A002315 -id:A002315 - cp's OEIS frontend

A046090 Consider all Pythagorean triples (X,X+1,Z) ordered by increasing Z; sequence gives X+1 values.

1, 4, 21, 120, 697, 4060, 23661, 137904, 803761, 4684660, 27304197, 159140520, 927538921, 5406093004, 31509019101, 183648021600, 1070379110497, 6238626641380, 36361380737781, 211929657785304, 1235216565974041, 7199369738058940, 41961001862379597, 244566641436218640

Offset: 0

Eric W. Weisstein

Solution to a*(a-1) = 2b*(b-1) in natural numbers: a = a(n), b = b(n) = A011900(n).
n such that n^2 = (1/2)*(n+floor(sqrt(2)*n*floor(sqrt(2)*n))). - Benoit Cloitre, Apr 15 2003
Place a(n) balls in an urn, of which b(n) = A011900(n) are red; draw 2 balls without replacement; 2*Probability(2 red balls) = Probability(2 balls); this is equivalent to the Pell equation A(n)^2-2*B(n)^2 = -1 with a(n) = (A(n)+1)/2; b(n) = (B(n)+1)/2; and the fundamental solution (7;5) and the solution (3;2) for the unit form. - Paul Weisenhorn, Aug 03 2010
Find base x in which repdigit yy has a square that is repdigit zzzz, corresponding to Diophantine equation zzzz_x = (yy_x)^2; then, solution z = a(n) with x = A002315(n) and y = A001653(n+1) for n >= 1 (see Maurice Protat reference). - Bernard Schott, Dec 21 2022

			For n=4: a(4)=697; b(4)=493; 2*binomial(493,2)=485112=binomial(697,2). - _Paul Weisenhorn_, Aug 03 2010

A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, pp. 122-125, 1964.
Maurice Protat, Des Olympiades à l'Agrégation, De zzzz_x = (yy_x)^2 à Pell-Fermat, Problème 23, pp. 52-54, Ellipses, Paris, 1997.

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
T. W. Forget and T. A. Larkin, Pythagorean triads of the form X, X+1, Z described by recurrence sequences, Fib. Quart., 6 (No. 3, 1968), 94-104.
L. J. Gerstein, Pythagorean triples and inner products, Math. Mag., 78 (2005), 205-213.
H. J. Hindin, Stars, hexes, triangular numbers and Pythagorean triples, J. Rec. Math., 16 (1983/1984), 191-193. (Annotated scanned copy)
Ron Knott, Pythagorean Triples and Online Calculators
S. Northshield, An Analogue of Stern's Sequence for Z[sqrt(2)], Journal of Integer Sequences, 18 (2015), #15.11.6.
Eric Weisstein's World of Mathematics, Pythagorean Triple
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

Other 2 sides are A001652 and A001653.
Cf. A001009, A001541, A001542, A011900, A046729, A053141, A084159, A084703, A156035.
See comments in A301383.
Cf. A001109, A001653, A002024, A002315, A029549.

a046090 n = a046090_list !! n
a046090_list = 1 : 4 : map (subtract 2)
   (zipWith (-) (map (* 6) (tail a046090_list)) a046090_list)
-- Reinhard Zumkeller, Jan 10 2012

m:=30; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((1-3*x)/((1-6*x+x^2)*(1-x)))); // G. C. Greubel, Jul 15 2018

Digits:=100: seq(round((1+(7+5*sqrt(2))*(3+2*sqrt(2))^(n-1))/2)/2, n=0..20); # Paul Weisenhorn, Aug 03 2010

Join[{1},#+1&/@With[{c=3+2Sqrt[2]},NestList[Floor[c #]+3&,3,20]]] (* Harvey P. Dale, Aug 19 2011 *)
LinearRecurrence[{7,-7,1},{1,4,21},25] (* Harvey P. Dale, Apr 13 2012 *)
a[n_] := (2-ChebyshevT[n, 3]+ChebyshevT[n+1, 3])/4; Array[a, 21, 0] (* Jean-François Alcover, Jul 10 2016, adapted from PARI *)

a(n)=(2-subst(poltchebi(abs(n))-poltchebi(abs(n+1)),x,3))/4

x='x+O('x^30); Vec((1-3*x)/((1-6*x+x^2)*(1-x))) \\ G. C. Greubel, Jul 15 2018

a(n) = (-1+sqrt(1+8*b(n)*(b(n)+1)))/2 with b(n) = A011900(n). [corrected by Michel Marcus, Dec 23 2022]
a(n) = 6*a(n-1) - a(n-2) - 2, n >= 2, a(0) = 1, a(1) = 4.
a(n) = (A(n+1) - 3*A(n) + 2)/4 with A(n) = A001653(n).
A001652(n) = -a(-1-n).
From Barry E. Williams, May 03 2000: (Start)
G.f.: (1-3*x)/((1-6*x+x^2)*(1-x)).
a(n) = partial sums of A001541(n). (End)
From Charlie Marion, Jul 01 2003: (Start)
A001652(n)*A001652(n+1) + a(n)*a(n+1) = A001542(n+1)^2 = A084703(n+1).
Let a(n) = A001652(n), b(n) = this sequence and c(n) = A001653(n). Then for k > j, c(i)*(c(k) - c(j)) = a(k+i) + ... + a(i+j+1) + a(k-i-1) + ... + a(j-i) + k - j. For n < 0, a(n) = -b(-n-1). Also a(n)*a(n+2k+1) + b(n)*b(n+2k+1) + c(n)*c(n+2k+1) = (a(n+k+1) - a(n+k))^2; a(n)*a(n+2k) + b(n)*b(n+2k) + c(n)*c(n+2k) = 2*c(n+k)^2. (End)
a(n) = 1/2 + ((1-2^(1/2))/4)*(3 - 2^(3/2))^n + ((1+2^(1/2))/4)*(3 + 2^(3/2))^n. - Antonio Alberto Olivares, Oct 13 2003
2*a(n) = 2*A084159(n) + 1 + (-1)^(n+1) = 2*A046729(n) + 1 - (-1)^(n+1). - Lekraj Beedassy, Jul 16 2004
a(n) = A001109(n+1) - A053141(n). - Manuel Valdivia, Apr 03 2010
From Paul Weisenhorn, Aug 03 2010: (Start)
a(n+1) = round((1+(7+5*sqrt(2))*(3+2*sqrt(2))^n)/2);
b(n+1) = round((2+(10+7*sqrt(2))*(3+2*sqrt(2))^n)/4) = A011900(n+1).
(End)
a(n)*(a(n)-1)/2 = b(n)*b(n+1) and 2*a(n) - 1 = b(n) + b(n+1), where b(n) = A001109. - Kenneth J Ramsey, Apr 24 2011
T(a(n)) = A011900(n)^2 + A001109(n), where T(n) is the n-th triangular number. See also A001653. - Charlie Marion, Apr 25 2011
a(0)=1, a(1)=4, a(2)=21, a(n) = 7*a(n-1) - 7*a(n-2) + a(n-3). - Harvey P. Dale, Apr 13 2012
Limit_{n->oo} a(n+1)/a(n) = 3 + 2*sqrt(2) = A156035. - Ilya Gutkovskiy, Jul 10 2016
a(n) = A001652(n)+1. - Dimitri Papadopoulos, Jul 06 2017
a(n) = (A002315(n) + 1)/2. - Bernard Schott, Dec 21 2022
E.g.f.: (exp(x) + exp(3*x)*(cosh(2*sqrt(2)*x) + sqrt(2)*sinh(2*sqrt(2)*x)))/2. - Stefano Spezia, Mar 16 2024
a(n) = A002024(A029549(n))+1. - Pontus von Brömssen, Sep 11 2024

Additional comments from Wolfdieter Lang

Comment moved to A001653 by Claude Morin, Sep 22 2023

A030221 Chebyshev even-indexed U-polynomials evaluated at sqrt(7)/2.

Original entry on oeis.org

1, 6, 29, 139, 666, 3191, 15289, 73254, 350981, 1681651, 8057274, 38604719, 184966321, 886226886, 4246168109, 20344613659, 97476900186, 467039887271, 2237722536169, 10721572793574, 51370141431701, 246129134364931, 1179275530392954, 5650248517599839

Offset: 0

Wolfdieter Lang

a(n) = L(n,-5)*(-1)^n, where L is defined as in A108299; see also A004253 for L(n,+5). - Reinhard Zumkeller, Jun 01 2005
General recurrence is a(n) = (a(1)-1)*a(n-1) - a(n-2), a(1) >= 4; lim_{n->oo} a(n) = x*(k*x+1)^n, k =(a(1)-3), x=(1+sqrt((a(1)+1)/(a(1)-3)))/2. Examples in OEIS: a(1)=4 gives A002878. a(1)=5 gives A001834. a(1)=6 gives the present sequence. a(1)=7 gives A002315. a(1)=8 gives A033890. a(1)=9 gives A057080. a(1)=10 gives A057081. - Ctibor O. Zizka, Sep 02 2008
The primes in this sequence are 29, 139, 3191, 15289, 350981, 1681651, ... - Ctibor O. Zizka, Sep 02 2008
Inverse binomial transform of A030240. - Philippe Deléham, Nov 19 2009
For positive n, a(n) equals the permanent of the (2n)X(2n) matrix with sqrt(7)'s along the main diagonal, and i's along the superdiagonal and the subdiagonal (i is the imaginary unit). - John M. Campbell, Jul 08 2011
The aerated sequence (b(n))n>=1 = [1, 0, 6, 0, 29, 0, 139, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -3, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047 for a connection with Chebyshev polynomials. - Peter Bala, Mar 22 2015
From Wolfdieter Lang, Oct 26 2020: (Start)
((-1)^n)*a(n) = X(n) = ((-1)^n)*(S(n, 5) + S(n-1, 5)) and Y(n) = X(n-1) gives all integer solutions (modulo sign flip between X and Y) of X^2 + Y^2 + 5*X*Y = +7, for n = -oo..+oo, with Chebyshev S polynomials (see A049310), with S(-1, x) = 0, and S(-n, x) = - S(n-2, x), for n >= 2.
This binary indefinite quadratic form of discriminant 21, representing 7, has only this family of proper solutions (modulo sign flip), and no improper ones.
This comment is inspired by a paper by Robert K. Moniot (private communication). See his Oct 04 2020 comment in A027941 related to the case of x^2 + y^2 - 3*x*y = -1 (special Markov solutions). (End)

			G.f. = 1 + 6*x + 29*x^2 + 139*x^3 + 666*x^4 + 3191*x^5 + 15289*x^6 + ...

G. C. Greubel, Table of n, a(n) for n = 0..1000
Marco Abrate, Stefano Barbero, Umberto Cerruti, and Nadir Murru, Polynomial sequences on quadratic curves, Integers, Vol. 15, 2015, #A38.
K. Andersen, L. Carbone, and D. Penta, Kac-Moody Fibonacci sequences, hyperbolic golden ratios, and real quadratic fields, Journal of Number Theory and Combinatorics, Vol 2, No. 3 pp 245-278, 2011. See Section 9.
K. Dilcher and K. B. Stolarsky, A Pascal-type triangle characterizing twin primes, Amer. Math. Monthly, 112 (2005), 673-681. (see page 678)
Alex Fink, Richard K. Guy, and Mark Krusemeyer, Partitions with parts occurring at most thrice, Contributions to Discrete Mathematics, Vol 3, No 2 (2008), pp. 76-114. See Section 13.
Taras Goy and Mark Shattuck, Determinants of Toeplitz-Hessenberg Matrices with Generalized Leonardo Number Entries, Ann. Math. Silesianae (2023). See p. 18.
Christian Kassel and Christophe Reutenauer, Pairs of intertwined integer sequences, arXiv:2507.15780 [math.NT], 2025. See p. 13.
Tanya Khovanova, Recursive Sequences.
Wolfdieter Lang, On polynomials related to powers of the generating function of Catalan's numbers, Fib. Quart. 38 (2000) 408-419. Eq.(44), rhs, m=6.
Ioana-Claudia Lazăr, Lucas sequences in t-uniform simplicial complexes, arXiv:1904.06555 [math.GR], 2019.
Donatella Merlini and Renzo Sprugnoli, Arithmetic into geometric progressions through Riordan arrays, Discrete Mathematics 340.2 (2017): 160-174.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences, Integers, Volume 12A (2012) The John Selfridge Memorial Volume.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (5,-1).

Cf. A004253, A004254, A100047, A054493 (partial sums), A049310, A003501 (first differences), A299109 (subsequence of primes).

I:=[1,6]; [n le 2 select I[n] else 5*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Mar 22 2015

A030221 := proc(n)
    option remember;
    if n <= 1 then
        op(n+1,[1,6]);
    else
        5*procname(n-1)-procname(n-2) ;
    end if;
end proc: # R. J. Mathar, Apr 30 2017

t[n_, k_?EvenQ] := I^k*Binomial[n-k/2, k/2]; t[n_, k_?OddQ] := -I^(k-1)*Binomial[n+(1-k)/2-1, (k-1)/2]; l[n_, x_] := Sum[t[n, k]*x^(n-k), {k, 0, n}]; a[n_] := (-1)^n*l[n, -5]; Table[a[n], {n, 0, 20}] (* Jean-François Alcover, Jul 05 2013, after Reinhard Zumkeller *)
a[ n_] := ChebyshevU[2 n, Sqrt[7]/2]; (* Michael Somos, Jan 22 2017 *)

{a(n) = simplify(polchebyshev(2*n, 2, quadgen(28)/2))}; /* Michael Somos, Jan 22 2017 */

[(lucas_number2(n,5,1)-lucas_number2(n-1,5,1))/3 for n in range(1,22)] # Zerinvary Lajos, Nov 10 2009

a(n) = 5*a(n-1) - a(n-2), a(-1)=-1, a(0)=1.
a(n) = U(2*n, sqrt(7)/2).
G.f.: (1+x)/(x^2-5*x+1).
a(n) = A004254(n) + A004254(n+1).
a(n) ~ (1/2 + (1/6)*sqrt(21))*((1/2)*(5 + sqrt(21)))^n. - Joe Keane (jgk(AT)jgk.org), May 16 2002
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i); then a(n) = (-1)^n*q(n, -7). - Benoit Cloitre, Nov 10 2002
A054493(2*n) = a(n)^2 for all n in Z. - Michael Somos, Jan 22 2017
a(n) = -a(-1-n) for all n in Z. - Michael Somos, Jan 22 2017
0 = -7 + a(n)*(+a(n) - 5*a(n+1)) + a(n+1)*(+a(n+1)) for all n in Z. - Michael Somos, Jan 22 2017
a(n) = S(n, 5) + S(n-1, 5) = S(2*n, sqrt(7)) (see above in terms of U), for n >= 0 with S(-1, 5) = 0, where the coefficients of the Chebyshev S polynomials are given in A049310. - Wolfdieter Lang, Oct 26 2020
From Peter Bala, May 16 2025: (Start)
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 1/a(n)) = 1/7 (telescoping series: 7/(a(n) - 1/a(n)) = 1/A004254(n+1) + 1/A004254(n)).
Product_{n >= 1} (a(n) + 1)/(a(n) - 1) = sqrt(7/3) (telescoping product: Product_{k = 1..n} ((a(k) + 1)/(a(k) - 1))^2 = 7/3 * (1 - 8/A231087(n+1))). (End)

A088165 NSW primes: NSW numbers that are also prime.

Original entry on oeis.org

7, 41, 239, 9369319, 63018038201, 489133282872437279, 19175002942688032928599, 123426017006182806728593424683999798008235734137469123231828679

Offset: 1

Christian Schroeder, Sep 21 2003

nonn

Next term a(9) is too large (99 digits) to include here. - Ray Chandler, Sep 21 2003
These primes are the prime RMS numbers (A140480): primes p such that (1+p^2)/2 is a square r^2. Then r is a Pell number, A000129. - T. D. Noe, Jul 01 2008
Also prime numerators with an odd index in A001333. - Ctibor O. Zizka, Aug 13 2008
r in the above note of T. D. Noe is a prime Pell number (A000129) with an odd index. - Ctibor O. Zizka, Aug 13 2008
General recurrence is a(n) = (a(1)-1)*a(n-1) - a(n-2), a(1) >= 4, lim_{n->infinity} a(n) = x*(k*x+1)^n, k = a(1)-3, x = (1+sqrt((a(1)+1)/(a(1)-3)))/2. Examples in the OEIS: a(1)=4 gives A002878, primes in it A121534. a(1)=5 gives A001834, primes in it A086386. a(1)=6 gives A030221, primes in it not in the OEIS {29, 139, 3191, ...}. a(1)=7 gives A002315, primes in it A088165. a(1)=8 gives A033890, primes in it not in the OEIS (do there exist any ?). a(1)=9 gives A057080, primes in it not in the OEIS {71, 34649, 16908641, ...}. a(1)=10 gives A057081, primes in it not in the OEIS {389806471, 192097408520951, ...}. - Ctibor O. Zizka, Sep 02 2008

Paulo Ribenboim, The New Book of Prime Number Records, 3rd edition, Springer-Verlag, New York, 1995, pp. 367-369.

Alois P. Heinz, Table of n, a(n) for n = 1..14
Morris Newman, Daniel Shanks, H. C. Williams, Simple groups of square order and an interesting sequence of primes, Acta Arith., 38 (1980/1981), pp. 129-140.
M. Newman, D. Shanks and L. L. Foster, Simple groups of square order (6176), The American Mathematical Monthly, Vol. 86, No. 4 (Apr., 1979), pp. 314-315.
The Prime Glossary, NSW numbers

Cf. A002315 (NSW numbers), A005850 (indices for NSW primes).

w=3+quadgen(32); forprime(p=2,1e3, if(ispseudoprime(t=imag((1+w)*w^p)), print1(t", "))) \\ Charles R Greathouse IV, Apr 29 2015

a(n) mod A005850(n) = 1. - Altug Alkan, Mar 17 2016

More terms from Ray Chandler, Sep 21 2003

A054320 Expansion of g.f.: (1 + x)/(1 - 10*x + x^2).

Original entry on oeis.org

1, 11, 109, 1079, 10681, 105731, 1046629, 10360559, 102558961, 1015229051, 10049731549, 99482086439, 984771132841, 9748229241971, 96497521286869, 955226983626719, 9455772314980321, 93602496166176491, 926569189346784589, 9172089397301669399, 90794324783669909401

Offset: 0

Ignacio Larrosa Cañestro, Feb 27 2000

Chebyshev's even-indexed U-polynomials evaluated at sqrt(3).
a(n)^2 is a star number (A003154).
Any k in the sequence has the successor 5*k + 2*sqrt(3(2*k^2 + 1)). - Lekraj Beedassy, Jul 08 2002
{a(n)} give the values of x solving: 3*y^2 - 2*x^2 = 1. Corresponding values of y are given by A072256(n+1). x + y = A001078(n+1). - Richard R. Forberg, Nov 21 2013
The aerated sequence (b(n))n>=1 = [1, 0, 11, 0, 109, 0, 1079, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -8, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047. - Peter Bala, Mar 22 2015

			a(1)^2 = 121 is the 5th star number (A003154).

G. C. Greubel, Table of n, a(n) for n = 0..1000
K. Andersen, L. Carbone, and D. Penta, Kac-Moody Fibonacci sequences, hyperbolic golden ratios, and real quadratic fields, Journal of Number Theory and Combinatorics, Vol 2, No. 3 pp 245-278, 2011. See Section 9.
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (I).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (II).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (III).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (IV).
Alex Fink, Richard K. Guy, and Mark Krusemeyer, Partitions with parts occurring at most thrice, Contrib. Discr. Math. 3 (2) (2008), pp. 76-114. See Section 13.
Tanya Khovanova, Recursive Sequences
Eric Weisstein's World of Mathematics, Star Number
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences, Integers, Volume 12A (2012) The John Selfridge Memorial Volume.
Index entries for linear recurrences with constant coefficients, signature (10,-1).
Index entries for sequences related to Chebyshev polynomials.

A member of the family A057078, A057077, A057079, A005408, A002878, A001834, A030221, A002315, A033890, A057080, A057081, A054320, which are the expansions of (1+x) / (1-kx+x^2) with k = -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. - Philippe Deléham, May 04 2004
Cf. A003154, A006061, A031138, A007667, A004189.
Cf. A138281. Cf. A100047.
Cf. A142238.

a:=[1,11];; for n in [3..30] do a[n]:=10*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jul 22 2019

I:=[1,11]; [n le 2 select I[n] else 10*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Mar 22 2015

CoefficientList[Series[(1+x)/(1-10x+x^2), {x,0,30}], x] (* Vincenzo Librandi, Mar 22 2015 *)
a[c_, n_] := Module[{},
   p := Length[ContinuedFraction[ Sqrt[ c]][[2]]];
   d := Numerator[Convergents[Sqrt[c], n p]];
   t := Table[d[[1 + i]], {i, 0, Length[d] - 1, p}];
   Return[t];
] (* Complement of A142238 *)
a[3/2, 20] (* Gerry Martens, Jun 07 2015 *)

a(n)=subst(poltchebi(n+1)-poltchebi(n),x,5)/4;

(a(n)-1)^2 + a(n)^2 + (a(n)+1)^2 = b(n)^2 + (b(n)+1)^2 = c(n), where b(n) is A031138 and c(n) is A007667.
a(n) = 10*a(n-1) - a(n-2).
a(n) = (sqrt(6) - 2)/4*(5 + 2*sqrt(6))^(n+1) - (sqrt(6) + 2)/4*(5 - 2*sqrt(6))^(n+1).
a(n) = U(2*(n-1), sqrt(3)) = S(n-1, 10) + S(n-2, 10) with Chebyshev's U(n, x) and S(n, x) := U(n, x/2) polynomials and S(-1, x) := 0. S(n, 10) = A004189(n+1), n >= 0.
6*a(n)^2 + 3 is a square. Limit_{n->oo} a(n)/a(n-1) = 5 + 2*sqrt(6). - Gregory V. Richardson, Oct 13 2002
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i), then (-1)^n*q(n, -12) = a(n). - Benoit Cloitre, Nov 10 2002
a(n) = L(n,-10)*(-1)^n, where L is defined as in A108299; see also A072256 for L(n,+10). - Reinhard Zumkeller, Jun 01 2005
From Reinhard Zumkeller, Mar 12 2008: (Start)
(sqrt(2) + sqrt(3))^(2*n+1) = a(n)*sqrt(2) + A138288(n)*sqrt(3);
a(n) = A138288(n) + A001078(n).
a(n) = A001079(n) + 3*A001078(n). (End)
a(n) = A142238(2n) = A041006(2n)/2 = A041038(2n)/4. - M. F. Hasler, Feb 14 2009
a(n) = sqrt(A006061(n)). - Zak Seidov, Oct 22 2012
a(n) = sqrt((3*A072256(n)^2 - 1)/2). - T. D. Noe, Oct 23 2012
(sqrt(3) + sqrt(2))^(2*n+1) - (sqrt(3) - sqrt(2))^(2*n+1) = a(n)*sqrt(8). - Bruno Berselli, Oct 29 2019
a(n) = A004189(n)+A004189(n+1). - R. J. Mathar, Oct 01 2021
E.g.f.: exp(5*x)*(2*cosh(2*sqrt(6)*x) + sqrt(6)*sinh(2*sqrt(6)*x))/2. - Stefano Spezia, May 16 2023
From Peter Bala, May 09 2025: (Start)
a(n) = Dir(n, 5), where Dir(n, x) denotes the n-th row polynomial of the triangle A244419.
a(n)^2 - 10*a(n)*a(n+1) + a(n+1)^2 = 12.
More generally, for arbitrary x, a(n+x)^2 - 10*a(n+x)*a(n+x+1) + a(n+x+1)^2 = 12 with a(n) := (sqrt(6) - 2)/4*(5 + 2*sqrt(6))^(n+1) - (sqrt(6) + 2)/4*(5 - 2*sqrt(6))^(n+1) as given above.
a(n+1/2) = sqrt(3) * A001078(n+1).
a(n+3/4) + a(n+1/4) = sqrt(6)*sqrt(sqrt(3) + 1) * A001078(n+1).
a(n+3/4) - a(n+1/4) = sqrt(sqrt(3) - 1) * A001079(n+1).
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 1/a(n)) = 1/12 (telescoping series: for n >= 1, 1/(a(n) - 1/a(n)) = 1/A004291(n) + 1/A004291(n+1)).
Product_{n >= 1} (a(n) + 1)/(a(n) - 1) = sqrt(3/2) (telescoping product: Product_{n = 1..k} ((a(n) + 1)/(a(n) - 1))^2 = 3/2 * (1 - 1/A171640(k+2))). (End)

Chebyshev comments from Wolfdieter Lang, Oct 31 2002

A075528 Triangular numbers that are half other triangular numbers.

Original entry on oeis.org

0, 3, 105, 3570, 121278, 4119885, 139954815, 4754343828, 161507735340, 5486508657735, 186379786627653, 6331426236682470, 215082112260576330, 7306460390622912753, 248204571168918457275, 8431648959352604634600, 286427860046819639119128

Offset: 0

Christian G. Bower, Sep 19 2002

This is the sequence of 1/2 the areas, x(n)*y(n)/2, of the ordered Pythagorean triples (x(n), y(n)=x(n)+1, z(n)) with x(0)=0, y(0)=1, z(0)=1, a(0)=0 and x(1)=3, y(1)=4, z(1)=5, a(1)=3. - George F. Johnson, Aug 24 2012

Colin Barker, Table of n, a(n) for n = 0..653
Martin V. Bonsangue, Gerald E. Gannon and Laura J. Pheifer, Misinterpretations can sometimes be a good thing, Math. Teacher, vol. 95, No. 6 (2002) pp. 446-449.
Harvey J. Hindin, Stars, hexes, triangular numbers and Pythagorean triples, J. Rec. Math., 16 (1983/1984), 191-193. (Annotated scanned copy)
Vladimir Pletser, Recurrent Relations for Multiple of Triangular Numbers being Triangular Numbers, arXiv:2101.00998 [math.NT], 2021.
Vladimir Pletser, Closed Form Equations for Triangular Numbers Multiple of Other Triangular Numbers, arXiv:2102.12392 [math.GM], 2021.
Vladimir Pletser, Triangular Numbers Multiple of Triangular Numbers and Solutions of Pell Equations, arXiv:2102.13494 [math.NT], 2021.
Vladimir Pletser, Congruence Properties of Indices of Triangular Numbers Multiple of Other Triangular Numbers, arXiv:2103.03019 [math.GM], 2021.
Vladimir Pletser, Searching for multiple of triangular numbers being triangular numbers, 2021.
Vladimir Pletser, Using Pell equation solutions to find all triangular numbers multiple of other triangular numbers, 2022.
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. A000129, A000217, A001652, A002315, A029546, A029549, A046090, A046729, A053141, A084159, A096979.

CoefficientList[ Series[ 3x/(1 - 35 x + 35 x^2 - x^3), {x, 0, 15}], x] (* Robert G. Wilson v, Jun 24 2011 *)

concat(0, Vec(3*x/((1-x)*(1-34*x+x^2)) + O(x^20))) \\ Colin Barker, Jun 18 2015

a(n) = 3*A029546(n-1) = A029549(n)/2.
G.f.: 3*x/((1-x)*(1-34*x+x^2)).
From George F. Johnson, Aug 24 2012: (Start)
a(n) = ((3+2*sqrt(2))^(2*n+1) + (3-2*sqrt(2))^(2*n+1) - 6)/64.
8*a(n)+1 = A000129(2*n+1)^2.
16*a(n)+1 = A002315(n)^2.
128*a(n)^2 + 24*a(n) + 1 is a perfect square.
a(n+1) = 17*a(n) + 3/2 + 3*sqrt((8*a(n)+1)*(16*a(n)+1))/2.
a(n-1) = 17*a(n) + 3/2 - 3*sqrt((8*a(n)+1)*(16*a(n)+1))/2.
a(n-1)*a(n+1) = a(n)*(a(n)-3); a(n+1) = 34*a(n) - a(n-1) + 3.
a(n+1) = 35*a(n) - 35*a(n-1) + a(n-2); a(n) = A096979(2*n)/2.
a(n) = A084159(n)*A046729(n)/4 = A001652(n)*A046090(n)/4.
Lim_{n->infinity} a(n)/a(n-1) =  17 + 12*sqrt(2).
Lim_{n->infinity} a(n)/a(n-2) = (17 + 12*sqrt(2))^2 = 577 + 408*sqrt(2).
Lim_{n->infinity} a(n)/a(n-r) = (17 + 12*sqrt(2))^r.
Lim_{n->infinity} a(n-r)/a(n) = (17 + 12*sqrt(2))^(-r) = (17 - 12*sqrt(2))^r.
(End)
a(n) = 34*a(n-1) - a(n-2) + 3, n >= 2. - R. J. Mathar, Nov 07 2015
a(n) = A000217(A053141(n)). - R. J. Mathar, Aug 16 2019
a(n) = (a(n-1)*(a(n-1)-3))/a(n-2) for n > 2. - Vladimir Pletser, Apr 08 2020
Sum_{n>=1} 1/a(n) = 2*(3 - 2*sqrt(2)). - Amiram Eldar, Dec 04 2024

A046176 Indices of square numbers that are also hexagonal.

Original entry on oeis.org

1, 35, 1189, 40391, 1372105, 46611179, 1583407981, 53789260175, 1827251437969, 62072759630771, 2108646576008245, 71631910824649559, 2433376321462076761, 82663163018885960315, 2808114166320660573949

Offset: 1

Eric W. Weisstein

Bisection (even part) of Chebyshev sequence with Diophantine property.
(3*b(n))^2 - 2*(2*a(n+1))^2 = 1 with companion sequence b(n) = A077420(n), n >= 0.
Sequence also refers to inradius of primitive Pythagorean triangles with consecutive legs, odd followed by even. - Lekraj Beedassy, Apr 23 2003
As n increases, this sequence is approximately geometric with common ratio r = lim_{n -> oo} a(n)/a(n-1) = (1 + sqrt(2))^4 = 17 + 12*sqrt(2). - Ant King, Nov 08 2011
Integers of the form sqrt((m+1)*(2*m+1)). The corresponding values of m form A078522. Subsequence of A284876. - Jonathan Sondow, Apr 07 2017

M. Rignaux, Query 2175, L'Intermédiaire des Mathématiciens, 24 (1917), 80.

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Tanya Khovanova, Recursive Sequences.
E. Kilic, Y. T. Ulutas, and N. Omur, A Formula for the Generating Functions of Powers of Horadam's Sequence with Two Additional Parameters, J. Int. Seq. 14 (2011) #11.5.6, table 4, k=1, t=3.
Serge Perrine, About the diophantine equation z^2 = 32y^2 - 16, SCIREA Journal of Mathematics (2019) Vol. 4, Issue 5, 126-139.
Maciej Ulas, On certain diophantine equations related to triangular and tetrahedral numbers, arXiv:0811.2477 [math.NT] (2008) , v_n in Theorem 5.6.
P. H. van der Kamp, Global classification of two-component..., Found. Comput. Math. 9 (5) (2009) 559-597 near Eq. (4.7).
Eric Weisstein's World of Mathematics, Hexagonal Square Number.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (34,-1).

Cf. A000129, A008844, A046177, A078522, A284876.

Cf. A001109, A001110 (partial sums).

List([0..20], n-> Lucas(2,-1, 4*n-2)[1]/2 ); # G. C. Greubel, Jan 13 2020

I:=[1,35]; [n le 2 select I[n] else 34*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Nov 22 2011

seq( simplify(ChebyshevU(2*(n-1), 3)), n = 1..20); # G. C. Greubel, Jan 13 2020

LinearRecurrence[{34, -1}, {1, 35}, 15] (* Ant King, Nov 08 2011 *)
Fibonacci[4*Range[20] -2, 2]/2 (* G. C. Greubel, Jan 13 2020 *)

vector(21, n, polchebyshev(2*(n-1), 2, 3) ) \\ G. C. Greubel, Jan 13 2020

[lucas_number1(4*n-2, 2,-1)/2 for n in (1..20)] # G. C. Greubel, Jan 13 2020

a(n) = 34*a(n-1) - a(n-2); a(0)=-1, a(1)=1.
a(n+1) = S(2*n, 6) = S(n, 34) + S(n-1, 34), n >= 1, with S(n, x) := U(n, x/2) Chebyshev's polynomials of the second kind. See A049310. S(n, 34) = A029547(n).
G.f.: x*(1+x)/(1-34*x+x^2).
a(n+1) = Sum_{k=0..n} (-1)^k*binomial(2*n-k, k)*6^(2*(n-k)), n >= 0.
a(n) = A001109(2n+1). - Lekraj Beedassy, Apr 23 2003
Define f(x,s) = s*x + sqrt((s^2-1)*x^2+1); f(0,s)=0. a(n) = f(f(a(n-1),3),3). - Marcos Carreira, Dec 27 2006
From Antonio Alberto Olivares, Mar 22 2008: (Start)
a(n) = (sqrt(2)/8)*(3 + 2*sqrt(2))*(17 + 12*sqrt(2))^(n-1) - (sqrt(2)/8)*(3 - 2*sqrt(2))*(17 - 12*sqrt(2))^(n-1).
a(n) = (sqrt(2)/8)*( (17+12*sqrt(2))^(n-1/2) - (17-12*sqrt(2))^(n-1/2) ).
a(n) = (sqrt(2)/8)*( (3+2*sqrt(2))^(2n-1) - (3-2*sqrt(2))^(2n-1) ).
a(n) = (sqrt(2)/8)*( (1+sqrt(2))^(4n-2) - (1-sqrt(2))^(4n-2) ).
a(n) = 35*a(n-1) - 35*a(n-2) + a(n-3). (End)
a(n+1) = 17*a(n) + 6*sqrt(8*a(n)^2+1) for n >= 0. - Richard Choulet, May 01 2009
a(n) = b such that (-1)^(n+1) * Integral_{x=-Pi/2..Pi/2} cos((2*n-1)*x)/(3-sin(x)) dx = c + b*log(2). - Francesco Daddi, Aug 01 2011
a(n) are the nonzero integer square roots of A227970. - Richard R. Forberg, Aug 01 2013
a(n) = y/5, where y are solutions to: y^2 = 2x^2 - x - 3. - Richard R. Forberg, Nov 24 2013
a(n) = sqrt((A078522(n)+1)*(2*A078522(n)+1)). - Jonathan Sondow, Apr 07 2017
a(n) = Pell(4*n-2)/2. - G. C. Greubel, Jan 13 2020
a(n) = A001653(n)*A002315(n). - Gerry Martens, Mar 23 2024

Chebyshev comments from Wolfdieter Lang, Nov 29 2002

A057080 Even-indexed Chebyshev U-polynomials evaluated at sqrt(10)/2.

Original entry on oeis.org

1, 9, 71, 559, 4401, 34649, 272791, 2147679, 16908641, 133121449, 1048062951, 8251382159, 64962994321, 511452572409, 4026657584951, 31701808107199, 249587807272641, 1965000650073929, 15470417393318791, 121798338496476399

Offset: 0

Wolfdieter Lang, Aug 04 2000

a(n) = L(n,-8)*(-1)^n, where L is defined as in A108299; see also A070997 for L(n,+8). - Reinhard Zumkeller, Jun 01 2005
General recurrence is a(n)=(a(1)-1)*a(n-1)-a(n-2), a(1)>=4, lim n->infinity a(n)= x*(k*x+1)^n, k =(a(1)-3), x=(1+sqrt((a(1)+1)/(a(1)-3)))/2. Examples in OEIS: a(1)=4 gives A002878. a(1)=5 gives A001834. a(1)=6 gives A030221. a(1)=7 gives A002315. a(1)=8 gives A033890. a(1)=9 gives A057080. a(1)=10 gives A057081. - Ctibor O. Zizka, Sep 02 2008
The primes in this sequence are 71, 34649, 16908641, 8251382159, 31701808107199,... - Ctibor O. Zizka, Sep 02 2008
The aerated sequence (b(n))n>=1 = [1, 0, 9, 0, 71, 0, 559, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -6, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047. - Peter Bala, Mar 22 2015
From Klaus Purath, May 06 2025: (Start)
Nonnegative solutions to the Diophantine equation 3*a(n)^2 - 5*b(n)^2 = -2. The corresponding b(n) are A070997(n). Note that (a(n)*a(n+2) - a(n+1)^2)/2 = -5 and (b(n)*b(n+2) - b(n+1)^2)/2 = 3.
(a(n) + b(n))/2 = (a(n+1) - b(n+1))/2 = A001090(n+1) = Lucas U(8,1). Also a(n)*b(n+1) - a(n+1)*b(n) = -2.
a(n) = (t(i+2n+1) - t(i))/(t(i+n+1) - t(i+n)) as long as t(i+n+1) - t(i+n) != 0 for integer i and n >= 0 where (t) is a sequence satisfying t(i+3) = 9*t(i+2) - 9*t(i+1) + t(i) or t(i+2) = 8*t(i+1) - t(i), regardless of the initial values and including this sequence itself. (End)

G. C. Greubel, Table of n, a(n) for n = 0..1000
Marco Abrate, Stefano Barbero, Umberto Cerruti, and Nadir Murru, Polynomial sequences on quadratic curves, Integers, Vol. 15, 2015, #A38.
K. Andersen, L. Carbone, and D. Penta, Kac-Moody Fibonacci sequences, hyperbolic golden ratios, and real quadratic fields, Journal of Number Theory and Combinatorics, Vol 2, No. 3 pp 245-278, 2011. See Section 9.
Alex Fink, Richard K. Guy, and Mark Krusemeyer, Partitions with parts occurring at most thrice, Contributions to Discrete Mathematics, Vol 3, No 2 (2008), pp. 76-114. See Section 13.
Tanya Khovanova, Recursive Sequences
Wolfdieter Lang, On polynomials related to powers of the generating function of Catalan's numbers, Fib. Quart. 38 (2000) 408-419. Eq.(44), rhs, m=10.
Donatella Merlini and Renzo Sprugnoli, Arithmetic into geometric progressions through Riordan arrays, Discrete Mathematics 340.2 (2017): 160-174.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences Integers, Volume 12A (2012) The John Selfridge Memorial Volume
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (8,-1).

Cf. A033890, A100047.

a:=[1,9];; for n in [3..30] do a[n]:=8*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Dec 06 2019

I:=[1,9]; [n le 2 select I[n] else 8*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Mar 22 2015

A057080 := proc(n)
    option remember;
    if n <= 1 then
        op(n+1,[1,9]);
    else
        8*procname(n-1)-procname(n-2) ;
    end if;
end proc: # R. J. Mathar, Apr 30 2017

CoefficientList[Series[(1+x)/(1-8x+x^2), {x, 0, 33}], x] (* Vincenzo Librandi, Mar 22 2015 *)

Vec((1+x)/(1-8*x+x^2) + O(x^30)) \\ Michel Marcus, Mar 22 2015

[(lucas_number2(n,8,1)-lucas_number2(n-1,8,1))/6 for n in range(1, 21)] # Zerinvary Lajos, Nov 10 2009

For all elements x of the sequence, 15*x^2 + 10 is a square. Lim. n-> Inf. a(n)/a(n-1) = 4 + sqrt(15). - Gregory V. Richardson, Oct 13 2002
a(n) = 8*a(n-1) - a(n-2), a(-1)=-1, a(0)=1.
a(n) = S(n, 8) + S(n-1, 8) = S(2*n, sqrt(10)) with S(n, x) := U(n, x/2), Chebyshev polynomials of 2nd kind, A049310. S(n, 8) = A001090(n).
G.f.: (1+x)/(1-8*x+x^2).
a(n) = ( ((4+sqrt(15))^(n+1) - (4-sqrt(15))^(n+1)) + ((4+sqrt(15))^n - (4-sqrt(15))^n) )/(2*sqrt(15)). - Gregory V. Richardson, Oct 13 2002
a(n) = sqrt((5*A070997(n)^2 - 2)/3) (cf. Richardson comment).
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i) then a(n) = (-1)^n*q(n,-10). - Benoit Cloitre, Nov 10 2002
a(n) = Jacobi_P(n,1/2,-1/2,4)/Jacobi_P(n,-1/2,1/2,1); - Paul Barry, Feb 03 2006
a(n+1) = 4*a(n) + sqrt(5*(3*a(n)^2 + 2)). - Richard Choulet, Aug 30 2007
In addition to the first formula above: In general, the following applies to all recurrences (a(n)) of the form (8,-1) with a(0) = 1 and arbritrary a(1): 15*a(n)^2 + y = b^2 where y = x^2 + 8*x + 1 and x = a(1) - 8. Also y = a(k+1)^2 - a(k)*a(k+1) for any k >=0. - Klaus Purath, May 06 2025
From Peter Bala, May 09 2025: (Start)
a(n) = Dir(n, 4), where Dir(n, x) denotes the n-th row polynomial of the triangle A244419.
a(n)^2 - 8*a(n)*a(n+1) + a(n+1)^2 = 10.
More generally, for arbitrary x, a(n+x)^2 - 8*a(n+x)*a(n+x+1) + a(n+x+1)^2 = 10 with a(n) := ( ((4+sqrt(15))^(n+1) - (4-sqrt(15))^(n+1)) + ((4+sqrt(15))^n - (4-sqrt(15))^n) )/(2*sqrt(15)) as given above.
a(n+1/2) = sqrt(10) * A001090(n+1).
a(n+3/4) + a(n+1/4) = sqrt(10)*sqrt(sqrt(10) + 2) * A001090(n+1).
a(n+3/4) - a(n+1/4) = sqrt((sqrt(40) - 4)/3) * A001091(n+1).
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 1/a(n)) = 1/10 (telescoping series: for n >= 1, 10/(a(n) - 1/a(n)) = 1/A001090(n) + 1/A001090(n+1)).
Product_{n >= 1} (a(n) + 1)/(a(n) - 1) = sqrt(5/3) (telescoping product: Product_{n = 1..k} ((a(n) + 1)/(a(n) - 1))^2 = 5/3 * (1 - 2/(1 + A001091(k+1)))). (End)

A005319 a(n) = 6*a(n-1) - a(n-2).

Original entry on oeis.org

0, 4, 24, 140, 816, 4756, 27720, 161564, 941664, 5488420, 31988856, 186444716, 1086679440, 6333631924, 36915112104, 215157040700, 1254027132096, 7309005751876, 42600007379160, 248291038523084, 1447146223759344, 8434586304032980

Offset: 0

N. J. A. Sloane

Solutions y of the equation 2x^2-y^2=2; the corresponding x values are given by A001541. - N-E. Fahssi, Feb 25 2008
The lower intermediate convergents to 2^(1/2) beginning with 4/3, 24/17, 140/99, 816/577, form a strictly increasing sequence; essentially, numerators=A005319 and denominators=A001541. - Clark Kimberling, Aug 26 2008
Numbers n such that (ceiling(sqrt(n*n/2)))^2 = 1 + n*n/2. - Ctibor O. Zizka, Nov 09 2009
All nonnegative solutions of the indefinite binary quadratic form X^2 + 4*X*Y -4*Y^2 of discriminant 32, representing -4 are (X(n), Y(n)) = (a(n), A001653(n+1)), for n >= 0. - Wolfdieter Lang, Jun 13 2018
Also the number of edge covers in the n-triangular snake graph. - Eric W. Weisstein, Jun 08 2019
All of the positive integer solutions of a*b+1=x^2, a*c+1=y^2, b*c+1=z^2, x+z=2*y, 0A001542(n), b=A005319(n), c=A001542(n+1), x=A001541(n), y=A001653(n+1), z=A002315(n) with 0Michael Somos, Jun 26 2022
a(n) is the sum of 4*n consecutive powers of the silver ratio 1+sqrt(2), starting at (1+sqrt(2))^(-2*n) and ending at (1+sqrt(2))^(2*n-1). - Greg Dresden and Ruxin Sheng, Jul 25 2024

			G.f. = 4*x + 24*x^2 + 140*x^3 + 816*x^4 + 4756*x^5 + ... - _Michael Somos_, Jun 26 2022

P. de la Harpe, Topics in Geometric Group Theory, Univ. Chicago Press, 2000, p. 160, middle display.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Marius A. Burtea, Table of n, a(n) for n = 0..100
K. Andersen, L. Carbone, and D. Penta, Kac-Moody Fibonacci sequences, hyperbolic golden ratios, and real quadratic fields, Journal of Number Theory and Combinatorics, Vol 2, No. 3 pp 245-278, 2011. See Section 9.
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
John M. Campbell, An Integral Representation of Kekulé Numbers, and Double Integrals Related to Smarandache Sequences, arXiv preprint arXiv:1105.3399 [math.GM], 2011.
Tanya Khovanova, Recursive Sequences
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Soumeya M. Tebtoub, Hacène Belbachir, and László Németh, Integer sequences and ellipse chains inside a hyperbola, Proceedings of the 1st International Conference on Algebras, Graphs and Ordered Sets (ALGOS 2020), hal-02918958 [math.cs], 17-18.
Eric Weisstein's World of Mathematics, Edge Cover
Eric Weisstein's World of Mathematics, Triangular Snake Graph
Index entries for linear recurrences with constant coefficients, signature (6,-1).

Cf. A000129, A001109, A001541, A001542, A001652, A001653, A046090, A046729, A090390.

a:=[0,4]; [n le 2 select a[n] else 6*Self(n-1) - Self(n-2):n in [1..22]]; // Marius A. Burtea, Sep 19 2019

LinearRecurrence[{6, -1}, {0, 4}, 22] (* Jean-François Alcover, Sep 26 2017 *)
Table[((3 + 2 Sqrt[2])^n - (3 - 2 Sqrt[2])^n)/Sqrt[2], {n, 20}] // Expand (* Eric W. Weisstein, Jun 08 2019 *)
CoefficientList[Series[(4 x)/(1 - 6 x + x^2), {x, 0, 20}], x] (* Eric W. Weisstein, Jun 08 2019 *)
a[ n_] := 4*ChebyshevU[n-1, 3]; (* Michael Somos, Jun 26 2022 *)

{a(n) = 4*polchebyshev(n-1, 2, 3)}; /* Michael Somos, Jun 26 2022 */

G.f.: 4*x / ( 1-6*x+x^2 ). - Simon Plouffe in his 1992 dissertation.
G.f. for signed version beginning with 1: (1+2*x+x^2)/(1+6*x+x^2).
For any term n of the sequence, 2*n^2 + 4 is a perfect square. Limit_{n->infinity} a(n)/a(n-1) = 3 + 2*sqrt(2). - Gregory V. Richardson, Oct 06 2002
a(n) = ((3+2*sqrt(2))^n - (3-2*sqrt(2))^n) / sqrt(2). - Gregory V. Richardson, Oct 06 2002
(-1)^(n+1) = A090390(n+1) + A001542(n+1) + A046729(n) - a(n) (conjectured). - Creighton Dement, Nov 17 2004
For n > 0, a(n) = A000129(n+1)^2 - A000129(n-1)^2; a(n) = A046090(n-1) + A001652(n); e.g., 816 = 120 + 696; a(n) = A001653(n) - A001653(n-1); e.g., 816 = 985 - 169. - Charlie Marion Jul 22 2005
a(n) = 4*A001109(n). - M. F. Hasler, Mar 2009
For n > 1, a(n) is the denominator of continued fraction [1,4,1,4,...,1,4] with (n-1) repetitions of 1,4. For the numerators, see A001653. - Greg Dresden, Sep 10 2019
1/a(n) - 1/a(n+1) = 1/(Pell(2*n+1) - 1/Pell(2*n+1)) for n >= 1, where Pell(n) = A000129(n). - Peter Bala, Aug 21 2022
E.g.f.: sqrt(2)*exp(3*x)*sinh(2*sqrt(2)*x). - Stefano Spezia, Nov 25 2022
a(n) = 2*A000129(2*n). - Tanya Khovanova and MIT PRIMES STEP senior group, Apr 17 2024

A008844 Squares of sequence A001653: y^2 such that x^2 - 2*y^2 = -1 for some x.

Original entry on oeis.org

1, 25, 841, 28561, 970225, 32959081, 1119638521, 38034750625, 1292061882721, 43892069261881, 1491038293021225, 50651409893459761, 1720656898084610641, 58451683124983302025, 1985636569351347658201, 67453191674820837076801, 2291422880374557112953025

Offset: 0

N. J. A. Sloane

Numbers simultaneously square and centered square. E.g., a(1)=25 because 25 is the fourth centered square number and the fifth square number. - Steven Schlicker, Apr 24 2007
Solutions to A007913(x)=A007913(2x-1). - Benoit Cloitre, Apr 07 2002
From Ant King, Nov 09 2011: (Start)
Indices of positive hexagonal numbers that are also perfect squares.
As n increases, this sequence is approximately geometric with common ratio r = lim_{n -> infinity} a(n)/a(n-1) = (1 + sqrt(2))^4 = 17 + 12 * sqrt(2).
(End)
Also indices of hexagonal numbers (A000384) which are also centered octagonal numbers (A016754). - Colin Barker, Jan 25 2015
Also positive integers x in the solutions to 4*x^2 - 8*y^2 - 2*x + 8*y - 2 = 0, the corresponding values of y being A253826. - Colin Barker, Jan 25 2015
Squares that are sum of two consecutive squares: y^2 = (k + 1)^2 + k^2 is equivalent to x^2 - 2*y^2 = -1 with x = 2*k + 1. - Jean-Christophe Hervé, Nov 11 2015
Squares in the main diagonal of the natural number array, A000027. - Clark Kimberling, Mar 12 2023

			From _Ravi Kumar Davala_, May 26 2013: (Start)
A001333(0)=1, A001333(4)=17, A001333(8)=577, A000129(0)=0, A000129(2)=2, A000129(4)=12, A000129(8)=408 so clearly
a(n+m)=A001333(4*m)*a(n)-(A000129(2*m))^2+A000129(4*m)*sqrt(2*a(n)^2-a(n)), with m=1,2 is true.
A002203(0)=2, A002203(4)=34, A002203(8)=1154 so clearly
a(n+m)=(1/2)*A002203(4*m)*a(n)-(A000129(2*m))^2+A000129(4*m)*sqrt(2*a(n)^2-a(n)) is true for m=1,2
a(n+1)*a(n-1) = (a(n)+4)^2 , with n=1 is 841*1=(25+4)^2, for n=2 , 28561*25=(841+4)^2.
(End)
1 = 1 + 0, 25 = 16 + 9, 841 = 29^2 = 21^2 + 20^2 = 441 + 400.

Albert H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 256.

Muniru A Asiru, Table of n, a(n) for n = 0..100
Daniel C. Fielder, Special integer sequences controlled by three parameters, Fibonacci Quarterly 6, 1968, 64-70.
M. A. Gruber, Artemas Martin, A. H. Bell, J. H. Drummond, A. H. Holmes, and H. C. Wilkes, Problem 47, Amer. Math. Monthly, 4 (1897), 25-28.
Thomas Koshy, Products Involving Reciprocals of Gibonacci Polynomials, The Fibonacci Quarterly, Vol. 60, No. 1 (2022), pp. 15-24.
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Circular Segment, Forum Geometricorum, Vol. 18 (2018), 47-55.
Steven C. Schlicker, Numbers Simultaneously Polygonal and Centered Polygonal, Mathematics Magazine, Vol. 84, No. 5, December 2011
Eric Weisstein's World of Mathematics, Hexagonal Square Number.
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. A000290, A001844, A007913, A000384, A046177, A016754, A253826.

a := [1, 25, 841];; for i in [4..10^2] do a[i] := 35*a[i-1] - 35*a[i-2] + a[i-3]; od; a;  # Muniru A Asiru, Jan 17 2018

I:=[1,25,841]; [n le 3 select I[n] else 35*Self(n-1)-35*Self(n-2)+Self(n-3): n in [1..20]]; // Vincenzo Librandi, Jan 20 2018

CP := n -> 1+1/2*4*(n^2-n): N:=10: u:=3: v:=1: x:=4: y:=1: k_pcp:=[1]: for i from 1 to N do tempx:=x; tempy:=y; x:=tempx*u+8*tempy*v: y:=tempx*v+tempy*u: s:=(y+1)/2: k_pcp:=[op(k_pcp),CP(s)]: end do: k_pcp; # Steven Schlicker, Apr 24 2007

LinearRecurrence[{35, -35, 1}, {1, 25, 841}, 15] (* Ant King, Nov 09 2011 *)
CoefficientList[Series[(1 - 10 x + x^2) / ((1 - x) (1 - 34 x + x^2)), {x, 0, 33}], x] (* Vincenzo Librandi, Jan 20 2018 *)

a(n)=if(n<0,0,sqr(subst(poltchebi(n+1)+poltchebi(n),x,3)/4))

vector(40, n, n--; (([5, 2; 2, 1]^n)[1, 1])^2) \\ Altug Alkan, Nov 11 2015

From Benoit Cloitre, Jan 19 2003: (Start)
a(n) = A078522(n) + 1.
a(n) = ceiling(A*B^n) where A = (3 + 2*sqrt(2))/8 and B = 17 + 12*sqrt(2). (End)
G.f.: (1-10x+x^2)/((1-x)(1-34x+x^2)).
a(n) = ceiling(A046176(n)/sqrt(2)). - Helge Robitzsch (hrobi(AT)math.uni-goettingen.de), Jul 28 2000
a(n+1) = 17*a(n) - 4 + 12*sqrt(2*a(n)^2 - a(n)). - Richard Choulet, Sep 14 2007
Define x(n) + y(n)*sqrt(8) = (4+sqrt(8))*(3+sqrt(8))^n, s(n) = (y(n)+1)/2; then a(n) = (1/2)*(2+4*(s(n)^2 - s(n))). - Steven Schlicker, Apr 24 2007
From Ant King, Nov 09 2011: (Start)
a(n) = 35*a(n-1) - 35*a(n-2) + a(n-3).
a(n) = 34*a(n-1) - a(n-2) - 8.
a(n) = 1/8 * ((1 + sqrt(2))^(4*n-2) + (1 - sqrt(2))^(4*n-2) + 2).
a(n) = ceiling((1/8) * (1 + sqrt(2))^(4*n-2)). (End)
From Ravi Kumar Davala, May 26 2013: (Start)
a(n+2) = 577*a(n) - 144 + 408*sqrt(2*a(n)^2 - a(n)).
a(n+m) = A001333(4*m)*a(n) - (A000129(2*m))^2 + A000129(4*m)*sqrt(2*a(n)^2 - a(n)).
a(n+m) = (1/2)*A002203(4*m)*a(n) - (A000129(2*m))^2 + A000129(4*m)*sqrt(2*a(n)^2 - a(n)).
a(n+1)*a(n-1) = (a(n)+4)^2. (End)
a(n) = A001652(n)^2 + A046090(n)^2. - César Aguilera, Jan 15 2018
Limit_{n -> infinity} a(n)/a(n-1) = A156164. - César Aguilera, Jan 28 2018
sqrt(2*a(n))-1 = A002315(n). - Ezhilarasu Velayutham, Apr 05 2019
4*a(n) = 1 +3*A077420(n). - R. J. Mathar, Mar 05 2024
Product_{n>=0} (1 + 4/a(n)) = 2*sqrt(2) + 3 (Koshy, 2022, section 3, p. 19). - Amiram Eldar, Jan 23 2025

Entry edited by N. J. A. Sloane, Sep 14 2007

A057084 Scaled Chebyshev U-polynomials evaluated at sqrt(2).

Original entry on oeis.org

1, 8, 56, 384, 2624, 17920, 122368, 835584, 5705728, 38961152, 266043392, 1816657920, 12404916224, 84706066432, 578409201664, 3949625081856, 26969727041536, 184160815677440, 1257528709087232, 8586943147278336

Offset: 0

Wolfdieter Lang, Aug 11 2000

From Kival Ngaokrajang, Dec 14 2014 (Start):
-2*a(n-1) is the irrational part of the integer in Q(sqrt 2) giving the length of a Levy C-curve variant L(n)=(2*(2- sqrt 2))^n at iteration step n. The length of this C-curve is an integer in the real quadratic number field Q(sqrt 2), namely L(n) = A(n)+B(n)*sqrt(2) with A(n) = A084130(n) and B(n) = -2*a(n-1). See the construction rule and the illustration in the links.
The fractal dimension of the Levy C-curve is 2, but for this modified case it is log(4)/log(2 + sqrt 2) = 1.1289527...
(End)
For lim_{n->oo} a(n+1)/a(n) = 2*(2 + sqrt(2)) = 6.82842... see A365823. - Wolfdieter Lang, Nov 15 2023

			The first pairs [A(n),B(n)] determining the length L(n) are : [1, 0], [4, -2], [24, -16], [160, -112], [1088, -768], [7424, -5248], [50688, -35840], [346112, -244736], [2363392, -1671168], [16138240, -11411456], ... _Kival Ngaokrajang_, Dec 14 2014

S. Falcon, Iterated Binomial Transforms of the k-Fibonacci Sequence, British Journal of Mathematics & Computer Science, 4 (22): 2014.

G. C. Greubel, Table of n, a(n) for n = 0..1000
A. F. Horadam, Special properties of the sequence W_n(a,b; p,q), Fib. Quart., 5.5 (1967), 424-434. Case n->n+1, a=0,b=1; p=8, q=-8.
W. Lang, On polynomials related to powers of the generating function of Catalan's numbers, Fib. Quart. 38 (2000) 408-419. Eqs.(38) and (45),lhs, m=8.
Kival Ngaokrajang, Illustration of construction rule and initial terms
Wikipedia, Lévy C curve
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (8,-8).

Cf. A001109, A002315, A049310, A084130, A109466, A251732, A251733, A365823.

Join[{a=1,b=8},Table[c=8*b-8*a;a=b;b=c,{n,60}]] (* Vladimir Joseph Stephan Orlovsky, Jan 19 2011 *)
LinearRecurrence[{8,-8},{1,8},30] (* Harvey P. Dale, Feb 07 2015 *)

x='x+O('x^50); Vec(1/(1-8*x+8*x^2)) \\ G. C. Greubel, Jul 03 2017

[lucas_number1(n,8,8) for n in range(1, 21)] # Zerinvary Lajos, Apr 23 2009

a(n) = 8*(a(n-1)-a(n-2)), a(-1)=0, a(0)=1.
a(n) = S(n, 2*sqrt(2))*(2*sqrt(2))^n with S(n, x) := U(n, x/2), Chebyshev's polynomials of the 2nd kind, A049310.
a(2*k) = A002315(k)*8^k, a(2*k+1) = A001109(k+1)*8^(k+1).
G.f.: 1/(1-8*x+8*x^2).
a(n) = Sum_{k, 0<=k<=n} A109466(n,k)*8^k. [Philippe Deléham, Oct 28 2008]
Binomial transform of A002315. [Al Hakanson (hawkuu(AT)gmail.com), Aug 15 2009]

cp's OEIS Frontend

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A030221 Chebyshev even-indexed U-polynomials evaluated at sqrt(7)/2.

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A088165 NSW primes: NSW numbers that are also prime.

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A054320 Expansion of g.f.: (1 + x)/(1 - 10*x + x^2).

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A075528 Triangular numbers that are half other triangular numbers.

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A046176 Indices of square numbers that are also hexagonal.

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