A260003 -id:A260003 - cp's OEIS frontend

A260006 a(n) = f(1,n,n), where f is the Sudan function defined in A260002.

0, 3, 12, 35, 90, 217, 504, 1143, 2550, 5621, 12276, 26611, 57330, 122865, 262128, 557039, 1179630, 2490349, 5242860, 11010027, 23068650, 48234473, 100663272, 209715175, 436207590, 905969637, 1879048164, 3892314083, 8053063650, 16642998241, 34359738336

Offset: 0

Natan Arie Consigli, Jul 23 2015

f(1,n,n) = 2^n*(n+2) - (n+2) = (2^n - 1)*(n+2).
To evaluate the Sudan function see A260002 and A260003.
The numbers are alternately even and odd because for even n (2^n-1)*(n+2) is even and (2^(n+1)-1)*(n+1+2) is odd.
From Enrique Navarrete, Oct 02 2021: (Start)
a(n-2) is the number of ways we can write [n] as the union of 2 sets of sizes i, j which intersect in exactly 1 element (1 < i, j < n;  i = j allowed), n>=2.
For n = 4, a(n-2) = 12 since [4] can be written as the unions:
{1,2} U {1,3,4}; {2,3} U {1,2,4}; {1,2} U {2,3,4}; {2,3} U {1,3,4};
{1,3} U {1,2,4}; {2,4} U {1,2,3}; {1,3} U {2,3,4}; {2,4} U {1,3,4};
{1,4} U {1,2,3}; {3,4} U {1,2,3}; {1,4} U {2,3,4}; {3,4} U {1,2,4}. (End)

			a(4) = (2^4 - 1)*(4 + 2) = 90.

Colin Barker, Table of n, a(n) for n = 0..1000
Wikipedia, Sudan function (see diagonal of "Values of F1(x, y)" table).
Index entries for linear recurrences with constant coefficients, signature (6,-13,12,-4).

Cf. A000295 (f(1,0,n)), A000325 (f(1,2,n)), A005408 (f(1,n,1) = 2n+1), A001787 (n*2^(n-1)), A079583 (f(1,1,n)), A123720 (f(1,4,n)), A133124 (f(1,3,n)).
Cf. A260002, A260003, A260004, A260005.
Cf. A000984, A002697, A303602.

[(2^n-1)*(n+2): n in [0..30]]; // Vincenzo Librandi, Aug 22 2015

Table[(2^n -1)(n+2), {n, 0, 30}] (* Michael De Vlieger, Aug 22 2015 *)
CoefficientList[Series[x(3 -6x +2x^2)/((1-x)^2 (1-2x)^2), {x, 0, 40}], x] (* Vincenzo Librandi, Aug 22 2015 *)
LinearRecurrence[{6,-13,12,-4},{0,3,12,35},40] (* Harvey P. Dale, Mar 04 2023 *)

vector(40, n, n--; (2^n-1)*(n+2)) \\ Michel Marcus, Jul 29 2015

concat(0, Vec(x*(3-6*x+2*x^2)/((1-x)^2*(1-2*x)^2) + O(x^40))) \\ Colin Barker, Jul 29 2015

[(n+2)*(2^n -1) for n in (0..30)] # G. C. Greubel, Dec 30 2021

a(n) = (2^n -1)*(n+2).
a(n) = 6*a(n-1) - 13*a(n-2) + 12*a(n-3) - 4*a(n-4) for n>3. - Colin Barker, Jul 29 2015
G.f.: x*(3 - 6*x + 2*x^2) / ((1-x)^2*(1-2*x)^2). - Colin Barker, Jul 29 2015
E.g.f.: 2*(x+1)*exp(2*x) - (x+2)*exp(x). - Robert Israel, Aug 23 2015
From Enrique Navarrete, Oct 02 2021: (Start)
a(n-2) = Sum_{j=2..n/2} binomial(n,j)*j, n even > 2.
a(n-2) = (Sum_{j=2..floor(n/2)} binomial(n,j)*j) + (1/2)*binomial(n, ceiling(n/2))*ceiling(n/2), n odd > 1. (End)
From Wolfdieter Lang, Nov 12 2021: (Start)
The previous bisection becomes for a(n):
a(2*k) = 2*(A002697(k+1) - (k+1)), and a(2*k+1) = A303602(k+1) - (2*k+3)*(2 - A000984(k+1))/2 = (2*k+3)*(4^(k+1) - 2)/2, for k >= 0. (End)

A260002 Sudan Numbers: a(n)= f(n,n,n) where f is the Sudan function.

Original entry on oeis.org

0, 3, 15569256417

Offset: 0

Natan Arie Consigli, Jul 12 2015

The Sudan function is the first discovered not primitive recursive function that is still totally recursive like the well-known three-argument (or two-argument) Ackermann function ack(a,b,c) (or ack(a,b)).
The Sudan function is defined as follows:
f(0,x,y) = x+y;
f(z,x,0) = x;
f(z,x,y) = f(z-1, f(z,x,y-1), f(z,x,y-1)+y).
Just as the three-argument (or two-argument) Ackermann numbers A189896 (or A046859) are defined to be the numbers that are the answer of ack(n,n,n) (or ack(n,n)) for some natural number n, the Sudan numbers are: a(n) = f(n,n,n).
a(3)> 2^(76*2^(76*2^(76*2^(76*2^76)))) so is too big to be included.

			a(1) = f(1,1,1) = f(0, f(1,1,0), f(1,1,0)+1) = f(0, 1, 2) = 1+2 = 3.

Wikipedia, Sudan Function, Primitive Recursive, Ackermann function.

Cf. A189896, A046859, A260003, A260004, A260005, A260006.

f[z_, x_, y_] := f[z, x, y] =
Piecewise[{{x + y, z == 0}, {x,
    z > 0 && y == 0}, {f[z - 1, f[z, x, y - 1], f[z, x, y - 1] + y],
    z > 0 && y > 0} }];
a[n_] := f[n,n,n]

f(z,x,y)=if(z,if(y,my(t=f(z,x,y-1)); f(z-1, t, t+y),x),x+y)
a(n)=f(n,n,n) \\ Charles R Greathouse IV, Jul 28 2015

A260005 a(n) = f(2,n,2), where f is the Sudan function defined in A260002.

Original entry on oeis.org

19, 10228, 15569256417, 5742397643169488579854258, 36681813266165915713665394441869800619098139628586701684547

Offset: 0

Natan Arie Consigli, Jul 23 2015

nonn

Naturally a subsequence of A260004.
See A260002-A260003 for the evaluation of the Sudan function.
Using f(2,n,2) = f(1, f(2,n,1), f(2,n,1)+2) = 2^(f(2,n,1)+2)*(f(2,n,1)+2)-f(2,n,1)-4 and f(2,n,1) = f(1, n, n+1) = 2^(n+1)*(n+2)-(n+3) we have:
a(n)=f(2,n,2)
    =f(1, 2^(n+1)*(n+2)-(n+3), 2^(n+1)*(n+2)-(n+3)+2)
    =2^(2^(n+1)*(n+2)-(n+3)+2)*(2^(n+1)*(n+2)-(n+3)+2)-2^(n+1)*(n+2)+(n+3)-4
    =2^(2^(n+1)*(n+2)-(n+1))*(2^(n+1)*(n+2)-(n+1))-2^(n+1)*(n+2)+(n-1).

			a(1) = f(2,1,2) = f(1,f(2,1,1),f(2,1,1)+2) = f(1,8,10) = 2^10*(8+2)-10-2 = 10228.

Natan Arie' Consigli, Table of n, a(n) for n = 0..6
Wikipedia, Sudan function (see 3rd line of "Values of F2(x, y)" table).

Cf. A048493 (f(2,n,1)), A260002, A260003, A260004, A260006.

[2^(2^(n+1)*(n+2)-(n+1))*(2^(n+1)*(n+2)-(n+1))-2^(n+1)*(n+2)+(n-1):n in [0..5]]; // Vincenzo Librandi, Jul 27 2015

Table[2^(2^(n + 1) (n + 2) - (n + 1)) (2^(n + 1) (n + 2) - (n + 1)) - 2^(n + 1) (n + 2) + (n - 1), {n, 0, 5}] (* Vincenzo Librandi, Jul 27 2015 *)

a(n) = 2^(2^(n+1)*(n+2)-(n+1))*(2^(n+1)*(n+2)-(n+1))-2^(n+1)*(n+2)+(n-1);
vector(10, n, a(n-1)) \\ Altug Alkan, Oct 01 2015

a(n) = 2^(2^(n+1)*(n+2)-(n+1))*(2^(n+1)*(n+2)-(n+1))-2^(n+1)*(n+2)+(n-1).

A260004 Values of f(2,x,y) in increasing order, for x>=0, y>0 where f is the Sudan function defined in A260002.

Original entry on oeis.org

1, 8, 19, 27, 74, 185, 440, 1015, 2294, 5109, 10228, 11252, 24563, 53234, 114673, 245744, 524271, 1114094, 2359277, 4980716, 10485739, 22020074, 46137321, 88080360, 96468968, 201326567, 419430374, 872415205, 1811939300, 15569256417

Offset: 1

Natan Arie Consigli, Jul 23 2015

nonn

This is a subsequence of A260003 since f(2,x,y)= f(1, f(2,x,y-1), f(2,x,y-1)+y).
In this list we suppose that y>0. If we take y=0, all the natural numbers would be in the sequence.
To evaluate the Sudan function see A260002-A260003.

			f(2,0,1) = f(1, f(2,0,0), f(2,0,0)+1) = f(1,0,1) = 2^1(0+2)-1-2 = 1;
f(2,0,2) = f(1, [f(2,0,1)], [f(2,0,1)+1+1]) = f(1,1,3) = 2^3(1+2)-3-2 = 19;

Cf. A048493 (f(2,n,1)), A260002, A260003, A260005 (f(2,n,2)), A260006.

cp's OEIS Frontend

A260006 a(n) = f(1,n,n), where f is the Sudan function defined in A260002.

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A260002 Sudan Numbers: a(n)= f(n,n,n) where f is the Sudan function.

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A260005 a(n) = f(2,n,2), where f is the Sudan function defined in A260002.

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A260004 Values of f(2,x,y) in increasing order, for x>=0, y>0 where f is the Sudan function defined in A260002.

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