Displaying 41-50 of 115 results found.
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0, 1, 4, 7, 24, 41, 140, 239, 816, 1393, 4756, 8119, 27720, 47321, 161564, 275807, 941664, 1607521, 5488420, 9369319, 31988856, 54608393, 186444716, 318281039, 1086679440, 1855077841, 6333631924, 10812186007, 36915112104, 63018038201, 215157040700
COMMENTS
Also, numerators of the lower principal and intermediate convergents to 2^(1/2). The lower principal and intermediate convergents to 2^(1/2), beginning with 1/1, 4/3, 7/5, 24/17, 41/29, form a strictly increasing sequence; essentially, numerators= A143608 and denominators= A079496. Sequence a(n) such that a(2*n) = sqrt(2* A001108(2*n)) and a(2*n+1) = sqrt( A001108(2*n+1)). For n > 0, a(n) divides A******(k+1,n+1)-A******(k,n+1) where A****** is any one of A182431, A182439, A182440, A182441 and k is any nonnegative integer. If p is a prime of the form 8*r +/- 3 then a(p+1) == 0 (mod p); if p is a prime of the form 8*r +/- 1 then a(p-1) == 0 (mod p).
Numbers n such that sqrt(floor(n^2/2 + 1)) is an integer. The integer square roots are given by A079496. - Richard R. Forberg, Aug 01 2013 Define a binary operation o on the real numbers by x o y = x*sqrt(1 + y^2) + y*sqrt(1 + x^2). The operation o is commutative and associative with identity 0. Then we have
a(2*n + 1) = 1 o 1 o ... o 1 (2*n + 1 terms) and
a(2*n) = sqrt(2)*(1 o 1 o ... o 1) (2*n terms). Cf. A084068. This is a fourth-order divisibility sequence. Indeed, a(2*n) = sqrt(2)*U(2*n) and a(2*n+1) = U(2*n+1), where U(n) is the Lehmer sequence [Lehmer, 1930] defined by the recurrence U(n) = 2*sqrt(2)*U(n-1) - U(n-2) with U(0) = 0 and U(1) = 1. The solution to the recurrence is U(n) = (1/2)*( (sqrt(2) + 1)^n - (sqrt(2) - 1)^n ). (End)
REFERENCES
Serge Lang, Introduction to Diophantine Approximations, Addison-Wesley, New York, 1966.
FORMULA
a(2*n) = (a(2*n - 1) + a(2*n + 1))/2.
a(2*n + 1) = (a(2*n) + a(2*n + 2))/4.
a(n) = 6*a(n-2) - a(n-4).
G.f.: x*(1 + 4*x + x^2)/((1 + 2*x - x^2)*(1 - 2*x - x^2)) = x*(1 + 4*x + x^2)/(1 - 6*x^2 + x^4). (End)
a(n) = (((-2 - sqrt(2) + (-1)^n * (-2+sqrt(2))) * ((-1+sqrt(2))^n - (1+sqrt(2))^n)))/(4*sqrt(2)). - Colin Barker, Mar 27 2016
MAPLE
option remember;
if n <= 3 then
op(n+1, [0, 1, 4, 7]) ;
else
6*procname(n-2)-procname(n-4) ;
end if;
MATHEMATICA
a = -4; b = -1; Reap[While[b<2000000000, t = 4*b-a; Sow[t]; a=b; b=t; t = 2*b-a; Sow[t]; a=b; b=t]][[2, 1]]
CoefficientList[Series[x*(1 + 4*x + x^2)/(1 - 6*x^2 + x^4), {x, 0, 30}], x] (* Wesley Ivan Hurt, Aug 24 2014 *)
PROG
(PARI) a(n)=([0, 1, 0, 0; 0, 0, 1, 0; 0, 0, 0, 1; -1, 0, 6, 0]^n*[0; 1; 4; 7])[1, 1] \\ Charles R Greathouse IV, Jun 11 2015 (PARI) concat(0, Vec(x*(1+4*x+x^2)/((1+2*x-x^2)*(1-2*x-x^2)) + O(x^50))) \\ Colin Barker, Mar 27 2016 (Magma) I:=[0, 1, 4, 7]; [n le 4 select I[n] else 6*Self(n-2) - Self(n-4): n in [1..30]]; // G. C. Greubel, Mar 27 2018
a(n) = 6*a(n-1) - a(n-2), n >= 2, a(0)=1, a(1)=4.
+10
16
1, 4, 23, 134, 781, 4552, 26531, 154634, 901273, 5253004, 30616751, 178447502, 1040068261, 6061962064, 35331704123, 205928262674, 1200237871921, 6995498968852, 40772755941191, 237641036678294, 1385073464128573
COMMENTS
This sequence gives one half of all positive solutions y = y1 = a(n) of the first class of the Pell equation x^2 - 2*y^2 = -7. For the corresponding x=x1 terms see A054490(n). Therefore it also gives one fourth of all positive solutions x = x1 of the first class of the Pell equation x^2 - 2*y^2 = 14, with the y=y1 terms given by A054490. - Wolfdieter Lang, Feb 26 2015
REFERENCES
A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 122-125, 194-196.
LINKS
Seyed Hassan Alavi, Ashraf Daneshkhah, Cheryl E Praeger, Symmetries of biplanes, arXiv:2004.04535 [math.GR], 2020. See Lemma 7.9 p. 21.
FORMULA
a(n) = ((4+sqrt(2))/8)*(3+2*sqrt(2))^(n-1) + ((4-sqrt(2))/8)*(3-2*sqrt(2))^(n-1). - Antonio Alberto Olivares, Mar 29 2008 Sequence satisfies -7 = f(a(n), a(n+1)) where f(u, v) = u^2 + v^2 - 6*u*v. - Michael Somos, Sep 28 2008 G.f.: (1 - 2*x) / (1 - 6*x + x^2). a(n) = (7 + a(n-1)^2) / a(n-2). - Michael Somos, Sep 28 2008 a(n) = S(n,6) - 2*S(n-1, 6), n >= 0, with the Chebyshev polynomials S(n, x) ( A049310) with S(-1, x) = 0 evaluated at x = 6. S(n, 6) = A001109(n-1). See the g.f. and the Pell comment above. - Wolfdieter Lang, Feb 26 2015 E.g.f.: exp(3*x)*(4*cosh(2*sqrt(2)*x) + sqrt(2)*sinh(2*sqrt(2)*x))/4. - Stefano Spezia, Apr 30 2020
EXAMPLE
n = 2: A054490(2)^2 - 2*(2*a(2))^2 = 65^2 - 2*(2*23)^2 = -7,
MAPLE
a[0]:=1: a[1]:=4: for n from 2 to 26 do a[n]:=6*a[n-1]-a[n-2] od: seq(a[n], n=0..20); # Zerinvary Lajos, Jul 26 2006
MATHEMATICA
LinearRecurrence[{6, -1}, {1, 4}, 30] (* Harvey P. Dale, Aug 06 2020 *)
PROG
(PARI) {a(n) = real((3 + 2*quadgen(8))^n * (1 + quadgen(8) / 4))} /* Michael Somos, Sep 28 2008 */ (PARI) {a(n) = polchebyshev(n, 1, 3) + polchebyshev(n-1, 2, 3)} /* Michael Somos, Sep 28 2008 */
Bisection of Chebyshev sequence T(n,3) (odd part) with Diophantine property.
+10
15
1, 33, 1121, 38081, 1293633, 43945441, 1492851361, 50713000833, 1722749176961, 58522759015841, 1988051057361633, 67535213191279681, 2294209197446147521, 77935577499977736033, 2647515425801796877601
COMMENTS
(3*a(n))^2 - 2*(2*b(n))^2 = 1 with companion sequence b(n)= A046176(n+1), n>=0 (special solutions of Pell equation).
FORMULA
a(n) = 34*a(n-1) - a(n-2), a(-1)=1, a(0)=1.
a(n) = T(2*n+1, 3)/3 = S(n, 34) - S(n-1, 34), with S(n, x) := U(n, x/2), resp. T(n, x), Chebyshev's polynomials of the second, resp. first, kind. See A049310 and A053120. S(-1, x)=0, S(n, 34)= A029547(n), T(n, 3)= A001541(n). G.f.: (1-x)/(1-34*x+x^2).
a(n) = sqrt(8* A046176(n+1)^2 + 1)/3. a(n) = (k^n)+(k^(-n))-a(n-1) = A003499(2*n)-a(n-1), where k = (sqrt(2)+1)^4 = 17+12*sqrt(2) and a(0)=1. - Charles L. Hohn, Apr 05 2011
MATHEMATICA
a[c_, n_] := Module[{},
p := Length[ContinuedFraction[ Sqrt[ c]][[2]]];
d := Denominator[Convergents[Sqrt[c], n p]];
t := Table[d[[1 + i]], {i, 0, Length[d] - 1, p}];
Return[t];
PROG
(Magma) I:=[1, 33]; [n le 2 select I[n] else 34*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Nov 22 2011 (Maxima) makelist(expand(((1+sqrt(2))^(4*n+2)+(1-sqrt(2))^(4*n+2))/6), n, 0, 14); /* _Bruno Berselli, Nov 22 2011 */
CROSSREFS
Cf. similar sequences listed in A238379. Similar sequences of the type cosh((2*n+1)*arccosh(k))/k are listed in A302329. This is the case k=3.
a(n) = 4*a(n-1) + 4*a(n-2), a(0)=1, a(1)=2.
+10
14
1, 2, 12, 56, 272, 1312, 6336, 30592, 147712, 713216, 3443712, 16627712, 80285696, 387653632, 1871757312, 9037643776, 43637604352, 210700992512, 1017354387456, 4912221519872, 23718303629312, 114522100596736, 552961616904192, 2669934870003712
COMMENTS
Original name was: Generalized Fibonacci sequence.
FORMULA
G.f.: (1-2*x)/(1-4*x-4*x^2).
a(n) = 4*a(n-1) + 4*a(n-2), a(0)=1, a(1)=2.
a(n) = (2 + 2*sqrt(2))^n/2 + (2 - 2*sqrt(2))^n/2.
E.g.f.: exp(2*x)*cosh(2*x*sqrt(2)).
G.f.: G(0)/2, where G(k)= 1 + 1/(1 - x*(4*k-2)/(x*(4*k+2) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 27 2013
MAPLE
a:=proc(n) option remember; if n=0 then 1 elif n=1 then 2 else
a := n -> (2*I)^n*ChebyshevT(n, -I):
MATHEMATICA
LinearRecurrence[{4, 4}, {1, 2}, 30] (* Harvey P. Dale, Mar 01 2018 *)
PROG
(PARI) a(n)=if(n<0, 0, polsym(4+4*x-x^2, n)[n+1]/2)
(Sage) [lucas_number2(n, 4, -4)/2 for n in range(0, 23)] # Zerinvary Lajos, May 14 2009 (Magma) [2^(n-1)*Evaluate(DicksonFirst(n, -1), 2): n in [0..40]]; // G. C. Greubel, Oct 13 2022
a(n) = a(-n) = 34*a(n-1) - a(n-2), and a(0)=1, a(1)=17.
+10
13
1, 17, 577, 19601, 665857, 22619537, 768398401, 26102926097, 886731088897, 30122754096401, 1023286908188737, 34761632124320657, 1180872205318713601, 40114893348711941777, 1362725501650887306817, 46292552162781456490001
COMMENTS
This sequence {a(n)} gives all the nonnegative integer solutions of the Pell equation a(n)^2 - 32*(3* A091761(n))^2 = +1. - Wolfdieter Lang, Mar 09 2019
FORMULA
a(n) = (r^n + 1/r^n)/2 with r = 17 + sqrt(17^2-1).
a(n) = T(n, 17) = T(2*n, 3) with T(n, x) Chebyshev's polynomials of the first kind. See A053120. T(n, 3)= A001541(n). G.f.: (1-17*x)/(1-34*x+x^2).
G.f.: (1 - 17*x / (1 - 288*x / (17 - x))). - Michael Somos, Apr 05 2019 a(n) = (a^n + b^n)/2 where a = 17 + 12*sqrt(2) and b = 17 - 12*sqrt(2); sqrt(a(n)-1)/4 = A001109(n). - James R. Buddenhagen, Dec 09 2011 a(n) = sqrt(1 + 32*9* A091761(n)^2), n >= 0. See one of the Pell comments above. - Wolfdieter Lang, Mar 09 2019
EXAMPLE
G.f. = 1 + 17*x + 577*x^2 + 19601*x^3 + 665857*x^4 + 22619537*x^5 + ...
PROG
(Sage) [lucas_number2(n, 34, 1)/2 for n in range(0, 15)] # Zerinvary Lajos, Jun 27 2008 (Magma) I:=[1, 17]; [n le 2 select I[n] else 34*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Dec 18 2011 (Maxima) makelist(expand(((17+sqrt(288))^n+(17-sqrt(288))^n))/2, n, 0, 15); /* Vincenzo Librandi, Dec 18 2011 */ (PARI) {a(n) = polchebyshev( n, 1, 17)}; /* Michael Somos, Apr 05 2019 */
Array of ((k^n)+(k^(-n)))/2 where k=(sqrt(x^2+1)+x)^2 for integers x>=1.
+10
13
1, 3, 1, 17, 9, 1, 99, 161, 19, 1, 577, 2889, 721, 33, 1, 3363, 51841, 27379, 2177, 51, 1, 19601, 930249, 1039681, 143649, 5201, 73, 1, 114243, 16692641, 39480499, 9478657, 530451, 10657, 99, 1, 665857, 299537289, 1499219281, 625447713, 54100801, 1555849, 19601, 129, 1
COMMENTS
Conjecture: Given function f(x, y)=(sqrt(x^2+y)+x)^2; and constant k=f(x, y); then for all integers x>=1 and y=[+-]1, k may be irrational, but ((k^n)+(k^(-n)))/2 always produces integer sequences; y=1 results shown here; y=-1 results are A188644. Also square array A(n,k), n >= 1, k >= 0, read by antidiagonals, where A(n,k) is Chebyshev polynomial of the first kind T_{k}(x), evaluated at x=2*n^2+1. - Seiichi Manyama, Jan 01 2019
FORMULA
A(n,k) = Sum_{j=0..k} binomial(2*k,2*j)*(n^2+1)^(k-j)*n^(2*j). - Seiichi Manyama, Jan 01 2019
EXAMPLE
Square array begins:
| 0 1 2 3 4
-----+---------------------------------------------
1 | 1, 3, 17, 99, 577, ...
2 | 1, 9, 161, 2889, 51841, ...
3 | 1, 19, 721, 27379, 1039681, ...
4 | 1, 33, 2177, 143649, 9478657, ...
5 | 1, 51, 5201, 530451, 54100801, ...
6 | 1, 73, 10657, 1555849, 227143297, ...
7 | 1, 99, 19601, 3880899, 768398401, ...
8 | 1, 129, 33281, 8586369, 2215249921, ...
9 | 1, 163, 53137, 17322499, 5647081537, ...
10 | 1, 201, 80801, 32481801, 13057603201, ...
11 | 1, 243, 118097, 57394899, 27893802817, ...
12 | 1, 289, 167041, 96549409, 55805391361, ...
13 | 1, 339, 229841, 155831859, 105653770561, ...
14 | 1, 393, 308897, 242792649, 190834713217, ...
15 | 1, 451, 406801, 366934051, 330974107201, ...
...
MATHEMATICA
max = 9; y = 1; t = Table[k = ((x^2 + y)^(1/2) + x)^2; ((k^n) + (k^(-n)))/2 // FullSimplify, {n, 0, max - 1}, {x, 1, max}]; Table[ t[[n - k + 1, k]], {n, 1, max}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jul 17 2013 *)
CROSSREFS
A188644 (f(x, y) as above with y=-1).
T(n,k)=Number of nXk arrays of occupancy after each element stays put or moves to some horizontal or antidiagonal neighbor, with no occupancy greater than 2
+10
13
1, 3, 1, 7, 17, 1, 17, 119, 99, 1, 41, 866, 2008, 577, 1, 99, 6328, 46105, 33873, 3363, 1, 239, 46211, 1010078, 2460824, 571358, 19601, 1, 577, 337274, 22181855, 162430981, 131347807, 9637322, 114243, 1, 1393, 2460918, 487335857, 10794156688
COMMENTS
Table starts
.1........3............7................17...................41
.1.......17..........119...............866.................6328
.1.......99.........2008.............46105..............1010078
.1......577........33873...........2460824............162430981
.1.....3363.......571358.........131347807..........26135748494
.1....19601......9637322........7010554450........4205264399891
.1...114243....162555929......374176705928......676612123702617
.1...665857...2741882383....19970973695923...108863121975329470
.1..3880899..46248186229..1065911268786396.17515379016294357354
.1.22619537.780082534460.56890876705125765
EXAMPLE
Some solutions for n=3 k=4
..1..2..0..1....2..1..2..0....2..1..0..2....2..1..1..1....2..2..1..2
..1..0..2..1....1..1..0..0....1..1..2..0....0..2..0..1....0..0..0..2
..1..1..2..0....1..2..1..1....1..0..2..0....1..1..0..2....2..0..1..0
a(n) = 6*a(n-1) - a(n-2), n >= 2, a(0)=1, a(1)=2.
+10
12
1, 2, 11, 64, 373, 2174, 12671, 73852, 430441, 2508794, 14622323, 85225144, 496728541, 2895146102, 16874148071, 98349742324, 573224305873, 3340996092914, 19472752251611, 113495517416752, 661500352248901
COMMENTS
The sequence {2*a(n+1)}_{n >= 0}, gives all positive solutions y = y2(n) = 2*a(n+1) of the second class of the Pell equation x^2 - 2*y^2 = -7. For the corresponding terms x = x2(n) see A255236(n). See A255236 for comments on the first class solutions and the relation to the Pell equation x^2 - 2*y^2 = 14. (End)
REFERENCES
A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 122-125, 194-196.
LINKS
Seyed Hassan Alavi, Ashraf Daneshkhah, Cheryl E Praeger, Symmetries of biplanes, arXiv:2004.04535 [math.GR], 2020. See x'(n) in Lemma 7.9 p. 21.
FORMULA
a(n) = 7*a(n-1) - 7*a(n-2) + a(n-3); a(n) = ((4-sqrt(2))/8)*(3+2*sqrt(2))^(n-1)+((4+sqrt(2))/8)*(3-2*sqrt(2))^(n-1). - Antonio Alberto Olivares, Mar 29 2008 Sequence satisfies -7 = f(a(n), a(n+1)) where f(u, v) = u^2 + v^2 - 6*u*v. - Michael Somos, Sep 28 2008 G.f.: (1 - 4*x) / (1 - 6*x + x^2). a(n) = (7 + a(n-1)^2) / a(n-2). - Michael Somos, Sep 28 2008 a(n) = S(n, 6) - 4*S(n-1, 6), n>=0, with the Chebyshev polynomials S(n, x) ( A049310), with S(-1, x) = 0, evaluated at x = 6. S(n, 6) = A001109(n-1). See the g.f. and the Pell equation comment above. a(n) = 6*a(n-1) - a(n-2), n >= 1, a(-1) = 4, a(0) = 1. (See the name.) (End)
a(n+1) = sqrt(( A255236(n)^2 + 7)/2)/2, n >= 0. E.g.f.: exp(3*x)*(4*cosh(2*sqrt(2)*x) - sqrt(2)*sinh(2*sqrt(2)*x))/4. - Stefano Spezia, May 01 2020
EXAMPLE
n = 2: a(3) = sqrt((181^2 + 7)/2)/2 = 64.
MAPLE
a[0]:=1: a[1]:=2: for n from 2 to 26 do a[n]:=6*a[n-1]-a[n-2] od: seq(a[n], n=0..20); # Zerinvary Lajos, Jul 26 2006
MATHEMATICA
Union[Flatten[NestList[{#[[2]], #[[3]], 6#[[3]]-#[[2]]}&, {1, 2, 11}, 25]]] (* Harvey P. Dale, Mar 04 2011 *) LinearRecurrence[{6, -1}, {1, 2}, 30] (* Harvey P. Dale, Jun 12 2017 *)
PROG
(PARI) {a(n) = real((3 + 2*quadgen(8))^n * (1 - quadgen(8) / 4))} /* Michael Somos, Sep 28 2008 */ (PARI) {a(n) = polchebyshev(n, 1, 3) - polchebyshev(n-1, 2, 3)} /* Michael Somos, Sep 28 2008 */
a(0) = a(1) = 1; thereafter a(2*n+1) = 2*a(2*n) - a(2*n-1), a(2*n) = 4*a(2*n-1) - a(2*n-2).
+10
12
1, 1, 3, 5, 17, 29, 99, 169, 577, 985, 3363, 5741, 19601, 33461, 114243, 195025, 665857, 1136689, 3880899, 6625109, 22619537, 38613965, 131836323, 225058681, 768398401, 1311738121, 4478554083, 7645370045, 26102926097, 44560482149
COMMENTS
a(1)=1, a(n) is the smallest integer > a(n-1) such that sqrt(2)*a(n) is closer and > to an integer than sqrt(2)*a(n-1) (i.e., a(n) is the smallest integer > a(n-1) such that frac(sqrt(2)*a(n)) < frac(sqrt(2)*a(n-1)).
n such that floor(sqrt(2)*n^2) = n*floor(sqrt(2)*n).
The sequence 1,1,3,5,17,... has g.f. (1+x-3x^2-x^3)/(1-6x^2+x^4); a(n) = Sum_{k=0..floor(n/2)} C(n,2k)*2^(n-k-floor((n+1)/2)); a(n) = -(sqrt(2)-1)^n*((sqrt(2)/8-1/4)*(-1)^n - sqrt(2)/8 - 1/4) - (sqrt(2)+1)^n*((sqrt(2)/8-1/4)*(-1)^n - sqrt(2)/8 - 1/4); a(2n) = A001541(n) = A001333(2n); a(2n+1) = A001653(n) = A000129(2n+1). - Paul Barry, Jan 22 2005 The lower principal and intermediate convergents to 2^(1/2), beginning with 1/1, 4/3, 7/5, 24/17, 41/29, form a strictly increasing sequence; essentially, numerators= A143608 and denominators= A079496. - Clark Kimberling, Aug 27 2008 This sequence is a particular case of the following situation: a(0)=1, a(1)=a, a(2)=b
with the recurrence relation a(n+3)=(a(n+2)*a(n+1)+q)/a(n) where q is given in Z to have Q=(a*b^2+q*b+a+q)/(a*b) itself in Z.
The g.f is f: f(z)=(1+a*z+(b-Q)*z^2+(a*b+q-a*Q)*z^3)/(1-Q*z^2+z^4); so we have the linear recurrence: a(n+4)=Q*a(n+2)-a(n).
The general form of a(n) is given by:
a(2*m)=sum((-1)^p*binomial(m-p,p)*Q^(m-2*p),p=0..floor(m/2))+(b-Q)*sum((-1)^p*binomial(m-1-p,p)*Q^(m-1-2*p),p=0..floor((m-1)/2)) and
a(2*m+1)=a*sum((-1)^p*binomial(m-p,p)*Q^(m-2*p),p=0..floor(m/2))+(a*b+q-a*Q)*sum((-1)^p*binomial(m-1-p,p)*Q^(m-1-2*p),p=0..floor((m-1)/2))
(End)
REFERENCES
Serge Lang, Introduction to Diophantine Approximations, Addison-Wesley, New York, 1966.
FORMULA
a(2n+1) - a(2n) = a(2n) - a(2n-1) = A001542(n). a(2n+1) = ceiling((2+sqrt(2))/4*(3+2*sqrt(2))^n), a(2n) = ceiling(1/2*(3+2*sqrt(2))^n).
G.f.: (1 + x - 3*x^2 - x^3)/(1 - 6*x^2 + x^4).
a(n) = (sqrt(2)*sqrt(2+(3-2*sqrt(2))^n+(3+2*sqrt(2))^n))/(2+sqrt(2)+(-1)^n*(-2+sqrt(2))). - Gerry Martens, Jun 06 2015 a(n) = 2^(n - 1)*H(n, n mod 2, 1/2) for n >= 3 where H(n, a, b) = hypergeom([a - n/2, b - n/2], [1 - n], -1). - Peter Luschny, Sep 03 2019 a(n) == Pell(n)^(-1) (mod Pell(n+1)) where Pell(n) = A000129(n), use the identity a(n)*Pell(n) - A084068(n-1)*Pell(n+1) = 1, taken modulo Pell(n+1). - Gary W. Adamson, Nov 21 2023
EXAMPLE
1 + x + 3*x^2 + 5*x^3 + 17*x^4 + 29*x^5 + 99*x^6 + 169*x^7 + 577*x^8 + ...
MAPLE
H := (n, a, b) -> hypergeom([a - n/2, b - n/2], [1 - n], -1):
a := n -> `if`(n < 3, [1, 1, 3][n+1], 2^(n - 1)*H(n, irem(n, 2), 1/2)):
PROG
(PARI) {a(n) = n = abs(n); 2^((4-n)\2) * real( (10 + 7 * quadgen(8)) / 2 * (2 + quadgen(8))^(n-3) ) } /* Michael Somos, Sep 03 2013 */ (PARI) {a(n) = polcoeff( (1 + x - 3*x^2 - x^3) / (1 - 6*x^2 + x^4) + x * O(x^abs(n)), abs(n))} /* Michael Somos, Sep 03 2013 */
Squares k such that 2*k+1 is also a square.
+10
12
0, 4, 144, 4900, 166464, 5654884, 192099600, 6525731524, 221682772224, 7530688524100, 255821727047184, 8690408031080164, 295218051329678400, 10028723337177985444, 340681375412721826704, 11573138040695364122500, 393146012008229658338304, 13355391270239113019379844
COMMENTS
A204576 is this sequence written in binary. (End)
a(n+1), n >= 0, is the perimeter squared (x(n) + y(n) + z(n))^2 of the ordered primitive Pythagorean triple (x(n), y(n) = x(n) + 1, z(n)). The first two terms are (x(0)=0, y(0)=1, z(0)=1), a(1) = 2^2, and (x(1)=3, y(1)=4, z(1)=5), a(2) = 12^2. - George F. Johnson, Nov 02 2012
FORMULA
G.f.: 4*x*(1+x)/((1-x)*(1-34*x+x^2)). - R. J. Mathar, Dec 15 2008 a(n) = ((17+12*sqrt(2))^n + (17-12*sqrt(2))^n - 2)/8.
a(n+1) = 17*a(n) + 4 + 12*sqrt(a(n)*(2*a(n) + 1)).
a(n-1) = 17*a(n) + 4 - 12*sqrt(a(n)*(2*a(n) + 1)).
a(n-1)*a(n+1) = (a(n) - 4)^2.
a(n+1) = 34*a(n) - a(n-1) + 8 for n>1, a(0)=0, a(1)=4.
a(n+1) = 35*a(n) - 35*a(n-1) + a(n-2) for n>0, a(0)=0, a(1)=4, a(2)=144.
a(n) + a(n+1) = 4*(6* A029549(n) + 1). Limit_{n -> infinity} a(n)/a(n-r) = (17+12*sqrt(2))^r. (End)
Product_{n>=2} (1 - 4/a(n)) = sqrt(2)/3 + 1/2 (Koshy, 2022, section 3, p. 19). - Amiram Eldar, Jan 23 2025
MATHEMATICA
b[n_]:= b[n]= If[n<2, n, 34*b[n-1] -b[n-2] +2]; (* b= A001110 *) a[n_]:= 4*b[n]; Table[a[n], {n, 0, 30}]
4*ChebyshevU[Range[-1, 30], 3]^2 (* G. C. Greubel, Aug 18 2022 *)
PROG
(Magma) [4*Evaluate(ChebyshevU(n), 3)^2: n in [0..30]]; // G. C. Greubel, Aug 18 2022 (SageMath) [4*chebyshev_U(n-1, 3)^2 for n in (0..30)] # G. C. Greubel, Aug 18 2022
CROSSREFS
Cf. similar sequences with closed form ((1 + sqrt(2))^(4*r) + (1 - sqrt(2))^(4*r))/8 + k/4: this sequence (k=-1), A076218 (k=3), A278310 (k=-5).
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