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)
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
KEYWORD
nonn,tabl
AUTHOR
Gary W. Adamson, Aug 04 2019
EXTENSIONS
Terms a(37) and beyond from Andrew Howroyd, Feb 20 2022
STATUS
approved