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.

Previous Showing 21-25 of 25 results.

A381812 Number of moves required to reach a position with the maximum number of heads in the game of blet with 2*n coins.

Original entry on oeis.org

1, 1, 2, 5, 3, 6, 11, 7, 10, 17, 11, 16, 25, 15, 22, 33, 21, 28, 41, 27, 34
Offset: 2

Views

Author

Pontus von Brömssen, Mar 08 2025

Keywords

Comments

See A075273 or Rodriguez Villegas, Sadun, and Voloch (2002) for the definition of blet.
Rodriguez Villegas, Sadun, and Voloch (2002) prove that the maximum number of heads achievable is A047206(n).

Examples

			For n = 5, a(5) = 5 moves are needed to go from HTHTHTHTHT to a sequence with the maximum number A047206(5) = 8 of heads: HTHTHTHTHT -> THHTHTHTHH -> THHTHTTHTH -> THHHTHTHTH -> THHHHTHHTH -> HHHHHTHHHT.
		

Crossrefs

Programs

  • Python
    # see linked program

Extensions

a(17)-a(18) from Michael S. Branicky, Mar 11 2025
a(19)-a(22) from Bert Dobbelaere, Mar 15 2025

A093545 Sorted mapping of A093544 onto the integers.

Original entry on oeis.org

0, 2, 1, 5, 7, 3, 10, 12, 4, 15, 17, 6, 20, 22, 8, 25, 27, 9, 30, 32, 11, 35, 37, 13, 40, 42, 14, 45, 47, 16, 50, 52, 18, 55, 57, 19, 60, 62, 21, 65, 67, 23, 70, 72, 24, 75, 77, 26, 80, 82, 28, 85, 87, 29, 90, 92, 31, 95, 97, 33, 100, 102, 34, 105, 107, 36, 110, 112, 38
Offset: 0

Views

Author

Ralf Stephan, Mar 31 2004

Keywords

Comments

As A093544 contains the odd numbers not of form 12k+9, we map from modulo 12 to modulo 5: 1->0, 3->1, 5->2, 7->3, 11->4.

Crossrefs

Cf. A047206, A340615 (inverse permutation), A014682.

Programs

  • Mathematica
    CoefficientList[Series[x (x^10 + 3 x^9 + 5 x^8 + x^7 + 5 x^6 + 5 x^5 + 2 x^4 + 5 x^3 + 5 x^2 + x + 2)/(1 - x^3 - x^9 + x^12), {x, 0, 68}], x] (* Michael De Vlieger, Mar 05 2021 *)
  • PARI
    a(n)=5*(A093544(n)\12)+if(A093544(n)%12==11,4,((A093544(n)%12)-1)/2)

Formula

a(3n) = 5n, a(3n+1) = 5n+2, a(3n+2) = A047206(n).
G.f.: x*(x^10 + 3*x^9 + 5*x^8 + x^7 + 5*x^6 + 5*x^5 + 2*x^4 + 5*x^3 + 5*x^2 + x + 2)/(1 - x^3 - x^9 + x^12).

A075274 Number of reachable arrangements of coins in the game Blet starting with 2n coins that achieve the maximum number of heads.

Original entry on oeis.org

2, 3, 2, 5, 4, 7, 16, 9, 25, 55, 24, 91, 196, 70, 336, 714, 216
Offset: 2

Views

Author

N. J. A. Sloane, Oct 12 2002

Keywords

Comments

Rodriguez Villegas, Sadun, and Voloch (2002) prove that the maximum number of heads achievable is A047206(n). - Pontus von Brömssen, Mar 08 2025

Examples

			For 4 coins, starting from HTHT we can reach THTT, HHTH, TTTH, THHH, and no others. Of these, two arrangements HHTH and THHH achieve the maximum of 3 heads, so a(2) = 2.
		

Crossrefs

Programs

  • Python
    # see linked program

Extensions

Offset corrected by Michel Marcus, Sep 30 2017
a(16)-a(18) from and name clarified by Sean A. Irvine, Feb 14 2025

A340615 a(n) = k/2 if k is even, otherwise (3k+1)/2, where k = n + floor((n+1)/5).

Original entry on oeis.org

0, 2, 1, 5, 8, 3, 11, 4, 14, 17, 6, 20, 7, 23, 26, 9, 29, 10, 32, 35, 12, 38, 13, 41, 44, 15, 47, 16, 50, 53, 18, 56, 19, 59, 62, 21, 65, 22, 68, 71, 24, 74, 25, 77, 80, 27, 83, 28, 86, 89, 30, 92, 31, 95, 98, 33, 101, 34, 104, 107, 36, 110, 37, 113, 116, 39, 119, 40, 122
Offset: 0

Views

Author

Thomas Scheuerle, Jan 13 2021

Keywords

Comments

A permutation of the nonnegative integers related to the Collatz function (A014682).
As this sequence is a permutation, it can be decomposed into cycles.
Known finite cycles so far: (0), (1 2), (3 5), (24 44 80 48 86 155 93 167 100 60 36 65 39 71 128 230 138 248 446 803 1445 867 520 312 187 112 67 40).
Except for the 0-cycle, it is a necessary (but not sufficient) condition for the smallest number in a cycle that it is in the intersection of A032766 and A047206 (as suggested by Sebastian Karlsson, Jan 15 2021).
If we generate new sequences by including a variable m into the definition of the original sequence, k = n+floor((n+1)/m*5), a(n) = k/2 if k is even, otherwise (3k+1)/2, then the sequences with m > 1 are not permutations and are in some sense intermediates between A014682 and the original sequence. This may aid in research related to Collatz conjecture as it may help to understand how cycles are modified or disappearing by increasing values of m.
Like any permutation this sequence can be written as a product of two involutions. Example:
a(n) = p(q(n)) with p = (0)(3 1)(5 2)(6 4)(11 7)(10 8)(26 9)(20 12)... and q = (0)(5 1)(3 2)(11 4)(8 6)(20 7)(47 9)(14 10)... does there exist a nice example with known sequences?
The difference between "Collatz function" A014682 and this sequence is that A014682 contains all numbers of the type 3*m + 2 twice (A016789). The modification "+floor((n+1)/5)" removes these duplicates from the sequence. The distances between the removed numbers and the predecessors that stay inside the sequence are of the form 4*m + 3 and found in A004767. Example: A014682(3) = 5 and A014682(10) = 5. 5 = A016789(1) and 10-3 = 7 = A004767(1).
A093545 is the inverse permutation: n = A093545(a(n)).

Crossrefs

Cf. A014682, A093545 (inverse).

Programs

  • MATLAB
    function a = A340615(max_n)
        for n = 1:max_n
            k = (n-1)+floor(((n-1)+1)/5);
            if floor(k/2) == k/2
                a(n) = k/2;
            else
                a(n) = (k*3+1)/2;
            end
        end
    end
  • Mathematica
    Array[If[EvenQ@ #, #/2, (3 # + 1)/2] &[# + Floor[(# + 1)/5]] &, 69, 0] (* Michael De Vlieger, Jan 13 2021 *)

Formula

a(n) = 2*a(n-5) - a(n-10).

A370753 Antidiagonal products of A319840.

Original entry on oeis.org

1, 1, 4, 36, 576, 12800, 360000, 12192768, 481890304, 21743271936, 1101996057600, 61952000000000, 3824628881965056, 257164113195565056, 18704075505689706496, 1462975070062038220800, 122444006400000000000000, 10918111308394619734065152, 1033255398127440061257744384
Offset: 0

Views

Author

Stefano Spezia, Jun 22 2024

Keywords

Comments

a(n) has trailing zeros iff n is congruent to 0 or 1 mod 5. Cf. A008851.
a(n) is a square iff n = 1 or congruent to {1, 3, 4} mod 5. Cf. A047206.
It appears that: (Start)
a(n) is a cube iff n = 0, 1, or is of the form (3*m - 4)^3 with m > 1 (A016791);
the only fourth powers in the sequence are 1 and a(9) = 21743271936 = 384^4;
the only fifth powers in the sequence are 1 and a(32) = 227200942336^5;
a(n) is a sixth power iff n = 0, 1, or is of the form (6*m - 10)^3 with m > 1;
the only seventh powers in the sequence are 1 and a(128) = 77458109039896212820250015287665035595218944^7. (End)

Crossrefs

Programs

  • Mathematica
    a[0]=a[1]=1; a[n_]:=n^2*2^(n-2)*(n-1)^(n-2); Array[a,19,0]

Formula

a(0) = a(1) = 1, and a(n) = n^2*2^(n-2)*(n - 1)^(n-2) for n > 1.
Previous Showing 21-25 of 25 results.