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 1-3 of 3 results.

A065883 Remove factors of 4 from n (i.e., write n in base 4, drop final zeros, then rewrite in decimal).

Original entry on oeis.org

1, 2, 3, 1, 5, 6, 7, 2, 9, 10, 11, 3, 13, 14, 15, 1, 17, 18, 19, 5, 21, 22, 23, 6, 25, 26, 27, 7, 29, 30, 31, 2, 33, 34, 35, 9, 37, 38, 39, 10, 41, 42, 43, 11, 45, 46, 47, 3, 49, 50, 51, 13, 53, 54, 55, 14, 57, 58, 59, 15, 61, 62, 63, 1, 65, 66, 67, 17, 69, 70, 71, 18, 73, 74, 75
Offset: 1

Views

Author

Henry Bottomley, Nov 26 2001

Keywords

Examples

			a(7)=7, a(14)=14, a(28)=a(4*7)=7, a(56)=a(4*14)=14, a(112)=a(4^2*7)=7.

Links

Crossrefs

Cf. A214392, A235127, A350091 (drop final 2's).
Remove other factors: A000265, A038502, A132739, A244414, A242603, A004151.

Programs

  • Maple
    A065883:= n -> n/4^floor(padic:-ordp(n,2)/2):
map(A065883, [$1..1000]); # Robert Israel, Dec 08 2015
  • Mathematica
    If[Divisible[#,4],#/4^IntegerExponent[#,4],#]&/@Range[80] (* Harvey P. Dale, Aug 31 2013 *)
  • PARI
    a(n)=n/4^valuation(n,4); \\ Joerg Arndt, Dec 09 2015
  • Python
    def A065883(n): return n>>((~n&n-1).bit_length()&-2) # Chai Wah Wu, Jul 09 2022

Formula

If n mod 4 = 0 then a(n) = a(n/4), otherwise a(n) = n.
Multiplicative with a(p^e) = 2^(e (mod 2)) if p = 2 and a(p^e) = p^e if p is an odd prime.
a(n) = n/4^A235127(n).
a(n) = A214392(n) if n mod 16 != 0. - Peter Kagey, Sep 02 2015
From Robert Israel, Dec 08 2015: (Start)
G.f.: x/(1-x)^2 - 3 Sum_{j>=1} x^(4^j)/(1-x^(4^j))^2.
G.f. satisfies G(x) = G(x^4) + x/(1-x)^2 - 4 x^4/(1-x^4)^2. (End)
Sum_{k=1..n} a(k) ~ (2/5) * n^2. - Amiram Eldar, Nov 20 2022
Dirichlet g.f.: zeta(s-1)*(4^s-4)/(4^s-1). - Amiram Eldar, Jan 04 2023

A214394 If n mod 6 = 0 then n/6 else n.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 1, 7, 8, 9, 10, 11, 2, 13, 14, 15, 16, 17, 3, 19, 20, 21, 22, 23, 4, 25, 26, 27, 28, 29, 5, 31, 32, 33, 34, 35, 6, 37, 38, 39, 40, 41, 7, 43, 44, 45, 46, 47, 8, 49, 50, 51, 52, 53, 9, 55, 56, 57, 58, 59, 10, 61
Offset: 0

Views

Author

Jeremy Gardiner, Jul 15 2012

Keywords

Examples

			a(36) = 36/6 = 6.

Links

Crossrefs

Cf. A026741, A051176, A214392.
Cf. A044102, A244414.

Programs

  • Mathematica
    Table[If[Mod[n, 6] == 0, n/6, n], {n, 0, 50}] (* G. C. Greubel, Oct 26 2017 *)
  • PARI
    a(n)=if(n%6,n,n/6) \\ G. C. Greubel, Oct 26 2017
  • PARI
    first(n) = my(res = vector(n, i, i-1)); forstep(i = 1, n, 6, res[i] \= 6); res \\ David A. Corneth, Oct 28 2017

Formula

a(n) = 2*a(n-6) - a(n-12). - G. C. Greubel, Oct 26 2017 [corrected by Georg Fischer, Mar 12 2020]
a(n) = A244414(n) when n is not a multiple of 36 (A044102). - Michel Marcus, Oct 28 2017
a(n) = floor(n/6) + sign(n mod 6) * (n - floor(n/6)). - Wesley Ivan Hurt, Oct 28 2017

A214630 a(n) is the reduced numerator of 1/4 - 1/A109043(n)^2 = (1 - 1/A026741(n)^2)/4.

Original entry on oeis.org

-1, 0, 0, 2, 3, 6, 2, 12, 15, 20, 6, 30, 35, 42, 12, 56, 63, 72, 20, 90, 99, 110, 30, 132, 143, 156, 42, 182, 195, 210, 56, 240, 255, 272, 72, 306, 323, 342, 90, 380, 399, 420, 110, 462, 483, 506, 132, 552, 575, 600, 156
Offset: 0

Views

Author

Paul Curtz, Jul 23 2012

Keywords

Comments

The unreduced fractions are -1/0, 0/4, 0/1, 8/36, 3/16, 24/100, 2/9, 48/196, 15/64, 80/324, 6/25, ... = c(n)/A061038(n), say.
Note that c(n)=A061037(n) + (period of length 2: repeat 0, 3).
c(n) is a permutation of A198442(n). The corresponding ranks are (the 0's have been swapped for convenience) 0,2,1,6,4,10,... = A145979(n-2).
Define the following sequences, satisfying the recurrence a(n) = 2*a(n-4) - a(n-8),
e(n) = -1, 0, 0, 2, 1, 4, 1, 6, 3, 8, 2, 10, 5, ... (after -1, a permutation of A004526(n) or mix A026741(n-1), 2*n),
f(n) = 1, 2, 1, 4, 3, 6, 2, 8, 5, 10, 3, 12, 7, ..., (another permutation of A004526(n+2) or mix A026741(n+1), 2*n+2).
f(n) - e(n) = periodic of period length 4: repeat 2, 2, 1, 2.
e(n) + f(n) = 0, 2, 1, 6, 4, 10, ... = A145979(n-2).
Then c(n) = e(n)*f(n).
Note that A061038(n) - 4*c(n) = periodic of period length 4: repeat 4, 4, 1, 4.
After division (by period 2: repeat 1, 4, A010685(n)), the reduced fractions of c(n) are -1/0, 0/1 ?, 0/4 ?, 2/9, 3/16, 6/25, 2/9, 12/49, 15/64, 20/81, 6/25, ... = a(n)/b(n).
Note that a(1+4*n) + a(2+4*n) + a(3+4*n) = 2,20,56,... = A002378(1+3*n) = A194767(3*n).
A061037(n-2) - a(n-2) = 0, -3, 0, -3, 0, 3, 0, 15, 0, 33, 0, 57, ... = Fip(n-2).
Fip(n-2)/3 = 0,-1,0,-1,0,1,0,5,0,11,0,19,0,29, .... Without 0's: A165900(n) (a Fibonacci polynomial); also -A110331(n+1) (Pell numbers).
g(n) = -1, 0, 0, 1, 1, 2, 1, 3, 3, 4, ... = mix A026741(n-1), n.
h(n) = 1, 1, 1, 2, 3, 3, 2, 4, 5, 5, ... = mix A026741(n+1), n+1.
h(n) - g(n) = (period 2: repeat 2, 1, 1, 1 = A177704(n-1)).
k(n) = 1, 1, 0, 2, 3, 3, 1, 4, 5, 5, ... = mix A174239(n), n+1.
l(n) = -1, 0, 1, 1, 1, 2, 2, 3, 3, 4, ... .
k(n) - l(n) = period 4: repeat 2, 1, -1, 1.
2) By the second formula in the definition, we take first 1 - 1/A026741(n)^2.
Hence, using a convention for the first fraction, -1/0, 0/1, 0/1, 8/9, 3/4, 24/25, 8/9, 48/49, 15/16, 80/81, 24/25, ... = (A005563(n-1) - A033996(n))/A168077(n) = q(n)/A168077(n).
For a(n), we divide by 4.
Note that A214297 is the reduced numerator of 1/4 - 1/A061038(n).
Note also that A168077(n) = A026741(n)^2.

Links

Crossrefs

Cf. A000466, A002378, A131723.

Programs

  • Magma
    m:=50; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((2*x^9+3*x^8+6*x^7+2*x^6+6*x^5+6*x^4+2*x^3-1)/((1-x)^3*(x+1)^3*(x^2+1)^3))); // G. C. Greubel, Sep 20 2018
  • Mathematica
    CoefficientList[Series[(2*x^9+3*x^8+6*x^7+2*x^6+6*x^5+6*x^4+2*x^3-1)/((1-x)^3*(x+1)^3*(x^2+1)^3), {x, 0, 50}], x] (* G. C. Greubel, Sep 20 2018 *)
LinearRecurrence[{0,0,0,3,0,0,0,-3,0,0,0,1},{-1,0,0,2,3,6,2,12,15,20,6,30},60] (* Harvey P. Dale, Jul 01 2019 *)
  • PARI
    Vec(-(2*x^9+3*x^8+6*x^7+2*x^6+6*x^5+6*x^4+2*x^3-1)/((x-1)^3*(x+ 1)^3*(x^2+1)^3) + O(x^100)) \\ Colin Barker, Jan 22 2015

Formula

a(4*n) = 4*n^2-1 = (2*n-1)*(2*n+1), a(2*n+1) = a(4*n+2) = n(n+1).
a(n)= A198442(n)/(period of length 4: repeat 1,1,4,1=A010121(n+2)).
a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12). Is this the shortest possible recurrence? See A214297.
a(n+2) - a(n-2) = 0, 2, 4, 6, 2, 10, 12, 14, 4, ... = 2*A214392(n). a(-2)=a(-1)=0=a(1)=a(2).
a(n+4) - a(n-4) = 0, 4, 2, 12, 16, 20, 6, 28, 32, 36,... = 2*A188167(n). a(-4)=3=a(4), a(-3)=2=a(3).
a(n) = g(n) * h(n).
a(n) = k(n) * l(n).
G.f.: -(2*x^9+3*x^8+6*x^7+2*x^6+6*x^5+6*x^4+2*x^3-1) / ((x-1)^3*(x+1)^3*(x^2+1)^3). - Colin Barker, Jan 22 2015
From Luce ETIENNE, Apr 08 2017: (Start)
a(n) = (13*n^2-28-3*(n^2+4)*(-1)^n+3*(n^2-4)*((-1)^((2*n-1+(-1)^n)/4)+(-1)^((2*n+1-(-1)^n)/4)))/64.
a(n) = (13*n^2-28-3*(n^2+4)*cos(n*Pi)+6*(n^2-4)*cos(n*Pi/2))/64. (End)

Extensions

Edited by N. J. A. Sloane, Aug 04 2012
Showing 1-3 of 3 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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