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 1-5 of 5 results.

A116642 A116641 in binary.

Original entry on oeis.org

1, 101, 111, 1011, 10011, 10111, 11101, 11111, 100011, 100101, 101111, 110001, 110101, 1000001, 1000011, 1001001, 1001111, 1010101, 1011001, 1100001, 1100101, 1100111, 1110111, 1111001, 1111101, 10000011, 10000101, 10001111
Offset: 0

Views

Author

Antti Karttunen, Feb 20 2006

Keywords

Crossrefs

Cf. a(n) = A007088(A116641(n)).

A116623 a(0)=1, a(2n) = a(n)+A000079(A000523(2n)), a(2n+1) = 3*a(n) + A000079(A000523(2n+1)+1).

Original entry on oeis.org

1, 5, 7, 19, 11, 29, 23, 65, 19, 49, 37, 103, 31, 85, 73, 211, 35, 89, 65, 179, 53, 143, 119, 341, 47, 125, 101, 287, 89, 251, 227, 665, 67, 169, 121, 331, 97, 259, 211, 601, 85, 223, 175, 493, 151, 421, 373, 1087, 79, 205, 157, 439, 133, 367, 319, 925, 121
Offset: 0

Views

Author

Antti Karttunen, Feb 20 2006. Proposed by Pierre Lamothe (plamothe(AT)aei.ca), May 21 2004

Keywords

Comments

Viewed as a binary tree, this is (1); 5; 7,19; 11,29,23,65; ... Related to the parity vectors of Collatz and Terras trajectories.

Links

Crossrefs

Cf. a(n) = A116640(A059893(n)). a(A000225(n)) = A001047(n+1). For n>= 1 a(A000079(n)) = A062709(n+1). A116641 gives the terms in ascending order and without duplicates.

Programs

  • Maple
    A116623 := proc(n)
    option remember;
    if n = 0 then
        1;
    elif type(n,'even') then
        procname(n/2)+2^A000523(n) ;
    else
        3*procname(floor(n/2))+2^(1+A000523(n)) ;
    end if;
end proc: # R. J. Mathar, Nov 28 2016
  • Mathematica
    a[n_] := a[n] = Which[n == 0, 1, EvenQ[n], a[n/2] + 2^Floor@Log2[n], True, 3a[Floor[n/2]] + 2^(1 + Floor@Log2[n])];
Table[a[n], {n, 0, 56}] (* Jean-François Alcover, Sep 01 2023 *)

A119733 Offsets of the terms of the nodes of the reverse Collatz function.

Original entry on oeis.org

0, 1, 2, 5, 4, 7, 10, 19, 8, 11, 14, 23, 20, 29, 38, 65, 16, 19, 22, 31, 28, 37, 46, 73, 40, 49, 58, 85, 76, 103, 130, 211, 32, 35, 38, 47, 44, 53, 62, 89, 56, 65, 74, 101, 92, 119, 146, 227, 80, 89, 98, 125, 116, 143, 170, 251, 152, 179, 206, 287, 260, 341, 422, 665, 64, 67
Offset: 0

Views

Author

William Entriken, Jun 14 2006

Keywords

Comments

Create a binary tree starting with x. To follow 0 from the root, apply f(x)=2x. To follow 1, apply g(x)=(2x-1)/3. For example, starting with x, the string 010 {also known as f(g(f(x)))}, you would get (8x-2)/3. These expressions represent the reverse Collatz function and will provide numbers whose Collatz path may include x. These expressions will all be of the form (2^a*x-b)/3^c. This sequence concerns b. What makes b interesting is that if you draw the tree, each level of the tree will have the same sequence of values for b. The root of the tree x, can be written as (2^0*x-0)/3^0, which has the first value for b. Each subsequent level contains twice as many values of b.
This sequence is 0 followed by a permutation of A213539, and therefore consists of 0 plus the elements of A116640 multiplied by 2^k, where k >= 0. E.g., 1, 5, 7, 19 becomes 0, 2^0*1, 2^1*1, 2^0*5, 2^2*1, 2^0*7, 2^1*5, 2^0*19, ... - Joe Slater, Dec 19 2016
When this sequence is arranged as an irregular triangle the sum of each row a(2^k)...a(2^(k+1)-1) equals A081039(2^k). The cumulative sum from a(0) to a(2^k-1) equals A002697(k). - Joe Slater, Apr 12 2018

Examples

			a(1) = 1 = 2 * 0 + 3^0 since 0 written in binary contains no 1's.

Links

Crossrefs

Cf. A002697, A081039, A116623, A116640, A116641, A213539, A226383.

Programs

  • Maple
    a:= proc(n) `if`(n=0, 0, `if`(irem(n, 2, 'r')=0, 0,
      3^add(i, i=convert(r, base, 2)))+2*a(r))
    end:
seq(a(n), n=0..127);  # Alois P. Heinz, Aug 13 2017
  • Mathematica
    a[0] := 0; a[n_?OddQ] := 2a[(n - 1)/2] + 3^Plus@@IntegerDigits[(n - 1)/2, 2]; a[n_?EvenQ] := 2a[n/2]; Table[a[n], {n, 0, 65}] (* Alonso del Arte, Apr 21 2011 *)
  • PARI
    a(n) = my(ret=0); if(n, for(i=0,logint(n,2), if(bittest(n,i), ret=3*ret+1<Kevin Ryde, Oct 22 2021
  • Perl
    # call with n to get 2^n values
$depth=shift; sub funct { my ($i, $b, $c) = @_; if ($i < $depth) { funct($i+1, $b*2, $c); funct($i+1, 2*$b+$c, $c*3); } else { print "$b, "; } } funct(0, 0, 1); print " ";
  • Python
    from sympy.core.cache import cacheit
@cacheit
def a(n): return 0 if n==0 else 2*a((n - 1)//2) + 3**bin((n - 1)//2).count('1') if n%2 else 2*a(n//2)
print([a(n) for n in range(131)]) # Indranil Ghosh, Aug 13 2017

Formula

a(0) = 0, a(2*n + 1) = 2*a(n) + 3^wt(n) = 2*a(n) + A048883(n), a(2*n) = 2*a(n), where wt(n) = A000120(n) = the number 1's in the binary representation of n.
a(k) = [z^k] 1 + (1/(1-z)) * Sum_{s=0..n-1} 2^s*z^(2^s)*(1 - z^(2^s)) * Product_{r=s+1..n-1} (1 + 3*z^(2^r)), for 0 < k <= 2^n-1. - Wolfgang Hintze, Jul 28 2017
a(n) = Sum_{i=0..k} 2^e[i] * 3^i where binary expansion n = 2^e[0] + 2^e[1] + ... + 2^e[k] with descending e[0] > e[1] > ... > e[k] (A272011). [Martín Chabrera lemma 6.1, adapting index i] - Kevin Ryde, Oct 22 2021

A213539 Variant of numbers for which there is at least one 3-smooth representation that is special of level k.

Original entry on oeis.org

1, 2, 4, 5, 7, 8, 10, 11, 14, 16, 19, 20, 22, 23, 28, 29, 31, 32, 35, 37, 38, 40, 44, 46, 47, 49, 53, 56, 58, 62, 64, 65, 67, 70, 73, 74, 76, 79, 80, 85, 88, 89, 92, 94, 97, 98, 101, 103, 106, 112, 116, 119, 121, 124, 125, 128, 130, 131, 133, 134, 140, 143, 146
Offset: 0

Views

Author

Kenneth Vollmar, Mar 03 2013

Keywords

Comments

These numbers are of the form 3^k*2^{a_0} + 3^{k-1}*2^{a_1} + ... + 3^1*2^{a_{k-1}} + 3^0*2^{a_k} in which every power 3^i appears, 0 <= i <= k, and where a_i satisfies 0 <= a_0 < a_1 < ... < a_k.
These values are those of sequence A116640 in addition to any multiple of two of elements of this sequence. - Kenneth Vollmar, Jun 05 2013

Examples

			n=19 has two 3-smooth representations that are special of level k. At k=1, 19 = 3^1*2^0 + 3^0*2^4. At k=2, 19 = 3^2*2^0 + 3^1*2^1 + 3^0*2^2.

References

  • Kenneth Vollmar, Recursive calculation of 3-smooth representations special of level k, To be submitted mid-2013.

Links

Crossrefs

Cf. A116623, A116640, A116641, A119733, A226383, A003586.

Extensions

Corrected a reference to another sequence and added cross references - Joe Slater, Dec 19 2016

A327283 Irregular triangle T(n,k) read by rows: "residual summands" in reduced Collatz sequences (see Comments for definition and explanation).

Original entry on oeis.org

1, 1, 5, 1, 5, 19, 73, 347, 1, 7, 29, 103, 373, 1631, 1, 5, 23, 133, 1, 11, 1, 5, 19, 65, 451, 1, 7, 53, 1, 5, 31, 125, 503, 2533, 1, 1, 5, 19, 185, 1, 7, 29, 151, 581, 2255, 10861, 1, 5, 23, 85, 287, 925
Offset: 1

Views

Author

Bob Selcoe, Sep 15 2019

Keywords

Comments

Let R_s be the reduced Collatz sequence (cf. A259663) starting with s and let R_s(k), k >= 0 be the k-th term in R_s. Then R_(2n-1)(k) = (3^k*(2n-1) + T(n,k))/2^j, where j is the total number of halving steps from R_(2n-1)(0) to R_(2n-1)(k). T(n,k) is defined here as the "residual summand".
The sequence without duplicates is a permutation of A116641.

Examples

			Triangle starts:
  1;
  1, 5;
  1;
  1, 5, 19, 73,  347;
  1, 7, 29, 103, 373, 1631;
  1, 5, 23, 133;
  1, 11;
  1, 5, 19, 65,  451;
  1, 7, 53;
  1, 5, 31, 125, 503, 2533;
  1;
  1, 5, 19, 185;
  1, 7, 29, 151, 581, 2255, 10861;
  ...
T(5,4)=103 because R_9(4) = 13; the number of halving steps from R_9(0) to R_9(4) is 6, and 13 = (81*9 + 103)/64.

Crossrefs

Cf. A116623, A116641, A259663.

Formula

T(n,k) = 2^j*R_(2n-1)(k) - 3^k*(2n-1), as defined in Comments.
T(n,1) = 1; for k>1: T(n,k) = 3*T(n,k-1) + 2^i, where i is the total number of halving steps from R_(2n-1)(0) to R_(2n-1)(k-1).
Showing 1-5 of 5 results.

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