A060728 -id:A060728 - 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

A188628 a(n) = smallest k such that k*2^n - 7 is a square.

Original entry on oeis.org

7, 4, 2, 1, 1, 1, 2, 1, 11, 11, 32, 16, 8, 4, 2, 1, 4006, 2003, 9284, 4642, 2321, 107566, 53783, 313702, 156851, 1364479, 1493338, 746669, 12145148, 6072574, 3036287, 107186842, 53593421, 323781196, 161890598, 80945299, 3501584548, 1750792274, 875396137

Offset: 0

Michel Lagneau, Apr 06 2011

nonn

From Michel Lagneau, Mar 04 2015: (Start)
The sequence is infinite. Proof by induction:
Given an integer n>0. Suppose there are an infinity of perfect squares of the form k*2^n - 7 = b(n)^2 for some integer k.
For all integers n, there exists an integer b(n) such that b(n)^2 == -7 (mod 2^n). If n <=3, b(n)=1 is appropriate. Suppose there exists b(n) such that b(n)^2 == -7 (mod 2^n) for some n >=3. We prove that b(n+1)^2 == -7 (mod 2^(n+1)). In the first case, we take b(n+1) = b(n) if k is even, and in the second case we take b(n+1) = 2^(n-1)- b(n). Then b(n+1)^2 = b(n)^2 - 2^n*b(n) + 2^(2n-2) == b(n)^2 - 2^n*b(n) (mod 2^(n+1)).
But b(n)^2 = k*2^n-7 => b(n)^2 - 2^n*b(n) = 2^n(k-b(n)) - 7 == - 7 (mod 2^(n+1)) because k - b(n) is even (k is odd and b(n) is odd).
With n>=3 and b(n) odd, the proof is complete.
Finally, we see that the sequence b(n) is unbounded because b(n)^2 >=2^n - 7 for all positive integers n. This completes the proof because, for all p >=n, b(p)^2 == -7 (mod 2^n).
The corresponding squares of the sequence are 1, 1, 1, 3^2, 5^2, 11^2, 11^2, 53^2, 75^2, 181^2, 181^2, 181^2, 181^2, 181^2, 181^2, 16203^2, 16203^2, 49333^2, 49333^2, 49333^2, 474955^2, ...
With the property:
b(2) + b(3) = 1+1 = 2^1;
b(3) + b(4) = 1+3 = 2^2;
b(4) + b(5) = 3+5 = 2^3;
b(5) + b(6) = 5+11 = 2^4;
b(7) + b(8) = 11+53 = 2^6;
b(8) + b(9) = 53+75=2^7;
b(9) + b(10) = 75+181=2^8;
b(15) + b(16) = 181+16203=2^14;
(End)
To get a(n), find the smallest term of A117619 that is divisible by 2^n, and divide it by 2^n. - Michel Marcus, Mar 05 2015
a(n)=1 for n=3, 4, 5, 7, 15; see A060728. - Michel Marcus, Mar 05 2015

Michel Lagneau, Proof

Cf. A117619 (n^2+7).

A188628 := proc(n) for k from 1 do if issqr(k*2^n-7) then return k; end if; end do:
end proc:
for n from 0 do printf("%d,\n",A188628(n)) ; end do; # R. J. Mathar, Apr 09 2011

f[n_] := Block[{k = 0}, While[!IntegerQ[Sqrt[k*2^n - 7]], k++]; k]; Table[f[n], {n, 0, 24}] (* Michael De Vlieger, Mar 04 2015 *)

a(n) = {k=1; while (! issquare(k*2^n - 7), k++); k;} \\ Michel Marcus, Mar 04 2015

b(1) = b(2) = b(3) = 1; b(n+1)= 2^(n-1)-b(n) or b(n+1) = b(n).

k*2^n - 7 = b(n)^2 => a(n) = k = (b(n)^2 + 7)/2^n if b(n) different from b(n-1) or a(n) = a(n-1)/2 if a(n-1) is even (or if b(n) = b(n-1)).

a(0) lowered from 8 to 7, a(25) from 2986676 to 1364479 by R. J. Mathar, Apr 09 2011

a(31)-a(38) from Michel Lagneau, Mar 04 2015

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

A227079 Solutions n to the Diophantine equation: n = (2x^2 - 1)^2 = (6y^2 - 5).

Original entry on oeis.org

1, 49, 289, 5041, 274266721

Offset: 0

Raphie Frank, Aug 08 2013

x = {1, 2, 3, 6, 91} = A180445(n).
y = {1, 3, 7, 29, 6761} = A227077(n).
(sqrt(2*sqrt(n) + 2) - 1)^2 is a Ramanujan-Nagell Square = {1, 9, 25, 121, 32761} = A227078(n).

J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 181, p. 56, Ellipses, Paris 2008.
L. J. Mordell, Diophantine Equations, Academic Press, NY, 1969, p. 205.

Richard K. Guy, The Strong Law of Small Numbers (example #29).
Eric Weisstein's World of Mathematics, Ramanujan's Square Equation

Cf. A060728, A180445, A227077, A227078.

cp's OEIS Frontend

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

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A188628 a(n) = smallest k such that k*2^n - 7 is a square.

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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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A227079 Solutions n to the Diophantine equation: n = (2*x^2 - 1)^2 = (6*y^2 - 5).

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A227079 Solutions n to the Diophantine equation: n = (2x^2 - 1)^2 = (6y^2 - 5).