A076046 -id:A076046 - cp's OEIS frontend

A060728 Numbers n such that Ramanujan's equation x^2 + 7 = 2^n has an integer solution.

3, 4, 5, 7, 15

Offset: 1

Lekraj Beedassy, Apr 25 2001

See A038198 for corresponding x. - Lekraj Beedassy, Sep 07 2004
Also numbers such that 2^(n-3)-1 is in A000217, i.e., a triangular number. - M. F. Hasler, Feb 23 2009
With respect to M. F. Hasler's comment above, all terms 2^(n-3) - 1 are known as the Ramanujan-Nagell triangular numbers (A076046). - Raphie Frank, Mar 31 2013
Interestingly enough, all the solutions correspond to noncomposite x, i.e., x = 1 for the first term, and primes 3, 5, 11, 181 for the following terms. - M. F. Hasler, Mar 11 2024

			The fifth and ultimate solution to Ramanujan's equation is obtained for the 15th power of 2, so that we have x^2 + 7 = 2^15 yielding x = 181.

J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 181, p. 56, Ellipses, Paris 2008.
J. Roberts, Lure of the Integers. pp. 90-91, MAA 1992.
Ian Stewart & David Tall, Algebraic Number Theory and Fermat's Last Theorem, 3rd Ed. Natick, Massachusetts (2002): 96-98.

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. [_M. F. Hasler_, Feb 23 2009]
Anonymous, Developing a general 2nd degree Diophantine Equation x^2 + p = 2^n
M. Beeler, R. W. Gosper and R. Schroeppel, HAKMEM: item 31: A Ramanujan Problem (R. Schroeppel)
Curtis Bright, Solving Ramanujan's Square Equation Computationally
Spencer De Chenne, The Ramanujan-Nagell Theorem: Understanding the Proof
T. Do, Developing A General 2nd Degree Diophantine Equation x^2 + p = 2^n
A. Engel, Problem-Solving Strategies. p. 126.
Gerry Myerson, Bibliography
T. Nagell, The Diophantine equation x^2 + 7 = 2^n, Ark. Mat. 4 (1961), no. 2-3, 185-187.
S. Ramanujan, Journal of the Indian Mathematical Society, Question 464(v,120)
Eric Weisstein's World of Mathematics, Ramanujan's Square Equation
Eric Weisstein's World of Mathematics, Diophantine Equation 2nd Powers
Wikipedia, Carmichael's Theorem
Wikipedia, Diophantine equation

Cf. A002249, A038198, A076046, A077020, A077021, A107920, A215795, A227078

[n: n in [0..100] | IsSquare(2^n-7)]; // Vincenzo Librandi, Jan 07 2014

ramaNagell[n_] := Reduce[x^2 + 7 == 2^n, x, Integers] =!= False; Select[ Range[100], ramaNagell] (* Jean-François Alcover, Sep 21 2011 *)

is(n)=issquare(2^n-7) \\ Anders Hellström, Dec 12 2015

a(n) = log_2(8*A076046(n) + 8) = log_2(A227078(n) + 7)

Empirically, a(n) = Fibonacci(c + 1) + 2 = ceiling[e^((c - 1)/2)] + 2 where {c} is the complete set of positive solutions to {n in N | 2 cos(2*Pi/n) is in Z}; c is in {1,2,3,4,6} (see A217290).

Added keyword "full", M. F. Hasler, Feb 23 2009

A002249 a(n) = a(n-1) - 2*a(n-2) with a(0) = 2, a(1) = 1.

Original entry on oeis.org

2, 1, -3, -5, 1, 11, 9, -13, -31, -5, 57, 67, -47, -181, -87, 275, 449, -101, -999, -797, 1201, 2795, 393, -5197, -5983, 4411, 16377, 7555, -25199, -40309, 10089, 90707, 70529, -110885, -251943, -30173, 473713, 534059, -413367, -1481485

Offset: 0

N. J. A. Sloane, Iwan Duursma

4*2^n = A002249(n)^2 + 7*A001607(n)^2. See A077020, A077021.
Among presented initial elements of the sequence a(n), the maximal increasing or decreasing subsequences have length either 3 or 4. - Roman Witula, Aug 21 2012
This is the Lucas Sequence V_n(P, Q) = V_n(1, 2). U_n(1, 2) = A107920(n). - Raphie Frank, Dec 25 2013
The only numbers that occur more than once are 1=a(1)=a(4) and -5=a(3)=a(9).  See Noam D. Elkies's posting in the Mathematics Stack Exchange link. - Robert Israel, Dec 21 2016

			We have a(2)-a(7) = a(5)-a(4) = a(6)+a(4) = a(11)-a(10) = a(12)+a(10)=10. Further the following relations: ((1+i*sqrt(7))/2)^4 + ((1-i*sqrt(7))/2)^4 = 1 and ((1+i*sqrt(7))/2)^8 + ((1-i*sqrt(7))/2)^8 = -31. - _Roman Witula_, Aug 21 2012
G.f. = 2 + x - 3*x^2 - 5*x^3 + x^4 + 11*x^5 + 9*x^6 - 13*x^7 - 31*x^8 + ...
From _Raphie Frank_, Dec 05 2015: (Start)
V_n(1, 2) = a(1*n) = ((a(1) + sqrt(-7))/2)^n + ((a(1) - sqrt(-7))/2)^n; a(1) = 1.
V_n(-3, 4) = a(2*n) = ((a(2) + sqrt(-7))/2)^n + ((a(2) - sqrt(-7))/2)^n; a(2) = -3.
V_n(-5, 8) = a(3*n) = ((a(3) + sqrt(-7))/2)^n + ((a(3) - sqrt(-7))/2)^n; a(3) = -5.
V_n(11, 32) = a(5*n) = ((a(5) + sqrt(-7))/2)^n + ((a(5) - sqrt(-7))/2)^n; a(5) = 11.
V_n(-181, 8192) = a(13*n) = ((a(13) + sqrt(-7))/2)^n + ((a(13) - sqrt(-7))/2)^n; a(13) = -181.
(End)

T. D. Noe, Table of n, a(n) for n = 0..500
Paul Barry, On a Central Transform of Integer Sequences, arXiv:2004.04577 [math.CO], 2020.
Mathematics Stack Exchange, An exotic sequence, March 2014.
Mihai Prunescu, On other two representations of the C-recursive integer sequences by terms in modular arithmetic, arXiv:2406.06436 [math.NT], 2024. See p. 18.
Mihai Prunescu and Lorenzo Sauras-Altuzarra, On the representation of C-recursive integer sequences by arithmetic terms, arXiv:2405.04083 [math.LO], 2024. See p. 17.
Mihai Prunescu and Joseph M. Shunia, On modular representations of C-recursive integer sequences, arXiv:2502.16928 [math.NT], 2025. See p. 6.
Wikipedia, Lucas Sequence.
Index entries for linear recurrences with constant coefficients, signature (1,-2).

Cf. A014551, A107920, A060728.

I:=[2,1]; [n le 2 select I[n] else Self(n-1)-2*Self(n-2): n in [1..50]]; // Vincenzo Librandi, Nov 29 2015

A002249 := proc(n) option remember; >if n = 1 then 1 elif n = 2 then -3 else A002249(n-1>)-2*A002249(n-2); fi; end;

LinearRecurrence[{1,-2}, {2,1}, 50] (* Roman Witula, Aug 21 2012 *)
a[ n_] := 2^(n/2) ChebyshevT[ n, 8^(-1/2)] 2; (* Michael Somos, Jun 02 2014 *)
a[ n_] := 2^Min[0, n] SeriesCoefficient[ (2 - x) / (1 - x + 2 x^2), {x, 0, Abs @ n}]; (* Michael Somos, Jun 02 2014 *)
Table[2 Re[((1 + I Sqrt[7])/2)^n], {n, 0, 40}] (* Jean-François Alcover, Jun 02 2017 *)

{a(n) = if( n<0, 2^n * a(-n), polsym(2 - x + x^2, n)[n+1])}; /* Michael Somos, Jun 02 2014 */

{a(n) = 2 * real( ((1 + quadgen(-28)) / 2)^n )}; /* Michael Somos, Jun 02 2014 */

x='x+O('x^100); Vec((2-x)/(1-x+2*x^2)) \\ Altug Alkan, Dec 04 2015

from sympy import sqrt, re, I
def a(n): return 2*re(((1 + I*sqrt(7))/2)**n)
print([a(n) for n in range(40)]) # Indranil Ghosh, Jun 02 2017

[lucas_number2(n,1,2) for n in range(0, 40)] # Zerinvary Lajos, Apr 30 2009

G.f.: (2-x)/(1-x+2x^2). - Michael Somos, Oct 18 2002
a(n) = trace(A^n) for the square matrix A=[1, -2; 1, 0]. - Paul Barry, Sep 05 2003
a(n) = 2^((n+2)/2)*cos(-n*acot(sqrt(7)/7)). - Paul Barry, Sep 06 2003
a(n) = (-1)^n*(2*A110512(n) - A001607(n)) = ((1 + i*sqrt(7))/2)^n + ((1 - i*sqrt(7))/2)^n. - Roman Witula, Aug 21 2012
G.f.: G(0), where G(k) = 1 + 1/(1 - x*(7*k+1)/(x*(7*k+8) + 2/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 03 2013
(a(A060728(n) - 2))^2 = (A107920(2*(A060728(n)) - 4))^2 = 2^(A060728(n)) - 7 = A227078(n), the Ramanujan-Nagell squares. - Raphie Frank, Dec 25 2013
a(n) = [x^n] ( (1 + x + sqrt(1 + 2*x - 7*x^2))/2 )^n for n >= 1. - Peter Bala, Jun 23 2015
a(n) = (A107920(n+1) + 2*A107920(n+2) - A107920(n+3))/2. - Raphie Frank, Nov 28 2015
V_n(P,Q) = a(k*n) = ((a(k) + sqrt(-7))/2)^n + ((a(k) - sqrt(-7))/2)^n for k is in {1, 2, 3, 5, 13} = (A060728(n) - 2), P is in {1, -3, -5, 11, -181} = a(k), and Q is in {2, 4, 8, 32, 8192} = 2^k = (2*A076046(n) + 2) = (A227078(n) - 7)/4. P^2 - 4*Q = -7. - Raphie Frank, Dec 05 2015
From Peter Bala, Nov 16 2022: (Start)
The Gauss congruences hold: a(n*p^k) == a(n*p^(k-1)) (mod p^k) for all positive integers n and k and all primes p.
A268924(n) == a(3^n) (mod 3^n). (End)

A215795 Numbers n such that 2^n-1 is a triangular number (A000217).

Original entry on oeis.org

0, 1, 2, 4, 12

Offset: 1

V. Raman, Aug 23 2012

Aside from a(2), all terms are even. Probably complete; no more terms up to 10^6. - Charles R Greathouse IV, Sep 07 2012
This sequence maps to the Ramanujan-Nagell squares (8*(2^n - 1) + 1) and is therefore complete. - Raphie Frank, Sep 10 2012
Define equivalence classes on a specified real interval with respect to the symmetric transitive closure of R(x,y) = "x is an integer multiple of y". If any equivalence class is finite (the conditions for which are given in A328129), then a smallest equivalence class has cardinality 1, 2, 4 or 12. - Peter Munn, Jun 02 2021

Eric Weisstein's World of Mathematics, Ramanujan's Square Equation

Cf. A000217, A060728, A038198.
Cf. A076046 (triangular numbers of the form 2^n - 1).
Cf. A060728 (a(n) + 3).
Cf. A038198 (sqrt(8*(2^n - 1)+1)).
Cf. A215797 ((sqrt(8*(2^n - 1)+1) - 1)/2).
Cf. A328129.

Select[Range[0,15],OddQ[Sqrt[8(2^#-1)+1]]&] (* Harvey P. Dale, Dec 13 2024 *)

is(n)=issquare(8<Charles R Greathouse IV, Sep 07 2012

Four cross-references to the Ramanujan-Nagell problem added by Raphie Frank, Sep 10 2012

A227078 The Ramanujan-Nagell squares: A038198(n)^2.

Original entry on oeis.org

1, 9, 25, 121, 32761

Offset: 0

Raphie Frank, Jun 30 2013

a(n) = (2*x - 1)^2 = (sqrt(2)*sqrt(sqrt(6*y^2 - 5) + 1) - 1)^2 = 2^(z + 3) - 7 for x, y, z are the solutions to two Diophantine equations noted by R. K. Guy: 2*x^2*(x^2 - 1) = 3*(y^2 - 1) & x*(x - 1)/2 = 2^z - 1 (see A180445). x = {1, 2, 3, 6, 91} = A180445(n), y = {1, 3, 7, 29, 6761} = A227077(n), and z = {0, 1, 2, 4, 12} = A215795(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.

Curtis Bright, Solving Ramanujan's Square Equation Computationally
Eric Weisstein's World of Mathematics, Ramanujan's Square Equation

Cf. A060728, A076046, A180445, A227077, A215795.

a(n) + 7 = 2^A060728(n).

(a(n) - 1)/8 = A076046(n).

A180445 All positive solutions, x, for each of two Diophantine equations noted by Richard K. Guy.

Original entry on oeis.org

1, 2, 3, 6, 91

Offset: 1

Jonathan Vos Post, Sep 05 2010

2*(x^2)*((x^2)-1) = 3*((y^2)-1) has only these five positive solutions.
x*(x-1)/2 = (2^z)-1 has only these five positive solutions.
Richard K. Guy notes, as Example 29: "True, but why the coincidence?"
Algebraically, y solutions = {1, 3, 7, 29, 6761} can be derived from x solutions as follows: y = sqrt(((2*x^2 - 1)^2 + 5)/6). From this relationship it becomes clear that the form (((2*x^2 - 1)^2 + 5)/6) can only be an integer square for x is in {1, 2, 3, 6, 91}. Thus, x and y solutions are also unique integer solutions to the following equivalency: (2x^2 - 1)^2  = 6y^2 - 5. From this relationship the following statement naturally follows: ((sqrt(6*y^2 - 5) + 1)/2 - sqrt((sqrt(6*(y^2) - 5) + 1)/2))/2 = (2^z - 1) = {0, 1, 3, 15, 4095} = A076046(n), the Ramanujan-Nagell triangular numbers; z = {0, 1, 2, 4, 12} = (A060728(n) - 3). - Raphie Frank, Jun 26 2013

R. K. Guy, editor, Western Number Theory Problems, 1985-12-21 & 23, Typescript, Jul 13 1986, Dept. of Math. and Stat., Univ. Calgary, 11 pages. Annotated scan of pages 1, 3, 7, 9, with permission. See Problem 85:08.
Richard K. Guy, The Strong Law of Small Numbers (example #29).
R. K. Guy, The strong law of small numbers. Amer. Math. Monthly 95 (1988), no. 8, 697-712. [Annotated scanned copy]

Cf. A076046, A060728.

x = sqrt((sqrt(6*(y^2) - 5) + 1)/2) = (sqrt(2^(z + 3) - 7) + 1)/2; y = {1, 3, 7, 29, 6761} and z = (A060728(n) - 3) = A215795(n) = {0, 1, 2, 4, 12}. - Raphie Frank, Jun 23 2013

A247024 Primes of the form x*(2^x - x) - 1.

Original entry on oeis.org

3, 47, 347, 10139, 396527668833598369303619003, 1624796301562061610805093487, 2637188343637273091841153207596203623407

Offset: 1

R. J. Cano, Sep 09 2014

nonn

Subsequence of A245014. a(n) represents the identity between (p + 4k^2 + 1) and (2n*4^k) for the least prime p defining A245014(k). Those first k where it occurs are: 1,2,3,5,41,42,62,183,357,407.
Consider the table of Stirling numbers of the second kind (A008277). The second column contains the numbers A000225, or 2^m - 1, and the first subdiagonal contains the triangular numbers. If a number appears in both sequences, we have the equation f(x) = 2^x - x^2 + x - 2 = 0 which has integer roots x = 1, 2, 3. Set g(x) = (x - 1)*(2^x - 1). Then it is found that the sum f(x) + g(x) for some even x defines this sequence and satisfies in common with A245014: Both sequences have three consecutive terms (those first) such that when they are represented in decimal the third term is the concatenation of the two terms preceding it.
Prime or PRP for x = 2, 4, 6, 10, 82, 84, 124, 366, 714, 814, 1584, 8938, 17812, 27054, 35380, 71358. - Jens Kruse Andersen, Sep 10 2014
The complete solution to the remark on Stirling2 numbers in a comment above is given in A076046. See also my Oct 08 2014 remark in the history. - Wolfdieter Lang, Oct 16 2014

R. J. Cano, Table of n, a(n) for n = 1..11

Cf. A000225, A008277, A245014.

[a: x in [0..450] | IsPrime(a) where a is x*(2^x - x)-1]; // Vincenzo Librandi, Oct 16 2014

A247024:=n->`if`(isprime(n*(2^n-n)-1),n*(2^n-n)-1,NULL): seq(A247024(n), n=1..10^2); # Wesley Ivan Hurt, Sep 13 2014

Select[Table[n(2^n - n) - 1, {n, 200}], PrimeQ] (* Alonso del Arte, Sep 09 2014 *)

listToVec(l:list,Comp,Omega=#l)=vector(Omega,i,l[i][Comp]);
fplusg(x)=x*(2^x-x)-1;
find_next_N_terms(N,resumeFrom:list=List(),verbose=0)= {
  my(s:list,q,j,F,N0);
  s=resumeFrom;
  q=#s;
  N0=q;
  if(q,j=s[q][1]);
  while(q

ABC2 $a*(2^$a-$a)-1
a: from 2 to 100000 step 2
Charles R Greathouse IV, Sep 18 2014

Edited: 'penultimate column' replaced by 'first subdiagonal'. - Wolfdieter Lang, Oct 16 2014

A211202 Positive numbers n such that Lambda_n = A002336(n) is divisible by n.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 12, 15, 16, 18, 20, 21, 22, 23, 24

Offset: 1

Raphie Frank, Feb 18 2013

nonn

Observations:
For all n in this sequence to n = 24, then y = Lambda_n/n follows form: y = (x^2 + x^k) - (floor[z^2/4]) or y = (x^2 + x^k) + (floor[z^2/4]); k = 1 or 2 and z = 0, 1, 3, 6 or 7. y (= A222786) gives the average number of spheres/dimension of the laminated lattice Kissing numbers in A222785.
e.g. Where T_x is the x-th triangular number = (1/2*(x^2 + x)), 2*T_x is the x-th pronic number = (x^2 + x) = floor[(2*x + 1)^2/4], and S_x is the x-th square = (x^2) = floor[(2*x)^2/4]:
For k = 1, z = 0 or 1, then n = {1, 4, 6, 8, 15, 20, 24}, x = {1, 2, 3, 5, 12, 29, 90}, and y = 2*T_x = {2, 6, 12, 30, 156, 870, 8190}.
For k = 2, z = 0 or 1, then n = {1, 5, 7, 23}, x = {1, 2, 3, 45}, and y = 2*T_x + 2*T_(-x) = 2*S_x = {2, 8, 18, 4050}.
For k = 1, z = 3, then n = {3, 7, 12, 16}, x = {2, 4, 7, 16}, and y = 2*T_x - 2*T_1 = {4, 18, 54, 270}.
For k = 1, z = 6, then n = {2, 18}, x = {3, 20}, and y = 2*T_x - S_3 = {3, 411}.
For k = 1, z = 7, then n = {5, 7, 8, 21}, x = {4, 5, 6, 36}, and y = 2*T_x - 2*T_3 = {8, 18, 30, 1320}.
For k = 1, z = 7, then n = {6, 7, 12, 22}, x = {0, 2, 6, 47}, and y = 2*T_x + 2*T_3 = {12, 18, 54, 2268}.
For the special case where k = 1 and z = 0 or 1, then all associated x values follow form (A216162(n) - A216162(n - 2)) [type 1] or (A216162(n) - A216162(n - 1)) [type II] for some n in N. Type II x values = {1, 2, 5, 90} (= A215797(n+1)) are associated with the positive Ramanujan-Nagell triangular numbers = {1, 3, 15, 4095} (= A076046(n+1)) by the formula 1/2*(x^2 + x) = T_x.

			Lambda_6/6 = 72/6 = 12, so 6 is in this sequence.
Lambda_12/12 = 648/12 = 54, so 12 is in this sequence.
Lambda_18/18 = 7398/18 = 411, so 18 is in this sequence.
Lambda_24/24 = 196560/24 = 8190, so 24 is in this sequence.
But...
Lambda_19/19 = 10668/19 = 561.47368..., so 19 is not in this sequence.

Cf. A222785, A222786, A002336, A216162

A218129 2^(((c - 2)^2 + (c - 2))/2) + n = a(n), where c are the positive solutions to {y in N | 2cos(2Pi/y) is in Z}; c = {1,2,3,4,6}.

Original entry on oeis.org

1, 2, 4, 11, 1028

Offset: 0

Raphie Frank, Oct 21 2012

The set {c} consists of the complete set of positive solutions to the short proof of the Crystallographic Restriction Theorem {1, 2, 3, 4, 6} (see A217290).
Let {V} = prime(a(n)) = {2, 3, 7, 31, 8191}. Then all elements of {V} follow form x^2 + x + 1 for some x in R; x = {(sqrt(5) - 1)/2, 1, 2, 5, 90}. (V + V(mod 2) - 2)/2 gives the complete set of Ramanujan-Nagell triangular numbers (A076046) = {0, 1, 3, 15, 4095} == (2^F_(c + 1) - 2)/2 (see A215929); F_n the n-th Fibonacci number (A000045).
Additionally, 2*V - 1 = {3, 5, 13, 61, 16381} is prime and, therefore, all elements of {V} are links in a Cunningham chain of the 2nd kind (see A005382).

			2^(((1 - 2)^2 + (1 - 2))/2) + 0 = 2^(a(-1) - 1) + 0 = 1 = a(0).
2^(((2 - 2)^2 + (2 - 2))/2) + 1 = 2^(a(0) - 1) + 1 = 2 = a(1).
2^(((3 - 2)^2 + (3 - 2))/2) + 2 = 2^(a(1) - 1) + 2 = 4 = a(2).
2^(((4 - 2)^2 + (4 - 2))/2) + 3 = 2^(a(2) - 1) + 3 = 11 = a(3).
2^(((6 - 2)^2 + (6 - 2))/2) + 4 = 2^(a(3) - 1) + 4 = 1028 = a(4).

Eric W. Weisstein, MathWorld: Ramanujan's Square Equation
Wikipedia, Crystallographic Restriction Theorem

Cf. A217290, A215929, A076046.

to n = 4, then a(n) = 2^(a(n - 1) - 1) + n; a(-1) = 1.

cp's OEIS Frontend

A060728 Numbers n such that Ramanujan's equation x^2 + 7 = 2^n has an integer solution.

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A002249 a(n) = a(n-1) - 2*a(n-2) with a(0) = 2, a(1) = 1.

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A215795 Numbers n such that 2^n-1 is a triangular number (A000217).

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A227078 The Ramanujan-Nagell squares: A038198(n)^2.

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A180445 All positive solutions, x, for each of two Diophantine equations noted by Richard K. Guy.

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A247024 Primes of the form x*(2^x - x) - 1.

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Magma

Maple

Mathematica

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Extensions

A211202 Positive numbers n such that Lambda_n = A002336(n) is divisible by n.

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A218129 2^(((c - 2)^2 + (c - 2))/2) + n = a(n), where c are the positive solutions to {y in N | 2cos(2Pi/y) is in Z}; c = {1,2,3,4,6}.