A338431 - OEIS

%I #6 Jan 22 2021 22:45:54

%S 1,1,1,3,3,4,6,10,8,13,15,15,21,26,21,36,36,33,45,49,42,64,66,58,72,

%T 89,71,99,105,80,120,136,105,151,137,129,171,188,147,190,210,165,231,

%U 247,184,274,276,228,288,295

%N Row length of irregular triangle A337939.

%F a(1) = 1, and for n >= 2, a(n) = Sum_{k=1..floor(n/2)} k = A000217(floor(n/2)) if b(n) := floor(n/2) - delta(n) = A219839(n) = 0, where delta(n) = A055034(n), and if b(n) > 0, i.e., n = n(j) = A111774(j), for j >= 1, then a(n) <  A000217(floor(n/2)), determined by a(n) = A000217(delta(n)) + R(n), with R(n) = Sum_{k = delta(n)+1..floor(n/2)} (1 + degree(S(k-1, x) evaluated with C(n, x) = 0)), where the C polynomial coefficients are given in A187360.

%e n = 12: b(12) = 6 - 4 = 2 =  A219839(12) > 0, hence A000217(4) = 10,  R(4) = (1 + 2) + (1 + 1) = 5 from degree(S(4, x)/C(12,x) = 1*x^2) = 2 and degree(S(5, x)/C(12, x) = 2*x) = 1. Hence a(12) = 10 + 5 = 15.

%Y Cf. A000217, A055034, A219839, A337939.

%K nonn,easy

%O 1,4

%A _Wolfdieter Lang_, Jan 15 2021