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 1-6 of 6 results.

A129194 a(n) = (n/2)^2*(3 - (-1)^n).

Original entry on oeis.org

0, 1, 2, 9, 8, 25, 18, 49, 32, 81, 50, 121, 72, 169, 98, 225, 128, 289, 162, 361, 200, 441, 242, 529, 288, 625, 338, 729, 392, 841, 450, 961, 512, 1089, 578, 1225, 648, 1369, 722, 1521, 800, 1681, 882, 1849, 968, 2025, 1058, 2209, 1152, 2401, 1250, 2601, 1352
Offset: 0

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Author

Paul Barry, Apr 02 2007

Keywords

Comments

The numerator of the integral is 2,1,2,1,2,1,...; the moments of the integral are 2/(n+1)^2. See 2nd formula.
The sequence alternates between twice a square and an odd square, A001105(n) and A016754(n).
Partial sums of the positive elements give the absolute values of A122576. - Omar E. Pol, Aug 22 2011
Partial sums of the positive elements give A212760. - Omar E. Pol, Dec 28 2013
Conjecture: denominator of 4/n - 2/n^2. - Wesley Ivan Hurt, Jul 11 2016
Multiplicative because both A000290 and A040001 are. - Andrew Howroyd, Jul 25 2018

References

  • G. Pólya and G. Szegő, Problems and Theorems in Analysis II (Springer 1924, reprinted 1976), Part Eight, Chap. 1, Sect. 7, Problem 73.

Crossrefs

Programs

Formula

G.f.: x*(1 + 2*x + 6*x^2 + 2*x^3 + x^4)/(1-x^2)^3.
a(n+1) = denominator((1/(2*Pi))*Integral_{t=0..2*Pi} exp(i*n*t)(-((Pi-t)/i)^2)), i=sqrt(-1).
a(n) = 3*a(n-2) - 3*a(n-4) + a(n-6) for n > 5. - Paul Curtz, Mar 07 2011
a(n) is the numerator of the coefficient of x^4 in the Maclaurin expansion of exp(-n*x^2). - Francesco Daddi, Aug 04 2011
O.g.f. as a Lambert series: x*Sum_{n >= 1} J_2(n)*x^n/(1 + x^n), where J_2(n) denotes the Jordan totient function A007434(n). See Pólya and Szegő. - Peter Bala, Dec 28 2013
From Ilya Gutkovskiy, Jul 11 2016: (Start)
E.g.f.: x*((2*x + 1)*sinh(x) + (x + 2)*cosh(x))/2.
Sum_{n>=1} 1/a(n) = 5*Pi^2/24. [corrected by Amiram Eldar, Sep 11 2022] (End)
a(n) = A000290(n) / A040001(n). - Andrew Howroyd, Jul 25 2018
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/24 (A222171). - Amiram Eldar, Sep 11 2022
From Peter Bala, Jan 16 2024: (Start)
a(n) = Sum_{1 <= i, j <= n} (-1)^(1 + gcd(i,j,n)) = Sum_{d | n} (-1)^(d+1) * J_2(n/d), that is, the Dirichlet convolution of the pair of multiplicative functions f(n) = (-1)^(n+1) and the Jordan totient function J_2(n) = A007434(n). Hence this sequence is multiplicative. Cf. A193356 and A309337.
Dirichlet g.f.: (1 - 2/2^s)*zeta(s-2). (End)
a(n) = Sum_{1 <= i, j <= n} (-1)^(n + gcd(i, n)*gcd(j, n)) = Sum_{d|n, e|n} (-1)^(n+e*d) * phi(n/d)*phi(n/e). - Peter Bala, Jan 22 2024

Extensions

More terms from Michel Marcus, Dec 28 2013

A105647 Lexicographically earliest sequence of increasing numbers whose digits satisfy the "Fractal Jump" rule using only the digits 2 and 5: keep the first digit "d" of the sequence, then jump over the next "d" digits and keep the digit "e" on which you have landed. Jump now over the next "e" digits and keep the digit "f" on which you have landed, etc. The succession "def..." of kept digits is the sequence itself.

Original entry on oeis.org

2, 5, 25, 52, 55, 222, 252, 255, 552, 555, 2222, 5222, 5252, 5255, 22222, 22252, 22522, 22525, 22555, 25222, 52222, 55222, 55522, 55525, 55552, 55555, 222522, 225222, 225225, 225522, 252222, 252225, 522222, 2225222, 2225252, 2225255, 5222225, 5222522, 5225222
Offset: 1

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Author

Keywords

Examples

			The sequence and the "kept" digits begin
  2, 5, 25, 52, 55, 222, 252, 255, 552, ...
  ^      ^           ^    ^         ^
  2      5           2    5         5
		

Crossrefs

Extensions

Corrected and extended by Tyler Busby, Jan 08 2023

A359385 The lexicographically earliest "Increasing Term Fractal Jump Sequence" that does not use the digit 0 in any terms.

Original entry on oeis.org

1, 2, 21, 22, 23, 112, 122, 132, 133, 134, 141, 221, 311, 2112, 2113, 3111, 21111, 31113, 31114, 31124, 31131, 34111, 41121, 42111, 43111, 111121, 111122, 112111, 112311, 131111, 211112, 211113, 1111311, 1111312, 3111311, 3111312, 4111131, 4111132, 4141111
Offset: 1

Views

Author

Tyler Busby, Dec 29 2022

Keywords

Comments

The rules of an "Increasing Term Fractal Jump Sequence" are described in A105647.
The digit zero is omitted as it can create situations where the next term can require a 0 in the first digit.

Examples

			The sequence and the "kept" digits begin
  1, 2, 21, 22, 23, 112, 122, 132, 133, ...
  ^     ^    ^      ^ ^    ^    ^    ^
  1     2    2      1 2    2    2    3
		

Crossrefs

A306279 Numbers congruent to 3 or 18 mod 22.

Original entry on oeis.org

3, 18, 25, 40, 47, 62, 69, 84, 91, 106, 113, 128, 135, 150, 157, 172, 179, 194, 201, 216, 223, 238, 245, 260, 267, 282, 289, 304, 311, 326, 333, 348, 355, 370, 377, 392, 399, 414, 421, 436, 443, 458, 465, 480, 487, 502, 509, 524, 531, 546, 553, 568
Offset: 1

Views

Author

Davis Smith, Feb 02 2019

Keywords

Crossrefs

Programs

  • Maple
    seq(seq(22*i+j, j=[3, 18]), i=0..200);
  • Mathematica
    Select[Range[200], MemberQ[{3, 18}, Mod[#, 22]] &]
    Flatten[Table[{22n + 3, 22n + 18}, {n, 0, 43}]] (* Alonso del Arte, Feb 18 2019 *)
  • PARI
    for(n=3, 678, if((n%22==3) || (n%22==18), print1(n, ", ")))
    
  • PARI
    vector(62,n,11*n-6+2*(-1)^n)
    
  • PARI
    Vec(x*(3 + 15*x + 4*x^2) / ((1 - x)^2*(1 + x)) + O(x^40)) \\ Colin Barker, Feb 07 2019
    
  • Scala
    (3 to 949 by 22).union(18 to 942 by 22).sorted // Alonso del Arte, Feb 18 2019

Formula

a(n) = 11*n - 6 + 2*(-1)^n.
a(n) = 11*n - A105398(n + 4).
A007310(a(n) + 1) = 11*A007310(n).
From Colin Barker, Feb 07 2019: (Start)
G.f.: x*(3 + 15*x + 4*x^2) / ((1 - x)^2*(1 + x)).
a(n) = a(n - 1) + a(n - 2) - a(n - 3) for n > 3. (End)
E.g.f.: 4 + (11*x - 6)*exp(x) + 2*exp(-x). - David Lovler, Sep 08 2022

A155158 Period 4: repeat [1, 5, 7, 3].

Original entry on oeis.org

1, 5, 7, 3, 1, 5, 7, 3, 1, 5, 7, 3, 1, 5, 7, 3, 1, 5, 7, 3, 1, 5, 7, 3, 1, 5, 7, 3, 1, 5, 7, 3, 1, 5, 7, 3, 1, 5, 7, 3, 1, 5, 7, 3, 1, 5, 7, 3, 1, 5, 7, 3, 1, 5, 7, 3, 1, 5, 7, 3, 1, 5, 7, 3, 1, 5, 7, 3, 1, 5, 7, 3, 1, 5, 7, 3, 1, 5, 7, 3, 1, 5, 7, 3, 1, 5
Offset: 0

Views

Author

Paul Curtz, Jan 21 2009

Keywords

Crossrefs

Programs

Formula

a(n) = A048473(n) mod 10.
First differences: a(n+1)-a(n) = (-1)^floor(n/2)*A010694(n+1).
Second differences: a(n+2)-2*a(n+1)+a(n) = (-1)^floor(1+n/2)*A010696(n).
Third differences: a(n+3)-3*a(n+2)+3*a(n+1)-a(n) = (-1)^floor((n+3)/2)*A105398(n).
G.f.: (1+4*x+3*x^2)/(1-x+x^2-x^3). - Colin Barker, Feb 28 2012
a(n) = a(n-1) - a(n-2) + a(n-3) for n>2. - Wesley Ivan Hurt, Jul 08 2016

A359611 The lexicographically earliest "Increasing Term Fractal Jump Sequence".

Original entry on oeis.org

1, 2, 20, 22, 100, 200, 201, 1000, 20000, 20001, 110000, 2000000, 2000001, 110100000, 200000000, 200000001, 1101001000000, 2000000000020, 2000000010101, 10100010000000, 20000000000002, 20020000000001, 101001010010000, 100000000200000000000000
Offset: 1

Views

Author

Tyler Busby, Jan 06 2023

Keywords

Comments

The rules of an "Increasing Term Fractal Jump Sequence" are described in A105647.
We define a "forced" digit in Fractal Jump Sequences as a digit that is required to be a specific value by a digit that occurred previously in the sequence. This is in opposition to digits that could have any value selected for them without breaking the Fractal Jump Sequence rules. In the diagram below, the digits with carets below them are the forced digits.
To find a(n), increment a(n-1) until all of the forced digits that will positionally occur in a(n) satisfy their forced values. Then, to avoid leading zeros in a(n+1), if there are forced zeros immediately following the candidate a(n), continue to increment until it is the same number of digits longer as there are consecutive forced zeros, and continue to increment until the candidate a(n) once again satisfies all forcing criteria (including the new zeros).
The only digits that appear in this sequence are 0, 1, and 2, even though no numerals are arbitrarily restricted from appearing.

Examples

			The sequence and the "kept"/"forced" digits begin
  1, 2, 20, 22, 100, 200, 201, 1000, 20000, ...
  ^     ^    ^    ^  ^    ^    ^ ^^  ^  ^^
  1     2    2    0  2    2    1 00  2  00
In the case of computing a(5), we have a 22 for a(4), so we would normally increment to 23, as there is nothing forcing the next two digits. However, since there is a 0 forcing the following digit, we must increment to the smallest number that satisfies this forced 0 (as we can't have leading zeros in a(6)).
		

Crossrefs

Showing 1-6 of 6 results.