A371763 - OEIS

%I #16 Apr 19 2024 16:52:15

%S 0,1,1,5,6,4,13,18,15,9,29,42,39,28,16,61,90,87,68,45,25,125,186,183,

%T 148,105,66,36,253,378,375,308,225,150,91,49,509,762,759,628,465,318,

%U 203,120,64,1021,1530,1527,1268,945,654,427,264,153,81

%N Triangle read by rows: Trace of the Akiyama-Tanigawa algorithm for powers x^2.

%C The Akiyama-Tanigawa is a sequence-to-sequence transformation AT := A -> B. If A(n) = 1/(n + 1) then B(n) are the Bernoulli numbers. Tracing the algorithm generates a triangle where the right edge is sequence A and the left edge is its transform B.

%C Here we consider the sequence A(n) = n^2 that is transformed into sequence B(n) = |A344920(n)|. The case A(n) = n^3 is A371764. Sequence [1, 1, 1, ...] generates A023531 and sequence [0, 1, 2, 3, ...] generates A193738.

%C In their general form, the AT-transforms of the powers are closely related to the poly-Bernoulli numbers A099594 and generate the rows of the array A371761.

%F T(n, k) = n^2 if n=k, otherwise (k + 1)*(2^(n - k)*(k + 2) - 3). - _Detlef Meya_, Apr 19 2024

%e Triangle starts:

%e 0:                  0

%e 1:               1,   1

%e 2:             5,   6,   4

%e 3:          13,  18,  15,   9

%e 4:        29,  42,  39,  28,  16

%e 5:      61,  90,  87,  68,  45,  25

%e 6:    125, 186, 183, 148, 105,  66, 36

%e 7:  253, 378, 375, 308, 225, 150, 91, 49

%p ATProw := proc(k, n) local m, j, A;

%p    for m from 0 by 1 to n do

%p       A[m] := m^k;

%p       for j from m by -1 to 1 do

%p          A[j - 1] := j * (A[j] - A[j - 1])

%p    od od; convert(A, list) end:

%p ATPtriangle := (p, len) -> local k;

%p      ListTools:-Flatten([seq(ATProw(p, k), k = 0..len)]):

%p ATPtriangle(2, 9);

%t T[n,k] := If[n==k, n^2, (k+1)*(2^(n-k)*(k+2)-3)]; Flatten[Table[T[n,k],{n,0,9},{k,0,n}]] (* _Detlef Meya_, Apr 19 2024 *)

%o (Python)

%o # See function ATPowList in A371761.

%o (Julia)

%o function ATPtriangle(k::Int, len::Int)

%o     A = Vector{BigInt}(undef, len)

%o     B = Vector{Vector{BigInt}}(undef, len)

%o     for n in 0:len-1

%o         A[n+1] = n^k

%o         for j = n:-1:1

%o             A[j] = j * (A[j+1] - A[j])

%o         end

%o         B[n+1] = A[1:n+1]

%o     end

%o     return B

%o end

%o for (n, row) in enumerate(ATPtriangle(2, 9))

%o     println("$(n-1): ", row)

%o end

%Y Family of triangles: A023531 (n=0), A193738 (n=1), this triangle (n=2), A371764 (n=3).

%Y Cf. A371761, A344920, A099594.

%K nonn,tabl

%O 0,4

%A _Peter Luschny_, Apr 15 2024