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Group the natural numbers so that the n-th group contains smallest set of consecutive numbers whose sum is a multiple of the sum for the previous group: (1), (2), (3, 4, 5), (6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18), (19, 20, ... Sequence gives sums of terms of the groups.
5

%I #10 Jan 28 2022 01:29:33

%S 1,2,12,156,4680,42120,43931160,1489793497920,89075729055377937600,

%T 51745303414377493336016875691726400,

%U 11120662689162548692119420650118842670724939176955384978536000

%N Group the natural numbers so that the n-th group contains smallest set of consecutive numbers whose sum is a multiple of the sum for the previous group: (1), (2), (3, 4, 5), (6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18), (19, 20, ... Sequence gives sums of terms of the groups.

%C Let n be the last term of a group with sum s. The end of the next group is computed by solving x^2 + x - n*(n+1) == 0 (mod 2*s) and picking the smallest solution x > n. For example to find the term after 12 (the (3,4,5) group), solve x^2 + x - 5*6 == 0 (mod 2*12). The solutions are x = {2, 5, 18, 24}; the end of the next group must be 18 and so the next term is Sum_{i=6..18} i = 156. - Jonathan Cross (jcross(AT)juggler.net), May 08 2004

%Y Cf. A075632, A075633, A075634, A079798.

%K nonn

%O 1,2

%A _Amarnath Murthy_, Sep 30 2002

%E More terms from Jonathan Cross (jcross(AT)juggler.net), May 08 2004