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A348647
Irregular table read by rows; the n-th row contains the lengths of the runs of consecutive terms with the same parity in the n-th row of Pascal's triangle (A007318).
1
1, 2, 1, 1, 1, 4, 1, 3, 1, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 8, 1, 7, 1, 2, 6, 2, 1, 1, 1, 5, 1, 1, 1, 4, 4, 4, 1, 3, 1, 3, 1, 3, 1, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 16, 1, 15, 1, 2, 14, 2, 1, 1, 1, 13, 1, 1, 1, 4, 12, 4, 1, 3, 1, 11, 1, 3, 1
OFFSET
0,2
COMMENTS
For any n >= 0, the n-th row:
- is palindromic,
- has A038573(n+1) terms,
- has leading term A006519(n+1),
- has central term A080079(n+1).
FORMULA
Sum_{k = 1..A038573(n+1)} T(n, k) = n+1.
T(n, 1) = A006519(n+1).
T(n, A048896(n)) = A080079(n+1).
T(2^k-1, 1) = 2^k for any k >= 0.
EXAMPLE
Triangle begins:
1;
2;
1, 1, 1;
4;
1, 3, 1;
2, 2, 2;
1, 1, 1, 1, 1, 1, 1;
8;
1, 7, 1;
2, 6, 2;
1, 1, 1, 5, 1, 1, 1;
4, 4, 4;
1, 3, 1, 3, 1, 3, 1;
2, 2, 2, 2, 2, 2, 2;
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1;
16;
...
PROG
(PARI) row(n) = { my (b=binomial(n)%2, r=[], p=1, w=1); for (k=2, #b, if (p==b[k], w++, r=concat(r, w); p=b[k]; w=1)); concat(r, w) }
CROSSREFS
KEYWORD
nonn,look,tabf
AUTHOR
Rémy Sigrist, Oct 27 2021
STATUS
approved