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.

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A087475 a(n) = n^2 + 4.

Original entry on oeis.org

4, 5, 8, 13, 20, 29, 40, 53, 68, 85, 104, 125, 148, 173, 200, 229, 260, 293, 328, 365, 404, 445, 488, 533, 580, 629, 680, 733, 788, 845, 904, 965, 1028, 1093, 1160, 1229, 1300, 1373, 1448, 1525, 1604, 1685, 1768, 1853, 1940, 2029, 2120, 2213, 2308, 2405, 2504
Offset: 0

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Author

Gary W. Adamson, Sep 09 2003

Keywords

Comments

Schroeder, p. 330, states "For positive n, these winding numbers are precisely those whose continued fraction expansion is periodic and has period length 1".
Positive X values of solutions to the equation X^3 - 4*X^2 = Y^2. To find Y values: b(n) = n*(n^2 + 4). - Mohamed Bouhamida, Nov 06 2007
From Artur Jasinski, Oct 03 2008: (Start)
General formula for cotangent recurrences type:
a(n+1) = a(n)^3 + 3*a(n) and a(1)=k is
a(n) = floor(((k + sqrt(k^2 + 4))/2)^(3^(n-1))). (End)
Given sequences of the form S(n) = N*S(n-1) + S(n-2) starting (1, N, ...), and having convergents with discriminant (N^2 + 4), S(p) == (a(n))^((p-1)/2) mod p, for n>0, p = odd prime. Example: with N = 2 we have the Pell series (1, 2, 5, 12, 29, 70, 169, ...) with P(7) = 169. Then 169 == 8^3 mod 7, with a(2) = 8. Cf. Schroeder, "Number Theory in Science and Communication", p. 90, for N = 1: F(p) == 5^((p-1)/2) mod p. - Gary W. Adamson, Feb 23 2009
The only two real solutions of the form f(x) = A*x^p with positive p that satisfy f^(n)(x) = f^[-1](x), x >= 0, n >= 1, with f^(n) the n-th derivative and f^[-1] the compositional inverse of f, are obtained for p = p1(n) = (n + sqrt(a(n)))/2 and p = p2(n) = (n - sqrt(a(n)))/2, n >= 1, and A = A(n) = (fallfac(p,n))^(-p/(p+1)), for p = p1(n) and p = p2(n), respectively. Here fallfac(x, k) := product(x - j, j = 0..k-1), the falling factorials. See the T. Koshy reference, pp. 263-264 (there is also a solution for negative p if n is even; see the corresponding comment in A002522). - Wolfdieter Lang, Oct 21 2010, Oct 28 2010
(n + sqrt(a(n)))/2 = [n;n,n,...], with the regular continued fraction with period length 1. For a simple proof see, e.g., the Schroeder reference, pp. 330-331. See also the first comment above.

Examples

			a(2) = 8, discriminant of algebraic representation of barover(2) = [2,2,2,...] = sqrt 2 - 1 = 0.41421356... = ((sqrt 8) - 2)/2. a(3) = 13, discriminant of barover(3) = [3,3,3,...] = 0.3027756... = ((sqrt 13) - 3)/2.

References

  • Manfred R. Schroeder, "Fractals, Chaos, Power Laws"; W.H. Freeman & Co, 1991, p. 330-331.
  • Manfred R. Schroeder, "Number Theory in Science and Communication", Springer Verlag, 5th ed., 2009. [From Gary W. Adamson, Feb 23 2009]
  • Thomas Koshy, "Fibonacci and Lucas Numbers with Applications", John Wiley and Sons, New York, 2001. [From Wolfdieter Lang, Oct 21 2010]

Links

Crossrefs

Cf. A001082, A002378, A002522, A005563, A028347, A036666, A046092, A053755, A062717, A155965, A155966, A156701, A156798.

Programs

Formula

n^2 + 4 are discriminant terms in the formula for Positive Silver Mean Constants, defined as barover(n), = (sqrt (n^2 + 4) - n)/2. Such constants barover(n) = C have the property: 1/C - C = n.
a(n) = A156701(n) / A053755(n). - Reinhard Zumkeller, Feb 13 2009
a(n) = A156798(n)/A002522(n). - Reinhard Zumkeller, Feb 16 2009
a(n) = a(n-1) + 2*n-1 (with a(0)=4). - Vincenzo Librandi, Nov 22 2010
G.f.: (4 - 7*x + 5*x^2)/(1 - x)^3. - Colin Barker, Jan 06 2012
a(n)^3 = A155965(n)^2 + A155966(n)^2. - Vincenzo Librandi, Feb 22 2012
From Amiram Eldar, Jul 13 2020: (Start)
Sum_{n>=0} 1/a(n) = (1 + 2*Pi*coth(2*Pi))/8.
Sum_{n>=0} (-1)^n/a(n) = (1 + 2*Pi*cosech(2*Pi))/8 = A371803. (End)
E.g.f.: exp(x)*(4 + x + x^2). - Stefano Spezia, Jul 08 2023
From Amiram Eldar, Feb 05 2024: (Start)
Product_{n>=0} (1 - 1/a(n)) = sqrt(3)*sinh(sqrt(3)*Pi)/(2*sinh(2*Pi)).
Product_{n>=0} (1 + 1/a(n)) = sqrt(5)*sinh(sqrt(5)*Pi)/(2*sinh(2*Pi)). (End)

A371799 Rectangular array, read by downward antidiagonals: row n shows the numbers m>1 in whose prime factorization p(1)^e(1)*p(2)^e(2)* ...*p(k)^e(k), all e(i) are <= 1 and the number of 0' s in {e(i)} is n-1.

Original entry on oeis.org

2, 6, 3, 30, 10, 5, 210, 15, 14, 7, 2310, 42, 21, 22, 11, 30030, 70, 35, 33, 26, 13, 510510, 105, 66, 55, 39, 34, 17, 9699690, 330, 110, 77, 65, 51, 38, 19, 223092870, 462, 154, 78, 91, 85, 57, 46, 23, 6469693230, 770, 165, 130, 102, 114, 95, 69, 58, 29
Offset: 1

Views

Author

Clark Kimberling, Apr 10 2024

Keywords

Examples

			15 = 2^0*3^1*51, so (e(1),e(2),e(3)) = (0,1,1), so 15 is in row 2
Corner:
 2   6   30  210  2310  30030  510510  9699690
 3   10  15   42    70    105     330      462
 5   14  21   35    66    110     154      165
 7   22  33   55    77     78     130      182
11   26  39   65    91    102     143      170
13   34  51   85   114    119     187      190
17   38  57   95   133    138     209      230
19   46  69  115   161    174     253      290
23   58  87  145   186    203     310      319

Crossrefs

Cf. A000040 (the primes, column 1), A002110 (row 1), A005117 (increasing sequence of all terms of the array), A340316, A371801, A371802, A371803, A371804.

Programs

  • Mathematica
    exps := Map[#[[2]] &, Sort[Join[#, Complement[Map[{Prime[#], 0} &, Range[PrimePi[Last[#][[1]]]]], Map[{#[[1]], 0} &, #]]]] &[FactorInteger[#]]] &;
m = Map[Transpose[#][[1]] &, GatherBy[Map[{#[[1]], Count[#[[2]], 0]} &,     Select[Map[{#, exps[#]} &, Range[2, 5000]], Max[#[[2]]] <= 1 &]], #[[2]] &]];
z = 12; row1 = Table[Apply[Times, Prime[Range[n]]], {n, 1, z}];
r = Join[{row1}, Table[Take[m[[n]], z], {n, 2, z}]];
Grid[r]  (* array *)
w[n_, k_] := r[[n]][[k]]
Table[w[n - k + 1, k], {n, z}, {k, n, 1, -1}] // Flatten
(* sequence *)(* Peter J. C. Moses, Mar 21 2024 *)
Showing 1-2 of 2 results.

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