A339359 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 [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, 10, 10, 5, 1, 1, 6, 6, 1, 1, 1, 5, 6, 1, 1, 7, 12, 7, 2, 2, 6, 11, 7, 1, 1, 8, 8, 3, 9, 9, 4, 8, 7, 2, 8, 8, 1, 1, 9, 16, 11, 12, 18, 13, 12, 15, 9, 10, 16, 9, 1, 1, 10, 10, 7, 7, 2, 2, 3, 3, 9, 9, 4, 4, 3, 3, 6, 14, 10, 1, 1, 7, 15, 10, 1

Offset

0,5

Comments

This sequence combines features of Pascal's triangle (A007318) and of A214365.

Rows 0 to 5 match that of Pascal's triangle, thereafter the values differ.

Every column is eventually periodic.

Links

Table of n, a(n) for n=0..90.

Example

The first rows are:
    1
    1, 1
    1, 2, 1
    1, 3, 3, 1
    1, 4, 6, 4, 1
    1, 5, 10, 10, 5, 1
    1, 6, 6, 1, 1, 1, 5, 6, 1
    1, 7, 12, 7, 2, 2, 6, 11, 7, 1
    1, 8, 8, 3, 9, 9, 4, 8, 7, 2, 8, 8, 1
    1, 9, 16, 11, 12, 18, 13, 12, 15, 9, 10, 16, 9, 1

Prog

(PARI) { r=[1]; for (n=0, 10, apply (v -> print1(v", "), r); d=concat(apply(digits, r)); r=vector(#d+1, k, if (k==1, d[1], k==#d+1, d[#d], d[k-1]+d[k]))) }

Crossrefs

See A339379 for a similar sequence.

Cf. A007318, A214365.

Sequence in context: A275198 A095145 A095144 * A144398 A034932 A374378

Adjacent sequences: A339356 A339357 A339358 * A339360 A339361 A339362

Keyword

nonn,base,tabf,easy

Author

Rémy Sigrist, Dec 02 2020

Status

approved