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.

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

Author

Bernard Schott, Aug 04 2020

Keywords

• nonn

Comments

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].

Examples

			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).
		

References

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

Links

• 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).

Crossrefs

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

Formula

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