This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A326793 #10 Jul 29 2019 12:00:28 %S A326793 1,1,2,5,12,30,75,188,472,1186,2981,7494,18842,47376,119126,299545, %T A326793 753220,1894018,4762640,11976010,30114592,75725485,190417684, %U A326793 478820320,1204031670,3027633300,7613224740,19144059492,48139261637,121050006438 %N A326793 The number of small Schröder paths such that the area between the path and the x-axis contains n up-triangles. %C A326793 We define two types of plane triangles - up-triangles with vertices at the integer lattice points (x, y), (x+1, y+1) and (x+2, y) and down-triangles with vertices at the integer lattice points (x, y), (x-1, y+1) and (x+1, y+1). The area beneath a small Schröder path may be decomposed in a unique manner into a collection of up- and down-triangles. This decomposition produces a triangle stack in the sense of A224704. Here we are counting triangle stacks containing n up-triangles. See the Links section for an illustration. %H A326793 P. Bala, <a href="/A326793/a326793.pdf">Illustration for a(3) = 5</a> %F A326793 O.g.f. as a continued fraction: (u marks up-triangles) %F A326793 A(u) = 1/(1 - u/(1 - u - u^2/(1 - u^2 - u^3/(1 - u^3 - u^4/(1 - u^4 - (...) ))))) = 1 + u + 2*u^2 + 5*u^3 + 12*u^4 + .... %F A326793 A(u) = 1/(1 - u/(1 - (u + u^2)/(1 - u^3/(1 - (u^2 + u^4)/(1 - u^5/(1 - (u^3 + u^6)/(1 - u^7/( (...) )))))))). %F A326793 A(u) = 1/(2 - (1 + u)/(2 - (1 + u^2)/(2 - (1 + u^3)/(2 - (...) )))). %F A326793 A(u) = N(u)/D(u), where N(u) = Sum_{n >= 0} u^(n^2+n)/ Product_{k = 1..n} ((1 - u^k)^2) and D(u) = Sum_{n >= 0} u^(n^2)/ Product_{k = 1..n} ((1 - u^k)^2). %F A326793 a(n) ~ c*d^n, where c = 0.29475 98606 22204 98206 41002 ..., d = 2.51457 96438 78729 18851 04371 .... %F A326793 Row sums of A326792. %Y A326793 Cf. A224704, A326792. %K A326793 nonn,easy %O A326793 0,3 %A A326793 _Peter Bala_, Jul 25 2019