A339379 Irregular triangle read by rows; the first row simply contains the value 1; given the succession of digits of the n-th row, say [d_0, ..., d_k], the (n+1)-th row is the succession of digits of [d_0, d_0+d_1, d_1+d_2, ..., d_{k-1}+d_k, d_k].

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 6, 4, 1, 1, 5, 1, 0, 1, 0, 5, 1, 1, 6, 6, 1, 1, 1, 5, 6, 1, 1, 7, 1, 2, 7, 2, 2, 6, 1, 1, 7, 1, 1, 8, 8, 3, 9, 9, 4, 8, 7, 2, 8, 8, 1, 1, 9, 1, 6, 1, 1, 1, 2, 1, 8, 1, 3, 1, 2, 1, 5, 9, 1, 0, 1, 6, 9, 1

Offset

0,5

Comments

This sequence combines features of Pascal's triangle (A007318) and of A093086.

Rows 0 to 4 match that of Pascal's triangle, thereafter the values differ.

Every column is eventually periodic.

Links

Alois P. Heinz, Rows n = 0..30, flattened

Example

The first rows are:
    1
    1, 1
    1, 2, 1
    1, 3, 3, 1
    1, 4, 6, 4, 1
    1, 5, 1, 0, 1, 0, 5, 1
    1, 6, 6, 1, 1, 1, 5, 6, 1
    1, 7, 1, 2, 7, 2, 2, 6, 1, 1, 7, 1
    1, 8, 8, 3, 9, 9, 4, 8, 7, 2, 8, 8, 1
    1, 9, 1, 6, 1, 1, 1, 2, 1, 8, 1, 3, 1, 2, 1, 5, 9, 1, 0, 1, 6, 9, 1

Prog

(PARI) { r = [1]; for (n=0, 9, apply (v -> print1 (v ", "), r); d = concat(apply(v -> if (v, digits(v), [0]), r)); r = concat(apply(v -> if (v, di
gits(v), [0]), vector(#d+1, k, if (k==1, d[k], k==#d+1, d[#d], d[k-1]+d[k]))))) }

Crossrefs

See A339359 for a similar sequence.

Cf. A007318, A093086.

Sequence in context: A008975 A140280 A140586 * A095143 A242312 A140279

Adjacent sequences: A339376 A339377 A339378 * A339380 A339381 A339382

Keyword

nonn,base,tabf,easy

Author

Rémy Sigrist, Dec 02 2020

Status

approved