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Branden, arXiv:1410.6601: sum_k f(k) x^k / k! = f(theta) e^x = C e^x x^m prod (1 + x/a_j), a_j > 0); f(k) has one sign for k >= 0, so coefficients of one sign stay of one sign; applied to f(n, .) for each n"},"verdict":"proved"},"notes":["the division remainder is removed by row annihilators of the shifted parent"],"parent_proof":{"attempts":[{"detail":"c and the seeds are nonzero, real-rooted and of one sign, so P_(n0 + k m + i) = c^m P_(n0 + i) is too","route":"periodic_product","success":true,"variant":"P_(n+k) = c(x) P_n"}],"conditional_on":[],"route":"periodic_product","rows_checked":10,"statement":"every parent row P_n, n >= 0, is zero or real-rooted with coefficients of one sign","status":"proved"},"rows_compared":16,"schema":"polytool.recurrence-preserver-image.v1","statement":"proved: Q_n = lambda(n) T_n(P_(n+0)) with T = f(n, theta) = 1 + (2) theta + (1) theta^2 for every n >= 0, the side conditions hold for every n, and every parent row P_n, n >= 0, is zero or real-rooted with coefficients of one sign; so every Q_n, n >= 0, is zero or real-rooted with roots in (-inf, 0]"},"kind":"preserver_image"}