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.

Showing 1-8 of 8 results.

A086485 Duplicate of A068310.

Original entry on oeis.org

3, 2, 15, 6, 35, 3, 7, 5, 11, 30, 143, 42, 195, 14, 255, 2, 323, 10, 399, 110, 483, 33, 23
Offset: 2

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Author

Keywords

A005563 a(n) = n*(n+2) = (n+1)^2 - 1.

Original entry on oeis.org

0, 3, 8, 15, 24, 35, 48, 63, 80, 99, 120, 143, 168, 195, 224, 255, 288, 323, 360, 399, 440, 483, 528, 575, 624, 675, 728, 783, 840, 899, 960, 1023, 1088, 1155, 1224, 1295, 1368, 1443, 1520, 1599, 1680, 1763, 1848, 1935, 2024, 2115, 2208, 2303, 2400, 2499, 2600
Offset: 0

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Author

Keywords

Comments

Erdős conjectured that n^2 - 1 = k! has a solution if and only if n is 5, 11 or 71 (when k is 4, 5 or 7).
Second-order linear recurrences y(m) = 2y(m-1) + a(n)*y(m-2), y(0) = y(1) = 1, have closed form solutions involving only powers of integers. - Len Smiley, Dec 08 2001
Number of edges in the join of two cycle graphs, both of order n, C_n * C_n. - Roberto E. Martinez II, Jan 07 2002
Let k be a positive integer, M_n be the n X n matrix m_(i,j) = k^abs(i-j) then det(M_n) = (-1)^(n-1)*a(k-1)^(n-1). - Benoit Cloitre, May 28 2002
Also numbers k such that 4*k + 4 is a square. - Cino Hilliard, Dec 18 2003
For each term k, the function sqrt(x^2 + 1), starting with 1, produces an integer after k iterations. - Gerald McGarvey, Aug 19 2004
a(n) mod 3 = 0 if and only if n mod 3 > 0: a(A008585(n)) = 2; a(A001651(n)) = 0; a(n) mod 3 = 2*(1-A079978(n)). - Reinhard Zumkeller, Oct 16 2006
a(n) is the number of divisors of a(n+1) that are not greater than n. - Reinhard Zumkeller, Apr 09 2007
Nonnegative X values of solutions to the equation X^3 + X^2 = Y^2. To find Y values: b(n) = n(n+1)(n+2). - Mohamed Bouhamida, Nov 06 2007
Sequence allows us to find X values of the equation: X + (X + 1)^2 + (X + 2)^3 = Y^2. To prove that X = n^2 + 2n: Y^2 = X + (X + 1)^2 + (X + 2)^3 = X^3 + 7*X^2 + 15X + 9 = (X + 1)(X^2 + 6X + 9) = (X + 1)*(X + 3)^2 it means: (X + 1) must be a perfect square, so X = k^2 - 1 with k>=1. we can put: k = n + 1, which gives: X = n^2 + 2n and Y = (n + 1)(n^2 + 2n + 3). - Mohamed Bouhamida, Nov 12 2007
From R. K. Guy, Feb 01 2008: (Start)
Toads and Frogs puzzle:
This is also the number of moves that it takes n frogs to swap places with n toads on a strip of 2n + 1 squares (or positions, or lily pads) where a move is a single slide or jump, illustrated for n = 2, a(n) = 8 by
T T - F F
T - T F F
T F T - F
T F T F -
T F - F T
- F T F T
F - T F T
F F T - T
F F - T T
I was alerted to this by the Holton article, but on consulting Singmaster's sources, I find that the puzzle goes back at least to 1867.
Probably the first to publish the number of moves for n of each animal was Edouard Lucas in 1883. (End)
a(n+1) = terms of rank 0, 1, 3, 6, 10 = A000217 of A120072 (3, 8, 5, 15). - Paul Curtz, Oct 28 2008
Row 3 of array A163280, n >= 1. - Omar E. Pol, Aug 08 2009
Final digit belongs to a periodic sequence: 0, 3, 8, 5, 4, 5, 8, 3, 0, 9. - Mohamed Bouhamida, Sep 04 2009 [Comment edited by N. J. A. Sloane, Sep 24 2009]
Let f(x) be a polynomial in x. Then f(x + n*f(x)) is congruent to 0 (mod f(x)); here n belongs to N. There is nothing interesting in the quotients f(x + n*f(x))/f(x) when x belongs to Z. However, when x is irrational these quotients consist of two parts, a) rational integers and b) integer multiples of x. The present sequence represents the non-integer part when the polynomial is x^2 + x + 1 and x = sqrt(2), f(x+n*f(x))/f(x) = A056108(n) + a(n)*sqrt(2). - A.K. Devaraj, Sep 18 2009
For n >= 1, a(n) is the number for which 1/a(n) = 0.0101... (A000035) in base (n+1). - Rick L. Shepherd, Sep 27 2009
For n > 0, continued fraction [n, 1, n] = (n+1)/a(n); e.g., [6, 1, 6] = 7/48. - Gary W. Adamson, Jul 15 2010
Starting (3, 8, 15, ...) = binomial transform of [3, 5, 2, 0, 0, 0, ...]; e.g., a(3) = 15 = (1*3 + 2*5 +1*2) = (3 + 10 + 2). - Gary W. Adamson, Jul 30 2010
a(n) is essentially the case 0 of the polygonal numbers. The polygonal numbers are defined as P_k(n) = Sum_{i=1..n} ((k-2)*i-(k-3)). Thus P_0(n) = 2*n-n^2 and a(n) = -P_0(n+2). See also A067998 and for the case k=1 A080956. - Peter Luschny, Jul 08 2011
a(n) is the maximal determinant of a 2 X 2 matrix with integer elements from {1, ..., n+1}, so the maximum determinant of a 2x2 matrix with integer elements from {1, ..., 5} = 5^2 - 1 = a(4) = 24. - Aldo González Lorenzo, Oct 12 2011
Using four consecutive triangular numbers t1, t2, t3 and t4, plot the points (0, 0), (t1, t2), and (t3, t4) to create a triangle. Twice the area of this triangle are the numbers in this sequence beginning with n = 1 to give 8. - J. M. Bergot, May 03 2012
Given a particle with spin S = n/2 (always a half-integer value), the quantum-mechanical expectation value of the square of the magnitude of its spin vector evaluates to = S(S+1) = n(n+2)/4, i.e., one quarter of a(n) with n = 2S. This plays an important role in the theory of magnetism and magnetic resonance. - Stanislav Sykora, May 26 2012
Twice the harmonic mean [H(x, y) = (2*x*y)/(x + y)] of consecutive triangular numbers A000217(n) and A000217(n+1). - Raphie Frank, Sep 28 2012
Number m such that floor(sqrt(m)) = floor(m/floor(sqrt(m))) - 2 for m > 0. - Takumi Sato, Oct 10 2012
The solutions of equation 1/(i - sqrt(j)) = i + sqrt(j), when i = (n+1), j = a(n). For n = 1, 2 + sqrt(3) = 3.732050.. = A019973. For n = 2, 3 + sqrt(8) = 5.828427... = A156035. - Kival Ngaokrajang, Sep 07 2013
The integers in the closed form solution of a(n) = 2*a(n-1) + a(m-2)*a(n-2), n >= 2, a(0) = 0, a(1) = 1 mentioned by Len Smiley, Dec 08 2001, are m and -m + 2 where m >= 3 is a positive integer. - Felix P. Muga II, Mar 18 2014
Let m >= 3 be a positive integer. If a(n) = 2*a(n-1) + a(m-2) * a(n-2), n >= 2, a(0) = 0, a(1) = 1, then lim_{n->oo} a(n+1)/a(n) = m. - Felix P. Muga II, Mar 18 2014
For n >= 4 the Szeged index of the wheel graph W_n (with n + 1 vertices). In the Sarma et al. reference, Theorem 2.7 is incorrect. - Emeric Deutsch, Aug 07 2014
If P_{k}(n) is the n-th k-gonal number, then a(n) = t*P_{s}(n+2) - s*P_{t}(n+2) for s=t+1. - Bruno Berselli, Sep 04 2014
For n >= 1, a(n) is the dimension of the simple Lie algebra A_n. - Wolfdieter Lang, Oct 21 2015
Finding all positive integers (n, k) such that n^2 - 1 = k! is known as Brocard's problem, (see A085692). - David Covert, Jan 15 2016
For n > 0, a(n) mod (n+1) = a(n) / (n+1) = n. - Torlach Rush, Apr 04 2016
Conjecture: When using the Sieve of Eratosthenes and sieving (n+1..a(n)), with divisors (1..n) and n>0, there will be no more than a(n-1) composite numbers. - Fred Daniel Kline, Apr 08 2016
a(n) mod 8 is periodic with period 4 repeating (0,3,0,7), that is a(n) mod 8 = 5/2 - (5/2) cos(n*Pi) - sin(n*Pi/2) + sin(3*n*Pi/2). - Andres Cicuttin, Jun 02 2016
Also for n > 0, a(n) is the number of times that n-1 occurs among the first (n+1)! terms of A055881. - R. J. Cano, Dec 21 2016
The second diagonal of composites (the only prime is number 3) from the right on the Klauber triangle (see Kival Ngaokrajang link), which is formed by taking the positive integers and taking the first 1, the next 3, the following 5, and so on, each centered below the last. - Charles Kusniec, Jul 03 2017
Also the number of independent vertex sets in the n-barbell graph. - Eric W. Weisstein, Aug 16 2017
Interleaving of A000466 and A033996. - Bruce J. Nicholson, Nov 08 2019
a(n) is the number of degrees of freedom in a triangular cell for a Raviart-Thomas or Nédélec first kind finite element space of order n. - Matthew Scroggs, Apr 22 2020
From Muge Olucoglu, Jan 19 2021: (Start)
For n > 1, a(n-2) is the maximum number of elements in the second stage of the Quine-McCluskey algorithm whose minterms are not covered by the functions of n bits. At n=3, we have a(3-2) = a(1) = 1*(1+2) = 3 and f(A,B,C) = sigma(0,1,2,5,6,7).
.
0 1 2 5 6 7
+---------------
*(0,1)| X X
(0,2)| X X
(1,5)| X X
*(2,6)| X X
*(5,7)| X X
(6,7)| X X
.
*: represents the elements that are covered. (End)
1/a(n) is the ratio of the sum of the first k odd numbers and the sum of the next n*k odd numbers. - Melvin Peralta, Jul 15 2021
For n >= 1, the continued fraction expansion of sqrt(a(n)) is [n; {1, 2n}]. - Magus K. Chu, Sep 09 2022
Number of diagonals parallel to an edge in a regular (2*n+4)-gon (cf. A367204). - Paolo Xausa, Nov 21 2023
For n >= 1, also the number of minimum cyclic edge cuts in the (n+2)-trapezohedron graph. - Eric W. Weisstein, Nov 21 2024
For n >= 1, a(n) is the sum of the interior angles of a polygon with n+2 sides, in radians, multiplied by (n+2)/Pi. - Stuart E Anderson, Aug 06 2025

Examples

			G.f. = 3*x + 8*x^2 + 15*x^3 + 24*x^4 + 35*x^5 + 48*x^6 + 63*x^7 + 80*x^8 + ...
		

References

  • E. R. Berlekamp, J. H. Conway and R. K. Guy, Winning Ways, Academic Press, NY, 2 vols., 1982, see index under Toads and Frogs Puzzle.
  • Martin Gardner, Perplexing Puzzles and Tantalizing Teasers, p. 21 (for "The Dime and Penny Switcheroo").
  • R. K. Guy, Unsolved Problems in Theory of Numbers, Section D25.
  • Derek Holton, Math in School, 37 #1 (Jan 2008) 20-22.
  • Edouard Lucas, Récréations Mathématiques, Gauthier-Villars, Vol. 2 (1883) 141-143.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

Formula

G.f.: x*(3-x)/(1-x)^3. - Simon Plouffe in his 1992 dissertation
a(n) = A000290(n+1) - 1.
A002378(a(n)) = A002378(n)*A002378(n+1); e.g., A002378(15)=240=12*20. - Charlie Marion, Dec 29 2003
a(n) = A067725(n)/3. - Zerinvary Lajos, Mar 06 2007
a(n) = Sum_{k=1..n} A144396(k). - Zerinvary Lajos, May 11 2007
a(n) = A134582(n+1)/4. - Zerinvary Lajos, Feb 01 2008
A143053(a(n)) = A000290(n+1), for n > 0. - Reinhard Zumkeller, Jul 20 2008
a(n) = Real((n+1+i)^2). - Gerald Hillier, Oct 12 2008
A053186(a(n)) = 2*n. - Reinhard Zumkeller, May 20 2009
a(n) = (n! + (n+1)!)/(n-1)!, n > 0. - Gary Detlefs, Aug 10 2009
a(n) = floor(n^5/(n^3+1)) with offset 1 (a(1)=0). - Gary Detlefs, Feb 11 2010
a(n) = a(n-1) + 2*n + 1 (with a(0)=0). - Vincenzo Librandi, Nov 18 2010
Sum_{n>=1} 1/a(n) = 3/4. - Mohammad K. Azarian, Dec 29 2010
a(n) = 2/(Integral_{x=0..Pi/2} (sin(x))^(n-1)*(cos(x))^3), for n > 0. - Francesco Daddi, Aug 02 2011
a(n) = A002378(n) + floor(sqrt(A002378(n))); pronic number + its root. - Fred Daniel Kline, Sep 16 2011
a(n-1) = A008833(n) * A068310(n) for n > 1. - Reinhard Zumkeller, Nov 26 2011
G.f.: U(0) where U(k) = -1 + (k+1)^2/(1 - x/(x + (k+1)^2/U(k+1))); (continued fraction, 3-step). - Sergei N. Gladkovskii, Oct 19 2012
a(n) = 15*C(n+4,3)*C(n+4,5)/(C(n+4,2)*C(n+4,4)). - Gary Detlefs, Aug 05 2013
a(n) = (n+2)!/((n-1)! + n!), n > 0. - Ivan N. Ianakiev, Nov 11 2013
a(n) = 3*C(n+1,2) - C(n,2) for n >= 0. - Felix P. Muga II, Mar 11 2014
a(n) = (A016742(n+1) - 4)/4 for n >= 0. - Felix P. Muga II, Mar 11 2014
a(-2 - n) = a(n) for all n in Z. - Michael Somos, Aug 07 2014
A253607(a(n)) = 1. - Reinhard Zumkeller, Jan 05 2015
E.g.f.: x*(x + 3)*exp(x). - Ilya Gutkovskiy, Jun 03 2016
For n >= 1, a(n^2 + n - 2) = a(n-1) * a(n). - Miko Labalan, Oct 15 2017
Sum_{n>=1} (-1)^(n+1)/a(n) = 1/4. - Amiram Eldar, Nov 04 2020
From Amiram Eldar, Feb 17 2021: (Start)
Product_{n>=1} (1 + 1/a(n)) = 2.
Product_{n>=1} (1 - 1/a(n)) = -sqrt(2)*sin(sqrt(2)*Pi)/Pi. (End)
a(n) = A000290(n+2) - n*2. See Bounded Squares illustration. - Leo Tavares, Oct 05 2021
From Leo Tavares, Oct 10 2021: (Start)
a(n) = A008585(n) + 2*A000217(n-1). See Trapezoids illustration.
2*A005563 = A054000(n+1). See Trapagons illustration.
a(n) = 2*A000217(n) + n. (End)
a(n) = (n+2)!!/(n-2)!! for n > 1. - Jacob Szlachetka, Jan 02 2022

Extensions

Partially edited by Joerg Arndt, Mar 11 2010
More terms from N. J. A. Sloane, Aug 01 2010

A008833 Largest square dividing n.

Original entry on oeis.org

1, 1, 1, 4, 1, 1, 1, 4, 9, 1, 1, 4, 1, 1, 1, 16, 1, 9, 1, 4, 1, 1, 1, 4, 25, 1, 9, 4, 1, 1, 1, 16, 1, 1, 1, 36, 1, 1, 1, 4, 1, 1, 1, 4, 9, 1, 1, 16, 49, 25, 1, 4, 1, 9, 1, 4, 1, 1, 1, 4, 1, 1, 9, 64, 1, 1, 1, 4, 1, 1, 1, 36, 1, 1, 25, 4, 1, 1, 1, 16, 81, 1, 1, 4, 1, 1, 1, 4, 1, 9, 1, 4, 1, 1, 1, 16, 1
Offset: 1

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Author

Keywords

Comments

The Dirichlet generating function of the arithmetic function of the largest t-th power dividing n is zeta(s)*zeta(t*s-t)/zeta(s*t), here with t=2 and in A008834 and A008835 with t=3 and t=4, respectively. - R. J. Mathar, Feb 19 2011

Crossrefs

Programs

  • Haskell
    a008833 n = head $ filter ((== 0) . (mod n)) $
       reverse $ takeWhile (<= n) $ tail a000290_list
    -- Reinhard Zumkeller, Nov 13 2011
    
  • Maple
    A008833 := proc(n)
        expand(numtheory:-nthpow(n,2)) ;
    end proc:
    seq(A008833(n), n=1..100) ;
  • Mathematica
    a[n_] := First[ Select[ Reverse[ Divisors[n]], IntegerQ[Sqrt[#]]&, 1]]; Table[a[n], {n, 1, 100}] (* Jean-François Alcover, Dec 12 2011 *)
    f[p_, e_] := p^(2*Floor[e/2]); a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Jul 07 2020 *)
  • PARI
    A008833(n)=n/core(n) \\ Michael B. Porter, Oct 17 2009
    
  • Python
    from sympy.ntheory.factor_ import core
    def A008833(n): return n//core(n) # Chai Wah Wu, Dec 30 2021

Formula

a(n) = A000188(n)^2 = n/A007913(n). Cf. A019554.
Multiplicative with a(p^e) = p^(2[e/2]). - David W. Wilson, Aug 01 2001
Dirichlet g.f.: zeta(s)*zeta(2s-2)/zeta(2s). - R. J. Mathar, Oct 31 2011
a(n) = A005563(n-1) / A068310(n) for n > 1. - Reinhard Zumkeller, Nov 26 2011
Sum_{k=1..n} a(k) ~ Zeta(3/2) * n^(3/2) / (3*Zeta(3)). - Vaclav Kotesovec, Feb 01 2019
a(A059897(n,k)) = A059897(a(n), a(k)). - Peter Munn, Nov 30 2019
From Ridouane Oudra, May 11 2025: (Start)
a(n) = Sum_{d|n} lambda(d)*d*psi(n/d), where lambda = A008836 and psi = A001615.
a(n) = lambda(n) * Sum_{d|n} lambda(d)*d*phi(n/d).
a(n) = A008836(n) * A358272(n). (End)

A067872 Least m > 0 for which m*n^2 + 1 is a square.

Original entry on oeis.org

3, 2, 7, 3, 23, 8, 47, 15, 79, 24, 119, 2, 167, 48, 3, 63, 287, 80, 359, 6, 88, 120, 527, 28, 623, 168, 727, 12, 839, 44, 959, 255, 216, 288, 8, 20, 1367, 360, 19, 77, 1679, 22, 1847, 30, 208, 528, 2207, 7, 2399, 624, 128, 42, 2807, 728, 696, 3, 160, 840, 3479, 11
Offset: 1

Views

Author

Lekraj Beedassy, Feb 25 2002

Keywords

Comments

Least m > 0 for which x^2 - m*y^2 = 1 has a solution with y = n.
For n > 1, a(n) <= n^2-2. - Chai Wah Wu, Jan 26 2016
Considering the greatest prime factor (gpf) of terms, two curves are evident. One has gpf ~ n^2 (see a(9949)), the other has gpf ~ n^2/7 (see a(9923)). See the gpf plot link. - Bill McEachen, Apr 13 2025

Examples

			a(4)=3, based on 3*4^2 + 1 = 7^2.
		

Crossrefs

Programs

  • Haskell
    a067872 n = (until ((== 1) . a010052 . (+ 1)) (+ nn) nn) `div` nn
                where nn = n ^ 2
    -- Reinhard Zumkeller, Jun 28 2013
    
  • Mathematica
    a[n_] := For[m=1, True, m++, If[IntegerQ[Sqrt[m*n^2+1]], Return[m]]]; Table[a[n], {n, 100}]
    lm[n_]:=Module[{m=1},While[!IntegerQ[Sqrt[m n^2+1]],m++];m]; Array[lm,60] (* Harvey P. Dale, Feb 24 2013 *)
  • Python
    def A067872(n):
        y, x, n2 = n*(n+2), 2*n+3, n**2
        m, r = divmod(y,n2)
        while r:
            y += x
            x += 2
            m, r = divmod(y,n2)
        return m # Chai Wah Wu, Jan 25 2016

Formula

For n a power of an odd prime, a(n) = n^2 - 2. For n twice a power of an odd prime, a(n) = (n/2)^2 - 1. - T. D. Noe, Sep 13 2007

Extensions

Edited by Dean Hickerson, Mar 19 2002

A128972 n^3 - 1 divided by its largest cube divisor.

Original entry on oeis.org

7, 26, 63, 124, 215, 342, 511, 91, 37, 1330, 1727, 2196, 2743, 3374, 4095, 614, 17, 254, 7999, 9260, 10647, 12166, 13823, 1953, 17575, 19682, 813, 24388, 26999, 29790, 32767, 4492, 39303, 42874, 46655, 1876, 54871, 59318, 63999, 8615, 74087, 79506
Offset: 2

Views

Author

Jonathan Vos Post, Apr 28 2007

Keywords

Comments

In other words, cubefree part of n^3-1, or cubefree kernel of n^3-1. Cube analog of A068310.

Examples

			a(9) = (9^3-1)/8 = (2^3 * 7 * 13)/(2^3) = 728/8 = 91.
a(10) = (10^3-1)/27 = (3^3 * 37)/(3^3) = 999/27 = 37.
a(18) = (18^3-1)/343 = (7^3 * 17)/(7^3) = 5831/343 = 17.
		

Crossrefs

Programs

  • Maple
    a:= n -> mul(f[1]^(f[2] mod 3), f = ifactors(n^3-1)[2]):
    seq(a(n),n=2..100); # Robert Israel, Sep 24 2014

Formula

a(n) = A062378(A068601(n)) = A062378(n^3-1).

Extensions

More terms from Carl R. White, Nov 09 2010

A128251 n^4 - 1 divided by its largest fourth power divisor.

Original entry on oeis.org

15, 5, 255, 39, 1295, 150, 4095, 410, 9999, 915, 20735, 1785, 38415, 3164, 65535, 5220, 104975, 8145, 159999, 12155, 234255, 17490, 331775, 24414, 456975, 33215, 614655, 44205, 809999, 57720, 1048575, 74120, 1336335, 93789, 1679615, 117135
Offset: 2

Views

Author

Jonathan Vos Post, May 03 2007

Keywords

Comments

In other words, biquadratefree part of n^4-1, or biquadratefree kernel of n^4-1. Fourth power analog of what A128972 is to cubes and A068310 is to squares. A046100 Biquadratefree numbers. A008835 Largest 4th power dividing n.

Examples

			a(3) = 5 because (3^4 - 1)/16 = 80/16 = (2^4 * 5)/(2^4) = 5.
a(5) = 39 because (5^4 - 1)/16 = 624/16 = (2^4 * 3 * 13)/(2^4) = 39.
a(7) = 150 because (7^4 - 1)/16 = 2400/16 = (2^5 * 3 * 5^2)/(2^4) = 150.
a(9) = 410 because (9^4 - 1)/16 = 6560/16 = (2^5 * 5 * 41)/(2^4) = 410.
a(63) = 61535 because (63^4 - 1)/256 = 15752960/256 = (2^8 * 5 * 31 * 397)/(2^8) = 61535.
		

Crossrefs

Formula

a(n) = (n^4 - 1)/A008835(n^4 - 1) = (A000583(n)-1)/A008835((A000583(n)-1)).

A354672 Numbers x with property that x is not the smallest possible value in the Pellian equation x^2 - D*y^2 = 1 with D = squarefree part of (x^2 - 1).

Original entry on oeis.org

7, 17, 26, 31, 49, 71, 97, 99, 127, 161, 199, 241, 244, 287, 337, 362, 391, 449, 485, 511, 577, 647, 721, 799, 846, 881, 967, 1057, 1151, 1249, 1351, 1457, 1567, 1681, 1799, 1921, 2024, 2047, 2177, 2311, 2449, 2591, 2737, 2887, 2889, 3041, 3199, 3361, 3363
Offset: 1

Views

Author

Herbert Kociemba, Jun 02 2022

Keywords

Comments

Alternatively numbers k such that A033314(k) <> A068310(k).
Conjecture: this sequence is equivalent to the sorted distinct values of cos(m*arccos(k)), where m and k are integers greater than 1. - Jennifer Buckley, Apr 23 2024

Examples

			a(2)=17. The squarefree part of 17^2 - 1 = 288 is D = 2. But the smallest possible solution to x^2 - 2*y^2 = 1 is not x = 17 but x = 3 (with y = 2).
15 is not a term: the squarefree part of 15^2 - 1 = 224 is D = 14 and x^2 - 14*y^2 = 1 has indeed the minimal solution x = 15 (and y = 4).
		

Programs

  • Mathematica
    squarefreepart[n_] :=
      Times @@ Power @@@ ({#[[1]], Mod[#[[2]], 2]} & /@ FactorInteger[n]);
    a = {}; NMAX = 3400; dict // Clear;
    For[n = 2, n <= NMAX, n++, s = squarefreepart[n^2 - 1];
    If[ ! IntegerQ[dict[s]], dict[s] = 1, AppendTo[a, n]]]; a

A358821 a(n) is the largest square dividing n^2 - 1.

Original entry on oeis.org

1, 4, 1, 4, 1, 16, 9, 16, 9, 4, 1, 4, 1, 16, 1, 144, 1, 36, 1, 4, 1, 16, 25, 16, 225, 4, 9, 4, 1, 64, 1, 64, 1, 36, 1, 36, 1, 16, 1, 16, 1, 4, 9, 4, 9, 16, 49, 400, 49, 100, 1, 36, 1, 144, 1, 16, 1, 4, 1, 4, 9, 64, 9, 64, 1, 4, 1, 4, 1, 144, 1, 144, 25, 4
Offset: 2

Views

Author

Darío Clavijo, Jan 04 2023

Keywords

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := p^(2*Floor[e/2]); a[n_] := Times @@ (f @@@ FactorInteger[n^2 - 1]); Array[a, 100, 2] (* Amiram Eldar, Jan 04 2023 *)
  • PARI
    a(n) = (n^2-1)/core(n^2-1); \\ Michel Marcus, Feb 19 2023
  • Python
    from sympy.ntheory.factor_ import core
    def a(n):
      return (n**2-1)//core(n**2-1)
    

Formula

a(n) = A008833(n^2 - 1).
a(n) = A008833(A005563(n-1)).
a(n) = A005563(n-1) / A068310(n).
a(n) = A005563(n-1) / A007913(A005563(n-1)).
a(n) = (n^2 - 1) / A068310(n).
a(n) = (n^2 - 1) / A007913(n^2 - 1).
Showing 1-8 of 8 results.