Displaying 1-10 of 37 results found.
19, 2089, 10009, 229771, 1100899, 2779769539, 33629651653051, 161129341280179, 3698955932459041, 406851522918841459, 23583148518971728185859, 720404932310119551295422832411, 4593030255749728350335873388896105689, 6111807570415039356543996013033519421475691
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.
MATHEMATICA
Select[LinearRecurrence[{5, -1}, {1, 4}, 200], PrimeQ] (* Paolo Xausa, Mar 19 2024 *)
a(n) = 5*a(n-1) - a(n-2) for n > 1, a(0) = 0, a(1) = 1.
(Formerly M3930)
+10
73
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
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 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 valus.
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)
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).
LINKS
A. F. Horadam, Pell Identities, Fib. Quart., Vol. 9, No. 3, 1971, pp. 245-252.
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. a(n) = Sum_{k=0..n-1} binomial(n+k, 2*k+1)*2^k. - Paul Barry, Nov 30 2004 a(n+1) = Sum_{k=0..n} Gegenbauer_C(n-k,k+1,2). - Paul Barry, Apr 21 2009 Product {n >= 1} (1 + 1/a(n)) = (1/3)*(3 + sqrt(21)). - Peter Bala, Dec 23 2012 Product {n >= 2} (1 - 1/a(n)) = (1/10)*(3 + sqrt(21)). - Peter Bala, Dec 23 2012 0 = -1 + 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) = 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)
EXAMPLE
G.f. = x + 5*x^2 + 24*x^3 + 115*x^4 + 551*x^5 + 2640*x^6 + 12649*x^7 + ...
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 *)
PROG
(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 (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
Triangle read by rows, 0 <= k <= n: T(n,k) = binomial(n-[(k+1)/2],[k/2])*(-1)^[(k+1)/2].
+10
58
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
COMMENTS
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); 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,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,2) = 1; L(n,-2) = A005408(n)*(-1)^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 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 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
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).
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
EXAMPLE
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;
...
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 *)
PROG
(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)
(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]
CROSSREFS
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
Chebyshev even-indexed U-polynomials evaluated at sqrt(7)/2.
+10
33
1, 6, 29, 139, 666, 3191, 15289, 73254, 350981, 1681651, 8057274, 38604719, 184966321, 886226886, 4246168109, 20344613659, 97476900186, 467039887271, 2237722536169, 10721572793574, 51370141431701, 246129134364931, 1179275530392954, 5650248517599839
COMMENTS
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 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 ((-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)
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) ~ (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 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
EXAMPLE
G.f. = 1 + 6*x + 29*x^2 + 139*x^3 + 666*x^4 + 3191*x^5 + 15289*x^6 + ...
MAPLE
option remember;
if n <= 1 then
op(n+1, [1, 6]);
else
5*procname(n-1)-procname(n-2) ;
end if;
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 *)
PROG
(Sage) [(lucas_number2(n, 5, 1)-lucas_number2(n-1, 5, 1))/3 for n in range(1, 22)] # Zerinvary Lajos, Nov 10 2009 (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 (PARI) {a(n) = simplify(polchebyshev(2*n, 2, quadgen(28)/2))}; /* Michael Somos, Jan 22 2017 */
Array T(k,n) read by antidiagonals. G.f.: x(1-x)/(1-kx+x^2), k>1.
+10
29
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
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}.
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).
EXAMPLE
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, ...
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 *)
PROG
(PARI) T(k, n)=polcoeff(x*(1-x)/(1-k*x+x*x), n)
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).
Expansion of (1 - x)/(1 - 36*x + x^2).
+10
29
1, 35, 1259, 45289, 1629145, 58603931, 2108112371, 75833441425, 2727895778929, 98128414600019, 3529895029821755, 126978092658983161, 4567681440693572041, 164309553772309610315, 5910576254362452399299, 212616435603275976764449
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.
323*a(n+1)^2 - ((a(n+2)-a(n))/2)^2 = 34.
Sum_{n>0} 1/(a(n) - 1/a(n)) = 1/34.
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).
MATHEMATICA
CoefficientList[Series[(1 - x)/(1 - 36 x + x^2), {x, 0, 20}], x] (* or *) LinearRecurrence[{36, -1}, {1, 35}, 20]
PROG
(Magma) [n le 2 select 35^(n-1) else 36*Self(n-1)-Self(n-2): n in [1..20]];
(Magma) R<x>:=PowerSeriesRing(Integers(), 20); Coefficients(R!( (1 - x)/(1 - 36*x + x^2))); // Marius A. Burtea, Jan 14 2020 (Sage)
m = 20; L.<x> = PowerSeriesRing(ZZ, m); f = (1-x)/(1-36*x+x^2)
print(f.coefficients())
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).
a(n) = 5*a(n-1) - a(n-2), with a(0) = 2, a(1) = 5.
(Formerly M1540)
+10
22
2, 5, 23, 110, 527, 2525, 12098, 57965, 277727, 1330670, 6375623, 30547445, 146361602, 701260565, 3359941223, 16098445550, 77132286527, 369562987085, 1770682648898, 8483850257405, 40648568638127, 194758992933230, 933146396028023, 4470972987206885
COMMENTS
Except for the first term, positive values of x (or y) satisfying x^2 - 5xy + y^2 + 21 = 0. - Colin Barker, Feb 08 2014
REFERENCES
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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.
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)
EXAMPLE
G.f. = 2 + 5*x + 23*x^2 + 110*x^3 + 527*x^4 + 2525*x^5 + ... - Michael Somos, Oct 25 2022
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 *)
PROG
(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 (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<x>:=PowerSeriesRing(Integers(), 25); Coefficients(R!((2-5*x)/(1-5*x+x^2))); // Marius A. Burtea, Jan 16 2020 (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
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.
+10
15
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
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
FORMULA
G.f. for row n: (1 + x)/(1 + n*x + x^2), n >= 1.
EXAMPLE
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
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}]]
PROG
(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 */
CROSSREFS
Cf. A285992, A299107, A299109, A088165, A117522, A299100, A299101, A113501, A269251, A269252, A269253, A269254, A294099, A298675, A298677, A298878, A299045, A299071. Cf. A094954 (unsigned version of this array, but missing the first row). Cf. Rows: A057078, A033999, A099496, A079935 (or A001835), A004253, A001653, A049685, A070997, A070998, A138288 (or A072256), ...
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.
+10
14
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
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
LINKS
W. W. Chao, Problem 2981, Crux Mathematicorum, 30 (2004), p. 430.
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
EXAMPLE
a(5) = 61 because (1 + a(4) + a(4)^2)/a(3) = (1 + 13 + 169)/3 = 61.
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 *)
PROG
(PARI) Vec(x*(1-5*x+3*x^2)/((1-x)*(1-5*x+x^2)) + O(x^30)) \\ Michel Marcus, Aug 03 2016 (Magma) [n le 2 select 1 else 5*Self(n-1)-Self(n-2)-1: n in [1..30]]; // Vincenzo Librandi, Dec 25 2018 (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
AUTHOR
M. Benito, O. Ciaurri and E. Fernandez (oscar.ciaurri(AT)dmc.unirioja.es), Jan 13 2005
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.
+10
14
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
COMMENTS
Mirror image of triangle in A121314.
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
EXAMPLE
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;
MATHEMATICA
m = 13;
DELTA[Join[{1, 0, 1}, Table[0, {m}]], Join[{0, 1}, Table[0, {m}]], m] // Flatten (* Jean-François Alcover, Feb 19 2020 *)
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