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 A339790 First coefficient of the lindep transform of sigma(n). 3
 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 3, 1, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 3, 2, 1, 2, 2, 4, 1, 3, 1, 1, 4, 2, 1, 2, 1, 1, 5, 1, 1, 4, 3, 1, 5, 2, 1, 1, 1, 2, 3, 1, 3, 1, 1, 1, 5, 1, 1, 3, 1, 2, 3, 1, 4, 1, 1, 3, 2, 2, 1, 3, 4, 2, 5, 1, 1, 5, 4, 6, 3, 2, 4, 3, 1, 4, 7, 6 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,9 COMMENTS If b(n) is a sequence of integers, we will call the "lindep transform" of b(n) the triplet of sequences (x(n), y(n), z(n)) such as: (i) x(n) >= 1 (ii) x(n) + abs (y(n)) + abs (z(n)) is minimal (iii) x(n)*b(n) + y(n)*n + z(n) = 0 (iv) if with the conditions (i), (ii), (iii) there exist several triplets (x(n), y(n), z(n)) we then choose the one with minimal y(n). We call x(n) the first coefficient of the lindep transform of b(n), y(n) the second and z(n) the third. As this corresponds to the lindep function of PARI/GP this transform is called "lindep transform". LINKS Benoit Cloitre, a(n)/sqrt(n) every 1000 up to 6.10^6 FORMULA Conjecture: a(n) << sqrt(n) with 0 < limsup n-->infty a(n)/sqrt(n) < infty exists (see graphic). Trivially liminf a(n)/sqrt(n) = 0 since for prime n we have a(n)=1. PROG (PARI) a(n)=(lindep([sigma(n), n, 1])*sign(lindep([sigma(n), n, 1])[1]))[1] CROSSREFS Cf. A000203, A339791, A339792. Sequence in context: A300826 A334926 A305936 * A334924 A211111 A074971 Adjacent sequences:  A339787 A339788 A339789 * A339791 A339792 A339793 KEYWORD nonn AUTHOR Benoit Cloitre, Dec 17 2020 STATUS approved

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Last modified September 17 06:16 EDT 2021. Contains 347478 sequences. (Running on oeis4.)