 2004

			a(0) = 2^11 - 11 = 2037.
a(1) = 2^12 - 11 = 4085.

T. Skolem, S. Chowla and D. J. Lewis, The diophantine equation 2^(n+2) - 7 = x^2 and related problems, Proc. Amer. Math. Soc., 10 (1959), 663-669.
Index entries for linear recurrences with constant coefficients, signature (3,-2).

Cf. A038198, A060728.

Table[2^(n + 11) - 11, {n, 0, 30}] (* Stefan Steinerberger, Mar 06 2006 *)

first(m)=vector(m,i,i--;2^(i + 11) - 11) \\ Anders Hellström, Aug 26 2015

From Colin Barker, May 11 2012: (Start)
a(n) = 3*a(n-1)-2*a(n-2).
G.f.: (2037-2026*x)/((1-x)*(1-2*x)). (End)

More terms from Stefan Steinerberger, Mar 06 2006

A322548 Integers x such that x^2 + 119 = 15*2^y.

Original entry on oeis.org

1, 11, 19, 29, 61, 701

Offset: 1

Tomohiro Yamada, Dec 14 2018

The exponents y of the corresponding powers of 2 are 3, 4, 5, 6, 8, 15.
The list gives all positive integers x such that x^2 + 119 = 15*2^y.
Yann Bugeaud proposed the problem to prove that there is an absolute constant C such that, for any positive integers D, k and a prime number p such that gcd(D, kp) = 1, the Diophantine equation x^2 + D = k*p^n has at most C integer solutions (x, n) (Problem 9 of the list of 22 open problems below).

			a(2) = 11: 11^2 + 119 = 240 = 15*2^4.

Jan-Hendrik Evertse, Some open problems about Diophantine equations, 22 problems posed at the Instructional conference and workshop "Solvability of Diophantine equations", May 7-16, 2007, Lorentz Center, Leiden.
Jörg Stiller, The Diophantine equation x^2 + 119 = 15*2^n has exactly six solutions, Rocky Mountain J. Math. 26 (1996), 295-298.

Cf. A038198 (All solutions to x^2 + 7 = 2^y).

s={}; Do[r = Solve[x^2 + 119 == 15*2^k && x >= 0, x, Integers]; If[Length[r]>0, AppendTo[s, x/.r[[1]]]], {k,1,15}]; s (* Amiram Eldar, Dec 15 2018 *)

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A248346 Primes of the form 2^x - y^2, with y^2 < 2^x.

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A336819 Odd values of D > 0 for which the generalized Ramanujan-Nagell equation x^2 + D = 2^m has two or more solutions in the positive integers.

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A336881 a(n) is the number of solutions (x, m) of the generalized Ramanujan-Nagell equation x^2 + n = 2^m, x > 0, m > 0, n > 0.

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A098808 a(n) = 2^(n + 11) - 11.

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A322548 Integers x such that x^2 + 119 = 15*2^y.

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