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%I #15 Nov 26 2020 23:21:39
%S 1,1,1,1,1,1,2,3,4,5,1,2,3,4,5,1,2,3,4,5,1,2,3,4,5,1,2,3,4,5,2,4,6,8,
%T 10,3,6,9,12,15,4,8,12,16,20,5,10,15,20,25,1,2,3,4,5,2,4,6,8,10,3,6,9,
%U 12,15,4,8,12,16,20,5,10,15,20,25,1,2,3,4,5,2,4,6,8,10,3,6,9,12,15,4,8,12,16,20,5,10,15,20,25
%N a(n) = a(floor(n / 5)) * (n mod 5 + 1); initial terms are 1.
%C If a(n) is arranged in a table with row lengths 5, then the first column is the transpose of the first row, followed the transpose of the second row, followed by the transpose of the third row, and so on. The remainder of each row (except the first) is an arithmetic progression whose start and step size equals the first entry of the row.
%C a(n) = O(n).
%C limsup_n a(n) = +oo.
%e a(10) = a(2) * 1 = 1.
%e a(13) = a(2) * 4 = 4.
%o (Python)
%o def a(n):
%o if n < 5:
%o return 1
%o q, r = divmod(n, 5)
%o return a(q) * (r + 1)
%o (PARI) a(n) = if (n < 5, 1, a(n\5)*(n % 5 + 1)); \\ _Michel Marcus_, Nov 26 2020
%Y Cf. A194459.
%Y Cf. A048896 (with 2 instead of 5, but shifted).
%K nonn
%O 0,7
%A _Robert Dougherty-Bliss_, Nov 25 2020
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