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 1-2 of 2 results.

A087289 a(n) = 2^(2*n+1) + 1.

Original entry on oeis.org

3, 9, 33, 129, 513, 2049, 8193, 32769, 131073, 524289, 2097153, 8388609, 33554433, 134217729, 536870913, 2147483649, 8589934593, 34359738369, 137438953473, 549755813889, 2199023255553, 8796093022209, 35184372088833, 140737488355329, 562949953421313, 2251799813685249
Offset: 0

Views

Author

W. Edwin Clark, Aug 29 2003

Keywords

Comments

Number of pairs of polynomials (f,g) in GF(2)[x] satisfying deg(f) <= n, deg(g) <= n and gcd(f,g) = 1.
An unpublished result due to Stephen Suen, David desJardins, and W. Edwin Clark. This is the case k = 2, q = 2 of their formula q^((n+1)*k) * (1 - 1/q^(k-1) + (q-1)/q^((n+1)*k)) for the number of ordered k-tuples (f_1, ..., f_k) of polynomials in GF(q)[x] such that deg(f_i) <= n for all i and gcd(f_1, ..., f_k) = 1.
Apparently the same as A084508 shifted left.
Terms in binary are palindromes of the form 1x1 where x is a string of 2*n zeros (A152577). - Brad Clardy, Sep 01 2011
For n > 0, a(n) is the number k such that the number of iterations of the map k -> (3k +1)/8 == 4 (mod 8) until reaching (3k +1)/8 <> 4 (mod 8) equals n. (see the Collatz problem: the start of the parity trajectory of a(n) is n times {100} = 100100100100...100abcd...). - Michel Lagneau, Jan 23 2012
An Engel expansion of 2 to the base 4 as defined in A181565, with the associated series expansion 2 = 4/3 + 4^2/(3*9) + 4^3/(3*9*33) + 4^4/(3*9*33*129) + .... Cf. A199561 and A207262. - Peter Bala, Oct 29 2013
For x = A083420(n), y = A000079(n+1), z = a(n) then x^2 + 2*y^2 = z^2. - Vincenzo Librandi, Jun 09 2014
A254046(n+1) is the 3-adic valuation of a(n). - Fred Daniel Kline, Jan 11 2017

Examples

			a(0) = 3 since there are three pairs, (0,1), (1,0) and (1,1) of polynomials (f,g) in GF(2)[x] of degree at most 0 such that gcd(f,g) = 1.
		

Crossrefs

Programs

  • Magma
    [2^(2*n+1) + 1: n in [0..30]]; // Vincenzo Librandi, May 16 2011
    
  • Mathematica
    Table[2^(2 n + 1) + 1, {n, 0, 20}] (* or *) 3 NestList[4 # - 1 &, 1, 20]
    (* or *) CoefficientList[Series[(3 - 6 x)/((1 - x) (1 - 4 x)), {x, 0, 20}], x] (* Michael De Vlieger, Mar 03 2017 *)
  • PARI
    a(n)=2^(2*n+1)+1 \\ Charles R Greathouse IV, Sep 24 2015

Formula

G.f.: (3-6*x)/((1-x)*(1-4*x)).
a(n) = 3*A007583(n).
a(n) = 4*a(n-1) - 3. - Lekraj Beedassy, Apr 29 2005
a(n) = A099393(n+1) - 2*A099393(n). - Brad Clardy, Sep 01 2011
a(n) = 2^(2*n + 1)*a(-1-n) for all n in Z. - Michael Somos, Jan 11 2017
a(n) = A283070(n) - 1. - Peter M. Chema, Mar 02 2017
From Elmo R. Oliveira, Feb 22 2025: (Start)
E.g.f.: exp(x)*(2*exp(3*x) + 1).
a(n) = 5*a(n-1) - 4*a(n-2). (End)

A095372 1+integers repeating "90" decimal digit pattern.

Original entry on oeis.org

1, 91, 9091, 909091, 90909091, 9090909091, 909090909091, 90909090909091, 9090909090909091, 909090909090909091, 90909090909090909091, 9090909090909090909091, 909090909090909090909091
Offset: 0

Views

Author

Labos Elemer, Jun 07 2004

Keywords

Comments

These numbers arise for example as divisors of several repunits (A002275).
The aerated sequence A(n) = [1, 0, 91, 0, 9091, 0, 909091,...] is a divisibility sequence, i.e., A(n) divides A(m) whenever n divides m. It is the case P1 = 0, P2 = -11^2, Q = 10 of the 3-parameter family of 4th-order linear divisibility sequences found by Williams and Guy. - Peter Bala, Aug 22 2019
Except for a(0) = 1, these terms M are such that 21 * M = 1M1, where 1M1 denotes the concatenation of 1, M and 1. Actually 21 is A329914(1) and a(1) = A329915(1) = 91, and the terms >=91 form the set {M_21}; for example, 21 * 909091 = 1(909091)1. - Bernard Schott, Dec 01 2019

Examples

			Digit-pattern P=[ab..z] repeating integers equal formally with P*(-1+10^(Ln))/(-1+10^L), where L is the length of pattern;
a(9) divides A002275(38) repunit. See A095371.
		

Crossrefs

Programs

  • Mathematica
    Table[1+90*(100^n-1)/99, {n, 0, 20}]

Formula

a(n) = 1 + 90*(-1 + 100^n)/99 = (10^(2*n+1) + 1)/11. - Rick L. Shepherd, Aug 01 2004
From Colin Barker, Jul 03 2013: (Start)
a(n) = 101*a(n-1) - 100*a(n-2).
G.f.: -(10*x-1)/((x-1)*(100*x-1)). (End)
E.g.f.: exp(x)*(1 + 10*(exp(99*x) - 1)/11). - Elmo R. Oliveira, Mar 15 2025
Showing 1-2 of 2 results.