|
Search: id:A075631
|
|
|
| A075631 |
|
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. |
|
+0
6
|
|
| 1, 2, 12, 156, 4680, 42120, 43931160, 1489793497920, 89075729055377937600, 51745303414377493336016875691726400, 11120662689162548692119420650118842670724939176955384978536000
(list; graph; listen)
|
|
|
OFFSET
|
1,2
|
|
|
COMMENT
|
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, i=6 to 18] = 156. - Jonathan Cross (jcross(AT)juggler.net), May 08 2004
|
|
CROSSREFS
|
Cf. A075632, A075633, A075634, A079798.
Sequence in context: A053549 A139383 A130529 this_sequence A127182 A012646 A012328
Adjacent sequences: A075628 A075629 A075630 this_sequence A075632 A075633 A075634
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
Amarnath Murthy (amarnath_murthy(AT)yahoo.com), Sep 30 2002
|
|
EXTENSIONS
|
More terms from Jonathan Cross (jcross(AT)juggler.net), May 08 2004
|
|
|
Search completed in 0.002 seconds
|
