A306309 - OEIS

%I #16 Feb 09 2019 11:12:45

%S 1,1,1,1,2,1,1,3,3,1,1,4,6,4,1,1,5,1,1,5,1,1,6,6,2,6,6,1,1,7,12,8,8,

%T 12,7,1,1,8,19,2,16,2,19,8,1,1,9,27,21,18,18,21,27,9,1,1,1,36,48,39,

%U 36,39,48,36,1,1,1,2,37,84,87,75,75,87,84,37,2,1

%N The "zeroless Pascal triangle" read by rows.

%C Left and right edges are all 1's, interior entries are obtained by removing zeros from the sum of the two numbers above them.

%C For any k >= 0 and n >= 0, let d_k(n) = T(n+k, k).

%C For any k >= 0, d_k is eventually periodic: by induction:

%C - for k = 0: for any n >= 0, d_0(n) = 1, hence d_0 is 1-periodic,

%C - suppose that the property is true for some k >= 0,

%C - d_k is eventually p_k-periodic, and so d_k is bounded, say by m_k,

%C - d_{k+1}(n+1) - d_{k+1}(n) = d_k(n+1) <= m_k,

%C - so the first difference of d_{k+1} is bounded by m_k,

%C - A004719 has arbitrary large gaps,

%C   and we can choose a range of m_k+1 terms that do not belong to A004719,

%C   say x_k..x_k+m_k (with x_k > 1),

%C - d_{k+1}(0) = 1 < x_k,

%C   and if d_{k+1}(n) < x_k, then d_{k+1)(n+1) < x_k,

%C   so d_{k+1} is bounded by x_k,

%C - let D_{k+1}(n) = d_{k+1}(n*p_k},

%C - D_{k+1} is bounded,

%C   so D_{k+1}(n + q_k) = D_{k+1}(n) for some n >= 0 and q_k > 0,

%C - we can assume that n*p_k is beyond the transient part of d_k,

%C - d_{k+1}(n*p_k + q_k*p_k + 1) = d_{k+1}(n*p_k+q_k*p_k) + d_k(n*p_k+q_k*p_k + 1)

%C                                = d_{k+1}(n*p_k)         + d_k(n*p_k + 1)

%C                                = d_{k+1}(n*p_k + 1),

%C - we can generalize: for any m >= n*p_k, d_{k+1}(m + q_k*p_k) = d_{k+1)(m),

%C - and d_{k+1} is (at least q_k*p_k-)periodic, QED.

%H Rémy Sigrist, <a href="/A306309/a306309.png">Colored representation of the first 1000 rows</a> (where the hue is function of T(n, k))

%H Rémy Sigrist, <a href="/A306309/a306309.gp.txt">PARI program for A306309</a>

%F T(n, 0) = T(n, n) = 1 for n >= 0.

%F T(n, k) = A004719(T(n-1, k-1) + T(n-1, k)) for n >= 0 and k = 1..n-1.

%F T(n, 1) = A177274(n-1) for any n > 0.

%e Triangle begins:

%e                     1

%e                   1   1

%e                 1   2   1

%e               1   3   3   1

%e             1   4   6   4   1

%e           1   5   1   1   5   1

%e         1   6   6   2   6   6   1

%e       1   7  12   8   8  12   7   1

%e     1   8  19   2  16   2  19   8   1

%e   1   9  27  21  18  18  21  27   9   1

%e ...

%o (PARI) See Links section.

%Y Cf. A004719, A007318, A177274.

%K nonn,tabl,base

%O 0,5

%A _Rémy Sigrist_, Feb 06 2019