{"id":2134,"date":"2016-02-18T06:00:08","date_gmt":"2016-02-18T04:00:08","guid":{"rendered":"http:\/\/www.glc.us.es\/~jalonso\/exercitium\/?p=2134"},"modified":"2016-05-02T13:09:03","modified_gmt":"2016-05-02T11:09:03","slug":"cantidad-de-numeros-pentanacci-impares","status":"publish","type":"post","link":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/cantidad-de-numeros-pentanacci-impares\/","title":{"rendered":"Cantidad de n\u00fameros Pentanacci impares"},"content":{"rendered":"<p>Los n\u00fameros de Pentanacci se definen mediante las ecuaciones<\/p>\n<pre lang=\"text\">\n   P(0) = 0\n   P(1) = 1\n   P(2) = 1\n   P(3) = 2\n   P(4) = 4\n   P(n) = P(n-1) + P(n-2) + P(n-3) + P(n-4) + P(n-5), si n > 4\n<\/pre>\n<p>Los primeros n\u00fameros de Pentanacci son<\/p>\n<pre lang=\"text\">\n  0, 1, 1, 2, 4, 8, 16, 31, 61, 120, 236, 464, 912, 1793, 3525, ...\n<\/pre>\n<p>Se obseeva que<\/p>\n<ul>\n<li>hasta P(5)  hay 1 impar: el 1 (aunque aparece dos veces); <\/li>\n<li>hasta P(7)  hay 2 impares distintos: 1 y 31;<\/li>\n<li>hasta P(10) hay 3 impares distintos: 1, 31 y 61;<\/li>\n<li>hasta P(15) hay 5 impares distintos: 1, 31 y 61, 1793 y 3525.<\/li>\n<\/ul>\n<p>Definir la funci\u00f3n<\/p>\n<pre lang=\"text\">\n   nPentanacciImpares :: Integer -> Integer\n<\/pre>\n<p>tal que (nPentanacciImpares n) es la cantidad de n\u00fameros impares distintos desde P(0) hasta P(n). Por ejemplo,<\/p>\n<pre lang=\"text\">\n   nPentanacciImpares 5                             ==  1\n   nPentanacciImpares 7                             ==  2\n   nPentanacciImpares 10                            ==  3\n   nPentanacciImpares 15                            ==  5\n   nPentanacciImpares 100                           ==  33\n   nPentanacciImpares 1000                          ==  333\n   nPentanacciImpares 10000                         ==  3333\n   nPentanacciImpares (10^(10^6)) `mod` (10^9)      ==  333333333\n   length (show (nPentanacciImpares2 (10^(10^6))))  ==  1000000\n<\/pre>\n<h4>Soluciones<\/h4>\n<pre lang=\"haskell\">\nimport Data.List (genericLength, genericTake)\n\n-- 1\u00aa definici\u00f3n \n-- =============\n\nnPentanacciImpares1 :: Integer -> Integer\nnPentanacciImpares1 0 = 0\nnPentanacciImpares1 1 = 1\nnPentanacciImpares1 n = \n    genericLength (filter odd (genericTake (n+1) pentanacci)) - 1\n\npentanacci :: [Integer]\npentanacci = p (0, 1, 1, 2, 4)\n    where p (a, b, c, d, e) = a : p (b, c, d, e, a + b + c + d + e)\n\n-- 2\u00aa definici\u00f3n\n-- =============\n\n-- \u03bb> map nPentanacciImpares1 [1..31]\n-- [1,1,1,1,1,1,2,3,3,3,3,3,4,5,5,5,5,5,6,7,7,7,7,7,8,9,9,9,9,9,10]\n-- \u03bb> [(head xs, length xs) | xs <- group (map nPentanacciImpares1 [1..37])]\n-- [(1,6),(2,1),(3,5),(4,1),(5,5),(6,1),(7,5),(8,1),(9,5),(10,1),(11,5),(12,1)]\n\nnPentanacciImpares2 :: Integer -> Integer\nnPentanacciImpares2 0 = 0\nnPentanacciImpares2 1 = 1\nnPentanacciImpares2 n = 2 * q + min r 2 - 1\n    where (q,r) = n `quotRem` 6\n\n-- 3\u00aa definici\u00f3n\n-- =============\n\nnPentanacciImpares3 :: Integer -> Integer\nnPentanacciImpares3 0 = 0\nnPentanacciImpares3 1 = 1\nnPentanacciImpares3 n | r == 5    = 2*q + 2 \n                      | otherwise = 2*q + 1 \n    where (q,r) = divMod (n-2) 6\n<\/pre>\n","protected":false},"excerpt":{"rendered":"<p>Los n\u00fameros de Pentanacci se definen mediante las ecuaciones P(0) = 0 P(1) = 1 P(2) = 1 P(3) = 2 P(4) = 4 P(n) = P(n-1) + P(n-2) + P(n-3) + P(n-4) + P(n-5), si n > 4 Los primeros n\u00fameros de Pentanacci son 0, 1, 1, 2, 4, 8, 16, 31, 61, 120,&#8230;<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"jetpack_post_was_ever_published":false,"_kad_post_transparent":"","_kad_post_title":"","_kad_post_layout":"","_kad_post_sidebar_id":"","_kad_post_content_style":"","_kad_post_vertical_padding":"","_kad_post_feature":"","_kad_post_feature_position":"","_kad_post_header":false,"_kad_post_footer":false,"_jetpack_newsletter_access":"","_jetpack_dont_email_post_to_subs":false,"_jetpack_newsletter_tier_id":0,"_jetpack_memberships_contains_paywalled_content":false,"footnotes":"","_jetpack_memberships_contains_paid_content":false},"categories":[4],"tags":[286,328,38,258,306,84,11,254,6],"jetpack_featured_media_url":"","jetpack_sharing_enabled":true,"_links":{"self":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts\/2134"}],"collection":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/comments?post=2134"}],"version-history":[{"count":3,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts\/2134\/revisions"}],"predecessor-version":[{"id":2397,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts\/2134\/revisions\/2397"}],"wp:attachment":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/media?parent=2134"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/categories?post=2134"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/tags?post=2134"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}