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%I #19 Dec 09 2020 01:16:44
%S 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,
%T 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,
%U 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
%N 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].
%C This sequence combines features of Pascal's triangle (A007318) and of A214365.
%C Rows 0 to 5 match that of Pascal's triangle, thereafter the values differ.
%C Every column is eventually periodic.
%e The first rows are:
%e 1
%e 1, 1
%e 1, 2, 1
%e 1, 3, 3, 1
%e 1, 4, 6, 4, 1
%e 1, 5, 10, 10, 5, 1
%e 1, 6, 6, 1, 1, 1, 5, 6, 1
%e 1, 7, 12, 7, 2, 2, 6, 11, 7, 1
%e 1, 8, 8, 3, 9, 9, 4, 8, 7, 2, 8, 8, 1
%e 1, 9, 16, 11, 12, 18, 13, 12, 15, 9, 10, 16, 9, 1
%o (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]))) }
%Y See A339379 for a similar sequence.
%Y Cf. A007318, A214365.
%K nonn,base,tabf,easy
%O 0,5
%A _Rémy Sigrist_, Dec 02 2020
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