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 1-4 of 4 results.

A371358 Number of binary strings of length n which have more 00 than 01 substrings.

Original entry on oeis.org

0, 0, 1, 2, 4, 10, 21, 42, 89, 184, 371, 758, 1546, 3122, 6315, 12782, 25780, 51962, 104759, 210934, 424404, 853806, 1716759, 3450158, 6932169, 13924260, 27959805, 56130762, 112662414, 226080318, 453595341, 909925794, 1825052601, 3660020992, 7339006091
Offset: 0

Views

Author

Robert P. P. McKone, Mar 19 2024

Keywords

Examples

			a(4) = 4: 0000, 0001, 1000, 1100.
a(5) = 10: 00000, 00001, 00010, 00011, 00100, 01000, 10000, 10001, 11000, 11100.

Links

Crossrefs

Cf. A163493 (equal 00 and 01), A371564 (more 01 than 00), A090129 (equal 01 and 10), A182027 (equal 00 and 11), A370048 (one more 00 than 01).
Cf. A000079(n-2) (more 01 than 10, for n>=2).

Programs

  • Maple
    b:= proc(n, l, t) option remember; `if`(n+t<1, 0, `if`(n=0, 1,
      add(b(n-1, i, t+`if`(l=0, (-1)^i, 0)), i=0..1)))
    end:
a:= n-> b(n, 2, 0):
seq(a(n), n=0..34);  # Alois P. Heinz, Mar 20 2024
  • Mathematica
    tup[n_] := Tuples[{0, 1}, n];
cou[lst_List] := Count[lst, {0, 0}] > Count[lst, {0, 1}];
par[lst_List] := Partition[lst, 2, 1];
a[n_] := Map[cou, Map[par, tup[n]]] // Boole // Total;
Monitor[Table[a[n], {n, 0, 18}], {n, Table[a[m], {m, 0, n - 1}]}]
  • PARI
    { a371358(n) = 2^(n-1) - sum(k=0, n\3, binomial(2*k,k) * (2*binomial(n-2*k,n-3*k) - binomial(n-2*k-1,n-3*k))) / 2; } \\ Max Alekseyev, May 01 2024

Formula

a(n) = 2^n - A163493(n) - A371564(n).
a(n) = ((4*n^2-15*n+7)*a(n-1) -(5*n^2-22*n+14)*a(n-2) +2*(3*n^2-14*n+10)*a(n-3) -4*(3*n^2-16*n+18)*a(n-4) +8*(n-2)*(n-4)*a(n-5)) / (n*(n-3)) for n>=5. - Alois P. Heinz, Mar 20 2024
For n >= 2, a(n) = 2*a(n-1) + A163493(n-1) - A163493(n-2) - A370048(n-2). - Max Alekseyev, Apr 30 2024
a(n) = 2^(n-1) - (1/2) * Sum_{k=0..floor(n/3)} binomial(2*k,k) * (2*binomial(n-2*k,n-3*k) - binomial(n-2*k-1,n-3*k)). - Max Alekseyev, May 01 2024
G.f. 1/(1-2*x)/2 - (1+x)/(2*sqrt(1-2*x+x^2-4*x^3+4*x^4)). - Max Alekseyev, Apr 30 2024

A370048 Number of binary strings of length n in which the number of substrings 00 is one more than that of substrings 01.

Original entry on oeis.org

0, 0, 1, 1, 2, 6, 10, 18, 40, 76, 141, 285, 558, 1066, 2097, 4121, 8000, 15660, 30763, 60171, 117918, 231690, 454816, 893208, 1756688, 3455580, 6799195, 13388587, 26375466, 51974798, 102470402, 202108730, 398756664, 787025260, 1553900235, 3068937675, 6062944710, 11981429394, 23683822694, 46828287038
Offset: 0

Views

Author

Max Alekseyev, Apr 30 2024

Keywords

Crossrefs

Cf. A000079, A090129, A163493, A182027, A371358, A371564.

Programs

  • PARI
    { a370048(n) = (n > 1) * sum(m=0,(n-1)\3, binomial(2*m,m+1) * binomial(n-1-2*m,m) + binomial(2*m+1,m) * binomial(n-2-2*m,m) ); }
  • Python
    from math import comb
def A370048(n): return 0 if n<2 else 1+sum((x:=comb((k:=m<<1),m+1)*comb(n-1-k,m))+x*(k+1)*(n-1-3*m)//(m*(n-1-k)) for m in range(1,(n+2)//3)) # Chai Wah Wu, May 01 2024

Formula

For n >= 2, a(n) = Sum_{m=0..floor((n-1)/3)} binomial(2*m,m+1) * binomial(n-1-2*m,m) + binomial(2*m+1,m) * binomial(n-2-2*m,m).
For n >= 4, a(n) = ( (n-2)*(2*n-1)*(n^2-n-4)*a(n-1) - (n^2-5*n+2)*(n^2+n-4)*a(n-2) + 2*(n-3)*n^2*(2*n-3)*a(n-3) - 4*(n-3)*(n-1)^2*n*a(n-4) ) / (n-2)^2 / (n-1) / (n+2).
a(n) = 2*A371358(n+1) - A371358(n+2) + A163493(n+1) - A163493(n).
G.f. ((1-x^2-2*x^3)*(1-2*x+x^2-4*x^3+4*x^4)^(-1/2) - 1 - x)/x^2/2, which can be expressed in terms of g.f. C(x) = (1-sqrt(1-4*x))/x/2 for Catalan number (A000108) as x*((x+1)*C(x^3/(1-x))-1)/(1-x-2*x^3*C(x^3/(1-x))).

A371570 Number of binary necklaces of length n which have more 01 than 00 substrings.

Original entry on oeis.org

0, 0, 2, 3, 6, 15, 29, 56, 118, 237, 467, 946, 1905, 3796, 7618, 15303, 30614, 61319, 122951, 246202, 492971, 987542, 1977560, 3959289, 7927969, 15873190, 31776708, 63614397, 127346134, 254908115, 510233309, 1021273672, 2044071894, 4091064805, 8187770675
Offset: 0

Views

Author

Robert P. P. McKone, Mar 28 2024

Keywords

Comments

A necklace may also be referred to as circular or cyclic strings.

Examples

			a(3) = 3: 011, 101, 110.
a(4) = 6: 0101, 0111, 1010, 1011, 1101, 1110.
a(5) = 15: 00101, 01001, 01010, 01011, 01101, 01111, 10010, 10100, 10101, 10110, 10111, 11010, 11011, 11101, 11110.

Crossrefs

Cf. A217464 (necklaces with equal 00 and 01), A371668 (necklaces with more 00 than 01).
Cf. A126869 (necklaces with equal 00 and 11, for n>=1), A058622 (necklaces with more 00 than 11).
Cf. A163493 (strings with equal 00 and 01), A371358 (strings with more 00 than 01), A371564 (strings with more 01 than 00).

Programs

  • Mathematica
    tup[n_] := Tuples[{0, 1}, n];
tupToNec[n_] := Map[Append[#, #[[1]]] &, tup[n]];
cou[lst_List] := Count[lst, {0, 1}] > Count[lst, {0, 0}];
par[lst_List] := Partition[lst, 2, 1];
a[0] = 0;
a[n_] := Map[cou, Map[par, tupToNec[n]]] // Boole // Total;
Monitor[Table[a[n], {n, 0, 18}], {n, Table[a[m], {m, 0, n - 1}]}]

Formula

a(n) = 2^n - A217464(n) - A371668(n).
a(n) = -(((n-3)*(n-2) - 8*(n-5)^2*(n-2)*a(n-5) + 4*(n*((3n-34)*n+117)-114)*a(n-4) + 2*(((32-3n)*n-95)*n+62)*a(n-3) + (((5n-52)*n+157)*n-114)*a(n-2) + (((39-4n)*n-103)*n+58)*a(n-1))/((n-6)*(n-3)*n)) for n>=7.

A371668 Number of binary necklaces of length n which have more 00 than 01 substrings.

Original entry on oeis.org

0, 1, 1, 1, 5, 11, 19, 43, 93, 181, 371, 771, 1547, 3121, 6357, 12821, 25805, 52123, 105031, 211243, 425215, 855457, 1719257, 3455153, 6942387, 13942111, 27993317, 56197117, 112785797, 226311535, 454043339, 910778203, 1826666093, 3663122277, 7344953123
Offset: 0

Views

Author

Robert P. P. McKone, Apr 02 2024

Keywords

Examples

			a(3) = 1: 000.
a(4) = 5: 0000, 0001, 0010, 0100, 1000.
a(5) = 11: 00000, 00001, 00010, 00011, 00100, 00110, 01000, 01100, 10000, 10001, 11000.

Crossrefs

Cf. A217464 (necklaces with equal 00 and 01), A371570 (necklaces with more 01 than 00).
Cf. A126869 (necklaces with equal 00 and 11, for n>=1), A058622 (necklaces with more 00 than 11).
Cf. A163493 (strings with equal 00 and 01), A371358 (strings with more 00 than 01), A371564 (strings with more 01 than 00).

Programs

  • Mathematica
    tup[n_] := Tuples[{0, 1}, n];
tupToNec[n_] := Map[Append[#, #[[1]]] &, tup[n]];
cou[lst_List] := Count[lst, {0, 0}] > Count[lst, {0, 1}];
par[lst_List] := Partition[lst, 2, 1];
a[0] = 0;
a[n_] := a[n] = Map[cou, Map[par, tupToNec[n]]] // Boole // Total;
Monitor[Table[a[n], {n, 0, 18}], {n, Table[a[m], {m, 0, n - 1}]}]

Formula

a(n) = 2^n - A217464(n) - A371570(n).
a(n) = (8*(n-7)*a(n-7) + 4*(29-5*n)*a(n-6) + (26*n-110)*a(n-5) + (77-23*n)*a(n-4) + (15*n-46)*a(n-3) + (22-10*n)*a(n-2) + 5*(n-1)*a(n-1))/n for n>=7.
Showing 1-4 of 4 results.

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

Content is available under The OEIS End-User License Agreement.