A018903 -id:A018903 - cp's OEIS frontend

A228352 Triangle read by rows, giving antidiagonals of an array of sequences representing the number of compositions of n when there are N types of ones (the sequences in the array begin (1, N, ...)).

1, 1, 1, 1, 2, 2, 1, 3, 5, 4, 1, 4, 10, 13, 8, 1, 5, 17, 34, 34, 16, 1, 6, 26, 73, 116, 89, 32, 1, 7, 37, 136, 314, 396, 233, 64, 1, 8, 50, 229, 712, 1351, 1352, 610, 128, 1, 9, 65, 358, 1418, 3728, 5813, 4616, 1597, 256, 1, 10, 82, 529, 2564, 8781, 19520, 25012, 15760, 4181, 512

Offset: 1

Gary W. Adamson, Aug 20 2013

The array sequence beginning (1, N, ...) is such that a(n) in the sequence represents the numbers of compositions of n when there are N types of ones.

			Array sequence beginning (1, 3, 10, 34, 116, ...) is the binomial transform of (1, 2, 5, 12, 70, ...) in A073133.
First few sequences in the array:
  1, 1,  2,  4,   8,  16, ...; = A011782
  1, 2,  5, 13,  34,  89, ...; = A001519
  1, 3, 10, 34, 116, 396, ...; = A007052
... followed by A018902, A018903, A018904, the latter beginning (1, 6, ...). First few rows of the triangle:
  1;
  1,  1;
  1,  2,  2;
  1,  3,  5,   4;
  1,  4, 10,  13,    8;
  1,  5, 17,  34,   34,   16;
  1,  6, 26,  73,  116,   89,    32;
  1,  7, 37, 136,  314,  396,   233,    64;
  1,  8, 50, 229,  712, 1351,  1352,   610,   128;
  1,  9, 65, 358, 1418, 3728,  5813,  4616,  1597,  256;
  1, 10, 82, 529, 2564, 8781, 19520, 25012, 15760, 4181, 512;
  ...

Alois P. Heinz, Rows n = 1..141, flattened

Cf. A073133, A011782, A001519, A007052, A018902, A018903, A018904.

A:= proc(N, n) option remember;
      `if`(n=0, 1, N*A(N, n-1) +add(A(N, n-j), j=2..n))
    end:
seq(seq(A(d-n, n), n=0..d-1), d=1..11); # Alois P. Heinz, Aug 20 2013

A[k_, n_] := A[k, n] = If[n == 0, 1, k*A[k, n-1] + Sum[A[k, n-j], {j, 2, n}]]; Table[A[d-n, n], {d, 1, 11}, {n, 0, d-1}] // Flatten (* Jean-François Alcover, May 27 2016, after Alois P. Heinz *)

Antidiagonals of an array in which a(n+2) = (N+1)*a(n+1) - (n-1)*a(n); with array sequences beginning (1, N, ...).
Array sequence beginning (1, N, ...) is the binomial transform of the sequence in A073133 beginning (1, (N-1), ...).
Given the first sequence of the array is (1, 1, 2, 4, 8, 16, ...), successive sequences are INVERT transforms of previous sequences.
Array sequence beginning (1, N, ...) is such that a(n), n>1 is N*(a) + a(n-1) + a(n-2) + a(n-3) + a(n-4) + ... + a(0).

A337129 Triangular array read by rows: T(n,0) = 2^n, T(n,k) = Sum_{i=n-k..n, j=0..i-n+k, i<>n or j<>k} T(i,j) for k > 0.

Original entry on oeis.org

1, 2, 3, 4, 6, 16, 8, 12, 32, 84, 16, 24, 64, 168, 440, 32, 48, 128, 336, 880, 2304, 64, 96, 256, 672, 1760, 4608, 12064, 128, 192, 512, 1344, 3520, 9216, 24128, 63168, 256, 384, 1024, 2688, 7040, 18432, 48256, 126336, 330752, 512, 768, 2048, 5376, 14080, 36864, 96512, 252672, 661504, 1731840

Offset: 0

Oboifeng Dira, Sep 14 2020

			The triangle  T(n,k) begins:
   n\k  0    1    2    3    4    5
   0:   1
   1:   2    3
   2:   4    6    16
   3:   8    12   32  84
   4:   16   24   64  168  440
   5:   32   48   128 336  880  2304
   ...
T(3,2) = ((3+sqrt(5))^3-(3-sqrt(5))^3)*(2)/(4*sqrt(5)) = (64*sqrt(5))/(2*sqrt(5)) = 32.

Andrew Howroyd, Table of n, a(n) for n = 0..1325 (rows 0..50)

Cf. A000079 (1st column), A069429 (diagonal), A018903 (row sums), A001906, A004171.

T := proc (n, k) if k = 0 and 0 <= n then 2^n elif 1 <= k and k <= n then round((((3+sqrt(5))^(k+1)-(3-sqrt(5))^(k+1))*(2^(n-k))/(4*sqrt(5)))) else 0 end if end proc:seq(print(seq(T(n, k), k=0..n)), n=0..9);

T[n_, 0] := 2^n;
T[n_, n_] := 2^(n-1) Fibonacci[2n+2];
T[n_, k_] /; 0Jean-François Alcover, Nov 13 2020 *)

T(n,k) = if (k == 0, 2^n, my(w=quadgen(5, 'w)); ((2*w+2)^(k+1)-(4-2*w)^(k+1))*(2^(n-k))/(4*(2*w-1))); \\ Michel Marcus, Sep 14 2020

Row(n)={Vecrev(polcoef((1-x*y)*(1-2*x*y)/((1-6*x*y+4*x^2*y^2)*(1-2*x)) + O(x*x^n), n))} \\ Andrew Howroyd, Sep 23 2020

T(n,0) = 2^n.
T(n,k) = ((3+sqrt(5))^(k+1)-(3-sqrt(5))^(k+1))*(2^(n-k))/(4*sqrt(5)) for 1<=k<=n.
T(n+1,n) = 2*T(n,n).
T(n+m,n) = 2^m*T(n,n), for m>=1.
T(n,n) = A069429(n) = 2^(n-1)*A001906(n+1) for n>=1.
T(2*n,n) = (1/2)*A099157(n+1) = A004171(n-1)*A001906(n+1) for n>=1.
G.f.: (1 - x*y)*(1 - 2*x*y)/((1 - 6*x*y + 4*x^2*y^2)*(1 - 2*x)). - Andrew Howroyd, Sep 23 2020

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A228352 Triangle read by rows, giving antidiagonals of an array of sequences representing the number of compositions of n when there are N types of ones (the sequences in the array begin (1, N, ...)).

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