A309495 Triangle read by rows, derived from A007318, row sums = the Bell Sequence.

1, 1, 1, 1, 2, 2, 1, 3, 5, 6, 1, 4, 9, 17, 21, 1, 5, 14, 34, 67, 82, 1, 6, 20, 58, 148, 290, 354, 1, 7, 27, 90, 275, 701, 1368, 1671, 1, 8, 35, 131, 460, 1411, 3579, 6986, 8536, 1, 9, 44, 182, 716, 2536, 7738, 19620, 38315, 46814, 1, 10, 54, 244, 1057, 4213, 14846, 45251, 114798, 224189, 273907

Offset

1,5

Comments

As described in A160185, we extract eigensequences of a rotated variant of Pascal's triangle:

1;

3, 1;

3, 2, 1;

1, 1, 1, 1;

Say, for these 4 columns, the eigensequence is (1, 4, 9, 15). Then preface the latter with a zero and take the first finite difference row, = (1, 3, 5, 6), fourth row of the triangle.

Links

Andrew Howroyd, Table of n, a(n) for n = 1..1275 (rows 1..50)

Formula

T(n,k) = A121207(n,k) - A121207(n, k-1) for k >= 2.

Example

Row 5 of A121207 is (1, 5, 14, 31, 52). Preface with a zero and take the first difference row:
     (0,  1,  5, 14, 31, 52)
  (..., 1,  4,  9, 17, 21) = row 5 of the triangle.
First few rows of the triangle:
  1;
  1, 1;
  1, 2,  2;
  1, 3,  5,  6;
  1, 4,  9, 17,  21;
  1, 5, 14, 34,  67,  82;
  1, 6, 20, 58, 148, 290, 354;
  ...

Prog

(PARI) \\ here U(n) is A121207.
U(n)={my(M=matrix(n, n)); for(n=1, n, M[n, 1]=1; for(k=1, n-1, M[n, k+1]=sum(j=1, k, M[n-j, k-j+1]*binomial(n-2, j-1)))); M}
T(n)={my(A=U(n+1)); vector(n, n, my(t=A[n+1, 2..n+1]); t-concat([0], t[1..n-1]))}
{ my(A=T(10)); for(n=1, #A, print(A[n])) } \\ Andrew Howroyd, Feb 20 2022

Crossrefs

Row sums are A000110.

Main diagonal is A032346.

Cf. A007318, A121207, A160185.

Sequence in context: A076038 A095788 A071944 * A080955 A340108 A340107

Adjacent sequences: A309492 A309493 A309494 * A309496 A309497 A309498

Keyword

nonn,tabl

Author

Gary W. Adamson, Aug 04 2019

Extensions

Terms a(37) and beyond from Andrew Howroyd, Feb 20 2022

Status

approved