A001541 -id:A001541 - cp's OEIS frontend

A090965 a(n) = 8a(n-1) - 4a(n-2), where a(0) = 1, a(1) = 4.

1, 4, 28, 208, 1552, 11584, 86464, 645376, 4817152, 35955712, 268377088, 2003193856, 14952042496, 111603564544, 833020346368, 6217748512768, 46409906716672, 346408259682304, 2585626450591744, 19299378566004736

Offset: 0

Philippe Deléham, Feb 29 2004

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (8,-4).

Cf. A001075.

Sum_{k>=0} A086645(n,k)*m^k for m = 0, 1, 2, 4 gives A000007, A081294, A001541, A083884.

a:=[1,4];; for n in [3..20] do a[n]:=8*a[n-1]-4*a[n-2]; od; a; # G. C. Greubel, Feb 03 2019

m:=20; R:=PowerSeriesRing(Integers(), m); Coefficients(R!( (1-4*x)/(1-8*x+4*x^2) )); // G. C. Greubel, Feb 03 2019

LinearRecurrence[{8,-4}, {1,4}, 20] (* G. C. Greubel, Feb 03 2019 *)

my(x='x+O('x^20)); Vec((1-4*x)/(1-8*x+4*x^2)) \\ G. C. Greubel, Feb 03 2019

[lucas_number2(n,8,4)/2 for n in range(0,21)] # Zerinvary Lajos, Jul 08 2008

a(n) = Sum_{k>=0} binomial(2*n, 2*k)*3^k = Sum_{k>=0} A086645(n, k)*3^k.
a(n) = 2^n*A001075(n).
G.f.: (1-4*x)/(1-8*x+4*x^2). - Philippe Deléham, Sep 07 2009
G.f.: G(0)/2, where G(k)= 1 + 1/(1 - x*(3*k-4)/(x*(3*k-1) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 28 2013
From Peter Bala, Feb 19 2022: (Start)
a(n) = Sum_{k = 0..floor(n/2)} 4^(n-2*k)*12^k*binomial(n,2*k).
a(n) = [x^n] (4*x + sqrt(1 + 12*x^2))^n.
G.f.: A(x) = 1 + x*B'(x)/B(x), where B(x) = 1/sqrt(1 - 8*x + 4*x^2) is the g.f. of A069835.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p and positive integers n and k. (End)

Corrected by T. D. Noe, Nov 07 2006

A123335 a(n) = -2*a(n-1) + a(n-2) for n>1, a(0)=1, a(1)=-1.

Original entry on oeis.org

1, -1, 3, -7, 17, -41, 99, -239, 577, -1393, 3363, -8119, 19601, -47321, 114243, -275807, 665857, -1607521, 3880899, -9369319, 22619537, -54608393, 131836323, -318281039, 768398401, -1855077841, 4478554083, -10812186007, 26102926097, -63018038201, 152139002499

Offset: 0

Philippe Deléham, Jun 27 2007

Inverse binomial transform of A077957.
The inverse of the g.f. is 3-x-2/(1+x) which generates 1, 1, -2, +2, -2, +2, ... (-2, +2 periodically continued). - Gary W. Adamson, Jan 10 2011
Pisano period lengths:  1, 1, 8, 4, 12, 8, 6, 4, 24, 12, 24, 8, 28, 6, 24, 8, 16, 24, 40, 12, ... - R. J. Mathar, Aug 10 2012
a(n) is the rational part of the Q(sqrt(2)) integer (sqrt(2) - 1)^n = a(n) + A077985(n-1)*sqrt(2), with A077985(-1) = 0. - Wolfdieter Lang, Dec 07 2014
3^n*a(n) = A251732(n) gives the rational part of the integer in Q(sqrt(2)) giving the length of a variant of Lévy's C-curve at iteration step n. - Wolfdieter Lang, Dec 07 2014
Define u(0) = 1/0, u(1) = -1/1, and u(n) = -(8 + 3*u(n-1)*u(n-2))/(3*u(n-1) + 2*u(n-2)) for n>1. Then u(n) = a(n)/A000219(n). - Michael Somos, Apr 19 2022

			G.f. = 1 - x + 3*x^2 - 7*x^3 + 17*x^4 - 41*x^5 + 99*x^6 + ... - _Michael Somos_, Apr 19 2022

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (-2,1).

Cf. A000129, A001333, A077985, A251732, A001541 (bisection), A002315 (bisection).

[Round(1/2*((-1-Sqrt(2))^n+(-1+Sqrt(2))^n)): n in [0..30]]; // G. C. Greubel, Oct 12 2017

a:= n-> (M-> M[2, 1]+M[2, 2])(<<2|1>, <1|0>>^(-n)):
seq(a(n), n=0..33);  # Alois P. Heinz, Jun 22 2021

LinearRecurrence[{-2,1},{1,-1},40] (* Harvey P. Dale, Nov 03 2011 *)

x='x+O('x^50); Vec((1+x)/(1+2*x-x^2)) \\ G. C. Greubel, Oct 12 2017

{a(n) = real((-1 + quadgen(8))^n)}; /* Michael Somos, Apr 19 2022 */

a(n) = (-1)^n*A001333(n).
G.f.: (1+x)/(1+2*x-x^2).
a(n) = A077985(n) + A077985(n-1). - R. J. Mathar, Mar 28 2011
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x*(2*k-1)/(x*(2*k+1) + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 19 2013
G.f.: 1/(1 + x/(1 + 2*x/(1 - x))). - Michael Somos, Apr 19 2022
E.g.f.: exp(-x)*cosh(sqrt(2)*x). - Stefano Spezia, Feb 01 2023

Corrected by N. J. A. Sloane, Oct 05 2008

A050795 Numbers n such that n^2 - 1 is expressible as the sum of two nonzero squares in at least one way.

Original entry on oeis.org

3, 9, 17, 19, 33, 35, 51, 73, 81, 99, 105, 129, 145, 147, 161, 163, 179, 195, 201, 233, 243, 273, 289, 291, 297, 339, 361, 387, 393, 451, 465, 467, 483, 489, 513, 521, 577, 579, 585, 611, 627, 649, 675, 721, 723, 739, 777, 801, 809, 819, 849, 883, 899, 915

Offset: 1

Patrick De Geest, Sep 15 1999

nonn

Analogous solutions exist for the sum of two identical squares z^2-1 = 2.r^2 (e.g. 99^2-1 = 2.70^2). Values of 'z' are the terms in sequence A001541, values of 'r' are the terms in sequence A001542.
Looking at a^2 + b^2 = c^2 - 1 modulo 4, we must have a and b even and c odd. Taking a = 2u, b = 2v and c = 2w - 1 and simplifying, we get u^2 + v^2 = w(w+1). - Franklin T. Adams-Watters, May 19 2008
If n is in this sequence, then so is n^(2^k), for all k >= 0. - Altug Alkan, Apr 13 2016

			E.g. 51^2 - 1 = 10^2 + 50^2 = 22^2 + 46^2 = 34^2 + 38^2.

Chai Wah Wu, Table of n, a(n) for n = 1..10000
E.-B. Escott, Query 2521, L'Intermédiaire des Mathématiciens, 10 (1903), 285. [Contains errors]
Index entries for sequences related to sums of squares

Cf. A050796, A050797, A001541, A001542, A001333.

Cf. A140612, A002378.

t={}; Do[i=c=1; While[iJayanta Basu, Jun 01 2013 *)
Select[Range@ 1000, Length[PowersRepresentations[#^2 - 1, 2, 2] /. {0, } -> Nothing] > 0 &] (* _Michael De Vlieger, Apr 13 2016 *)

select( {is_A050795(n)=#qfbsolve(Qfb(1,0,1),n^2-1,2)}, [1..999]) \\ M. F. Hasler, Mar 07 2022

from itertools import islice, count
from sympy import factorint
def A050795_gen(startvalue=2): # generator of terms >= startvalue
    for k in count(max(startvalue,2)):
        if all(map(lambda d: d[0] % 4 != 3 or d[1] % 2 == 0, factorint(k**2-1).items())):
            yield k
A050795_list = list(islice(A050795_gen(),20)) # Chai Wah Wu, Mar 07 2022

a(n) = 2*A140612(n) + 1. - Franklin T. Adams-Watters, May 19 2008

{k : A025426(k^2-1)>0}. - R. J. Mathar, Mar 07 2022

A204517 Square root of floor[A055859(n)/7].

Original entry on oeis.org

0, 0, 0, 1, 3, 6, 17, 48, 96, 271, 765, 1530, 4319, 12192, 24384, 68833, 194307, 388614, 1097009, 3096720, 6193440, 17483311, 49353213, 98706426, 278635967, 786554688, 1573109376, 4440692161, 12535521795, 25071043590, 70772438609, 199781794032, 399563588064

Offset: 1

M. F. Hasler, Jan 15 2012

nonn

M. F. Hasler, Truncated squares, OEIS wiki, Jan 16 2012
Index to sequences related to truncating digits of squares.

See also A031149=sqrt(A023110) (base 10), A204502=sqrt(A204503) (base 9), A204514=sqrt(A055872) (base 8), A204516=sqrt(A055859) (base 7), A204518=sqrt(A055851) (base 6), A204520=sqrt(A055812) (base 5), A004275=sqrt(A055808) (base 4), A001075=sqrt(A055793) (base 3), A001541=sqrt(A055792) (base 2).

b=7;for(n=1,2e9,issquare(n^2\b) & print1(sqrtint(n^2\b),","))

A204517(n)=polcoeff((x^4 + 3*x^5 + 6*x^6 + x^7)/(1 - 16*x^3 + x^6+O(x^n)),n)

A204517(n) = sqrt(floor(A204516(n)^2/7)).

G.f. = (x^4 + 3*x^5 + 6*x^6 + x^7)/(1 - 16*x^3 + x^6)

A322836 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where A(n,k) is Chebyshev polynomial of the first kind T_{n}(x), evaluated at x=k.

Original entry on oeis.org

1, 1, 0, 1, 1, -1, 1, 2, 1, 0, 1, 3, 7, 1, 1, 1, 4, 17, 26, 1, 0, 1, 5, 31, 99, 97, 1, -1, 1, 6, 49, 244, 577, 362, 1, 0, 1, 7, 71, 485, 1921, 3363, 1351, 1, 1, 1, 8, 97, 846, 4801, 15124, 19601, 5042, 1, 0, 1, 9, 127, 1351, 10081, 47525, 119071, 114243, 18817, 1, -1

Offset: 0

Seiichi Manyama, Dec 28 2018

			Square array begins:
   1, 1,    1,     1,      1,      1,       1, ...
   0, 1,    2,     3,      4,      5,       6, ...
  -1, 1,    7,    17,     31,     49,      71, ...
   0, 1,   26,    99,    244,    485,     846, ...
   1, 1,   97,   577,   1921,   4801,   10081, ...
   0, 1,  362,  3363,  15124,  47525,  120126, ...
  -1, 1, 1351, 19601, 119071, 470449, 1431431, ...

Seiichi Manyama, Antidiagonals n = 0..139, flattened
Wikipedia, Chebyshev polynomials.
Index entries for sequences related to Chebyshev polynomials.

Mirror of A101124.
Columns 0-20 give A056594, A000012, A001075, A001541, A001091, A001079, A023038, A011943(n+1), A001081, A023039, A001085, A077422, A077424, A097308, A097310, A068203, A322888, A056771, A322889, A078986, A322890.
Rows 0-10 give A000012, A001477, A056220, A144129, A144130, A243131, A243132, A243133, A243134, A243135, A243136.
Main diagonal gives A115066.
Cf. A323182 (Chebyshev polynomial of the second kind).

Table[ChebyshevT[n-k, k], {n, 0, 10}, {k, n, 0, -1}] // Flatten (* Amiram Eldar, Dec 28 2018 *)

T(n,k) = polchebyshev(n,1,k);
matrix(7, 7, n, k, T(n-1,k-1)) \\ Michel Marcus, Dec 28 2018

T(n, k) = round(cos(n*acos(k)));\\ Seiichi Manyama, Mar 05 2021

T(n, k) = if(n==0, 1, n*sum(j=0, n, (2*k-2)^j*binomial(n+j, 2*j)/(n+j))); \\ Seiichi Manyama, Mar 05 2021

A(0,k) = 1, A(1,k) = k and A(n,k) = 2 * k * A(n-1,k) - A(n-2,k) for n > 1.
A(n,k) = cos(n*arccos(k)). - Seiichi Manyama, Mar 05 2021
A(n,k) = n * Sum_{j=0..n} (2*k-2)^j * binomial(n+j,2*j)/(n+j) for n > 0. - Seiichi Manyama, Mar 05 2021

A075114 Perfect powers n such that 2n + 1 is a perfect power; the value of y^b in the solution of the Diophantine equation x^a - 2y^b = 1.

Original entry on oeis.org

4, 121, 144, 4900, 166464, 5654884, 192099600, 6525731524, 221682772224, 7530688524100, 255821727047184, 8690408031080164, 295218051329678400, 10028723337177985444, 340681375412721826704

Offset: 1

Zak Seidov, Oct 11 2002

Note that the first ten numbers in this sequence are all squares. Except for 121, these squares are the y^2 in the Pell equation x^2 - 2y^2 = 1, whose solutions (x,y) are in sequences A001541 and A001542. The equation x^a - 2y^b = 1 is very similar to Catalan's equation x^a - y^b = 1, which has only one solution. Bennett shows that the equation x^2 - 2y^b = 1 has no solutions for b>2. Hence all the terms in this sequence are squares and solutions other than the Pell solutions must satisfy x^a - 2y^2 = 1 for a>2. The one known solution is 3^5 - 2*11^2 = 1. Are there any others? - T. D. Noe, Mar 29 2006

M. A. Bennett, Products of Consecutive Integers, Bull. London Math. Soc. 36 (2004), 683-694

Cf. A001597.

Cf. A117547 (square root of terms).

pp = Select[ Range[10^8], Apply[ GCD, Last[ Transpose[ FactorInteger[ # ]]]] > 1 & ]; Select[pp, Apply[GCD, Last[ Transpose[ FactorInteger[2# + 1]]]] > 1 & ]
lim=10^14; lst={}; k=2; While[n=Floor[lim^(1/k)]; n>1, lst=Join[lst,Range[2,n]^k]; k++ ]; lst=Union[lst]; Intersection[lst,(lst-1)/2] (*T. D. Noe, Mar 29 2006 *)

Empirical G.f.: x*(117*x^4-4091*x^3+3951*x^2+19*x-4) / ((x-1)*(x^2-34*x+1)). - Colin Barker, Dec 21 2012

Extended by Robert G. Wilson v, Oct 15 2002
More terms from T. D. Noe, Mar 29 2006
More terms from T. D. Noe, Nov 19 2006

A348692 Triangle whose n-th row lists the integers m such that A000178(n) / m! is a square, where A000178(n) = n$ = 1!2!...*n! is the superfactorial of n; if there is no such m, then n-th row = 0.

Original entry on oeis.org

1, 2, 0, 2, 0, 0, 0, 3, 4, 0, 0, 0, 6, 0, 8, 9, 0, 8, 9, 0, 7, 0, 10, 0, 0, 0, 12, 0, 0, 0, 14, 0, 0, 0, 15, 16, 0, 18, 0, 18, 0, 0, 0, 20, 0, 0, 0, 22, 0, 0, 0, 24, 25, 0, 0, 0, 26, 0, 0, 0, 28, 0, 0, 0, 30, 0, 32, 0, 32, 0, 0, 0, 34, 0, 0, 0, 35, 36, 0, 0, 0, 38, 0, 0, 0, 40

Offset: 1

Bernard Schott, Oct 30 2021

This sequence is the generalization of a problem proposed during the 17th Tournament of Towns (Spring 1996) and also during the first stage of the Moscow Mathematical Olympiad (1995-1996); the problem asked the question for n = 100 (see Andreescu-Gelca reference, Norman Do link, and Examples section).
Exhaustive results coming from Mabry-McCormick's link and adapted for OEIS:
-> n$ (A000178) is never a square if n > 1.
-> There is no solution if n is odd > 1, hence row(2q+1) = 0 when q > 0.
-> When n is even and there is a solution, then m belongs to {n/2 - 2, n/2 - 1, n/2, n/2 + 1, n/2 + 2}.
-> If 4 divides n (A008536), m = n/2 is always a solution because
    (n$) / (n/2)! = ( 2^(n/4) * Product_{j=1..n/2} ((2j-1)!) )^2.
-> For other cases, see Formula section.
-> When n is even, there are 0, 1 or 2 solutions, so, the maximal length of a row is 2.
-> It is not possible to get more than three consecutive 0 terms, and three consecutive 0 terms correspond to three consecutive rows such that (n, n+1, n+2) = (4u+1, 4u+2, 4u+3) for some u >= 1.

			For n = 4, 4$ / 3! = 48, 4$ / 4! = 12 but 4$ / 2! = 12^2, hence, m = 2.
For n = 8, 8$ / 2! is not a square, but m_1 = 3 because 8$ / 3! = 29030400^2 and m_2 = 4 because 8$ / 4! = 14515200^2.
For n = 14, m_1 = 8 because 14$ / 8! = 1309248519599593818685440000000^2 and m_2 = 9 because 14$ / 9! = 436416173199864606228480000000^2.
For n = 16, m_1 = 8 because 16$ / 8! = 6848282921689337839624757371207680000000000^2 and m_2 = 9 because 16$ / 9! = 2282760973896445946541585790402560000000000^2.
For n = 18, m = 7 because 18$ / 7! = 29230177671473293820176594405114531928195727360000000000000^2 and there is no other solution.
For n = 100, m = 50, unique solution to the Olympiad problems.
Triangle begins:
    1;
    2;
    0;
    2;
    0;
    0;
    0;
    8,  9;
    0;
    ...

Titu Andreescu and Rǎzvan Gelca, Putnam and Beyond, New York, Springer, 2007, problem 725, pp. 253 and 686.
Peter J. Taylor and A. M. Storozhev, Tournament of Towns 1993-1997, Book 4, Tournament 17, Spring 1996, O Level, Senior questions, Australian Mathematics Trust, 1998, problem 3, p. 96.

Diophante, A1963 - Le vilain petit canard (in French).
Norman Do, Factorial fun, Puzzle Corner 13, Gaz. Aust. Math. Soc. 36, 2009, 176-179, page 178.
Rick Mabry and Laura McCormick, Square products of punctured sequences of factorials, Gaz. Aust. Math. Soc., 2009, pages 346-352.
Tournament of Towns, Tournament 17, 1995-1996, Spring 1996, O Level, Senior questions, question 3 (in Russian).
Index to sequences related to Olympiads and other Mathematical competitions.

Cf. A000178, A001541, A008536, A035008, A060626, A139098.

sf(n)=prod(k=2, n, k!); \\ A000178
row(n) = my(s=sf(n)); Vec(select(issquare, vector(n, k, s/k!), 1));
lista(nn) = {my(list = List()); for (n=1, nn, my(r=row(n)); if (#r, for (k=1, #r, listput(list, r[k])), listput(list, 0));); Vec(list);} \\ Michel Marcus, Oct 30 2021

When there are two such integers m, then m_1 < m_2.
If n = 8*q^2 (A139098), then m_1 = n/2 - 1 = 4q^2-1 (see example for n=8).
If n = 8q*(q+1) (A035008), then m_2 = n/2 + 1 = (2q+1)^2 (see example for n=16).
if n = 4q^2 - 2 (A060626), then m_1 = n/2 + 1 = 2q^2 (see example for n=14).
If n = 2q^2, q>1 in A001541, then m = n/2 - 2 = q^2-2 (see example for n=18).
If n = 2q^2-4, q>1 in A001541, then m_2 = n/2 + 2 = q^2 (see example for n=14).

A182189 a(n) = 6*a(n-1) - a(n-2) - 4 with n > 1, a(0)=1, a(1)=3.

Original entry on oeis.org

1, 3, 13, 71, 409, 2379, 13861, 80783, 470833, 2744211, 15994429, 93222359, 543339721, 3166815963, 18457556053, 107578520351, 627013566049, 3654502875939, 21300003689581, 124145519261543, 723573111879673, 4217293152016491, 24580185800219269, 143263821649299119

Offset: 0

Kenneth J Ramsey, Apr 17 2012

If p is a prime of the form 8*n +- 1 then a(p) == 3 (mod p); if p is a prime of the form 8*n +- 3 then a(p) == -1 (mod p).

The terms a(n) > 1 satisfy a(n)^5 + b(n)^5 = c(n)^5 + d(n)^5 where b(n) = a(n) - 2, c(n) = (a(n)-1) + i*ceiling((a(n)-1)*sqrt(2)), and d(n) is the conjugate of c(n), where i is the imaginary unit. Note that Re(c(n)) is A001542(n) and Im(d(n)) is A001541(n). - Pedro Caceres, Dec 30 2017

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

Cf. A000129, A001541, A001542, A182190, A182431.

I:=[1,3]; [n le 2 select I[n] else 6*Self(n-1)-Self(n-2)-4: n in [1..41]]; // Bruno Berselli, Jun 07 2012

m = -11;n = -1; c = 0;
list3 = Reap[While[c < 22, t = 6 n - m - 4; Sow[t];m = n; n = t;c++]][[2,1]]
CoefficientList[Series[(1-4*x-x^2)/((1-x)*(1-6*x+x^2)),{x,0,40}],x] (* Vincenzo Librandi, Jul 26 2012 *)
1 + Fibonacci[2*Range[0, 40], 2] (* G. C. Greubel, May 24 2021 *)

my(x='x+O('x^40)); Vec((1-4*x-x^2)/((1-x)*(1-6*x+x^2))) \\ Altug Alkan, Dec 30 2017

[1 + lucas_number1(2*n,2,-1) for n in (0..40)] # G. C. Greubel, May 24 2021

G.f.: (1-4*x-x^2)/((1-x)*(1-6*x+x^2)). - Bruno Berselli, Jun 07 2012

a(n) = 1 + A000129(2*n). - G. C. Greubel, May 24 2021

A204512 Square roots of [A055872/8]: Their square written in base 8, with some digit appended, is again a square.

Original entry on oeis.org

0, 0, 0, 1, 2, 6, 12, 35, 70, 204, 408, 1189, 2378, 6930, 13860, 40391, 80782, 235416, 470832, 1372105, 2744210, 7997214, 15994428, 46611179, 93222358, 271669860, 543339720, 1583407981, 3166815962, 9228778026, 18457556052, 53789260175, 107578520350

Offset: 1

M. F. Hasler, Jan 15 2012

Base-8 analog of A031150. The square of the terms (= truncated squares A055872) are listed in A204504.

See also A031149=sqrt(A023110) (base 10), A204502=sqrt(A204503) (base 9), A204514=sqrt(A055872) (base 8), A204516=sqrt(A055859) (base 7), A204518=sqrt(A055851) (base 6), A204520=sqrt(A055812) (base 5), A004275=sqrt(A055808) (base 4), A001075=sqrt(A055793) (base 3), A001541=sqrt(A055792) (base 2).

CoefficientList[Series[(x^4 (1+2x))/(1-6x^2+x^4),{x,0,40}],x] (* Harvey P. Dale, Nov 30 2020 *)

b=8;for(n=1,1e7,issquare(n^2\b) & print1(sqrtint(n^2\b)","))

a(n)=polcoeff((2*x^5 + x^4)/(x^4 - 6*x^2 + 1+O(x^n)),n)

G.f. = x^4*(1 + 2*x)/(1 - 6*x^2 + x^4)

A237599 Positive integers k such that x^2 - 6xy + y^2 + k = 0 has integer solutions.

Original entry on oeis.org

4, 7, 8, 16, 23, 28, 31, 32, 36, 47, 56, 63, 64, 68, 71, 72, 79, 92, 100, 103, 112, 119, 124, 127, 128, 136, 144, 151, 164, 167, 175, 184, 188, 191, 196, 199, 200, 207, 223, 224, 239, 248, 252, 256, 263, 271, 272, 279, 284, 287, 288, 292, 311, 316, 324, 328

Offset: 1

Colin Barker, Feb 10 2014

nonn

Nonnegative numbers of the form 8x^2 - y^2. - Jon E. Schoenfield, Jun 03 2022

			4 is in the sequence because x^2 - 6xy + y^2 + 4 = 0 has integer solutions, for example (x, y) = (1, 5).

N. J. A. Sloane et al., Binary Quadratic Forms and OEIS (Index to related sequences, programs, references)

Cf. A001653 (k = 4), A006452 (k = 7), A001541 (k = 8), A075870 (k = 16), A156066 (k = 23), A217975 (k = 28), A003499 (k = 32), A075841 (k = 36), A077443 (k = 56).
Cf. A031363, A084917, A237351, A237606, A237609, A237610.
For primes see A007522 and A141175.
For a list of sequences giving numbers and/or primes represented by binary quadratic forms, see the "Binary Quadratic Forms and OEIS" link.

cp's OEIS Frontend

A090965 a(n) = 8*a(n-1) - 4*a(n-2), where a(0) = 1, a(1) = 4.

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A123335 a(n) = -2*a(n-1) + a(n-2) for n>1, a(0)=1, a(1)=-1.

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A050795 Numbers n such that n^2 - 1 is expressible as the sum of two nonzero squares in at least one way.

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A204517 Square root of floor[A055859(n)/7].

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A322836 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where A(n,k) is Chebyshev polynomial of the first kind T_{n}(x), evaluated at x=k.

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A075114 Perfect powers n such that 2n + 1 is a perfect power; the value of y^b in the solution of the Diophantine equation x^a - 2y^b = 1.

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A348692 Triangle whose n-th row lists the integers m such that A000178(n) / m! is a square, where A000178(n) = n$ = 1!*2!*...*n! is the superfactorial of n; if there is no such m, then n-th row = 0.

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A090965 a(n) = 8a(n-1) - 4a(n-2), where a(0) = 1, a(1) = 4.

A348692 Triangle whose n-th row lists the integers m such that A000178(n) / m! is a square, where A000178(n) = n$ = 1!2!...*n! is the superfactorial of n; if there is no such m, then n-th row = 0.