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
Examples
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
Links
- Paolo Xausa, Table of n, a(n) for n = 0..15582 (rows 0..30 of triangle, flattend).
Programs
-
Mathematica
NestList[Map[Total, Partition[Flatten[IntegerDigits[#]], 2, 1, {-1, 1}, 0]] &, {1}, 10] (* Paolo Xausa, Aug 19 2025 *) -
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]))) }
Comments