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A339379
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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].
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2
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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
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OFFSET
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0,5
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COMMENTS
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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.
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LINKS
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EXAMPLE
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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
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PROG
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(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]))))) }
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CROSSREFS
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See A339359 for a similar sequence.
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KEYWORD
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nonn,base,tabf,easy
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AUTHOR
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STATUS
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approved
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