A348647 - OEIS

%I #8 Oct 27 2021 15:26:08

%S 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,

%T 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,

%U 15,1,2,14,2,1,1,1,13,1,1,1,4,12,4,1,3,1,11,1,3,1

%N 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).

%C For any n >= 0, the n-th row:

%C - is palindromic,

%C - has A038573(n+1) terms,

%C - has leading term A006519(n+1),

%C - has central term A080079(n+1).

%H Rémy Sigrist, <a href="/A348647/b348647.txt">Table of n, a(n) for n = 0..6304</a> (rows for n = 0..254, flattened)

%H <a href="/index/Pas#Pascal">Index entries for triangles and arrays related to Pascal's triangle</a>

%F Sum_{k = 1..A038573(n+1)} T(n, k) = n+1.

%F T(n, 1) = A006519(n+1).

%F T(n, A048896(n)) = A080079(n+1).

%F T(2^k-1, 1) = 2^k for any k >= 0.

%e Triangle begins:

%e     1;

%e     2;

%e     1, 1, 1;

%e     4;

%e     1, 3, 1;

%e     2, 2, 2;

%e     1, 1, 1, 1, 1, 1, 1;

%e     8;

%e     1, 7, 1;

%e     2, 6, 2;

%e     1, 1, 1, 5, 1, 1, 1;

%e     4, 4, 4;

%e     1, 3, 1, 3, 1, 3, 1;

%e     2, 2, 2, 2, 2, 2, 2;

%e     1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1;

%e     16;

%e     ...

%o (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) }

%Y Cf. A006519, A007318, A038573, A047999, A048896, A080079.

%K nonn,look,tabf

%O 0,2

%A _Rémy Sigrist_, Oct 27 2021