A151975
The number of ways one can flip seven consecutive tails (or heads) when flipping a coin n times.
Original entry on oeis.org
0, 0, 0, 0, 0, 0, 0, 1, 3, 8, 20, 48, 112, 256, 576, 1279, 2811, 6126, 13256, 28512, 61008, 129952, 275712, 582913, 1228551, 2582048, 5412984, 11321744, 23631056, 49229312, 102377216, 212560127, 440668919, 912310222, 1886316324, 3895528632, 8035861664
Offset: 0
a(0)=0 means that there are no cases of seven consecutive tails (or heads) in zero coin flips. Likewise, a(1)=a(2)=...=a(6)=0. a(7)=1 since there is exactly one case of seven consecutive tails in seven coin flips.
- Colin Barker, Table of n, a(n) for n = 0..1000
- Benjamin E. Merkel, Probabilities of Consecutive Events in Coin Flipping, OhioLINK, 2011
- Index entries for linear recurrences with constant coefficients, signature (3,-1,-1,-1,-1,-1,-1,-2).
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N=66; x='x+O('x^N);
gf = (1-x)/(1-2*x); /* A011782(n): compositions of n */
gf -= 1/(1 - (x+x^2+x^3+x^4+x^5+x^6+x^7)); /* A066178(n): compositions of n into parts <=7 */
v151975=Vec(gf + 'a0); v151975[1]=0; /* kludge to get all terms */
v151975 /* show terms */
/* Joerg Arndt, Aug 06 2012 */
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concat(vector(7), Vec(x^7/((2*x-1)*(x^7+x^6+x^5+x^4+x^3+x^2+x-1)) + O(x^100))) \\ Colin Barker, Oct 16 2015
A341050
Cube array read by upward antidiagonals ignoring zero and empty terms: T(n, k, r) is the number of n-ary strings of length k, containing r consecutive 0's.
Original entry on oeis.org
1, 1, 1, 3, 1, 1, 3, 1, 5, 8, 1, 1, 3, 1, 5, 8, 1, 7, 21, 19, 1, 1, 3, 1, 5, 8, 1, 7, 21, 20, 1, 9, 40, 81, 43, 1, 1, 3, 1, 5, 8, 1, 7, 21, 20, 1, 9, 40, 81, 47, 1, 11, 65, 208, 295, 94, 1, 1, 3, 1, 5, 8, 1, 7, 21, 20, 1, 9, 40, 81, 48, 1, 11, 65, 208, 297, 107, 1, 13, 96, 425, 1024, 1037, 201
Offset: 2
For n = 5, k = 6 and r = 4, there are 65 strings: {000000, 000001, 000002, 000003, 000004, 000010, 000011, 000012, 000013, 000014, 000020, 000021, 000022, 000023, 000024, 000030, 000031, 000032, 000033, 000034, 000040, 000041, 000042, 000043, 000044, 010000, 020000, 030000, 040000, 100000, 100001, 100002, 100003, 100004, 110000, 120000, 130000, 140000, 200000, 200001, 200002, 200003, 200004, 210000, 220000, 230000, 240000, 300000, 300001, 300002, 300003, 300004, 310000, 320000, 330000, 340000, 400000, 400001, 400002, 400003, 400004, 410000, 420000, 430000, 440000}
The first seven slices of the tetrahedron (or pyramid) are:
-----------------Slice 1-----------------
1
-----------------Slice 2-----------------
1
1 3
-----------------Slice 3-----------------
1
1 3
1 5 8
-----------------Slice 4-----------------
1
1 3
1 5 8
1 7 21 19
-----------------Slice 5-----------------
1
1 3
1 5 8
1 7 21 20
1 9 40 81 43
-----------------Slice 6-----------------
1
1 3
1 5 8
1 7 21 20
1 9 40 81 47
1 11 65 208 295 94
-----------------Slice 7-----------------
1
1 3
1 5 8
1 7 21 20
1 9 40 81 48
1 11 65 208 297 107
1 13 96 425 1024 1037 201
Cf.
A005408,
A003215,
A005917,
A022521,
A022522,
A022523,
A022524,
A022525,
A022526,
A022527,
A022528,
A022529,
A022530,
A022531,
A022532,
A022533,
A022534,
A022535,
A022536,
A022537,
A022538,
A022539,
A022540 (k=x, r=1, where x is the x-th Nexus Number).
Cf.
A000567 [(k=4, r=2),(k=5, r=3),(k=6, r=4),...,(k=x, r=x-2)].
Cf.
A103532 [(k=6, r=3),(k=7, r=4),(k=8, r=5),...,(k=x, r=x-3)].
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m[r_, n_] := Normal[With[{p = 1/n}, SparseArray[{Band[{1, 2}] -> p, {i_, 1} /; i <= r -> 1 - p, {r + 1, r + 1} -> 1}]]]; T[n_, k_, r_] := MatrixPower[m[r, n], k][[1, r + 1]]*n^k; DeleteCases[Transpose[PadLeft[Reverse[Table[T[n, k, r], {k, 2, 8}, {r, 2, k}, {n, 2, r}], 2]], 2 <-> 3], 0, 3] // Flatten
A167826
a(n) is the number of n-tosses having a run of 3 or more heads and a run of 3 or more tails for a fair coin.
Original entry on oeis.org
0, 0, 0, 0, 0, 2, 8, 26, 74, 194, 482, 1152, 2674, 6068, 13524, 29704, 64460, 138482, 294988, 623834, 1311086, 2740666, 5702270, 11815752, 24395678, 50209572, 103048168, 210965064, 430938832, 878534170
Offset: 1
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b[1] = 0; b[2] = 1; b[3] = 1; b[n_]: = b[n-1] + b[n-2] + b[n-3]; Table[2^n - 2*(Sum[b[n + 3 - i], {i, 1, 3}] - Fibonacci[n + 1]), {n, 1, 30}]
LinearRecurrence[{4, -3, -3, 0, 3, 2}, {0, 0, 0, 0, 0, 2}, 50] (* G. C. Greubel, Jun 27 2016 *)
A369580
a(n) := f(n, n), where f(0,0) = 1/3, f(0,k) = 0 and f(k,0) = 3^(k-1) if k > 0, and f(n, m) = f(n, m-1) + f(n-1, m) + 3*f(n-1, m-1) otherwise.
Original entry on oeis.org
2, 16, 138, 1216, 10802, 96336, 861114, 7708416, 69072354, 619380496, 5557080938, 49879087296, 447852531986, 4022246329936, 36132550233498, 324645166734336, 2917340834679234, 26219438520320016, 235672871308226634, 2118552629658530496, 19046140604787232242, 171241206828437556816
Offset: 1
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def lis(n):
table = [[0]*(n+1) for _ in range(n+1)]
table[1][1] = 2
for i in range(1, n+1) :
table[i][0] = 3**(i-1)
for i in range(1, n+1) :
for j in range(1, n+1) :
if (i == 1 and j == 1) :
continue
table[i][j] = table[i][j-1] + table[i-1][j] + 3*table[i-1][j-1]
return [int(table[i][i]) for i in range(1, n+1)]
A265725
Number of binary strings of length n having at least one run of length at least 4.
Original entry on oeis.org
0, 0, 0, 0, 2, 6, 16, 40, 94, 214, 476, 1040, 2242, 4782, 10112, 21232, 44318, 92046, 190364, 392264, 805746, 1650518, 3372816, 6877656, 13998142, 28442918, 57707324, 116925600, 236630274, 478372062, 966145664, 1949583456, 3930972094, 7920443038, 15948482236
Offset: 0
For n=5 there are 6 such strings: 00000, 00001, 01111, and their complements.
Comments