A076445 -id:A076445 - cp's OEIS frontend

A001075 a(0) = 1, a(1) = 2, a(n) = 4*a(n-1) - a(n-2).

1, 2, 7, 26, 97, 362, 1351, 5042, 18817, 70226, 262087, 978122, 3650401, 13623482, 50843527, 189750626, 708158977, 2642885282, 9863382151, 36810643322, 137379191137, 512706121226, 1913445293767, 7141075053842, 26650854921601, 99462344632562, 371198523608647

Offset: 0

N. J. A. Sloane

Chebyshev's T(n,x) polynomials evaluated at x=2.
x = 2^n - 1 is prime if and only if x divides a(2^(n-2)).
Any k in the sequence is succeeded by 2*k + sqrt{3*(k^2 - 1)}. - Lekraj Beedassy, Jun 28 2002
For all elements x of the sequence, 12*x^2 - 12 is a square. Lim_{n -> infinity} a(n)/a(n-1) = 2 + sqrt(3) = (4 + sqrt(12))/2 which preserves the kinship with the equation "12*x^2 - 12 is a square" where the initial "12" ends up appearing as a square root. - Gregory V. Richardson, Oct 10 2002
This sequence gives the values of x in solutions of the Diophantine equation x^2 - 3*y^2 = 1; the corresponding values of y are in A001353. The solution ratios a(n)/A001353(n) are obtained as convergents of the continued fraction expansion of sqrt(3): either as successive convergents of [2;-4] or as odd convergents of [1;1,2]. - Lekraj Beedassy, Sep 19 2003 [edited by Jon E. Schoenfield, May 04 2014]
a(n) is half the central value in a list of three consecutive integers, the lengths of the sides of a triangle with integer sides and area. - Eugene McDonnell (eemcd(AT)mac.com), Oct 19 2003
a(3+6*k) - 1 and a(3+6*k) + 1 are consecutive odd powerful numbers. See A076445. - T. D. Noe, May 04 2006
The intermediate convergents to 3^(1/2), beginning with 3/2, 12/7, 45/26, 168/97, comprise a strictly increasing sequence; essentially, numerators=A005320, denominators=A001075. - Clark Kimberling, Aug 27 2008
The upper principal convergents to 3^(1/2), beginning with 2/1, 7/4, 26/15, 97/56, comprise a strictly decreasing sequence; numerators=A001075, denominators=A001353. - Clark Kimberling, Aug 27 2008
a(n+1) is the Hankel transform of A000108(n) + A000984(n) = (n+2)*Catalan(n). - Paul Barry, Aug 11 2009
Also, numbers such that floor(a(n)^2/3) is a square: base 3 analog of A031149, A204502, A204514, A204516, A204518, A204520, A004275, A001541. - M. F. Hasler, Jan 15 2012
Pisano period lengths:  1, 2, 2, 4, 3, 2, 8, 4, 6, 6, 10, 4, 12, 8, 6, 8, 18, 6, 5, 12, ... - R. J. Mathar, Aug 10 2012
Except for the first term, positive values of x (or y) satisfying x^2 - 4*x*y + y^2 + 3 = 0. - Colin Barker, Feb 04 2014
Except for the first term, positive values of x (or y) satisfying x^2 - 14*x*y + y^2 + 48 = 0. - Colin Barker, Feb 10 2014
A triangle with row sums generating the sequence can be constructed by taking the production matrix M. Take powers of M, extracting the top rows.
M =
  1, 1, 0, 0, 0, 0, ...
  2, 0, 3, 0, 0, 0, ...
  2, 0, 0, 3, 0, 0, ...
  2, 0, 0, 0, 3, 0, ...
  2, 0, 0, 0, 0, 3, ...
  ...
The triangle generated from M is:
   1,
   1,  1,
   3,  1, 3,
  11,  3, 3, 9,
  41, 11, 9, 9, 27,
   ...
The left border is A001835 and row sums are (1, 2, 7, 26, 97, ...). - Gary W. Adamson, Jul 25 2016
Even-indexed terms are odd while odd-indexed terms are even. Indeed, a(2*n) = 2*(a(n))^2 - 1 and a(2*n+1) = 2*a(n)*a(n+1) - 2. - Timothy L. Tiffin, Oct 11 2016
For each n, a(0) divides a(n), a(1) divides a(2n+1), a(2) divides a(4*n+2), a(3) divides a(6*n+3), a(4) divides a(8*n+4), a(5) divides a(10n+5), and so on. Thus, a(k) divides a((2*n+1)*k) for each k > 0 and n >= 0. A proof of this can be found in Bhargava-Kedlaya-Ng's first solution to Problem A2 of the 76th Putnam Mathematical Competition. Links to the exam and its solutions can be found below. - Timothy L. Tiffin, Oct 12 2016
From Timothy L. Tiffin, Oct 21 2016: (Start)
If any term a(n) is a prime number, then its index n will be a power of 2. This is a consequence of the results given in the previous two comments. See A277434 for those prime terms.
a(2n) == 1 (mod 6) and a(2*n+1) == 2 (mod 6). Consequently, each odd prime factor of a(n) will be congruent to 1 modulo 6 and, thus, found in A002476.
a(n) == 1 (mod 10) if n == 0 (mod 6), a(n) == 2 (mod 10) if n == {1,-1} (mod 6), a(n) == 7 (mod 10) if n == {2,-2} (mod 6), and a(n) == 6 (mod 10) if n == 3 (mod 6). So, the rightmost digits of a(n) form a repeating cycle of length 6: 1, 2, 7, 6, 7, 2. (End)
a(A298211(n)) = A002350(3*n^2). - A.H.M. Smeets, Jan 25 2018
(2 + sqrt(3))^n = a(n) + A001353(n)*sqrt(3), n >= 0; integers in the quadratic number field Q(sqrt(3)). - Wolfdieter Lang, Feb 16 2018
Yong Hao Ng has shown that for any n, a(n) is coprime with any member of A001834 and with any member of A001835. - René Gy, Feb 26 2018
Positive numbers k such that 3*(k-1)*(k+1) is a square. - Davide Rotondo, Oct 25 2020
a(n)*a(n+1)-1 = a(2*n+1)/2 = A001570(n) divides both a(n)^6+1 and a(n+1)^6+1. In other words, for k = a(2*n+1)/2, (k+1)^6 has divisors congruent to -1 modulo k (cf. A350916). - Max Alekseyev, Jan 23 2022

			2^6 - 1 = 63 does not divide a(2^4) = 708158977, therefore 63 is composite. 2^5 - 1 = 31 divides a(2^3) = 18817, therefore 31 is prime.
G.f. = 1 + 2*x + 7*x^2 + 26*x^3 + 97*x^4 + 362*x^5 + 1351*x^6 + 5042*x^7 + ...

Serge Lang, Introduction to Diophantine Approximations, Addison-Wesley, New York, 1966.
Eugene McDonnell, "Heron's Rule and Integer-Area Triangles", Vector 12.3 (January 1996) pp. 133-142.
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.

Indranil Ghosh, Table of n, a(n) for n = 0..1745 (terms 0..200 from T. D. Noe)
Christian Aebi and Grant Cairns, Lattice Equable Parallelograms, arXiv:2006.07566 [math.NT], 2020.
Christian Aebi and Grant Cairns, Less than Equable Triangles on the Eisenstein lattice, arXiv:2312.10866 [math.CO], 2023.
Krassimir T. Atanassov and Anthony G. Shannon, On intercalated Fibonacci sequences, Notes on Number Theory and Discrete Mathematics (2020) Vol. 26, No. 3, 218-223.
C. Banderier and D. Merlini, Lattice paths with an infinite set of jumps, FPSAC02, Melbourne, 2002.
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
H. Brocard, Notes élémentaires sur le problème de Peel [sic], Nouvelle Correspondance Mathématique, 4 (1878), 337-343.
Yurii S. Bystryk, Vitalii L. Denysenko, and Volodymyr I. Ostryk, Lune and Lens Sequences, ResearchGate preprint, 2024. See pp. 30, 43, 56.
Chris Caldwell, Primality Proving, Arndt's theorem.
J. B. Cosgrave and K. Dilcher, A role for generalized Fermat numbers, Math. Comp., 2016.
G. Dresden and Y. Li, Periodic Weighted Sums of Binomial Coefficients, arXiv:2210.04322 [math.NT], 2022.
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242.
Margherita Maria Ferrari and Norma Zagaglia Salvi, Aperiodic Compositions and Classical Integer Sequences, Journal of Integer Sequences, Vol. 20 (2017), Article 17.8.8.
R. K. Guy, Letter to N. J. A. Sloane concerning A001075, A011943, A094347 [Scanned and annotated letter, included with permission]
Kiran S. Kedlaya, The 76th William Lowell Putnam Mathematical Competition, Dec 05 2015.
Kiran S. Kedlaya, Solutions to the 76th William Lowell Putnam Mathematical Competition, Dec 05 2015.
Tanya Khovanova, Recursive Sequences
Clark Kimberling, Best lower and upper approximates to irrational numbers, Elemente der Mathematik, 52 (1997) 122-126.
Pablo Lam-Estrada, Myriam Rosalía Maldonado-Ramírez, José Luis López-Bonilla, and Fausto Jarquín-Zárate, The sequences of Fibonacci and Lucas for each real quadratic fields Q(Sqrt(d)), arXiv:1904.13002 [math.NT], 2019.
Eugene McDonnell, Heron's Rule and Integer-Area Triangles, At Play With J, 2010.
Valcho Milchev and Tsvetelina Karamfilova, Domino tiling in grid - new dependence, arXiv:1707.09741 [math.HO], 2017.
Yong Hao Ng, A partition in three classes of the set of all prime numbers?, Mathematics Stack Exchange.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
F. V. Waugh and M. W. Maxfield, Side-and-diagonal numbers, Math. Mag., 40 (1967), 74-83.
Index entries for sequences related to Chebyshev polynomials.
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (4,-1).

Bisections are A011943 and A094347.

Cf. A001353, A065918, A071954, A001571, A001834, A003500, A016064, A082840, A026150, A277434.

a001075 n = a001075_list !! n
a001075_list =
   1 : 2 : zipWith (-) (map (4 *) $ tail a001075_list) a001075_list
-- Reinhard Zumkeller, Aug 11 2011

I:=[1, 2]; [n le 2 select I[n] else 4*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Dec 19 2017

A001075 := proc(n)
    orthopoly[T](n,2) ;
end proc:
seq(A001075(n),n=0..30) ; # R. J. Mathar, Apr 14 2018

Table[ Ceiling[(1/2)*(2 + Sqrt[3])^n], {n, 0, 24}]
CoefficientList[Series[(1-2*x) / (1-4*x+x^2), {x, 0, 24}], x] (* Jean-François Alcover, Dec 21 2011, after Simon Plouffe *)
LinearRecurrence[{4,-1},{1,2},30] (* Harvey P. Dale, Aug 22 2015 *)
Round@Table[LucasL[2n, Sqrt[2]]/2, {n, 0, 20}] (* Vladimir Reshetnikov, Sep 15 2016 *)
ChebyshevT[Range[0, 20], 2] (* Eric W. Weisstein, May 26 2017 *)
a[ n_] := LucasL[2*n, x]/2 /. x->Sqrt[2]; (* Michael Somos, Sep 05 2022 *)

{a(n) = subst(poltchebi(abs(n)), x, 2)};

{a(n) = real((2 + quadgen(12))^abs(n))};

{a(n) = polsym(1 - 4*x + x^2, abs(n))[1 + abs(n)]/2};

a(n)=polchebyshev(n,1,2) \\ Charles R Greathouse IV, Nov 07 2016

my(x='x+O('x^30)); Vec((1-2*x)/(1-4*x+x^2)) \\ G. C. Greubel, Dec 19 2017

[lucas_number2(n,4,1)/2 for n in range(0, 25)] # Zerinvary Lajos, May 14 2009

def a(n):
    Q = QuadraticField(3, 't')
    u = Q.units()[0]
    return (u^n).lift().coeffs()[0]  # Ralf Stephan, Jun 19 2014

G.f.: (1 - 2*x)/(1 - 4*x + x^2). - Simon Plouffe in his 1992 dissertation
E.g.f.: exp(2*x)*cosh(sqrt(3)*x).
a(n) = 4*a(n-1) - a(n-2) = a(-n).
a(n) = (S(n, 4) - S(n-2, 4))/2 = T(n, 2), with S(n, x) := U(n, x/2), S(-1, x) := 0, S(-2, x) := -1. U, resp. T, are Chebyshev's polynomials of the second, resp. first, kind. S(n-1, 4) = A001353(n), n >= 0. See A049310 and A053120.
a(n) = A001353(n+2) - 2*A001353(n+1).
a(n) = sqrt(1 + 3*A001353(n)) (cf. Richardson comment, Oct 10 2002).
a(n) = 2^(-n)*Sum_{k>=0} binomial(2*n, 2*k)*3^k = 2^(-n)*Sum_{k>=0} A086645(n, k)*3^k. - Philippe Deléham, Mar 01 2004
a(n) = ((2 + sqrt(3))^n + (2 - sqrt(3))^n)/2; a(n) = ceiling((1/2)*(2 + sqrt(3))^(n)).
a(n) = cosh(n * log(2 + sqrt(3))).
a(n) = Sum_{k=0..floor(n/2)} binomial(n, 2*k)*2^(n-2*k)*3^k. - Paul Barry, May 08 2003
a(n+2) = 2*a(n+1) + 3*Sum_{k>=0} a(n-k)*2^k. - Philippe Deléham, Mar 03 2004
a(n) = 2*a(n-1) + 3*A001353(n-1). - Lekraj Beedassy, Jul 21 2006
a(n) = left term of M^n * [1,0] where M = the 2 X 2 matrix [2,3; 1,2]. Right term = A001353(n). Example: a(4) = 97 since M^4 * [1,0] = [A001075(4), A001353(4)] = [97, 56]. - Gary W. Adamson, Dec 27 2006
Binomial transform of A026150: (1, 1, 4, 10, 28, 76, ...). - Gary W. Adamson, Nov 23 2007
First differences of A001571. - N. J. A. Sloane, Nov 03 2009
Sequence satisfies -3 = f(a(n), a(n+1)) where f(u, v) = u^2 + v^2 - 4*u*v. - Michael Somos, Sep 19 2008
a(n) = Sum_{k=0..n} A201730(n,k)*2^k. - Philippe Deléham, Dec 06 2011
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x*(3*k - 4)/(x*(3*k - 1) - 2/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 28 2013
a(n) = Sum_{k=0..n} A238731(n,k). - Philippe Deléham, Mar 05 2014
a(n) = (-1)^n*(A125905(n) + 2*A125905(n-1)), n > 0. - Franck Maminirina Ramaharo, Nov 11 2018
a(n) = (tan(Pi/12)^n + tan(5*Pi/12)^n)/2. - Greg Dresden, Oct 01 2020
From Peter Bala, Aug 17 2022: (Start)
a(n) = (1/2)^n * [x^n] ( 4*x + sqrt(1 + 12*x^2) )^n.
The g.f. A(x) satisfies A(2*x) = 1 + x*B'(x)/B(x), where B(x) = 1/sqrt(1 - 8*x + 4*x^2) is the g.f. of A069835.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p >= 3 and positive integers n and k.
Sum_{n >= 1} 1/(a(n) - (3/2)/a(n)) = 1.
Sum_{n >= 1} (-1)^(n+1)/(a(n) + (1/2)/a(n)) = 1/3.
Sum_{n >= 1} 1/(a(n)^2 - 3/2) = 1 - 1/sqrt(3). (End)
a(n) = binomial(2*n, n) + 2*Sum_{k > 0} binomial(2*n, n+2*k)*cos(k*Pi/3). - Greg Dresden, Oct 11 2022
2*a(n) + 2^n = 3*Sum_{k=-n..n} (-1)^k*binomial(2*n, n+6*k). - Greg Dresden, Feb 07 2023

More terms from James Sellers, Jul 10 2000

Chebyshev comments from Wolfdieter Lang, Oct 31 2002

A056169 Number of unitary prime divisors of n.

Original entry on oeis.org

0, 1, 1, 0, 1, 2, 1, 0, 0, 2, 1, 1, 1, 2, 2, 0, 1, 1, 1, 1, 2, 2, 1, 1, 0, 2, 0, 1, 1, 3, 1, 0, 2, 2, 2, 0, 1, 2, 2, 1, 1, 3, 1, 1, 1, 2, 1, 1, 0, 1, 2, 1, 1, 1, 2, 1, 2, 2, 1, 2, 1, 2, 1, 0, 2, 3, 1, 1, 2, 3, 1, 0, 1, 2, 1, 1, 2, 3, 1, 1, 0, 2, 1, 2, 2, 2, 2, 1, 1, 2, 2, 1, 2, 2, 2, 1, 1, 1, 1, 0, 1, 3, 1, 1, 3

Offset: 1

Labos Elemer, Jul 27 2000

The zeros of this sequences are the powerful numbers (A001694). There are no arbitrarily long subsequences with a given upper bound; for example, every sequence of 4 values includes one divisible by 2 but not 4, so there are no more than 3 consecutive zeros. Similarly, there can be no more than 23 consecutive values with none divisible by both 2 and 3 but neither 4 nor 9 (so a(n) >= 2), etc. In general, this gives an upper bound that is a (relatively) small multiple of the k-th primorial number (prime(k)#). One suspects that the actual upper bounds for such subsequences are quite a bit lower; e.g., Erdős conjectured that there are no three consecutive powerful numbers. - Franklin T. Adams-Watters, Aug 08 2006
In particular, for every A048670(k)*A002110(k) consecutive terms, at least one is greater than or equal to k. - Charlie Neder, Jan 03 2019
Following Catalan's conjecture (which became Mihăilescu's theorem in 2002), the first case of two consecutive zeros in this sequence is for a(8) and a(9), because 8 = 2^3 and 9 = 3^2, and there are no other consecutive zeros for consecutive powers. However, there are other pairs of consecutive zeros at powerful numbers (A001694, A060355). The next example is a(288) = a(289) = 0, because 288 = 2^5 * 3^2 and 289 = 17^2, then also a(675) and a(676). - Bernard Schott, Jan 06 2019
a(2k-1) is the number of primes p such that p || x + y and p^2 || x^(2k-1) + y^(2k-1) for some positive integers x and y. For any positive integers x, y and k > 1, there is no prime p such that p || x + y and p^2 || x^(2k) + y^(2k). - Jinyuan Wang, Apr 08 2020

			9 = 3^2 so a(9) = 0; 10 = 2 * 5 so a(10) = 2; 11 = 11^1 so a(11) = 1.

T. D. Noe, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, Catalan's Conjecture.
Index entries for sequences computed from exponents in factorization of n.

Cf. A001221, A001694, A002110, A034444, A056170, A055231, A076445, A162642, A275812, A295659, A295662, A295664, A001694.
See also A060355.
Cf. A077761, A085548.

a056169 = length . filter (== 1) . a124010_row
-- Reinhard Zumkeller, Sep 10 2013

a:= n-> nops(select(i-> i[2]=1, ifactors(n)[2])):
seq(a(n), n=1..120);  # Alois P. Heinz, Mar 27 2017

Join[{0},Table[Count[Transpose[FactorInteger[n]][[2]],1],{n,2,110}]] (* Harvey P. Dale, Mar 15 2012 *)
Table[DivisorSum[n, 1 &, And[PrimeQ@ #, CoprimeQ[#, n/#]] &], {n, 105}] (* Michael De Vlieger, Nov 28 2017 *)

a(n)=my(f=factor(n)[,2]); sum(i=1,#f,f[i]==1) \\ Charles R Greathouse IV, Apr 29 2015

from sympy import factorint
def a(n):
    f=factorint(n)
    return 0 if n==1 else sum(1 for i in f if f[i]==1)
print([a(n) for n in range(1, 101)]) # Indranil Ghosh, Jun 19 2017

;; With memoization-macro definec.
(definec (A056169 n) (if (= 1 n) 0 (+ (if (= 1 (A067029 n)) 1 0) (A056169 (A028234 n))))) ;; Antti Karttunen, Nov 28 2017

A prime factor of n is unitary iff its exponent is 1 in prime factorization of n. In general, gcd(p, n/p) = 1 or = p.
Additive with a(p^e) = 1 if e = 1, 0 otherwise.
a(n) = #{k: A124010(n,k) = 1, k = 1..A001221}. - Reinhard Zumkeller, Sep 10 2013
From Antti Karttunen, Nov 28 2017: (Start)
a(1) = 0; for n > 1, a(n) = A063524(A067029(n)) + a(A028234(n)).
a(n) = A001221(A055231(n)) = A001222(A055231(n)).
a(n) = A001221(n) - A056170(n) = A001221(n) - A001221(A000188(n)).
a(n) = A001222(n) - A275812(n).
a(n) = A162642(n) - A295662(n).
a(n) <= A162642(n) <= a(n) + A295659(n).
a(n) <= A295664(n).
(End)
Sum_{k=1..n} a(k) ~ n * (log(log(n)) + B - C), where B is Mertens's constant (A077761) and C = Sum_{p prime} (1/p^2) = 0.452247... (A085548). - Amiram Eldar, Sep 28 2023

A001091 a(n) = 8*a(n-1) - a(n-2); a(0) = 1, a(1) = 4.

Original entry on oeis.org

1, 4, 31, 244, 1921, 15124, 119071, 937444, 7380481, 58106404, 457470751, 3601659604, 28355806081, 223244789044, 1757602506271, 13837575261124, 108942999582721, 857706421400644, 6752708371622431, 53163960551578804

Offset: 0

N. J. A. Sloane

a(15+30k)-1 and a(15+30k)+1 are consecutive odd powerful numbers. The first pair is 13837575261124 +- 1. See A076445. - T. D. Noe, May 04 2006
This sequence gives the values of x in solutions of the Diophantine equation x^2 - 15*y^2 = 1. The corresponding y values are in A001090. - Vincenzo Librandi, Nov 12 2010 [edited by Jon E. Schoenfield, May 04 2014]
The square root of 15*(n^2-1) at those numbers = 5 * A136325. - Richard R. Forberg, Nov 22 2013
For the above Diophantine equation x^2-15*y^2=1, x + y = A105426. - Richard R. Forberg, Nov 22 2013
a(n) solves for x in the Diophantine equation floor(3*x^2/5)= y^2. The corresponding y solutions are provided by A136325(n).  x + y = A070997(n). - Richard R. Forberg, Nov 22 2013
Except for the first term, values of x (or y) in the solutions to x^2 - 8xy + y^2 + 15 = 0. - Colin Barker, Feb 05 2014

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) - From N. J. A. Sloane, May 30 2012
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).

T. D. Noe, Table of n, a(n) for n=0..200
H. Brocard, Notes élémentaires sur le problème de Peel [sic], Nouvelle Correspondance Mathématique, 4 (1878), 337-343.
Tanya Khovanova, Recursive Sequences
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (8,-1).

Cf. A001090, A090965, A098269, A322836 (column 4).

a:=[1,4];; for n in [3..20] do a[n]:=8*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Aug 26 2019

R:=PowerSeriesRing(Integers(), 20); Coefficients(R!( (1-4*x)/(1-8*x+x^2) )); // G. C. Greubel, Aug 26 2019

LinearRecurrence[{8,-1},{1,4},20] (* Harvey P. Dale, May 01 2014 *)

PARI
```
a(n)=subst(poltchebi(n),x,4)
```

a(n)=n=abs(n); polcoeff((1-4*x)/(1-8*x+x^2)+x*O(x^n),n) /* Michael Somos, Jun 07 2005 */

def A001091_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( (1-4*x)/(1-8*x+x^2) ).list()
A001091_list(20) # G. C. Greubel, Aug 26 2019

G.f.: A(x) = (1-4*x)/(1-8*x+x^2). - Simon Plouffe in his 1992 dissertation
For all elements x of the sequence, 15*(x^2 -1) is a square. Limit_{n -> infinity} a(n)/a(n-1) = 4 + sqrt(15). - Gregory V. Richardson, Oct 11 2002
a(n) = sqrt(15*((A001090(n))^2)+1).
a(n) = ((4+sqrt(15))^n + (4-sqrt(15))^n)/2.
a(n) = 4*S(n-1, 8) - S(n-2, 8) = (S(n, 8) - S(n-2, 8))/2, n>=1; S(n, x) := U(n, x/2) with Chebyshev's polynomials of the 2nd kind, A049310, with S(-1, x) := 0 and S(-2, x) := -1.
a(n) = T(n, 4) with Chebyshev's polynomials of the first kind; see A053120.
a(n)=a(-n). - Ralf Stephan, Jun 06 2005
a(n)*a(n+3) - a(n+1)*a(n+2) = 120. - Ralf Stephan, Jun 06 2005
From Peter Bala, Feb 19 2022: (Start)
a(n) = Sum_{k = 0..floor(n/2)} 4^(n-2*k)*15^k*binomial(n,2*k).
a(n) = [x^n] (4*x + sqrt(1 + 15*x^2))^n.
The g.f. A(x) satisfies A(2*x) = 1 + x*B'(x)/B(x), where B(x) = 1/sqrt(1 - 16*x + 4*x^2) is the g.f. of A098269.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p >= 3 and positive integers n and k. (End)
From Peter Bala, Aug 17 2022: (Start)
Sum_{n >= 1} 1/(a(n) - (5/2)/a(n)) = 1/3.
Sum_{n >= 1} (-1)^(n+1)/(a(n) + (3/2)/a(n)) = 1/5.
Sum_{n >= 1} 1/(a(n)^2 - 5/2) =  1/3 - 1/sqrt(15). (End)
a(n) = A001090(n+1)-4*A001090(n). - R. J. Mathar, Dec 12 2023

More terms from Larry Reeves (larryr(AT)acm.org), Aug 25 2000

Chebyshev comments from Wolfdieter Lang, Oct 31 2002

A060355 Numbers k such that k and k+1 are powerful numbers.

Original entry on oeis.org

8, 288, 675, 9800, 12167, 235224, 332928, 465124, 1825200, 11309768, 384199200, 592192224, 4931691075, 5425069447, 13051463048, 221322261600, 443365544448, 865363202000, 8192480787000, 11968683934831, 13325427460800, 15061377048200, 28821995554247

Offset: 1

Jason Earls, Apr 01 2001

nonn

"Erdős conjectured in 1975 that there do not exist three consecutive powerful integers." - Guy
See Guy for Erdős's conjecture and statement that this sequence is infinite. - Jud McCranie, Oct 13 2002
It is easy to see that this sequence is infinite: if k is in the sequence, so is 4*k*(k+1). - Franklin T. Adams-Watters, Sep 16 2009
The first of a run of three consecutive powerful numbers (conjectured to be empty) are just those in this sequence and A076445. - Charles R Greathouse IV, Nov 16 2012
Jaroslaw Wroblewski (see Prime Puzzles link) shows that there are infinitely many terms k in this sequence such that neither k nor k+1 is a square. - Charles R Greathouse IV, Nov 19 2012
Paul Erdős wrote of meeting Kurt Mahler in 1936: "I almost immediately posed him the following problem: ... are there infinitely many consecutive powerful numbers? Mahler immediately answered: Trivially, yes! x^2 - 8y^2 = 1 has infinitely many solutions. I was a bit crestfallen since I felt that I should have thought of this myself." - Jonathan Sondow, Feb 08 2015
Of the first 39 terms k, only 7 are such that neither k nor k+1 is a square. - Jon E. Schoenfield, Jun 12 2024

			1825200 belongs to this sequence because both 1825200 = 2^4 * 3^3 * 5^2 * 13^2 and 1825201 = 7^2 * 193^2 = 1351^2 are powerful numbers. - _Labos Elemer_, May 03 2001

J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 288, pp. 74, Ellipses, Paris 2008.
R. K. Guy, Unsolved Problems in Number Theory, B16.
P. Shiu, On the number of square-full integers between successive squares, Volume 27, Issue 2 (December 1980), pp. 171-178.

Donovan Johnson, Table of n, a(n) for n = 1..39 (terms < 10^22)
C. K. Caldwell, Powerful Numbers.
P. Erdős, Some personal and mathematical reminiscences of Kurt Mahler, Austral. Math. Soc. Gaz., 16 (1) (1989), 1-2.
Jérôme Germoni, Nombres puissants au bac S, Images des Mathématiques, CNRS, 2018 (in French).
J. J. O'Connor and E. F. Robertson, Biography of Kurt Mahler.
Carlos Rivera, Problem 53. Powerful numbers revisited, The Prime Puzzles & Problems Connection.
Eric Weisstein's World of Mathematics, Powerful numbers.

Primitive elements are in A199801.
Cf. A001694, A060859.
Cf. A076446 (first differences of A001694).

a060355 n = a060355_list !! (n-1)
a060355_list = map a001694 $ filter ((== 1) . a076446) [1..]
-- Reinhard Zumkeller, Jun 03 2015, Nov 30 2012

f[n_]:=First[Union[Last/@FactorInteger[n]]];Select[Range[2000000],f[#]>1&&f[#+1]>1&] (* Vladimir Joseph Stephan Orlovsky, Jan 29 2012 *)
SequencePosition[Table[If[Min[FactorInteger[n][[;;,2]]]>1,1,0],{n,11310000}],{1,1}][[;;,1]] (* The program generates the first 10 terms of the sequence. *) (* Harvey P. Dale, Mar 27 2024 *)

is(n)=ispowerful(n)&&ispowerful(n+1) \\ Charles R Greathouse IV, Nov 16 2012

def A060355(n):
    a = sloane.A001694
    return a.is_powerful(n) and a.is_powerful(n+1)
[n for n in (1..333333) if A060355(n)] # Peter Luschny, Feb 08 2015

Corrected and extended by Jud McCranie, Jul 08 2001
More terms from Jud McCranie, Oct 13 2002
a(22)-a(23) from Donovan Johnson, Jul 29 2011

A062739 Odd powerful numbers.

Original entry on oeis.org

1, 9, 25, 27, 49, 81, 121, 125, 169, 225, 243, 289, 343, 361, 441, 529, 625, 675, 729, 841, 961, 1089, 1125, 1225, 1323, 1331, 1369, 1521, 1681, 1849, 2025, 2187, 2197, 2209, 2401, 2601, 2809, 3025, 3087, 3125, 3249, 3267, 3375, 3481, 3721, 3969, 4225

Offset: 1

Labos Elemer, Jul 12 2001

nonn

Smallest term of this sequence not also in A075109 is 675, followed by 1125. - Alonso del Arte, Nov 22 2011

			Consecutive-odd examples from Sentance: {25,27},{70225,70227},{189750625,189750627}

R. K. Guy, Unsolved Problems in Number Theory, B16

Zak Seidov, Table of n, a(n) for n = 1..10000
W. A. Sentance, Occurrences of consecutive odd powerful numbers, Amer. Math. Monthly, 88 (1981), 272-274.

Cf. A001694, A060355, A060859, A060860, A082695.

Cf. A076445 (consecutive odd powerful numbers).

Powerful[n_Integer] := (n ==1) || Min[Transpose[FactorInteger[n]][[2]]]>=2; Select[Range[5000],OddQ[ # ]&&Powerful[ # ]&] (* T. D. Noe, May 04 2006 *)
Join[{1},Select[Range[3,4301,2],Min[FactorInteger[#][[All,2]]]>1&]] (* Harvey P. Dale, Jan 08 2021 *)

It is not true that a(n) = A001694(2n-1).

Sum_{n>=1} 1/a(n) = (2/3) * Sum_{n>=1} 1/A001694(n) = 2*zeta(2)*zeta(3)/(3*zeta(6)) = (2/3) * A082695 = 1.2957309... - Amiram Eldar, Jun 23 2020

Checked by T. D. Noe, May 04 2006

A001081 a(n) = 16*a(n-1) - a(n-2).

Original entry on oeis.org

1, 8, 127, 2024, 32257, 514088, 8193151, 130576328, 2081028097, 33165873224, 528572943487, 8424001222568, 134255446617601, 2139663144659048, 34100354867927167, 543466014742175624, 8661355881006882817

Offset: 0

N. J. A. Sloane

Chebyshev's polynomials T(n,x) evaluated at x=8.
The a(n) give all (unsigned, integer) solutions of Pell equation a(n)^2 - 63*b(n)^2 = +1 with b(n)= A077412(n-1), n>=1 and b(0)=0.
Also gives solutions to the equation x^2-1=floor(x*r*floor(x/r)) where r=sqrt(7). - Benoit Cloitre, Feb 14 2004
a(7+14k)-1 and a(7+14k)+1 are consecutive odd powerful numbers. The first pair is 130576328+-1. See A076445. - T. D. Noe, May 04 2006
a(n)^2 - 7 * A001080(n)^2 = 1 (this property is equivalent to the second comment). - Vincenzo Librandi, Feb 17 2013
a(n+3)*a(n) - a(n+2)*a(n+1) = 16*63. - Bruno Berselli, Feb 18 2013

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) - From N. J. A. Sloane, May 30 2012
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).
V. Thébault, Les Récréations Mathématiques. Gauthier-Villars, Paris, 1952, p. 281.

T. D. Noe, Table of n, a(n) for n = 0..200
H. Brocard, Notes élémentaires sur le problème de Peel [sic], Nouvelle Correspondance Mathématique, 4 (1878), 337-343.
M. Davis, One equation to rule them all, Trans. New York Acad. Sci. Ser. II, 30 (1968), 766-773.
Tanya Khovanova, Recursive Sequences
Pablo Lam-Estrada, Myriam Rosalía Maldonado-Ramírez, José Luis López-Bonilla, Fausto Jarquín-Zárate, The sequences of Fibonacci and Lucas for each real quadratic fields Q(Sqrt(d)), arXiv:1904.13002 [math.NT], 2019.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
Mihai Prunescu, On other two representations of the C-recursive integer sequences by terms in modular arithmetic, arXiv:2406.06436 [math.NT], 2024. See p. 17.
Mihai Prunescu and Lorenzo Sauras-Altuzarra, On the representation of C-recursive integer sequences by arithmetic terms, arXiv:2405.04083 [math.LO], 2024. See p. 16.
Mihai Prunescu and Joseph M. Shunia, On modular representations of C-recursive integer sequences, arXiv:2502.16928 [math.NT], 2025. See p. 5.
N. J. Wildberger, Pell's equation without irrational numbers, J. Int. Seq. 13 (2010), 10.4.3, Section 5.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (16,-1).

Cf. A090727, A001080, A010727.

I:=[1, 8]; [n le 2 select I[n] else 16*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Feb 17 2013

LinearRecurrence[{16, -1}, {1, 8}, 30]
CoefficientList[Series[(1-8*x)/(1-16*x+x^2), {x, 0, 30}], x] (* G. C. Greubel, Dec 20 2017 *)
Table[LucasL[n, 16*I]*(-I)^n/2, {n,0,30}] (* G. C. Greubel, Jun 06 2019 *)

Vec((1-8*x)/(1-16*x+x^2)+O(x^30)) \\ Charles R Greathouse IV, Jul 02 2013

[lucas_number2(n,16,1)/2 for n in range(0,30)] # Zerinvary Lajos, Jun 26 2008

G.f.: (1-8*x)/(1-16*x+x^2). - Simon Plouffe in his 1992 dissertation.
For all members x of the sequence, 7*x^2 - 7 is a square. Limit_{n->infinity} a(n)/a(n-1) = 8 + 3*sqrt(7). - Gregory V. Richardson, Oct 13 2002
a(n) = T(n, 8) = (S(n, 16)-S(n-2, 16))/2, with S(n, x) := U(n, x/2) and T(n), resp. U(n, x), are Chebyshev's polynomials of the first, resp. second, kind. See A053120 and A049310. S(-2, x) := -1, S(-1, x) := 0, S(n, 16)= A077412(n).
a(n) = ((8 + 3*sqrt(7))^n + (8 - 3*sqrt(7))^n)/2.
a(n) = sqrt(63*A077412(n-1)^2 + 1), n>=1, (cf. Richardson comment).
a(n) = 16*a(n-1) - a(n-2) with a(1)=1 and a(2)=8. - Sture Sjöstedt, Nov 18 2011
a(n) = A077412(n) - 8*A077412(n-1). - R. J. Mathar, Jul 22 2017
a(n) = (-i)^n*Lucas(n, 16*i)/2, where i = sqrt(-1). - G. C. Greubel, Jun 06 2019

Chebyshev and Pell comments from Wolfdieter Lang, Nov 08 2002

A023038 a(n) = 12*a(n-1) - a(n-2).

Original entry on oeis.org

1, 6, 71, 846, 10081, 120126, 1431431, 17057046, 203253121, 2421980406, 28860511751, 343904160606, 4097989415521, 48831968825646, 581885636492231, 6933795669081126, 82623662392481281, 984550153040694246, 11731978174095849671, 139799187936109501806

Offset: 0

David W. Wilson

From Wolfdieter Lang, Nov 08 2002: (Start)
Chebyshev's polynomials T(n,x) evaluated at x=6.
a(n+1) give all (nontrivial, integer) solutions of Pell equation a(n+1)^2 - 35*b(n)^2 = +1 with b(n)=A004191(n), n>=0. (End)
a(35+70k)-1 and a(35+70k)+1 are consecutive odd powerful numbers. The first pair is 23101441813552306872262673994181386126 +- 1. See A076445. - T. D. Noe, May 04 2006
Numbers n such that 35*(n^2-1) is a square. - Vincenzo Librandi, Nov 19 2010
Except for the first term, positive values of x (or y) satisfying x^2 - 12xy + y^2 + 35 = 0. - Colin Barker, Feb 09 2014

T. D. Noe, Table of n, a(n) for n = 0..200
Tanya Khovanova, Recursive Sequences
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (12,-1).

Cf. A087800.

I:=[1, 6]; [n le 2 select I[n] else 12*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Dec 19 2017

A023038:=n->round(((6+sqrt(35))^n + (6-sqrt(35))^n)/2); seq(A023038(n), n=0..30); # Wesley Ivan Hurt, Feb 03 2014

Table[Round[((6 + Sqrt[35])^n + (6 - Sqrt[35])^n)/2], {n, 0, 30}] (* Wesley Ivan Hurt, Feb 03 2014 *)
nn = 20; CoefficientList[Series[(1 - 6*x)/(1 - 12*x + x^2), {x, 0, nn}], x] (* T. D. Noe, Feb 05 2014 *)

x='x+O('x^30); Vec((1-6*x)/(1-12*x+x^2)) \\ G. C. Greubel, Dec 19 2017

a(n) = T(n, 6) = (S(n, 12)-S(n-2, 12))/2 with S(n, x) := U(n, x/2) and T(n), resp. U(n, x), are Chebyshev's polynomials of the first, resp. second, kind. See A053120 and A049310. S(-2, x) := -1, S(-1, x) := 0, S(n, 12)=A004191(n).
a(n) = ((6+sqrt(35))^n + (6-sqrt(35))^n)/2.
G.f.: (1-6*x)/(1-12*x+x^2).
a(n)*a(n+3) - a(n+1)*a(n+2) = 420. - Ralf Stephan, Jun 06 2005

A001085 a(n) = 20*a(n-1) - a(n-2).

Original entry on oeis.org

1, 10, 199, 3970, 79201, 1580050, 31521799, 628855930, 12545596801, 250283080090, 4993116004999, 99612037019890, 1987247624392801, 39645340450836130, 790919561392329799, 15778745887395759850, 314783998186522867201, 6279901217843061584170

Offset: 0

N. J. A. Sloane

From Wolfdieter Lang, Nov 08 2002: (Start)
Chebyshev's polynomials T(n,x) evaluated at x=10.
The a(n) give all (unsigned, integer) solutions of Pell equation a(n)^2 - 99*b(n)^2 = +1 with b(n) = A075843(n), n >= 0. (End)
a(11+22k) - 1 and a(11+22k) + 1 are consecutive odd powerful numbers. The first pair is 99612037019890 +- 1. See A076445. - T. D. Noe, May 04 2006
This sequence gives the values of x in solutions of the Diophantine equation x^2 - 11*y^2 = 1. The corresponding y values are in A001084.
Except for the first term, positive values of x (or y) satisfying x^2 - 20xy + y^2 + 99 = 0. - Colin Barker, Feb 18 2014

			G.f. = 1 + 10*x + 199*x^2 + 3970*x^3 + 79201*x^4 + 1580050*x^5 + 31521799*x^6 + ...

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) - From N. J. A. Sloane, May 30 2012
"Questions D'Arithmétique", Question 3686, Solution by H.L. Mennessier, Mathesis, 65(4, Supplement) 1956, pp. 1-12.
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).

T. D. Noe, Table of n, a(n) for n = 0..200
H. Brocard, Notes élémentaires sur le problème de Peel, Nouvelle Correspondance Mathématique, 4 (1878), 161-169.
Tanya Khovanova, Recursive Sequences
Pablo Lam-Estrada, Myriam Rosalía Maldonado-Ramírez, José Luis López-Bonilla, and Fausto Jarquín-Zárate, The sequences of Fibonacci and Lucas for each real quadratic fields Q(Sqrt(d)), arXiv:1904.13002 [math.NT], 2019.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (20,-1).

Cf. A090728, A001084.

a:=[1,10];; for n in [3..30] do a[n]:=20*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jun 06 2019

I:=[1,10]; [n le 2 select I[n] else 20*Self(n-1)-Self(n-2): n in [1..30]]; // G. C. Greubel, Dec 20 2017

LinearRecurrence[{20, -1}, {1, 10}, 30] (* T. D. Noe, Dec 19 2011 *)
a[ n_] := ChebyshevT[ n, 10]; (* Michael Somos, May 27 2014 *)
a[ n_] := ((10 + Sqrt[99])^n + (10 - Sqrt[99])^n) / 2 // Simplify; (* Michael Somos, May 27 2014 *)
a[ n_] := With[{m = Abs @ n}, SeriesCoefficient[ (1 - 10 x) / (1 - 20 x + x^2), {x, 0, m}]]; (* Michael Somos, May 27 2014 *)
Table[LucasL[n, 20*I]*(-I)^n/2, {n,0,30}] (* G. C. Greubel, Jun 06 2019 *)

{a(n) = n=abs(n); polsym( 1 - 20*x + x^2, n) [n+1] / 2}; /* Michael Somos, May 27 2014 */

my(x='x+O('x^30)); Vec((1-10*x)/(1-20*x+x^2)) \\ G. C. Greubel, Dec 20 2017

[lucas_number2(n,20,1)/2 for n in range(0,20)] # Zerinvary Lajos, Jun 27 2008

For all members x of the sequence, 11*x^2 - 11 is a square. Limit_{n->infinity} a(n)/a(n-1) = 10 + 3*sqrt(11). - Gregory V. Richardson, Oct 13 2002
a(n) = T(n, 10) = (S(n, 20)-S(n-2, 20))/2, with S(n, x) := U(n, x/2) and T(n), resp. U(n, x), are Chebyshev's polynomials of the first, resp. second, kind. See A053120 and A049310. S(-2, x) := -1, S(-1, x) := 0, S(n-1, 20)= A075843(n).
G.f.: (1-10*x)/(1-20*x+x^2). - Simon Plouffe in his 1992 dissertation
a(n) = (((10+3*sqrt(11))^n + (10-3*sqrt(11))^n))/2.
a(n) = sqrt(99*A075843(n)^2 + 1), (cf. Richardson comment).
a(n) = (-i)^n*Lucas(n, 20*i)/2, where i = sqrt(-1) and Lucas(n, x) is the Lucas polynomial. - G. C. Greubel, Jun 06 2019

cp's OEIS Frontend

A001075 a(0) = 1, a(1) = 2, a(n) = 4*a(n-1) - a(n-2).

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A056169 Number of unitary prime divisors of n.

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A001091 a(n) = 8*a(n-1) - a(n-2); a(0) = 1, a(1) = 4.

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A060355 Numbers k such that k and k+1 are powerful numbers.

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A062739 Odd powerful numbers.

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