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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A355982 Primes in A004253.

Original entry on oeis.org

19, 2089, 10009, 229771, 1100899, 2779769539, 33629651653051, 161129341280179, 3698955932459041, 406851522918841459, 23583148518971728185859, 720404932310119551295422832411, 4593030255749728350335873388896105689, 6111807570415039356543996013033519421475691
Offset: 1

Views

Author

R. J. Mathar, Jul 22 2022

Keywords

Comments

Related to the search of solutions of pair of congruences p^2 == -3 (mod q), q^2 == -3 (mod p).
Cosgrave and Dilcher list indices of these primes 2, 5, 6, 8, 9, 14, 20, 21, 23, 26, ... up to 25000 (assuming offset 0 in A004253), including probable primes.

Crossrefs

Cf. A004253.

Programs

  • Mathematica
    Select[LinearRecurrence[{5, -1}, {1, 4}, 200], PrimeQ] (* Paolo Xausa, Mar 19 2024 *)

Formula

A000040 INTERSECT A004253.

Extensions

More terms from Jinyuan Wang, Jul 22 2022

A004254 a(n) = 5*a(n-1) - a(n-2) for n > 1, a(0) = 0, a(1) = 1.

Original entry on oeis.org

0, 1, 5, 24, 115, 551, 2640, 12649, 60605, 290376, 1391275, 6665999, 31938720, 153027601, 733199285, 3512968824, 16831644835, 80645255351, 386394631920, 1851327904249, 8870244889325, 42499896542376, 203629237822555, 975646292570399, 4674602225029440, 22397364832576801
Offset: 0

Views

Author

Keywords

Comments

Nonnegative values of y satisfying x^2 - 21*y^2 = 4; values of x are in A003501. - Wolfdieter Lang, Nov 29 2002
a(n) is equal to the permanent of the (n-1) X (n-1) Hessenberg matrix with 5's along the main diagonal, i's along the superdiagonal and the subdiagonal (i is the imaginary unit), and 0's everywhere else. - John M. Campbell, Jun 09 2011
For n >= 1, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,2,3,4}. - Milan Janjic, Jan 25 2015
From Klaus Purath, Jul 26 2024: (Start)
For any three consecutive terms (x, y, z), y^2 - xz = 1 always applies.
a(n) = (t(i+2n) - t(i))/(t(i+n+1) - t(i+n-1)) where (t) is any recurrence t(k) = 4t(k-1) + 4t(k-2) - t(k-3) or t(k) = 5t(k-1) - t(k-2) without regard to initial values.
In particular, if the recurrence (t) of the form (4,4,-1) has the same three initial values as the current sequence, a(n) = t(n) applies.
a(n) = (t(k+1)*t(k+n) - t(k)*t(k+n+1))/(y^2 - xz) where (t) is any recurrence of the current family with signature (5,-1) and (x, y, z) are any three consecutive terms of (t), for integer k >= 0. (End)

Examples

			G.f. = x + 5*x^2 + 24*x^3 + 115*x^4 + 551*x^5 + 2640*x^6 + 12649*x^7 + ...
		

References

  • F. A. Haight, On a generalization of Pythagoras' theorem, pp. 73-77 of J. C. Butcher, editor, A Spectrum of Mathematics. Auckland University Press, 1971.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Partial sums of A004253.
Cf. A000027, A001906, A001353, A003501, A030221. a(n) = sqrt((A003501(n)^2 - 4)/21).
First differences of a(n) are in A004253, partial sums in A089817.
Cf. A004253.
INVERT transformation yields A001109. - R. J. Mathar, Sep 11 2008

Programs

  • Magma
    [ n eq 1 select 0 else n eq 2 select 1 else 5*Self(n-1)-Self(n-2): n in [1..30] ]; // Vincenzo Librandi, Aug 19 2011
  • Maple
    A004254:=1/(1-5*z+z**2); # Simon Plouffe in his 1992 dissertation
  • Mathematica
    a[n_]:=(MatrixPower[{{1,3},{1,4}},n].{{1},{1}})[[2,1]]; Table[a[n],{n,0,40}] (* Vladimir Joseph Stephan Orlovsky, Feb 19 2010 *)
    a[ n_] := ChebyshevU[2 n - 1, Sqrt[7]/2] / Sqrt[7]; (* Michael Somos, Jan 22 2017 *)
  • PARI
    {a(n) = subst(4*poltchebi(n+1) - 10*poltchebi(n), x, 5/2) / 21}; /* Michael Somos, Dec 04 2002 */
    
  • PARI
    {a(n) = imag((5 + quadgen(84))^n) / 2^(n-1)}; /* Michael Somos, Dec 04 2002 */
    
  • PARI
    {a(n) = polchebyshev(n - 1, 2, 5/2)}; /* Michael Somos, Jan 22 2017 */
    
  • PARI
    {a(n) = simplify( polchebyshev( 2*n - 1, 2, quadgen(28)/2) / quadgen(28))}; /* Michael Somos, Jan 22 2017 */
    
  • Sage
    [lucas_number1(n,5,1) for n in range(27)] # Zerinvary Lajos, Jun 25 2008
    

Formula

G.f.: x/(1-5*x+x^2).
a(n) = (((5+sqrt(21))/2)^n-((5-sqrt(21))/2)^n)/sqrt(21). - Barry E. Williams, Aug 29 2000
a(n) = S(2*n-1, sqrt(7))/sqrt(7) = S(n-1, 5); S(n, x)=U(n, x/2), Chebyshev polynomials of 2nd kind, A049310.
A003501(n) = sqrt(21*a(n)^2 + 4).
a(n) = Sum_{k=0..n-1} binomial(n+k, 2*k+1)*2^k. - Paul Barry, Nov 30 2004
[A004253(n), a(n)] = [1,3; 1,4]^n * [1,0]. - Gary W. Adamson, Mar 19 2008
a(n+1) = Sum_{k=0..n} Gegenbauer_C(n-k,k+1,2). - Paul Barry, Apr 21 2009
a(n+1) = Sum_{k=0..n} A101950(n,k)*4^k. - Philippe Deléham, Feb 10 2012
From Peter Bala, Dec 23 2012: (Start)
Product {n >= 1} (1 + 1/a(n)) = (1/3)*(3 + sqrt(21)).
Product {n >= 2} (1 - 1/a(n)) = (1/10)*(3 + sqrt(21)). (End)
From Michael Somos, Jan 22 2017: (Start)
A054493(2*n - 1) = 7 * a(n)^2 for all n in Z.
a(n) = -a(-n) for all n in Z.
0 = -1 + a(n)*(+a(n) - 5*a(n+1)) + a(n+1)*(+a(n+1)) for all n in Z. (End)
Limit_{n->oo} a(n+1)/a(n) = (5 + sqrt(21))/2 = A107905. - Wolfdieter Lang, Nov 15 2023
From Klaus Purath, Jul 26 2024: (Start)
a(n) = 4(a(n-1) + a(n-2)) - a(n-3).
a(n) = 6(a(n-1) - a(n-2)) + a(n-3).
In general, for all sequences of the form U(n) = P*U(n-1) - U(n-2) the following applies:
U(n) = (P-1)*U(n-1) + (P-1)*U(n-2) - U(n-3).
U(n) = (P+1)*U(n-1) - (P+1)*U(n-2) + U(n-3). (End)
a(n) = (5*a(n-1)+sqrt(21*a(n-1)^2+4))/2 for n>0. - Alexandru Petrescu, Apr 15 2025
From Peter Bala, May 22 2025: (Start)
Product_{n >= 1} ((a(2*n) + 1)/(a(2*n) - 1))^2 = 7/3.
Product_{n >= 1} ((a(2*n+1) + 1)/(a(2*n+1) - 1))^2 = 25/21.
The o.g.f. A(x) satisfies A(x) + A(-x) + 10*A(x)*A(-x) = 0. The o.g.f. for A097778 equals -1/x * A(sqrt(x))*A(-sqrt(x)). (End)
E.g.f.: 2*exp(5*x/2)*sinh(sqrt(21)*x/2)/sqrt(21). - Stefano Spezia, Jul 02 2025

A108299 Triangle read by rows, 0 <= k <= n: T(n,k) = binomial(n-[(k+1)/2],[k/2])*(-1)^[(k+1)/2].

Original entry on oeis.org

1, 1, -1, 1, -1, -1, 1, -1, -2, 1, 1, -1, -3, 2, 1, 1, -1, -4, 3, 3, -1, 1, -1, -5, 4, 6, -3, -1, 1, -1, -6, 5, 10, -6, -4, 1, 1, -1, -7, 6, 15, -10, -10, 4, 1, 1, -1, -8, 7, 21, -15, -20, 10, 5, -1, 1, -1, -9, 8, 28, -21, -35, 20, 15, -5, -1, 1, -1, -10, 9, 36, -28, -56, 35, 35, -15, -6, 1, 1, -1, -11, 10, 45, -36, -84, 56, 70
Offset: 0

Views

Author

Reinhard Zumkeller, Jun 01 2005

Keywords

Comments

Matrix inverse of A124645.
Let L(n,x) = Sum_{k=0..n} T(n,k)*x^(n-k) and Pi=3.14...:
L(n,x) = Product_{k=1..n} (x - 2*cos((2*k-1)*Pi/(2*n+1)));
Sum_{k=0..n} T(n,k) = L(n,1) = A010892(n+1);
Sum_{k=0..n} abs(T(n,k)) = A000045(n+2);
abs(T(n,k)) = A065941(n,k), T(n,k) = A065941(n,k)*A087960(k);
T(2*n,k) + T(2*n+1,k+1) = 0 for 0 <= k <= 2*n;
T(n,0) = A000012(n) = 1; T(n,1) = -1 for n > 0;
T(n,2) = -(n-1) for n > 1; T(n,3) = A000027(n)=n for n > 2;
T(n,4) = A000217(n-3) for n > 3; T(n,5) = -A000217(n-4) for n > 4;
T(n,6) = -A000292(n-5) for n > 5; T(n,7) = A000292(n-6) for n > 6;
T(n,n-3) = A058187(n-3)*(-1)^floor(n/2) for n > 2;
T(n,n-2) = A008805(n-2)*(-1)^floor((n+1)/2) for n > 1;
T(n,n-1) = A008619(n-1)*(-1)^floor(n/2) for n > 0;
T(n,n) = L(n,0) = (-1)^floor((n+1)/2);
L(n,1) = A010892(n+1); L(n,-1) = A061347(n+2);
L(n,2) = 1; L(n,-2) = A005408(n)*(-1)^n;
L(n,3) = A001519(n); L(n,-3) = A002878(n)*(-1)^n;
L(n,4) = A001835(n+1); L(n,-4) = A001834(n)*(-1)^n;
L(n,5) = A004253(n); L(n,-5) = A030221(n)*(-1)^n;
L(n,6) = A001653(n); L(n,-6) = A002315(n)*(-1)^n;
L(n,7) = A049685(n); L(n,-7) = A033890(n)*(-1)^n;
L(n,8) = A070997(n); L(n,-8) = A057080(n)*(-1)^n;
L(n,9) = A070998(n); L(n,-9) = A057081(n)*(-1)^n;
L(n,10) = A072256(n+1); L(n,-10) = A054320(n)*(-1)^n;
L(n,11) = A078922(n+1); L(n,-11) = A097783(n)*(-1)^n;
L(n,12) = A077417(n); L(n,-12) = A077416(n)*(-1)^n;
L(n,13) = A085260(n);
L(n,14) = A001570(n); L(n,-14) = A028230(n)*(-1)^n;
L(n,n) = A108366(n); L(n,-n) = A108367(n).
Row n of the matrix inverse (A124645) has g.f.: x^floor(n/2)*(1-x)^(n-floor(n/2)). - Paul D. Hanna, Jun 12 2005
From L. Edson Jeffery, Mar 12 2011: (Start)
Conjecture: Let N=2*n+1, with n > 2. Then T(n,k) (0 <= k <= n) gives the k-th coefficient in the characteristic function p_N(x)=0, of degree n in x, for the n X n tridiagonal unit-primitive matrix G_N (see [Jeffery]) of the form
G_N=A_{N,1}=
(0 1 0 ... 0)
(1 0 1 0 ... 0)
(0 1 0 1 0 ... 0)
...
(0 ... 0 1 0 1)
(0 ... 0 1 1),
with solutions phi_j = 2*cos((2*j-1)*Pi/N), j=1,2,...,n. For example, for n=3,
G_7=A_{7,1}=
(0 1 0)
(1 0 1)
(0 1 1).
We have {T(3,k)}=(1,-1,-2,1), while the characteristic function of G_7 is p(x) = x^3-x^2-2*x+1 = 0, with solutions phi_j = 2*cos((2*j-1)*Pi/7), j=1,2,3. (End)
The triangle sums, see A180662 for their definitions, link A108299 with several sequences, see the crossrefs. - Johannes W. Meijer, Aug 08 2011
The roots to the polynomials are chaotic using iterates of the operation (x^2 - 2), with cycle lengths L and initial seeds returning to the same term or (-1)* the seed. Periodic cycle lengths L are shown in A003558 such that for the polynomial represented by row r, the cycle length L is A003558(r-1). The matrices corresponding to the rows as characteristic polynomials are likewise chaotic [cf. Kappraff et al., 2005] with the same cycle lengths but substituting 2*I for the "2" in (x^2 - 2), where I = the Identity matrix. For example, the roots to x^3 - x^2 - 2x + 1 = 0 are 1.801937..., -1.246979..., and 0.445041... With 1.801937... as the initial seed and using (x^2 - 2), we obtain the 3-period trajectory of 8.801937... -> 1.246979... -> -0.445041... (returning to -1.801937...). We note that A003558(2) = 3. The corresponding matrix M is: [0,1,0; 1,0,1; 0,1,1,]. Using seed M with (x^2 - 2*I), we obtain the 3-period with the cycle completed at (-1)*M. - Gary W. Adamson, Feb 07 2012

Examples

			Triangle begins:
  1;
  1,  -1;
  1,  -1,  -1;
  1,  -1,  -2,   1;
  1,  -1,  -3,   2,   1;
  1,  -1,  -4,   3,   3,  -1;
  1,  -1,  -5,   4,   6,  -3,  -1;
  1,  -1,  -6,   5,  10,  -6,  -4,   1;
  1,  -1,  -7,   6,  15, -10, -10,   4,   1;
  1,  -1,  -8,   7,  21, -15, -20,  10,   5,  -1;
  1,  -1,  -9,   8,  28, -21, -35,  20,  15,  -5,  -1;
  1,  -1, -10,   9,  36, -28, -56,  35,  35, -15,  -6,   1;
  ...
		

References

  • Friedrich L. Bauer, 'De Moivre und Lagrange: Cosinus eines rationalen Vielfachen von Pi', Informatik Spektrum 28 (Springer, 2005).
  • Jay Kappraff, S. Jablan, G. Adamson, & R. Sazdonovich: "Golden Fields, Generalized Fibonacci Sequences, & Chaotic Matrices"; FORMA, Vol 19, No 4, (2005).

Crossrefs

Cf. A049310, A039961, A124645 (matrix inverse).
Triangle sums (see the comments): A193884 (Kn11), A154955 (Kn21), A087960 (Kn22), A000007 (Kn3), A010892 (Fi1), A134668 (Fi2), A078031 (Ca2), A193669 (Gi1), A001519 (Gi3), A193885 (Ze1), A050935 (Ze3). - Johannes W. Meijer, Aug 08 2011
Cf. A003558.

Programs

  • Haskell
    a108299 n k = a108299_tabl !! n !! k
    a108299_row n = a108299_tabl !! n
    a108299_tabl = [1] : iterate (\row ->
       zipWith (+) (zipWith (*) ([0] ++ row) a033999_list)
                   (zipWith (*) (row ++ [0]) a059841_list)) [1,-1]
    -- Reinhard Zumkeller, May 06 2012
  • Maple
    A108299 := proc(n,k): binomial(n-floor((k+1)/2), floor(k/2))*(-1)^floor((k+1)/2) end: seq(seq(A108299 (n,k), k=0..n), n=0..11); # Johannes W. Meijer, Aug 08 2011
  • 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]; Table[t[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* Jean-François Alcover, May 16 2013 *)
  • PARI
    {T(n,k)=polcoeff(polcoeff((1-x*y)/(1-x+x^2*y^2+x^2*O(x^n)),n,x)+y*O(y^k),k,y)} (Hanna)
    

Formula

T(n,k) = binomial(n-floor((k+1)/2),floor(k/2))*(-1)^floor((k+1)/2).
T(n+1, k) = if sign(T(n, k-1))=sign(T(n, k)) then T(n, k-1)+T(n, k) else -T(n, k-1) for 0 < k < n, T(n, 0) = 1, T(n, n) = (-1)^floor((n+1)/2).
G.f.: A(x, y) = (1 - x*y)/(1 - x + x^2*y^2). - Paul D. Hanna, Jun 12 2005
The generating polynomial (in z) of row n >= 0 is (u^(2*n+1) + v^(2*n+1))/(u + v), where u and v are defined by u^2 + v^2 = 1 and u*v = z. - Emeric Deutsch, Jun 16 2011
From Johannes W. Meijer, Aug 08 2011: (Start)
abs(T(n,k)) = A065941(n,k) = abs(A187660(n,n-k));
T(n,n-k) = A130777(n,k); abs(T(n,n-k)) = A046854(n,k) = abs(A066170(n,k)). (End)

Extensions

Corrected and edited by Philippe Deléham, Oct 20 2008

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

Views

Author

Keywords

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)

A094954 Array T(k,n) read by antidiagonals. G.f.: x(1-x)/(1-kx+x^2), k>1.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 5, 1, 1, 4, 11, 13, 1, 1, 5, 19, 41, 34, 1, 1, 6, 29, 91, 153, 89, 1, 1, 7, 41, 169, 436, 571, 233, 1, 1, 8, 55, 281, 985, 2089, 2131, 610, 1, 1, 9, 71, 433, 1926, 5741, 10009, 7953, 1597, 1, 1, 10, 89, 631, 3409, 13201, 33461, 47956, 29681
Offset: 1

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Author

Ralf Stephan, May 31 2004

Keywords

Comments

Also, values of polynomials with coefficients in A098493 (see Fink et al.). See A098495 for negative k.
Number of dimer tilings of the graph S_{k-1} X P_{2n-2}.

Examples

			1,1,1,1,1,1,1,1,1,1,1,1,1,1, ...
1,2,5,13,34,89,233,610,1597, ...
1,3,11,41,153,571,2131,7953, ...
1,4,19,91,436,2089,10009,47956, ...
1,5,29,169,985,5741,33461,195025, ...
1,6,41,281,1926,13201,90481,620166, ...
		

Crossrefs

Rows are first differences of rows in array A073134.
Rows 2-14 are A000012, A001519, A079935/A001835, A004253, A001653, A049685, A070997, A070998, A072256, A078922, A077417, A085260, A001570. Other rows: A007805 (k=18), A075839 (k=20), A077420 (k=34), A078988 (k=66).
Columns include A028387. Diagonals include A094955, A094956. Antidiagonal sums are A094957.

Programs

  • Mathematica
    max = 14; row[k_] := Rest[ CoefficientList[ Series[ x*(1-x)/(1-k*x+x^2), {x, 0, max}], x]]; t = Table[ row[k], {k, 2, max+1}]; Flatten[ Table[ t[[k-n+1, n]], {k, 1, max}, {n, 1, k}]] (* Jean-François Alcover, Dec 27 2011 *)
  • PARI
    T(k,n)=polcoeff(x*(1-x)/(1-k*x+x*x),n)

Formula

Recurrence: T(k, 1) = 1, T(k, 2) = k-1, T(k, n) = kT(k, n-1) - T(k, n-2).
For n>3, T(k, n) = [k(k-2) + T(k, n-1)T(k, n-2)] / T(k, n-3).
T(k, n+1) = S(n, k) - S(n-1, k) = U(n, k/2) - U(n-1, k/2), with S, U = Chebyshev polynomials of second kind.
T(k+2, n+1) = Sum[i=0..n, k^(n-i) * C(2n-i, i)] (from comments by Benoit Cloitre).

A238379 Expansion of (1 - x)/(1 - 36*x + x^2).

Original entry on oeis.org

1, 35, 1259, 45289, 1629145, 58603931, 2108112371, 75833441425, 2727895778929, 98128414600019, 3529895029821755, 126978092658983161, 4567681440693572041, 164309553772309610315, 5910576254362452399299, 212616435603275976764449
Offset: 0

Views

Author

Bruno Berselli, Feb 25 2014

Keywords

Comments

First bisection of A041611.

Crossrefs

Cf. similar sequences with g.f. (1-x)/(1-k*x+x^2): A122367 (k=3), A079935 (k=4), A004253 (k=5), A001653 (k=6), A049685 (k=7), A070997 (k=8), A070998 (k=9), A138288 (k=10), A078922 (k=11), A077417 (k=12), A085260 (k=13), A001570 (k=14), A160682 (k=15), A157456 (k=16), A161595 (k=17). From 18 to 38, even k only, except k=27 and k=31: A007805 (k=18), A075839 (k=20), A157014 (k=22), A159664 (k=24), A153111 (k=26), A097835 (k=27), A159668 (k=28), A157877 (k=30), A111216 (k=31), A159674 (k=32), A077420 (k=34), this sequence (k=36), A097315 (k=38).

Programs

  • Magma
    [n le 2 select 35^(n-1) else 36*Self(n-1)-Self(n-2): n in [1..20]];
    
  • Magma
    R:=PowerSeriesRing(Integers(), 20); Coefficients(R!( (1 - x)/(1 - 36*x + x^2))); // Marius A. Burtea, Jan 14 2020
    
  • Mathematica
    CoefficientList[Series[(1 - x)/(1 - 36 x + x^2), {x, 0, 20}], x] (* or *) LinearRecurrence[{36, -1}, {1, 35}, 20]
  • PARI
    a(n)=([0,1; -1,36]^n*[1;35])[1,1] \\ Charles R Greathouse IV, May 10 2016
  • Sage
    m = 20; L. = PowerSeriesRing(ZZ, m); f = (1-x)/(1-36*x+x^2)
    print(f.coefficients())
    

Formula

G.f.: (1 - x)/(1 - 36*x + x^2).
a(n) = a(-n-1) = 36*a(n-1) - a(n-2).
a(n) = ((19-sqrt(323))/38)*(1+(18+sqrt(323))^(2*n+1))/(18+sqrt(323))^n.
a(n+1) - a(n) = 34*A144128(n+1).
323*a(n+1)^2 - ((a(n+2)-a(n))/2)^2 = 34.
Sum_{n>0} 1/(a(n) - 1/a(n)) = 1/34.
See also Tanya Khovanova in Links field:
a(n) = 35*a(n-1) + 34*Sum_{i=0..n-2} a(i).
a(n+2)*a(n) - a(n+1)^2 = 36-2 = 34 = 34*1,
a(n+3)*a(n) - a(n+1)*a(n+2) = 36*(36-2) = 1224 = 34*36.
Generalizing:
a(n+4)*a(n) - a(n+1)*a(n+3) = 44030 = 34*1295,
a(n+5)*a(n) - a(n+1)*a(n+4) = 1583856 = 34*46584,
a(n+6)*a(n) - a(n+1)*a(n+5) = 56974786 = 34*1675729, etc.,
where 1, 36, 1295, 46584, 1675729, ... is the sequence A144128, which is the second bisection of A041611.
a(n)^2 - 36*a(n)*a(n+1) + a(n+1)^2 + 34 = 0 (see comments by Colin Barker in similar sequences).

A003501 a(n) = 5*a(n-1) - a(n-2), with a(0) = 2, a(1) = 5.

Original entry on oeis.org

2, 5, 23, 110, 527, 2525, 12098, 57965, 277727, 1330670, 6375623, 30547445, 146361602, 701260565, 3359941223, 16098445550, 77132286527, 369562987085, 1770682648898, 8483850257405, 40648568638127, 194758992933230, 933146396028023, 4470972987206885
Offset: 0

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Author

Keywords

Comments

Positive values of x satisfying x^2 - 21*y^2 = 4; values of y are in A004254. - Wolfdieter Lang, Nov 29 2002
Except for the first term, positive values of x (or y) satisfying x^2 - 5xy + y^2 + 21 = 0. - Colin Barker, Feb 08 2014

Examples

			G.f. = 2 + 5*x + 23*x^2 + 110*x^3 + 527*x^4 + 2525*x^5 + ... - _Michael Somos_, Oct 25 2022
		

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • GAP
    a:=[2,5];; for n in [4..30] do a[n]:=5*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jan 16 2020
  • Magma
    I:=[2,5]; [n le 2 select I[n] else 5*Self(n-1) -Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 16 2020
    
  • Magma
    R:=PowerSeriesRing(Integers(), 25); Coefficients(R!((2-5*x)/(1-5*x+x^2))); // Marius A. Burtea, Jan 16 2020
    
  • Maple
    seq( simplify(2*ChebyshevT(n, 5/2)), n=0..30); # G. C. Greubel, Jan 16 2020
  • Mathematica
    a[0]=2; a[1]=5; a[n_]:= 5a[n-1] -a[n-2]; Table[a[n], {n,0,30}] (* Robert G. Wilson v, Jan 30 2004 *)
    LinearRecurrence[{5,-1},{2,5},30] (* Harvey P. Dale, May 12 2019 *)
    2*ChebyshevT[Range[0, 30], 5/2] (* G. C. Greubel, Jan 16 2020 *)
    a[ n_] := LucasL[n, 5*I]/I^n; (* Michael Somos, Oct 25 2022 *)
  • PARI
    {a(n) = subst(poltchebi(n),x,5/2)*2};
    
  • PARI
    {a(n) = polchebyshev(n,1,5/2)*2 }; /* Michael Somos, Oct 25 2022 */
    
  • Sage
    [lucas_number2(n,5,1) for n in range(37)] # Zerinvary Lajos, Jun 25 2008
    

Formula

a(n) = 5*S(n-1, 5) - 2*S(n-2, 5) = S(n, 5) - S(n-2, 5) = 2*T(n, 5/2), with S(n, x)=U(n, x/2), S(-1, x)=0, S(-2, x)=-1. U(n, x), resp. T(n, x), are Chebyshev's polynomials of the second, resp. first, kind. S(n-1, 5) = A004254(n), n>=0.
G.f.: (2-5*x)/(1-5*x+x^2). - Simon Plouffe in his 1992 dissertation.
a(n) ~ (1/2*(5 + sqrt(21)))^n. - Joe Keane (jgk(AT)jgk.org), May 16 2002
a(n) = ap^n + am^n, with ap=(5+sqrt(21))/2 and am=(5-sqrt(21))/2.
a(n) = sqrt(4 + 21*A004254(n)^2).
From Peter Bala, Jan 06 2013: (Start)
Let F(x) = Product_{n=0..inf} (1 + x^(4*n+1))/(1 + x^(4*n+3)). Let alpha = 1/2*(5 - sqrt(21)). This sequence gives the simple continued fraction expansion of 1 + F(alpha) = 2.19827 65373 95327 17782 ... = 2 + 1/(5 + 1/(23 + 1/(110 + ...))).
Also F(-alpha) = 0.79824 49142 28050 93561 ... has the continued fraction representation 1 - 1/(5 - 1/(23 - 1/(110 - ...))) and the simple continued fraction expansion 1/(1 + 1/((5-2) + 1/(1 + 1/((23-2) + 1/(1 + 1/((110-2) + 1/(1 + ...))))))).
F(alpha)*F(-alpha) has the simple continued fraction expansion 1/(1 + 1/((5^2-4) + 1/(1 + 1/((23^2-4) + 1/(1 + 1/((110^2-4) + 1/(1 + ...))))))).
(End)
a(n) = (A217787(k+3n) + A217787(k-3n))/A217787(k) for k>=3n. - Bruno Berselli, Mar 25 2013

Extensions

Chebyshev comments from Wolfdieter Lang, Oct 31 2002

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

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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

A101368 The sequence solves the following problem: find all the pairs (i,j) such that i divides 1+j+j^2 and j divides 1+i+i^2. In fact, the pairs (a(n),a(n+1)), n>0, are all the solutions.

Original entry on oeis.org

1, 1, 3, 13, 61, 291, 1393, 6673, 31971, 153181, 733933, 3516483, 16848481, 80725921, 386781123, 1853179693, 8879117341, 42542407011, 203832917713, 976622181553, 4679277990051, 22419767768701, 107419560853453, 514678036498563, 2465970621639361, 11815175071698241, 56609904736851843, 271234348612560973
Offset: 1

Views

Author

M. Benito, O. Ciaurri and E. Fernandez (oscar.ciaurri(AT)dmc.unirioja.es), Jan 13 2005

Keywords

Comments

Also, integers m such that 21*(3*m-1)^2 - 48 is a square. - Max Alekseyev, May 23 2022
a(n) is prime exactly for n = 3, 4, 5, 8, 16, 20, 22, 23, 58, 302, 386, 449, 479, 880 up to 1000. - Tomohiro Yamada, Dec 23 2018
Similarly, positive integers m,k with m|(1+k+^2) and k|(1-m+m^2) are consecutive terms of A061646, where m has an even index. - Max Alekseyev, May 23 2022

Examples

			a(5) = 61 because (1 + a(4) + a(4)^2)/a(3) = (1 + 13 + 169)/3 = 61.
		

Crossrefs

Programs

  • GAP
    a:=[1,1];; for n in [3..30] do a[n]:=5*a[n-1]-a[n-2]-1; od; Print(a); # Muniru A Asiru, Dec 28 2018
  • Magma
    [n le 2 select 1 else 5*Self(n-1)-Self(n-2)-1: n in [1..30]]; // Vincenzo Librandi, Dec 25 2018
    
  • Maple
    seq(coeff(series(x*(1-5*x+3*x^2)/((1-x)*(1-5*x+x^2)),x,n+1), x, n), n = 1 .. 30); # Muniru A Asiru, Dec 28 2018
  • Mathematica
    Rest@ CoefficientList[Series[x (1 - 5 x + 3 x^2)/((1 - x) (1 - 5 x + x^2)), {x, 0, 28}], x] (* or *)
    RecurrenceTable[{a[n] == (1 + a[n - 1] + a[n - 1]^2)/a[n - 2], a[1] == a[2] == 1}, a, {n, 1, 28}] (* or *)
    RecurrenceTable[{a[n] == 5 a[n - 1] - a[n - 2] - 1, a[1] == a[2] == 1}, a, {n, 1, 28}] (* or *)
    LinearRecurrence[{6, -6, 1}, {1, 1, 3}, 28] (* Michael De Vlieger, Aug 28 2016 *)
  • PARI
    Vec(x*(1-5*x+3*x^2)/((1-x)*(1-5*x+x^2)) + O(x^30)) \\ Michel Marcus, Aug 03 2016
    
  • PARI
    a(n)=([0,1,0;0,0,1;1,-6,6]^n*[3;1;1])[1,1] \\ Charles R Greathouse IV, Aug 28 2016
    

Formula

Recurrence: a(1)=a(2)=1 and a(n+1)=(1+a(n)+a(n)^2)/a(n-1) for n>2.
G.f.: x(1 - 5x + 3x^2) / [(1-x)(1 - 5x + x^2)]; a(n) = 2 * A089817(n-3) + 1, n>2. - Conjectured by Ralf Stephan, Jan 14 2005, proved by Max Alekseyev, Aug 03 2006
a(n) = 6a(n-1)-6a(n-2)+a(n-3), a(n) = 5a(n-1)-a(n-2)-1. - Floor van Lamoen, Aug 01 2006
a(n) = (4/3 - (2/7)*sqrt(21))*((5 + sqrt(21))/2)^n + (4/3 + (2/7)*sqrt(21))*((5 - sqrt(21))/2)^n + 1/3. - Floor van Lamoen, Aug 04 2006
For n>1, a(n) = (2 * A004253(n-1) + 1) / 3. - Max Alekseyev, May 23 2022

A165253 Triangle T(n,k), read by rows given by [1,0,1,0,0,0,0,0,0,...] DELTA [0,1,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 3, 1, 0, 1, 6, 5, 1, 0, 1, 10, 15, 7, 1, 0, 1, 15, 35, 28, 9, 1, 0, 1, 21, 70, 84, 45, 11, 1, 0, 1, 28, 126, 210, 165, 66, 13, 1, 0, 1, 36, 210, 462, 495, 286, 91, 15, 1, 0, 1, 45, 330, 924, 1287, 1001, 455, 120, 17, 1, 0, 1, 55, 495, 1716, 3003, 3003, 1820, 680
Offset: 0

Views

Author

Philippe Deléham, Sep 10 2009

Keywords

Comments

Mirror image of triangle in A121314.

Examples

			Triangle begins:
  1;
  1,    0;
  1,    1,    0;
  1,    3,    1,    0;
  1,    6,    5,    1,    0;
  1,   10,   15,    7,    1,    0;
  1,   15,   35,   28,    9,    1,    0;
  1,   21,   70,   84,   45,   11,    1,    0;
  1,   28,  126,  210,  165,   66,   13,    1,    0;
  1,   36,  210,  462,  495,  286,   91,   15,    1,    0,
  1,   45,  330,  924, 1287, 1001,  455,  120,   17,    1,    0;
		

Crossrefs

Programs

  • Mathematica
    m = 13;
    (* DELTA is defined in A084938 *)
    DELTA[Join[{1, 0, 1}, Table[0, {m}]], Join[{0, 1}, Table[0, {m}]], m] // Flatten (* Jean-François Alcover, Feb 19 2020 *)

Formula

T(0,0)=1, T(n,k) = binomial(n-1+k,2k) for n >= 1.
Sum {k=0..n} T(n,k)*x^k = A000012(n), A001519(n), A001835(n), A004253(n), A001653(n), A049685(n-1), A070997(n-1), A070998(n-1), A072256(n), A078922(n), A077417(n-1), A085260(n), A001570(n) for x = 0,1,2,3,4,5,6,7,8,9,10,11,12 respectively.
Sum_{k=0..n} T(n,k)*x^(n-k) = A000007(n), A001519(n), A047849(n), A165310(n), A165311(n), A165312(n), A165314(n), A165322(n), A165323(n), A165324(n) for x= 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 respectively. - Philippe Deléham, Sep 26 2009
T(n,k) = 2*T(n-1,k) + T(n-1,k-1) - T(n-2,k), T(0,0)=T(1,0)=1, T(1,1)=0. - Philippe Deléham, Feb 18 2012
G.f.: (1-x-y*x)/((1-x)^2-y*x). - Philippe Deléham, Feb 19 2012
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