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Bisection of A001353. Indices of square numbers which are also octagonal.
+20
21
1, 15, 209, 2911, 40545, 564719, 7865521, 109552575, 1525870529, 21252634831, 296011017105, 4122901604639, 57424611447841, 799821658665135, 11140078609864049, 155161278879431551, 2161117825702177665, 30100488280951055759, 419245718107612602961, 5839339565225625385695
COMMENTS
Chebyshev S-sequence with Diophantine property.
4*b(n)^2 - 3*a(n)^2 = 1 with b(n) = A001570(n), n>=0. As n increases, this sequence is approximately geometric with common ratio r = lim(n -> oo, a(n)/a(n-1)) = (2 + sqrt(3))^2 = 7 + 4 * sqrt(3). - Ant King, Nov 15 2011
REFERENCES
R. L. Graham, D. E. Knuth, and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 329.
J. D. E. Konhauser et al., Which Way Did the Bicycle Go?, MAA 1996, p. 104.
FORMULA
a(n) = 14*a(n-1) - a(n-2) for n>1.
a(n) = S(n, 14) + S(n-1, 14) = S(2*n, 4) with S(n, x) := U(n, x/2) Chebyshev's polynomials of the second kind. See A049310. S(-1, x) = 0, S(n, 14) = A007655(n+1) and S(n, 4) = A001353(n+1). G.f.: x*(1+x)/(1-14*x+x^2).
a(n) = (ap^(2*n+1) - am^(2*n+1))/(ap - am) with ap := 2+sqrt(3) and am := 2-sqrt(3).
a(n+1) = Sum_{k=0..n} (-1)^k*binomial(2*n-k, k)*16^(n-k), n>=0.
a(n) = sqrt((4* A001570(n-1)^2 - 1)/3). a(n) ~ 1/6*sqrt(3)*(2 + sqrt(3))^(2*n-1). - Joe Keane (jgk(AT)jgk.org), May 15 2002
4*a(n+1) = ( A001834(n))^2 + 4*( A001835(n+1))^2 - ( A001835(n))^2. E.g. 4*a(3) = 4*209 = 19^2 + 4*11^2 - 3^2 = ( A001834(2))^2 + 4*( A001835(3))^2 - ( A001835(2))^2. Generating floretion: 'i + 2'j + 3'k + i' + 2j' + 3k' + 4'ii' + 3'jj' + 4'kk' + 3'ij' + 3'ji' + 'jk' + 'kj' + 4e. - Creighton Dement, Dec 04 2004 a(n) = f(a(n-1),7) + f(a(n-2),7), where f(x,s) = s*x + sqrt((s^2-1)*x^2+1); f(0,s)=0. - Marcos Carreira, Dec 27 2006 a(n) = 1/6 * sqrt(3) * ( (tan(5*Pi/12)) ^ (2n-1) - (tan(Pi/12)) ^ (2n-1) ).
a(n) = floor (1/6 * sqrt(3) * (tan(5*Pi/12)) ^ (2n-1)). (End)
E.g.f.: 1 - exp(7*x)*(3*cosh(4*sqrt(3)*x) - 2*sqrt(3)*sinh(4*sqrt(3)*x))/3. - Stefano Spezia, Dec 12 2022
MAPLE
seq(coeff(series((1+x)/(1-14*x+x^2), x, n+1), x, n), n = 0..30); # G. C. Greubel, Dec 06 2019
MATHEMATICA
LinearRecurrence[{14, - 1}, {1, 15}, 17] (* Ant King, Nov 15 2011 *) CoefficientList[Series[(1+x)/(1-14x+x^2), {x, 0, 30}], x] (* Vincenzo Librandi, Jun 17 2014 *)
PROG
(Sage) [(lucas_number2(n, 14, 1)-lucas_number2(n-1, 14, 1))/12 for n in range(1, 18)] # Zerinvary Lajos, Nov 10 2009 (PARI) isok(n) = ispolygonal(n^2, 8); \\ Michel Marcus, Jul 09 2017 (Magma) I:=[1, 15]; [n le 2 select I[n] else 14*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Dec 06 2019 (GAP) a:=[1, 15];; for n in [3..30] do a[n]:=14*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Dec 06 2019
EXTENSIONS
Incorrect recurrence relation deleted by Ant King, Nov 15 2011
0, -1, -4, -15, -56, -209, -780, -2911, -10864, -40545, -151316, -564719, -2107560, -7865521, -29354524, -109552575, -408855776, -1525870529, -5694626340, -21252634831, -79315912984, -296011017105, -1104728155436, -4122901604639, -15386878263120, -57424611447841
FORMULA
G.f.: -x/(1-4*x+x^2).
a(n) = 4*a(n-1) - a(n-2); a(0)=0, a(1)=-1.
MAPLE
a[0]:=0: a[1]:=-1: for n from 2 to 27 do a[n]:=4*a[n-1]-a[n-2] od: seq(a[n], n=0..27);
MATHEMATICA
LinearRecurrence[{4, -1}, {0, -1}, 30] (* Harvey P. Dale, Nov 01 2019 *)
PROG
(PARI) x='x+O('x^30); Vec(-x/(1-4*x+x^2)) \\ G. C. Greubel, Feb 05 2018 (Magma) I:=[0, -1]; [n le 2 select I[n] else 4*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Feb 05 2018
1, 8, 45, 224, 1045, 4680, 20377, 86912, 364905, 1513160, 6211909, 25290720, 102251773, 410963336, 1643288625, 6541692416, 25939798993, 102503274120, 403800061789, 1586318259680, 6216231359205, 24304019419592, 94826736906697, 369285078314880, 1435615286196025
COMMENTS
Derivative of Chebyshev polynomials of the first kind evaluated at x=2.
FORMULA
G.f.: x*(1 - x)*(1 + x) / (1 - 4*x + x^2)^2.
a(n) = (((-(2-sqrt(3))^n + (2+sqrt(3))^n)*n)) / (2*sqrt(3)).
a(n) = 8*a(n-1) - 18*a(n-2) + 8*a(n-3) - a(n-4) for n>4.
(End)
MATHEMATICA
Table[ D[ ChebyshevT[n, x], x] /. x -> 2, {n, 25}]
CoefficientList[Series[-x(x^2 - 1)/(x^2 - 4x + 1)^2, {x, 0, 24}], x] (* Robert G. Wilson v, Aug 07 2018 *)
PROG
(PARI) Vec(x*(1 - x)*(1 + x) / (1 - 4*x + x^2)^2 + O(x^40)) \\ Colin Barker, Jul 28 2018 (PARI) a(n) = subst(deriv(polchebyshev(n)), x, 2); \\ Michel Marcus, Jul 29 2018
Odd primes p that divide a Lucas quotient studied by H. C. Williams: A001353(p - (3/p))/p, where (3/p) is a Jacobi symbol.
+20
4
COMMENTS
The condition for an odd prime p to be a member of this sequence is that p^2 divides A001353(p - (3/p)). Neither this quotient, nor the Lucas sequence U(4, 1) on which it is based, has a common name; but its fundamental discriminant of 3 places it between the quotient based on the Pell sequence U(2, -1) with discriminant 2 ( A000129), and that based on the Fibonacci sequence U(1, -1) with discriminant 5 ( A000045). Values of p dividing the Pell quotient will be found under A238736, while for the Fibonacci quotient it is known that there is no such p < 9.7*10^14. The interest in this family of number-theoretic quotients derives from H. C. Williams, "Some formulas concerning the fundamental unit of a real quadratic field," p. 440, which proves a formula connecting the present quotient with the Fermat quotient base 2 ( A007663), the Fermat quotient base 3 ( A146211), and the harmonic number H(floor(p/12)) (see the Formula section below). As is well known, the vanishing of each of these Fermat quotients is a necessary condition for the failure of the first case of Fermat's Last Theorem (see discussions under A001220 and A014127); and a corresponding result concerning this type of harmonic number was proved by Dilcher and Skula. Thus, the vanishing mod p of the quotient based on U(4, 1) is also a necessary condition for the failure of the first case of Fermat's Last Theorem. The pioneering computation for this quotient appears to be that of Elsenhans and Jahnel, "The Fibonacci sequence modulo p^2," p. 5, who report 103 as the only value of a(n) < 10^9. Extending the search to p < 2.5*10^10 has found only one further solution, 2297860813.
Let LucasQuotient(p) = A001353(p - (3/p))/p, q_2 = (2^(p-1) - 1)/p = A007663(p) be the corresponding Fermat quotient of base 2, q_3 = (3^(p-1) - 1)/p = A146211(p) be the corresponding Fermat quotient of base 3, H(floor(p/12)) be a harmonic number. Then Williams (1991) shows that 6*(3/p)*LucasQuotient(p) == -6*q_2 - 3*q_3 - 2*H(floor(p/12)) (mod p). Also with an initial 2, primes p such that p^2 divides A001353(p - Kronecker(12,p)) (note that 12 is the discriminant of the characteristic polynomial of A001353, x^2 - 4x + 1). - Jianing Song, Jul 28 2018
EXAMPLE
LucasQuotient(103) = 103*851367555454046677501642274766916900879231854719584128208.
MATHEMATICA
The following criteria are equivalent:
PrimeQ[p] &&
Mod[(MatrixPower[{{1, 2}, {1, 3}}, p-JacobiSymbol[3, p]-1].{{1}, {1}})[[2, 1]], p^2]==0
PrimeQ[p] && Mod[Last[LinearRecurrence[{4, -1}, {0, 1}, p-JacobiSymbol[3, p]+1]], p^2]==0
PROG
(PARI) isprime(p) && (Mod([2, 2; 1, 0], p^2)^(p-kronecker(3, p)))[2, 1]==0 \\ This test, which was used to find the second member of this sequence, is based on the test for A238736 devised by Charles R Greathouse IV
Smallest n such that A001353(a(n)) == 0 (mod n), i.e., x= A001075(a(n)) and y= A001353(a(n)) is the fundamental solution of the Pell equation x^2 - 3*(n*y)^2 = 1.
+20
4
1, 2, 3, 2, 3, 6, 4, 4, 9, 6, 5, 6, 6, 4, 3, 8, 9, 18, 5, 6, 12, 10, 11, 12, 15, 6, 27, 4, 15, 6, 16, 16, 15, 18, 12, 18, 18, 10, 6, 12, 7, 12, 11, 10, 9, 22, 23, 24, 28, 30, 9, 6, 9, 54, 15, 4, 15, 30, 29, 6, 30, 16, 36, 32, 6, 30, 17, 18, 33, 12, 7, 36, 18, 18, 15
COMMENTS
The fundamental solution of the Pell equation x^2 - 3*(n*y)^2 = 1 is the smallest solution of x^2 - 3*y^2 = 1 satisfying y == 0 (mod n).
For primes p > 2, 2^p-1 is a Mersenne prime if and only if a(2^p-1) = 2^(p-1). For example, a(7) = 4, a(31) = 16, a(127) = 64, but a(2047) = 495 < 1024. - Jianing Song, Jun 02 2022
REFERENCES
Michael J. Jacobson, Jr. and Hugh C. Williams, Solving the Pell Equation, Springer, 2009, pages 1-17.
FORMULA
a(n) <= n.
if n | m then a(n) | a(m).
a(3^m) = 3^m and a(2*3^m) = 2*3^m for m>=0.
In general: if p is prime and p == 3 (mod 4) then: a(n) = n iff n = p^m or n = 2*p^m, for m>=0.
a(p) | (p- A091338(p))/2 for p is an odd prime > 3. a(p^e) = a(p)*p^(e-r) for e >= r, where r is the largest number such that a(p^r) = a(p). r can be greater than 1, for p = 2, 103, 2297860813 (Cf. A238490). If gcd(m,n) = 1, then a(m*n) = lcm(a(m),a(n)). (End)
PROG
(Python)
xf, yf = 2, 1
x, n = 2*xf, 0
while n < 20000:
n = n+1
y1, y0, i = 0, yf, 1
while y0%n != 0:
y1, y0, i = y0, x*y0-y1, i+1
print(n, i)
Odd composite integers m such that A001353(m)^2 == 1 (mod m).
+20
4
35, 65, 91, 209, 455, 533, 595, 629, 679, 901, 923, 989, 1001, 1241, 1295, 1495, 1547, 1729, 1769, 1855, 1961, 1991, 2015, 2345, 2431, 2509, 2555, 2639, 2701, 2795, 2911, 3007, 3059, 3367, 3439, 3535, 3869, 3977, 4277, 4823, 5249, 5291, 5551, 5719, 5777
COMMENTS
For a, b integers, the generalized Lucas sequence is defined by the relation U(n+2)=a*U(n+1)-b*U(n) and U(0)=0, U(1)=1.
This sequence satisfies the relation U(p)^2 == 1 for p prime and b=1,-1.
The composite numbers with this property may be called weak generalized Lucas pseudoprimes of parameters a and b.
The current sequence is defined for a=4 and b=1.
REFERENCES
D. Andrica and O. Bagdasar, Recurrent Sequences: Key Results, Applications and Problems. Springer (2020).
MATHEMATICA
Select[Range[3, 6000, 2], CompositeQ[#] && Divisible[ChebyshevU[#-1, 2]*ChebyshevU[#-1, 2] - 1, #] &]
1, 14, 161, 1792, 19809, 218638, 2412353, 26614784, 293628097, 3239445006, 35739069409, 394290020096, 4349990523425, 47991114171406, 529460241815169, 5841251080892416, 64443392518654337, 710969410782059534
COMMENTS
Proposed by R. Guy in the seqfan list Mar 28 2009.
With an offset of 1, this sequence is the case P1 = 14, P2 = 32, Q = 1 of the 3 parameter family of 4th-order linear divisibility sequences found by Williams and Guy. - Peter Bala, Apr 27 2014
FORMULA
a(n) = 14*a(n-1) -34*a(n-2) +14*a(n-3) -a(n-4).
G.f.: (1-x^2)/(1-14*x+34*x^2-14*x^3+x^4).
The following remarks assume an offset of 1.
a(n) = (1/sqrt(17))*( T(n,(7 + sqrt(17))/2) - T(n,(7 - sqrt(17))/2) ), where T(n,x) is the Chebyshev polynomial of the first kind.
a(n) = the bottom left entry of the 2 X 2 matrix T(n,M), where M is the 2 X 2 matrix [0, -8; 1, 7].
a(n) = U(n-1,1/2*(4 + sqrt(2)))*U(n-1,1/2*(4 - sqrt(2))), where U(n,x) is the Chebyshev polynomial of the second kind.
See the remarks in A100047 for the general connection between Chebyshev polynomials of the first kind and 4th-order linear divisibility sequences. (End)
MAPLE
seq(simplify( ChebyshevU(n, (4+sqrt(2))/2)*ChebyshevU(n, (4-sqrt(2))/2) ), n = 0 .. 20); # G. C. Greubel, Dec 24 2019
MATHEMATICA
CoefficientList[Series[(1-x^2)/(1-14x+34x^2-14x^3+x^4), {x, 0, 20}], x] (* Vincenzo Librandi, Apr 28 2014 *) Table[Simplify[ChebyshevU[n, (4+Sqrt[2])/2]*ChebyshevU[n, (4-Sqrt[2])/2]], {n, 0, 20}] (* G. C. Greubel, Dec 24 2019 *)
PROG
(Magma) I:=[1, 14, 161, 1792]; [n le 4 select I[n] else 14*Self(n-1)-34*Self(n-2) +14*Self(n-3)-Self(n-4): n in [1..30]]; // Vincenzo Librandi, Apr 28 2014 (PARI) vector(21, n, round(polchebyshev(n-1, 2, (4+sqrt(2))/2)*polchebyshev(n-1, 2, (4-sqrt(2))/2)) ) \\ G. C. Greubel, Dec 24 2019 (Sage) [round(chebyshev_U(n, (4+sqrt(2))/2)*chebyshev_U(n, (4-sqrt(2))/2)) for n in (0..20)] # G. C. Greubel, Dec 24 2019 (GAP) a:=[1, 14, 161, 1792];; for n in [5..20] do a[n]:=14*a[n-1]-34*a[n-2] +14*a[n-3] -a[n-4]; od; a; # G. C. Greubel, Dec 24 2019
Odd composite integers m such that U(m)^2 == 1 (mod m) and V(m) == 4 (mod m), where U(m)= A001353(m) and V(m)= A003500(m) are the m-th generalized Lucas and Pell-Lucas numbers of parameters a=4 and b=1, respectively.
+20
3
209, 455, 901, 923, 989, 1295, 1729, 1855, 2015, 2345, 2639, 2701, 2795, 2911, 3007, 3439, 3535, 4823, 5291, 5719, 6061, 6767, 6989, 7421, 8569, 9503, 9869, 10439, 10609, 11041, 11395, 11951, 13133, 13529, 13735, 13871, 14701, 14839, 15505, 15841, 17119, 17815
COMMENTS
For a, b integers, the following sequences are defined:
generalized Lucas sequences by U(n+2)=a*U(n+1)-b*U(n) and U(0)=0, U(1)=1,
generalized Pell-Lucas sequences by V(n+2)=a*V(n+1)-b*V(n) and V(0)=2, V(1)=a.
These satisfy the identities U(p)^2 == 1 and V(p)==a (mod p) for p prime and b=1,-1.
These numbers may be called weak generalized Lucas-Bruckner pseudoprimes of parameters a and b.The current sequence is defined for a=4 and b=1.
MATHEMATICA
Select[Range[3, 10000, 2], CompositeQ[#] && Divisible[2*ChebyshevT[#, 2] - 4, #] && Divisible[ChebyshevU[#-1, 2]*ChebyshevU[#-1, 2] - 1, #] &]
CROSSREFS
Similar sequences: A337627 (a=4, b=-1).
Sum of cubes of first n terms of Lucas sequence U(4,1) ( A001353) divided by sum of their first powers.
+20
2
1, 13, 172, 2356, 32661, 454329, 6325816, 88099144, 1227032521, 17090245381, 238035989412, 3315412063548, 46177727142301, 643172746439665, 8958240642814960, 124772195953666576, 1737852501591502353, 24205162822158610557, 337134426993071036956, 4695676815022772628676, 65402340983109050660389
COMMENTS
For a Lucas sequence U(k,1), the sum of the cubes of the first n terms is divisible by the sum of the first n terms. This sequence corresponds to the case of k=4.
FORMULA
a(n) = 19*a(n-1)-76*a(n-2)+76*a(n-3)-19*a(n-4)+a(n-5) for n>5.
G.f.: x*(1-6*x+x^2) / ((1-x)*(1-14*x+x^2)*(1-4*x+x^2)).
(End)
MATHEMATICA
CoefficientList[Series[(1 - 6 x + x^2)/((1 - x) (1 - 14 x + x^2) (1 - 4 x + x^2)), {x, 0, 33}], x] (* Vincenzo Librandi, Dec 09 2015 *)
PROG
(PARI) Vec(x*(1-6*x+x^2)/((1-x)*(1-14*x+x^2)*(1-4*x+x^2)) + O(x^30)) \\ Colin Barker, Dec 08 2015 (Magma) I:=[1, 13, 172, 2356, 32661]; [n le 5 select I[n] else 19*Self(n-1)-76*Self(n-2)+76*Self(n-3)-19*Self(n-4)+Self(n-5): n in [1..30]]; // Vincenzo Librandi, Dec 09 2015
1, 4, 15, 14, 209, 13, 2911, 194, 2703, 181, 564719, 193, 7865521, 2521, 34945, 37634, 1525870529, 2701, 21252634831, 37441, 6779137, 489061, 4122901604639, 37633, 274758906449, 6811741, 19726764303, 7263361, 11140078609864049, 40321, 155161278879431551
COMMENTS
A prime p is called a unique-period prime in base b if there is no other prime q such that the period length of 1/q is equal to that of 1/p. If q = a(2p) = A001353(2*p)/(4* A001353(p)) = ((2 + sqrt(3))^p + (2 - sqrt(3))^p)/4 is prime (this happens for p = 3, 5, 7, 11, 13, 17, 19, 79, 151, 199, 233, 251, 317, ...), where p is an odd prime, then q is a unique-period prime in base b = (sqrt(12*q^2 - 3) - 1)/2 (1/q has period length 3) as well as in base b' = (sqrt(12*q^2 - 3) + 1)/2 (1/q has period length 6). For example, a(6) = 13 is prime, so 13 is the only prime whose reciprocal has period length 3 in base 22 and the only prime whose reciprocal has period length 6 in base 23. Compare: If q = A000129(p) = A008555(p), then q is a unique-period prime in base b = sqrt(2*q^2 - 1) (1/q has period length 4). By Lucas-Lehmer test, p is a Mersenne prime > 3 if and only if the smallest k such that p divides a(k) is k = (p - 1)/2.
For primes p, p^2 divides a(k) for some k if and only if p = 2 or p is in A238490. If p > 2, the only possible values for k are the divisors of (p - Legendre(3,p))/2 (e.g., 103^2 divides a(52) = 53028360515521 = 103^2 * 4998431569). Conjecturally there must be infinitely many primes p such that a(p) is prime, but no such p is known. [By the formula below, there is no such p. - Jianing Song, Oct 31 2024]
FORMULA
Product_{d|n} a(d) = A001353(n), that is, a(n) = A001353(n)/(Product_{d<n, d|n} a(d)). Equivalently, a(n) = Product_{d|n} A001353(d)^mu(n/d), where mu = A008683.
EXAMPLE
For n = 8 we have: a(1) = A001353(1), a(1)*a(2) = A001353(2), a(1)*a(2)*a(4) = A001353(4), a(1)*a(2)*a(4)*a(8) = A001353(8). The solution is a(1) = 1, a(2) = 4, a(4) = 14 and a(8) = 194.
PROG
(PARI) b(n) = if(n==1, [1], my(v=vector(n)); v[1]=1; v[2]=4; for(i=3, n, v[i]=4*v[i-1]-v[i-2]); v)
a(n) = my(d=divisors(n)); prod(i=1, #d, (b(n)[d[i]])^moebius(n/d[i]))
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