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A348175 Irregular table T(n,k) read by rows: T(n,k) = T(n-1,k/2) when k is even and 3*T(n-1,(k-1)/2) + 2^(n-1) when k is odd. T(0,0) = 0 and 0 <= k <= 2^n-1.

0, 0, 1, 0, 2, 1, 5, 0, 4, 2, 10, 1, 7, 5, 19, 0, 8, 4, 20, 2, 14, 10, 38, 1, 11, 7, 29, 5, 23, 19, 65, 0, 16, 8, 40, 4, 28, 20, 76, 2, 22, 14, 58, 10, 46, 38, 130, 1, 19, 11, 49, 7, 37, 29, 103, 5, 31, 23, 85, 19, 73, 65, 211

Offset: 0

Ryan Brooks, Oct 04 2021

			n\k 0  1  2  3  4  5  6  7
0   0
1   0  1
2   0  2  1  5
3   0  4  2 10  1  7  5 19

Cf. A001047 (right diagonal), A002697 (row sums), A119733.

Cf. A133457 (binary exponents).

T[0, 0] = 0; T[n_, k_] := T[n, k] = If[EvenQ[k], T[n - 1, k/2], 3*T[n - 1, (k - 1)/2] + 2^(n - 1)]; Table[T[n, k], {n, 0, 5}, {k, 0, 2^n - 1}] // Flatten (* Amiram Eldar, Oct 11 2021 *)

T(n, k) = if ((n==0) && (k==0), 0, if (k%2, 3*T(n-1,(k-1)/2) + 2^(n-1), T(n-1,k/2)));
tabf(nn) = for (n=0, nn, for (k=0, 2^n-1, print1(T(n,k), ", ")); print); \\ Michel Marcus, Oct 18 2021

T(n,k) = my(ret=0); for(i=0,n-1, if(bittest(k,n-1-i), ret=3*ret+1<Kevin Ryde, Oct 22 2021

T(n,k) = T(n-1,k/2) for k being even.
T(n,k) = 3*T(n-1,(k-1)/2) + 2^(n-1) for k being odd.
T(n,k) = 2*T(n-1,k) for 0 <= k <= 2^(n-1) - 1.
T(n,k) = Sum_{i=0..r} 2^(n-1-e[i]) * 3^i where binary expansion k = 2^e[0] + 2^e[1] + ... + 2^e[r] with ascending e[0] < e[1] < ... < e[r] (A133457). - Kevin Ryde, Oct 22 2021

A344747 a(n) = (1/6)*(3^n + (-2)^n - 1).

Original entry on oeis.org

0, 2, 3, 16, 35, 132, 343, 1136, 3195, 10012, 29183, 89256, 264355, 799892, 2386023, 7185376, 21501515, 64613772, 193622863, 581305496, 1743042675, 5230875652, 15689131703, 47074385616, 141209175835, 423655489532, 1270910544543, 3812843481736, 11438306748995

Offset: 1

Ryan Brooks, Jun 14 2021

			a(4) = (1/6)*(3^4 + (-2)^4 - 1) = (1/6)*(81+16-1) = 16.

Index entries for linear recurrences with constant coefficients, signature (2,5,-6).

Potentially related to A094554, A094555, and A094556 (which have the same recurrence).

LinearRecurrence[{2, 5, -6}, {0, 2, 3}, 30] (* Greg Dresden, Jun 19 2021 *)

G.f.: x*(2 - x)/((1 - x)*(1 + 2*x)*(1 - 3*x)). - Andrew Howroyd, Jun 15 2021

a(n) = A094554(n-1) + 2*A094556(n-1). - Greg Dresden, Jun 19 2021

A334754 The size of partitions of the decimal digits of Pi, starting directly after the decimal point, such that each partition contains the maximum number of digits possible while also avoiding any repeated digits. A digit must be in a partition if the current partition does not contain the current digit.

Original entry on oeis.org

2, 5, 2, 4, 3, 5, 3, 2, 6, 2, 5, 4, 1, 6, 7, 2, 10, 5, 5, 4, 4, 5, 2, 4, 6, 2, 6, 5, 7, 1, 5, 2, 3, 4, 3, 7, 2, 2, 1, 7, 5, 7, 1, 3, 1, 4, 3, 4, 3, 3, 6, 3, 7, 4, 2, 5, 4, 4, 4, 7, 4, 3, 5, 4, 5, 5, 5, 4, 6, 2, 5, 6, 5, 5, 2, 2, 2, 4, 2, 4, 1, 6, 4

Offset: 1

Ryan Brooks, May 10 2020

Assuming digits are random, the expected value for the size of the partitions is 3.66021568 = Sum_{k=1..10} k^2*9!/(10^k*(10-k)!).

			Pi=3.1415926535897932384626433... => ignore lead 3 and partition as such: 0.|14|15926|53|5897|932|38462|643|3... => 2,5,2,4,3,5,3,...

Sean A. Irvine, Java program (github)

Cf. A000796 (Pi). Essentially the same as A104807.

F(v)={my(L=List(), S=Set()); for(i=1, #v, if(setsearch(S, v[i]), listput(L,#S); S=Set()); S=setunion(S,[v[i]])); Vec(L)}
{ localprec(10^3); my(t=Pi-3); F(digits(floor(t*10^precision(t)))) } \\ Andrew Howroyd, Aug 10 2020

A334578 Double subfactorials: a(n) = (-1)^floor(n/2) * n!! * Sum_{i=0..floor(n/2)} (-1)^i/(n-2*i)!!.

Original entry on oeis.org

1, 1, 1, 2, 5, 11, 29, 76, 233, 685, 2329, 7534, 27949, 97943, 391285, 1469144, 6260561, 24975449, 112690097, 474533530, 2253801941, 9965204131, 49583642701, 229199695012, 1190007424825, 5729992375301, 30940193045449, 154709794133126, 866325405272573

Offset: 0

Ryan Brooks, May 06 2020

nonn

			a(5) = (5*3*1)*(1/(1) - 1/(3*1) + 1/(5*3*1)) = 15-5+1 = 11.

Alois P. Heinz, Table of n, a(n) for n = 0..807

Even bisection gives A000354.

Cf. A000166, A006882.

a:= proc(n) option remember; `if`(n<2, [0$2, 1$2][n+3],
      (n-1)*a(n-2)+(n-2)*a(n-4))
    end:
seq(a(n), n=0..32);  # Alois P. Heinz, May 06 2020

RecurrenceTable[{a[0] == 1, a[1] == 1, a[n] == n a[n-2] + (-1)^Floor[n/2]}, a, {n, 0, 32}] (* Jean-François Alcover, Nov 27 2020 *)

a(n) = n*a(n-2) + (-1)^floor(n/2).
a(2n) = A000354(n).
From Ryan Brooks, Oct 25 2020: (Start)
a(2n)/A006882(2n) ~ 1/sqrt(e) = A092605.
a(2n+1)/A006882(2n+1) ~ sqrt(Pi/(2*e))*erfi(1/sqrt(2)) = A306858. (End)

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User: Ryan Brooks

A348175 Irregular table T(n,k) read by rows: T(n,k) = T(n-1,k/2) when k is even and 3*T(n-1,(k-1)/2) + 2^(n-1) when k is odd. T(0,0) = 0 and 0 <= k <= 2^n-1.

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A344747 a(n) = (1/6)*(3^n + (-2)^n - 1).

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A334578 Double subfactorials: a(n) = (-1)^floor(n/2) * n!! * Sum_{i=0..floor(n/2)} (-1)^i/(n-2*i)!!.

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