A076445 -id:A076445 - cp's OEIS frontend

A097310 Chebyshev T-polynomials T(n,14) with Diophantine property.

1, 14, 391, 10934, 305761, 8550374, 239104711, 6686381534, 186979578241, 5228741809214, 146217791079751, 4088869408423814, 114342125644787041, 3197490648645613334, 89415396036432386311, 2500433598371461203374, 69922725358364481308161, 1955335876435834015425134

Offset: 0

Wolfdieter Lang, Aug 31 2004

a(n)^2 - 195 b(n)^2 = +1 with b(n):=A097311(n) gives all nonnegative solutions of this Pell equation.
a(195+390k)-1 and a(195+390k)+1 are consecutive odd powerful numbers. See A076445. - T. D. Noe, May 04 2006
Except for the first term, positive values of x (or y) satisfying x^2 - 28xy + y^2 + 195 = 0. - Colin Barker, Feb 23 2014

Vincenzo Librandi, Table of n, a(n) for n = 0..700
Tanya Khovanova, Recursive Sequences
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (28,-1).

Cf. A090249, A097311.

LinearRecurrence[{28,-1},{1,14},20] (* Harvey P. Dale, Jan 29 2014 *)
CoefficientList[Series[(1 - 14 x)/(1 - 28 x + x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Feb 24 2014 *)

Vec((1-14*x)/(1-28*x+x^2) + O(x^100)) \\ Colin Barker, Feb 23 2014

[lucas_number2(n,28,1)/2 for n in range(0,16)] # Zerinvary Lajos, Jun 27 2008

a(n) = 28*a(n-1) - a(n-2), a(-1):= 14, a(0)=1.
a(n) = T(n, 14)= (S(n, 28)-S(n-2, 28))/2 = S(n, 28)-14*S(n-1, 28) with T(n, x), resp. S(n, x), Chebyshev's polynomials of the first, resp.second, kind. See A053120 and A049310. S(n, 28)=A097311(n).
a(n) = (ap^n + am^n)/2 with ap := 14+sqrt(195) and am := 14-sqrt(195).
a(n) = sum(((-1)^k)*(n/(2*(n-k)))*binomial(n-k, k)*(2*14)^(n-2*k), k=0..floor(n/2)), n>=1.
G.f.: (1-14*x)/(1-28*x+x^2).
a(n) = sqrt(1 + 195*A097311(n)^2), n>=0.

More terms from Colin Barker, Feb 23 2014

A363190 Odd powerful numbers (A062739) k such that the next powerful number after k is also odd.

Original entry on oeis.org

25, 121, 225, 343, 1089, 1323, 2187, 2197, 3025, 3087, 3249, 5929, 6125, 6859, 7803, 8575, 9261, 10125, 11881, 11907, 14161, 15125, 16641, 16807, 19683, 19773, 21025, 22707, 25921, 27889, 29241, 29791, 30375, 33275, 36125, 41067, 42849, 44217, 45125, 45369, 49729

Offset: 1

Amiram Eldar, May 21 2023

A076445 is a subsequence if there are no three consecutive integers that are powerful numbers (A001694).

			25 = 5^2 is a term since it is an odd powerful number and the next powerful number, 27 = 3^3, is also odd.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A001694, A062739, A076445, A363189, A363191, A363192.

With[{pow = Select[Range[10^5], # == 1 || Min[FactorInteger[#][[;; , 2]]] > 1 &]}, pow[[Select[Range[Length[pow] - 1], OddQ[pow[[#]]] && OddQ[pow[[#+1]]] &]]]]

lista(kmax) = {my(c = 0); for(k = 1, kmax, if(ispowerful(k), c++; if(k%2, print1(c, ", ")))); }

A238799 a(0) = 1, a(n+1) = 2a(n)^3 + 3a(n).

Original entry on oeis.org

1, 5, 265, 37220045, 103124220135120334842385, 2193370648451279691104497113491599222165108730278225579497595691360405

Offset: 0

Arkadiusz Wesolowski, Mar 05 2014

a(6) has 209 digits and is too large to include.
Except for the first term, this is a subsequence of A175180.
The squares larger than 1 are in A076445.
If we define u(0) = 1 , u(n+1) = (u(n)/3)*(u(n)^2+9) / (u(n)^2 + 1), then u(n) = a(n) / A378683(n) ; this is Halley's method to calculate sqrt(3). - Robert FERREOL, Dec 21 2024

Wikipedia, Halley's method.

Cf. A175180, A378683.

RecurrenceTable[{a[0] == 1, a[n] == 2*a[n - 1]^3 + 3*a[n - 1]}, a[n], {n, 5}]
NestList[2#^3+3#&,1,5] (* Harvey P. Dale, Mar 22 2023 *)

a=1; print1(a, ", "); for(n=1, 5, b=2*a^3+3*a; print1(b, ", "); a=b);

{ A238799(n) = my(q=Mod(x,x^2-3)); lift( (1+q)*(2+q)^((3^n-1)/2) + (1-q)*(2-q)^((3^n-1)/2) )/2; } \\ Max Alekseyev, Sep 04 2018

a(n) = sqrt(2) * sinh( 3^n * arcsinh(1/sqrt(2)) ) = (1+sqrt(3))/2 * (2+sqrt(3))^((3^n-1)/2) + (1-sqrt(3))/2 * (2-sqrt(3))^((3^n-1)/2). - Max Alekseyev, Sep 04 2018

a(n) = ((1 + sqrt(3))^(3^n) + (1 - sqrt(3))^(3^n))/2^((3^n+1)/2) = A002531(3^n) = A080040(3^n)/2^((3^n+1)/2). - Robert FERREOL, Nov 19 2024

A118894 Numbers m such that the Pell equation x^2-m*y^2=1 has fundamental solution with x even.

Original entry on oeis.org

3, 7, 11, 15, 19, 23, 27, 31, 35, 43, 47, 51, 59, 63, 67, 71, 75, 79, 83, 87, 91, 99, 103, 107, 115, 119, 123, 127, 131, 135, 139, 143, 151, 159, 163, 167, 171, 175, 179, 187, 191, 195, 199, 211, 215, 219, 223, 227, 231, 235, 239, 243, 247, 251, 255, 263, 267

Offset: 1

T. D. Noe, May 04 2006

nonn

Numbers m such that A002350(m) is even. These m can be used to generate consecutive odd powerful numbers, as in A076445. As shown by Lang, the solution of Pell's equation is greatly simplified by Chebyshev polynomials of the first kind T(n,x), which is illustrated in A001075 for the case m=3. In that case, the solutions are x=T(n,2), for integer n>0. For any m in this sequence, let E(k)=T(m+2mk,A002350(m)). Then E(k)-1 and E(k)+1 are consecutive odd powerful numbers for k=0,1,2,...

Wolfdieter Lang, Chebyshev Polynomials and Certain Quadratic Diophantine Equations
H. W. Lenstra Jr., Solving the Pell equation, Notices AMS, 49 (2002), 182-192.

Cf. A001075, A001091, A023038, A001081, A001085, A077424, A097310 (x solutions for m=3, 15, 35, 63, 99, 143, 195).

A175180 Numbers k such that k^2 + 2 is powerful in the sense of A001694.

Original entry on oeis.org

5, 265, 13775, 716035, 9980583, 37220045, 1934726305

Offset: 1

Michel Lagneau, Mar 01 2010

This sequence is infinite (F. Luca in De Koninck).
The values of k^2 are a subset of A076445, so 23 terms of the sequence are known from there. - R. J. Mathar, Mar 05 2010
Together with 1, supersequence of A238799. - Arkadiusz Wesolowski, Mar 06 2014
From Amiram Eldar, Feb 23 2024: (Start)
a(8) <= 100568547815.
A041042(2*k) is a term for all k >= 0 (since 3^3 * A041043(n)^2 - A041042(n)^2 = -1 if n is odd and 2 if n is even). (End)

			5 is in the sequence because 5^2 + 2 = 3^3 is powerful.
265 is in the sequence because 265^2 + 2 = 51^2*3^3 is powerful.
13775 is in the sequence because 13775^2 + 2 = 2651^2 * 3^3 is powerful.

Jean-Marie De Koninck, Ces nombres qui nous fascinent, Entry 265, p. 71, Ellipses, Paris, 2008.

Jean-Marie De Koninck, Those Fascinating Numbers, Amer. Math. Soc., 2009, page 66.

Cf. A001694, A041042, A041043, A076445, A238799.

q[n_] := AllTrue[FactorInteger[n^2+2][[;;, 2]], # > 1 &]; Select[Range[10^6], q] (* Amiram Eldar, Feb 23 2024 *)

is(n)=ispowerful(n^2+2) \\ Charles R Greathouse IV, Feb 04 2013

Examples rephrased by R. J. Mathar, Feb 24 2010, Mar 05 2010

a(7) from Amiram Eldar, Feb 23 2024

A365983 Even numbers k such that k^2 - 1 is a powerful number.

Original entry on oeis.org

26, 70226, 130576328, 189750626, 512706121226, 13837575261124, 99612037019890, 1385331749802026

Offset: 1

Jud McCranie, Sep 24 2023

This sequence is a subsequence of A060860 (the even terms) and a supersequence of A094835. All the terms of A094835 are in this sequence, but 130576328 is not in A094835. A094835 also shows that this sequence is infinite.
Terms A076445(n)+1 are terms of this sequence because A076445(n) and A076445(n)+2 are powerful and (A076445(n)+1)^2-1 = A076445(n) * (A076445(n)+2), which is also powerful.
a(n) - 1 is an odd powerful number (A062739). - Amiram Eldar, Feb 23 2024

			26^2 - 1 = 675 = 3^3 * 5^2 is powerful.
130576328^2 - 1 = 17050177433963583 = 3^2 * 7^3 * 13^2 * 293^2 * 617^2, whose exponents are all greater than 1, so it is powerful.

Jean-Marie De Koninck, Those Fascinating Numbers, American Mathematical Society, 2009, entries 70226 and 485.

Index entries for sequences related to powerful numbers.

Cf. A000466, A001694, A062739, A094835, A060860, A076445.

seq[max_] := Module[{p = Union[Flatten[Table[i^2*j^3, {j, 1, max^(1/3), 2}, {i, 1, Sqrt[max/j^3], 2}]]], i}, i = Position[Differences[p], 2] // Flatten; Sqrt[p[[i]]*(p[[i]] + 2) + 1]]; seq[10^10] (* Amiram Eldar, Feb 23 2024 *)

isok(k) = !(k%2) && ispowerful(k^2-1); \\ Michel Marcus, Sep 25 2023

a(5)-a(8) from Amiram Eldar, Feb 23 2024

cp's OEIS Frontend

A097310 Chebyshev T-polynomials T(n,14) with Diophantine property.

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A363190 Odd powerful numbers (A062739) k such that the next powerful number after k is also odd.

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A238799 a(0) = 1, a(n+1) = 2*a(n)^3 + 3*a(n).

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A118894 Numbers m such that the Pell equation x^2-m*y^2=1 has fundamental solution with x even.

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A175180 Numbers k such that k^2 + 2 is powerful in the sense of A001694.

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A365983 Even numbers k such that k^2 - 1 is a powerful number.

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A238799 a(0) = 1, a(n+1) = 2a(n)^3 + 3a(n).