cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A001834 a(0) = 1, a(1) = 5, a(n) = 4*a(n-1) - a(n-2).

Original entry on oeis.org

1, 5, 19, 71, 265, 989, 3691, 13775, 51409, 191861, 716035, 2672279, 9973081, 37220045, 138907099, 518408351, 1934726305, 7220496869, 26947261171, 100568547815, 375326930089, 1400739172541, 5227629760075, 19509779867759, 72811489710961, 271736178976085
Offset: 0

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Author

Keywords

Comments

Sequence also gives values of x satisfying 3*y^2 - x^2 = 2, the corresponding y being given by A001835(n+1). Moreover, quadruples(p, q, r, s) satisfying p^2 + q^2 + r^2 = s^2, where p = q and r is either p+1 or p-1, are termed nearly isosceles Pythagorean and are given by p = {x + (-1)^n}/3, r = p-(-1)^n, s = y for n > 1. - Lekraj Beedassy, Jul 19 2002
a(n)= A002531(1+2*n). - Anton Vrba (antonvrba(AT)yahoo.com), Feb 14 2007
361 written in base A001835(n+1) - 1 is the square of a(n). E.g., a(12) = 2672279, A001835(13) - 1 = 1542840. We have 361_(1542840) = 3*1542840 + 6*1542840 + 1 = 2672279^2. - Richard Choulet, Oct 04 2007
The lower principal convergents to 3^(1/2), beginning with 1/1, 5/3, 19/11, 71/41, comprise a strictly increasing sequence; numerators=A001834, denominators=A001835. - Clark Kimberling, Aug 27 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 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 A299109. a(1) = 7 gives A002315, primes in it A088165. a(1) = 8 gives A033890, primes in it not in OEIS (do there exist any?). a(1) = 9 gives A057080, primes in {71, 34649, 16908641, ...}. a(1) = 10 gives A057081, primes in it {389806471, 192097408520951, ...}. - Ctibor O. Zizka, Sep 02 2008
Inverse binomial transform of A030192. - Philippe Deléham, Nov 19 2009
For positive n, a(n) equals the permanent of the (2*n) X (2*n) tridiagonal matrix with sqrt(6)'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
x-values in the solution to 3x^2 + 6 = y^2 (see A082841 for the y-values). - Sture Sjöstedt, Nov 25 2011
Pisano period lengths: 1, 1, 2, 4, 3, 2, 8, 4, 6, 3, 10, 4, 12, 8, 6, 8, 18, 6, 5, 12, ... - R. J. Mathar, Aug 10 2012
The aerated sequence (b(n))A100047%20for%20a%20connection%20with%20Chebyshev%20polynomials.%20-%20_Peter%20Bala">{n>=1} = [1, 0, 5, 0, 19, 0, 71, 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 = -2, 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
Yong Hao Ng has shown that for any n, a(n) is coprime with any member of A001835 and with any member of A001075. - René Gy, Feb 26 2018
From Wolfdieter Lang, Oct 15 2020: (Start)
((-1)^n)*a(n) = X(n) = (-1)^n*(S(n, 4) + S(n-1, 4)) and Y(n) = X(n-1) gives all integer solutions (modulo sign flip between X and Y) of X^2 + Y^2 + 4*X*Y = +6, 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 12, representing 6, 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)
Floretion Algebra Multiplication Program, FAMP Code: A001834 = (4/3)vesseq[ - .25'i + 1.25'j - .25'k - .25i' + 1.25j' - .25k' + 1.25'ii' + .25'jj' - .75'kk' + .75'ij' + .25'ik' + .75'ji' - .25'jk' + .25'ki' - .25'kj' + .25e], apart from initial term

Examples

			G.f. = 1 + 5*x + 19*x^2 + 71*x^3 + 265*x^4 + 989*x^5 + 3691*x^6 + ...
		

References

  • Bastida, Julio R. Quadratic properties of a linearly recurrent sequence. Proceedings of the Tenth Southeastern Conference on Combinatorics, Graph Theory and Computing (Florida Atlantic Univ., Boca Raton, Fla., 1979), pp. 163--166, Congress. Numer., XXIII-XXIV, Utilitas Math., Winnipeg, Man., 1979. MR0561042 (81e:10009)
  • Leonhard Euler, (E388) Vollstaendige Anleitung zur Algebra, Zweiter Theil, reprinted in: Opera Omnia. Teubner, Leipzig, 1911, Series (1), Vol. 1, p. 375.
  • Serge Lang, Introduction to Diophantine Approximations, Addison-Wesley, New York, 1966.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • P.-F. Teilhet, Reply to Query 2094, L'Intermédiaire des Mathématiciens, 10 (1903), 235-238.

Crossrefs

A bisection of sequence A002531.
Cf. A001352, A001835, A086386 (prime members).
Cf. A026150.
a(n)^2+1 = A094347(n+1).

Programs

  • Haskell
    a001834 n = a001834_list !! (n-1)
    a001834_list = 1 : 5 : zipWith (-) (map (* 4) $ tail a001834_list) a001834_list
    -- Reinhard Zumkeller, Jan 23 2012
    
  • Magma
    I:=[1,5]; [n le 2 select I[n] else 4*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Mar 22 2015
  • Maple
    f:=n->((1+sqrt(3))^(2*n+1)+(1-sqrt(3))^(2*n+1))/2^(n+1); # N. J. A. Sloane, Nov 10 2009
  • Mathematica
    a[0] = 1; a[1] = 5; a[n_] := a[n] = 4a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 25}] (* Robert G. Wilson v, Apr 24 2004 *)
    Table[Expand[((1+Sqrt[3])^(2*n+1)+(1+Sqrt[3])^(2*n+1))/2^(n+1)],{n, 0, 20}] (* Anton Vrba, Feb 14 2007 *)
    LinearRecurrence[{4, -1}, {1, 5}, 50] (* Sture Sjöstedt, Nov 27 2011 *)
    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 A002531 *)
    a[3, 20] (* Gerry Martens, Jun 07 2015 *)
    Round@Table[LucasL[2n+1, Sqrt[2]]/Sqrt[2], {n, 0, 20}] (* Vladimir Reshetnikov, Sep 15 2016 *)
  • PARI
    {a(n) = real( (2 + quadgen(12))^n * (1 + quadgen(12)) )}; /* Michael Somos, Sep 19 2008 */
    
  • PARI
    {a(n) = subst( polchebyshev(n-1, 2) + polchebyshev(n, 2), x, 2)}; /* Michael Somos, Sep 19 2008 */
    
  • SageMath
    [(lucas_number2(n,4,1)-lucas_number2(n-1,4,1))/2 for n in range(1, 27)] # Zerinvary Lajos, Nov 10 2009
    

Formula

a(n) = ((1 + sqrt(3))^(2*n + 1) + (1 - sqrt(3))^(2*n + 1))/2^(n + 1). - N. J. A. Sloane, Nov 10 2009
a(n) = (1/2) * ((1 + sqrt(3))*(2 + sqrt(3))^n + (1 - sqrt(3))*(2 - sqrt(3))^n). - Dean Hickerson, Dec 01 2002
From Mario Catalani, Apr 11 2003: (Start)
With a = 2 + sqrt(3), b = 2 - sqrt(3): a(n) = (1/sqrt(2))(a^(n + 1/2) - b^(n + 1/2)).
a(n) - a(n-1) = A003500(n).
a(n) = sqrt(1 + 12*A061278(n) + 12*A061278(n)^2). (End)
a(n) = ((1 + sqrt(3))^(2*n + 1) + (1 - sqrt(3))^(2*n + 1))/2^(n + 1). - Anton Vrba, Feb 14 2007
G.f.: (1 + x)/((1 - 4*x + x^2)). Simon Plouffe in his 1992 dissertation.
a(n) = S(2*n, sqrt(6)) = S(n, 4) + S(n-1, 4); S(n, x) := U(n, x/2), Chebyshev polynomials of 2nd kind, A049310. S(n, 4) = A001353(n).
For all members x of the sequence, 3*x^2 + 6 is a square. Limit_{n->infinity} a(n)/a(n-1) = 2 + sqrt(3). - Gregory V. Richardson, Oct 10 2002
a(n) = 2*A001571(n) + 1. - Bruce Corrigan (scentman(AT)myfamily.com), Nov 04 2002
Let q(n, x) = Sum_{i=0..n} x^(n - i)*binomial(2*n - i, i); then (-1)^n*q(n, -6) = a(n). - Benoit Cloitre, Nov 10 2002
a(n) = 2^(-n)*Sum_{k>=0} binomial(2*n + 1, 2*k)*3^k; see A091042. - Philippe Deléham, Mar 01 2004
a(n) = floor(sqrt(3)*A001835(n+1)). - Philippe Deléham, Mar 03 2004
a(n+1) - 2*a(n) = 3*A001835(n+1). Using the known relation A001835(n+1) = sqrt((a(n)^2 + 2)/3) it follows that a(n+1) - 2*a(n) = sqrt(3*(a(n)^2 + 2)). Therefore a(n+1)^2 + a(n)^2 - 4*a(n+1)*a(n) - 6 = 0. - Creighton Dement, Apr 18 2005
a(n) = L(n,-4)*(-1)^n, where L is defined as in A108299; see also A001835 for L(n,+4). - Reinhard Zumkeller, Jun 01 2005
a(n) = Jacobi_P(n, 1/2, -1/2, 2)/Jacobi_P(n, -1/2, 1/2, 1). - Paul Barry, Feb 03 2006
Equals binomial transform of A026150 starting (1, 4, 10, 28, 76, ...) and double binomial transform of (1, 3, 3, 9, 9, 27, 27, 81, 81, ...). - Gary W. Adamson, Nov 30 2007
Sequence satisfies 6 = f(a(n), a(n+1)) where f(u, v) = u^2 + v^2 - 4*u*v. - Michael Somos, Sep 19 2008
a(-1-n) = -a(n). - Michael Somos, Sep 19 2008
From Franck Maminirina Ramaharo, Nov 11 2018: (Start)
a(n) = (-1)^n*(5*A125905(n) + A125905(n+1)).
E.g.f.: exp(2*x)*(cosh(sqrt(3)*x) + sqrt(3)*sinh(sqrt(3)*x)). (End)
a(n) = A061278(n+1) - A061278(n-1) for n>=2. - John P. McSorley, Jun 20 2020
From Peter Bala, May 09 2025: (Start)
a(n) = Dir(n, 2), where Dir(n, x) denotes the n-th row polynomial of the triangle A244419.
a(n) - 2*a(n-1) = 3 * A001835(n) for n >= 1.
For arbitrary x, a(n+x)^2 - 4*a(n+x)*a(n+x+1) + a(n+x+1)^2 = 6 with a(n) := (1/2) * ((1 + sqrt(3))*(2 + sqrt(3))^n + (1 - sqrt(3))*(2 - sqrt(3))^n) as above. The particular case x = 0 is noted above,
a(n+1/2) = sqrt(6) * A001353(n+1).
a(n+3/4) + a(n+1/4) = sqrt(6*sqrt(6) + 12) * A001353(n+1).
a(n+3/4) - a(n+1/4) = sqrt(2*sqrt(6) - 4) * A001075(n+1).
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 1/a(n)) = 1/6 (telescoping series: for n >= 1, 1/(a(n) - 1/a(n)) = 1/A001352(n) + 1/A001352(n+1)).
Product_{n >= 1} (a(n) + 1)/(a(n) - 1) = sqrt(3) (telescoping product: Product_{n = 1..k} ((a(n) + 1)/(a(n) - 1))^2 = 3*(1 - 2/A102206(k))). (End)

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

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Comments

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)

Examples

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

Crossrefs

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

Programs

  • Magma
    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
    
  • Maple
    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
  • Mathematica
    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 *)
  • PARI
    {a(n) = simplify(polchebyshev(2*n, 2, quadgen(28)/2))}; /* Michael Somos, Jan 22 2017 */
  • Sage
    [(lucas_number2(n,5,1)-lucas_number2(n-1,5,1))/3 for n in range(1,22)] # Zerinvary Lajos, Nov 10 2009
    

Formula

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)

A298675 Rectangular array A: first differences of row entries of array A294099, read by antidiagonals.

Original entry on oeis.org

1, 2, -1, 3, 2, -2, 4, 7, 2, -1, 5, 14, 18, 2, 1, 6, 23, 52, 47, 2, 2, 7, 34, 110, 194, 123, 2, 1, 8, 47, 198, 527, 724, 322, 2, -1, 9, 62, 322, 1154, 2525, 2702, 843, 2, -2, 10, 79, 488, 2207, 6726, 12098, 10084, 2207, 2, -1, 11, 98, 702, 3842, 15127, 39202, 57965, 37634, 5778, 2, 1
Offset: 1

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Comments

From a problem in A269254. For detailed theory, see [Hone].
From Charles L. Hohn, Sep 28 2024: (Start)
For rows n >= 3, values x >= 3 where (x^2-4)/(n^2-4) is a square.
For rows n >= 3, Lim_{k->oo}(T(n, k+1)/T(n, k)) = (sqrt(n^2-4)+n)/2. (End)

Examples

			Array begins:
   1 -1  -2   -1     1      2       1       -1        -2         -1
   2  2   2    2     2      2       2        2         2          2
   3  7  18   47   123    322     843     2207      5778      15127
   4 14  52  194   724   2702   10084    37634    140452     524174
   5 23 110  527  2525  12098   57965   277727   1330670    6375623
   6 34 198 1154  6726  39202  228486  1331714   7761798   45239074
   7 47 322 2207 15127 103682  710647  4870847  33385282  228826127
   8 62 488 3842 30248 238142 1874888 14760962 116212808  914941502
   9 79 702 6239 55449 492802 4379769 38925119 345946302 3074591599
  10 98 970 9602 95050 940898 9313930 92198402 912670090 9034502498
		

Crossrefs

Programs

  • Mathematica
    t[n_, 0] := 2; t[n_, 1] := n; t[n_, k_] := n*t[n, k - 1] - t[n, k - 2]; Table[t[n, k], {n, 10}, {k, 10}] // Grid

Formula

A(n,k) = T_k(n), n >= 1, k >= 1, where T_j(x) = x*T_{j-1}(x) - T_{j-2}(x), j >= 2, T_0(x) = 2, T_1(x) = x, (dilated Chebyshev polynomials of the first kind).

A269254 To find a(n), define a sequence by s(k) = n*s(k-1) - s(k-2), with s(0) = 1, s(1) = n + 1; then a(n) is the smallest index k such that s(k) is prime, or -1 if no such k exists.

Original entry on oeis.org

1, 1, 2, 1, 2, 1, -1, 2, 2, 1, 2, 1, 2, -1, 2, 1, 3, 1, 2, 2, 2, 1, -1, 2, 6, 2, 3, 1, 3, 1, 2, 9, 9, -1, 2, 1, 6, 2, 2, 1, 2, 1, 5, 2, 2, 1, -1, 2, 5, 2, 9, 1, 2, 2, 2, 2, 6, 1, 2, 1, 14, -1, 5, 2, 2, 1, 5, 2, 3, 1, 6, 1, 8, 3, 6, 2, 3, 1, -1, 3, 18, 1, 2, 3, 2, 2, 3, 1, 2, 9, 3, 5, 2, 2, 96, 1, 3, -1, 5, 1, 2, 1, 2, 15, 14, 1, 44, 1, 3, -1
Offset: 1

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Author

Arkadiusz Wesolowski, Jul 09 2016

Keywords

Comments

The s(k) sequences can be viewed in A294099, where they appear as rows. - Peter Munn, Aug 31 2020
For n >= 3, a(n) is that positive integer k yielding the smallest prime of the form (x^y - 1/x^y)/(x - 1/x), where x = (sqrt(n+2) +- sqrt(n-2))/2 and y = 2*k + 1, or -1 if no such k exists.
Every positive term belongs to A005097.
When n=7, the sequence {s(k)} is A033890, which is Fibonacci(4i+2), and since x|y <=> F_x|F_y, and 2i+1|4i+2, A033890 is never prime, and so a(7)=-1. For the other -1 terms below 100, see the theorem below and the Klee link - N. J. A. Sloane, Oct 20 2017 and Oct 22 2017
Theorem (Brad Klee): For all n > 2, a(n^2 - 2) = -1. See Klee link for a proof. - L. Edson Jeffery, Oct 22 2017
Theorem (Based on work of Hans Havermann, L. Edson Jeffery, Brad Klee, Don Reble, Bob Selcoe, and N. J. A. Sloane) a(110) = -1. [For proof see link. - N. J. A. Sloane, Oct 23 2017]
From Bob Selcoe, Oct 24 2017, edited by N. J. A. Sloane, Oct 27 2017: (Start)
Suppose n = m^2 - 2, where m >= 3, and let j = m-2, with j >= 1.
For this value of n, the sequence s(k) satisfies s(k) = (c(k) + d(k))*(c(k) - d(k)), where c(0) = 1, d(0) = 0; and for k >= 1: c(k) = (j+2)*c(k-1) - d(k-1), and d(k) = c(k-1). So (as Brad Klee already proved) a(n) = -1 .
We have s(0) = 1 and s(1) = n+1 = j^2 + 4j + 3. In general, the coefficients of s(k) when expanded in powers of j are given by the (4k+2)-th row of A011973 (the triangle of coefficients of Fibonacci polynomials) in reverse order. For example, s(2) = j^4 + 8j^3 + 21j^2 + 20j + 5, s(3) = j^6 + 12j^5 + 55j^4 + 120j^3 + 126j^2 + 56j + 7, etc.
Perhaps the above comments could be generalized to apply to a(110) or to other n for which a(n) = -1?
(End)
For detailed theory, see [Hone]. - L. Edson Jeffery, Feb 09 2018

Examples

			Let b(k) be the recursive sequence defined by the initial conditions b(0) = 1, b(1) = 16, and the recursive equation b(k) = 15*b(k-1) - b(k-2). a(15) = 2 because b(2) = 239 is the smallest prime in b(k).
Let c(k) be the recursive sequence defined by the initial conditions c(0) = 1, c(1) = 18, and the recursive equation c(k) = 17*c(k-1) - c(k-2). a(17) = 3 because c(3) = 5167 is the smallest prime in c(k).
		

Crossrefs

Programs

  • Magma
    lst:=[]; for n in [1..85] do if n gt 2 and IsSquare(n+2) then Append(~lst, -1); else a:=n+1; c:=1; t:=1; if IsPrime(a) then Append(~lst, t); else repeat b:=n*a-c; c:=a; a:=b; t+:=1; until IsPrime(a); Append(~lst, t); end if; end if; end for; lst;
    
  • Mathematica
    kmax = 100;
    a[1] = a[2] = 1;
    a[n_ /; IntegerQ[Sqrt[n+2]]] = -1;
    a[n_] := Module[{s}, s[0] = 1; s[1] = n+1; s[k_] := s[k] = n s[k-1] - s[k-2]; For[k=1, k <= kmax, k++, If[PrimeQ[s[k]], Return[k]]]; Print["For n = ", n, ", k = ", k, " exceeds the limit kmax = ", kmax]; -1];
    Array[a, 110] (* Jean-François Alcover, Aug 05 2018 *)
  • PARI
    allocatemem(2^30);
    default(primelimit,(2^31)+(2^30));
    s(n,k) = if(0==k,1,if(1==k,(1+n),((n*s(n,k-1)) - s(n,k-2))));
    A269254(n) = { my(k=1); if((n>2)&&issquare(2+n),-1,while(!isprime(s(n,k)),k++);(k)); }; \\ Antti Karttunen, Oct 20 2017

Formula

If n is prime then a(n-1) = 1.

Extensions

a(86)-a(94) from Antti Karttunen, Oct 20 2017
a(95)-a(109) appended by L. Edson Jeffery, Oct 22 2017

A294099 Rectangular array read by (upward) antidiagonals: A(n,k) = Sum_{j=0..k} (-1)^floor(j/2)*binomial(k-floor((j+1)/2), floor(j/2))*n^(k-j), n >= 1, k >= 0.

Original entry on oeis.org

1, 1, 2, 1, 3, 1, 1, 4, 5, -1, 1, 5, 11, 7, -2, 1, 6, 19, 29, 9, -1, 1, 7, 29, 71, 76, 11, 1, 1, 8, 41, 139, 265, 199, 13, 2, 1, 9, 55, 239, 666, 989, 521, 15, 1, 1, 10, 71, 377, 1393, 3191, 3691, 1364, 17, -1, 1, 11, 89, 559, 2584, 8119, 15289, 13775, 3571, 19, -2
Offset: 1

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Author

Keywords

Comments

This array is used to compute A269254: A269254(n) = least k such that A(n,k) is a prime, or -1 if no such k exists.
For detailed theory, see [Hone]. - L. Edson Jeffery, Feb 09 2018
The array can be extended to k<0 with A(n, k) = -A(n, -k-1) for all k in Z. - Michael Somos, Jun 19 2023

Examples

			Array begins:
  1   2    1    -1     -2      -1        1         2          1          -1
  1   3    5     7      9      11       13        15         17          19
  1   4   11    29     76     199      521      1364       3571        9349
  1   5   19    71    265     989     3691     13775      51409      191861
  1   6   29   139    666    3191    15289     73254     350981     1681651
  1   7   41   239   1393    8119    47321    275807    1607521     9369319
  1   8   55   377   2584   17711   121393    832040    5702887    39088169
  1   9   71   559   4401   34649   272791   2147679   16908641   133121449
  1  10   89   791   7030   62479   555281   4935050   43860169   389806471
  1  11  109  1079  10681  105731  1046629  10360559  102558961  1015229051
		

Crossrefs

Programs

  • Mathematica
    (* Array: *)
    Grid[Table[LinearRecurrence[{n, -1}, {1, 1 + n}, 10], {n, 10}]]
    (* Array antidiagonals flattened (gives this sequence): *)
    A294099[n_, k_] := Sum[(-1)^(Floor[j/2]) Binomial[k - Floor[(j + 1)/2], Floor[j/2]] n^(k - j), {j, 0, k}]; Flatten[Table[A294099[n - k, k], {n, 11}, {k, 0, n - 1}]]
  • PARI
    {A(n, k) = sum(j=0, k, (-1)^(j\2)*binomial(k-(j+1)\2, j\2)*n^(k-j))}; /* Michael Somos, Jun 19 2023 */

Formula

A(n,0) = 1, A(n,1) = n + 1, A(n,k) = n*A(n,k-1) - A(n,k-2), n >= 1, k >= 2.
G.f. for row n: (1 + x)/(1 - n*x + x^2), n >= 1.
A(n, k) = B(-n, k) where B = A299045. - Michael Somos, Jun 19 2023

A298677 a(n) = 110*a(n-1) - a(n-2) for n >= 2, a(0)=1, a(1)=111.

Original entry on oeis.org

1, 111, 12209, 1342879, 147704481, 16246150031, 1786928798929, 196545921732159, 21618264461738561, 2377812544869509551, 261537761671184312049, 28766775971285404815839, 3164083819079723345430241, 348020453322798282592510671, 38279085781688731361830743569, 4210351415532437651518789281919
Offset: 0

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Author

Keywords

Comments

Sequence {s_k(110)} of A269254.
The sequence contains no primes; see A269254 for a proof by N. J. A. Sloane.
For detailed theory, see [Hone]. - L. Edson Jeffery, Feb 09 2018

Crossrefs

Programs

  • Mathematica
    s[0, n_] := 1; s[1, n_] := n + 1; s[k_, n_] := n*s[k - 1, n] - s[k - 2, n]; Table[s[k, 110], {k, 0, 15}]
    LinearRecurrence[{110, -1}, {1, 111}, 15]
    CoefficientList[Series[(1 + x)/(1 - 110*x + x^2), {x, 0, 14}], x]
  • PARI
    Vec((1 + x)/(1 - 110*x + x^2) + O(x^20)) \\ Colin Barker, Jan 25 2018

Formula

G.f.: (1 + x)/(1 - 110*x + x^2).
a(n) = (1/18)*((55+12*sqrt(21))^(-n)*(9-2*sqrt(21) + (9+2*sqrt(21))*(55+12*sqrt(21))^(2*n))). - Colin Barker, Jan 25 2018

A117522 Numbers k such that L(2*k + 1) is prime, where L(m) is a Lucas number.

Original entry on oeis.org

2, 3, 5, 6, 8, 9, 15, 18, 20, 23, 26, 30, 35, 39, 56, 156, 176, 251, 306, 308, 431, 548, 680, 2393, 2396, 2925, 3870, 4233, 5345, 6125, 6981, 7224, 9734, 17724, 18389, 22253, 25584, 28001, 40835, 44924, 47411, 70028, 74045, 79760, 91544, 96600, 101333, 172146, 193716, 221804, 266138, 287109, 308393, 315590, 318875, 325910, 346073, 450828, 525924
Offset: 1

Views

Author

Parthasarathy Nambi, Apr 26 2006

Keywords

Comments

For n = 24..43, we can only claim that L(2*a(n) + 1) is a probable prime. Sequence arises in a study of A269254; for detailed theory, see [Hone]. - L. Edson Jeffery, Feb 09 2018

Examples

			If k = 56, then L(2*k + 1) is a prime with twenty-four digits.
		

Crossrefs

Extensions

Values beyond 680 from L. Edson Jeffery, et al., Feb 02 2018
a(44)-a(56) from Robert Price, Jun 12 2025
a(57)-a(59) (using data in A001606) from Alois P. Heinz, Jun 12 2025

A269253 Smallest prime in the sequence s(k) = n*s(k-1) - s(k-2), with s(0) = 1, s(1) = n + 1 (or -1 if no such prime exists).

Original entry on oeis.org

2, 3, 11, 5, 29, 7, -1, 71, 89, 11, 131, 13, 181, -1, 239, 17, 5167, 19, 379, 419, 461, 23, -1, 599, 251894449, 701, 20357, 29, 25171, 31, 991, 36002209323169, 47468744103199, -1, 1259, 37, 2625505273, 1481, 1559, 41, 1721, 43, 150103799, 1979, 2069, 47, -1, 2351, 287762399
Offset: 1

Views

Author

Arkadiusz Wesolowski, Jul 09 2016

Keywords

Comments

For n >= 3, smallest prime of the form (x^y - 1/x^y)/(x - 1/x), where x = (sqrt(n+2) +/- sqrt(n-2))/2 and y is an odd positive integer, or -1 if no such prime exists.
When n=7, the sequence {s(k)} is A033890, which is Fibonacci(4i+2), and since x|y <=> F_x|F_y, and 2i+1|4i+2, A033890 is never prime, and so a(7) = -1. What is the proof for the other entries that are -1? Answer: See the Comments in A269254. - N. J. A. Sloane, Oct 22 2017
For detailed theory, see [Hone]. - L. Edson Jeffery, Feb 09 2018

Crossrefs

Programs

  • Magma
    lst:=[]; for n in [1..49] do if n gt 2 and IsSquare(n+2) then Append(~lst, -1); else a:=n+1; c:=1; if IsPrime(a) then Append(~lst, a); else repeat b:=n*a-c; c:=a; a:=b; until IsPrime(a); Append(~lst, a); end if; end if; end for; lst;
  • Mathematica
    terms = 172;
    kmax = 120;
    a[n_] := Module[{s, k}, s[k_] := s[k] = n s[k-1] - s[k-2]; s[0] = 1; s[1] = n+1; For[k = 1, k <= kmax, k++, If[PrimeQ[s[k]], Return[s[k]]]]];
    Array[a, terms] /. Null -> -1 (* Jean-François Alcover, Aug 30 2018 *)

Formula

If n is prime then a(n-1) = n.

A299045 Rectangular array: A(n,k) = Sum_{j=0..k} (-1)^floor(j/2)*binomial(k-floor((j+1)/2), floor(j/2))*(-n)^(k-j), n >= 1, k >= 0, read by antidiagonals.

Original entry on oeis.org

1, 1, 0, 1, -1, -1, 1, -2, 1, 1, 1, -3, 5, -1, 0, 1, -4, 11, -13, 1, -1, 1, -5, 19, -41, 34, -1, 1, 1, -6, 29, -91, 153, -89, 1, 0, 1, -7, 41, -169, 436, -571, 233, -1, -1, 1, -8, 55, -281, 985, -2089, 2131, -610, 1, 1, 1, -9, 71, -433, 1926, -5741, 10009, -7953, 1597, -1, 0
Offset: 1

Views

Author

Keywords

Comments

This array is used to compute A269252: A269252(n) = least k such that |A(n,k)| is a prime, or -1 if no such k exists.
For detailed theory, see [Hone].
The array can be extended to k<0 with A(n, k) = -A(n, -k-1) for all k in Z. - Michael Somos, Jun 19 2023

Examples

			Array begins:
1   0  -1     1     0      -1       1         0        -1           1
1  -1   1    -1     1      -1       1        -1         1          -1
1  -2   5   -13    34     -89     233      -610      1597       -4181
1  -3  11   -41   153    -571    2131     -7953     29681     -110771
1  -4  19   -91   436   -2089   10009    -47956    229771    -1100899
1  -5  29  -169   985   -5741   33461   -195025   1136689    -6625109
1  -6  41  -281  1926  -13201   90481   -620166   4250681   -29134601
1  -7  55  -433  3409  -26839  211303  -1663585  13097377  -103115431
1  -8  71  -631  5608  -49841  442961  -3936808  34988311  -310957991
1  -9  89  -881  8721  -86329  854569  -8459361  83739041  -828931049
		

Crossrefs

Cf. A094954 (unsigned version of this array, but missing the first row).

Programs

  • Mathematica
    (* Array: *)
    Grid[Table[LinearRecurrence[{-n, -1}, {1, 1 - n}, 10], {n, 10}]]
    (*Array antidiagonals flattened (gives this sequence):*)
    A299045[n_, k_] := Sum[(-1)^(Floor[j/2]) Binomial[k - Floor[(j + 1)/2], Floor[j/2]] (-n)^(k - j), {j, 0, k}]; Flatten[Table[A299045[n - k, k], {n, 11}, {k, 0, n - 1}]]
  • PARI
    {A(n, k) = sum(j=0, k, (-1)^(j\2)*binomial(k-(j+1)\2, j\2)*(-n)^(k-j))}; /* Michael Somos, Jun 19 2023 */

Formula

G.f. for row n: (1 + x)/(1 + n*x + x^2), n >= 1.
A(n, k) = B(-n, k) where B = A294099. - Michael Somos, Jun 19 2023

A299071 Union_{odd primes p, n >= 3} {T_p(n)}, where T_m(x) = x*T_{m-1}(x) - T_{m-2}(x), m >= 2, T_0(x) = 2, T_1(x) = x (dilated Chebyshev polynomials of the first kind).

Original entry on oeis.org

18, 52, 110, 123, 198, 488, 702, 724, 843, 970, 1298, 1692, 2158, 2525, 3330, 4048, 4862, 5778, 6726, 6802, 7940, 9198, 10084, 10582, 13752, 15550, 17498, 19602, 21868, 24302, 26910, 29698, 30248, 32672, 35838, 39603, 42770, 46548, 50542
Offset: 1

Views

Author

Keywords

Comments

From a problem in A269254. For detailed theory, see [Hone].
Sequence avoids numbers of the form T_p(T_2(j)).

Crossrefs

Programs

  • Mathematica
    maxT = 55000; maxn = 12;
    T[0][] = 2; T[1][x] = x;
    T[m_][x_] := T[m][x] = x T[m-1][x] - T[m-2][x];
    TT = Table[T[p][n], {p, Prime[Range[2, maxn]]}, {n, 3, Prime[maxn]}] // Flatten // Union // Select[#, # <= maxT&]&;
    avoid = Table[T[p][T[2][n]], {p, Prime[Range[2, maxn]]}, {n, 3, Prime[maxn] }] // Flatten // Union // Select[#, # <= maxT&]&;
    Complement[TT, avoid] (* Jean-François Alcover, Nov 03 2018 *)
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