{"group":{"id":1,"name":"Community","lockable":false,"created_at":"2012-01-18T18:02:15.000Z","updated_at":"2025-12-14T01:33:56.000Z","description":"Problems submitted by members of the MATLAB Central community.","is_default":true,"created_by":161519,"badge_id":null,"featured":false,"trending":false,"solution_count_in_trending_period":0,"trending_last_calculated":"2025-12-14T00:00:00.000Z","image_id":null,"published":true,"community_created":false,"status_id":2,"is_default_group_for_player":false,"deleted_by":null,"deleted_at":null,"restored_by":null,"restored_at":null,"description_opc":null,"description_html":null,"published_at":null},"problems":[{"id":44836,"title":"Iccanobif numbers 1","description":"There are a lot of problems in Cody that deal with Fibonacci numbers (1, 1, 2, 3, 5, 8, 13, 21 etc...) so let's turn things around a bit.  Literally.  This problem deals with the Iccanobif sequence.  Instead of adding the previous two numbers in the sequence, reverse the digits of the previous two numbers and add those together to get the next number in the sequence.\r\n\r\nIt starts out exactly the same as the Fibonacci (1, 1, 2, 3, 5, 8, 13) as all of the one digit numbers equal themselves when reversed.  However, instead of 8+13=21 for the next term, reverse the digits in 13 to get 31. The next term is now 8+31, or 39.  The next term after that is 31+93 (13 and 39 reversed) to get 124, and so on.  Unlike the starndard Fibonacci sequence, the Iccanobif sequence can actually decrease in value between terms.\r\n\r\nYou will be given a number, and you will be asked to calculate that term in the Iccanobif sequence.\r\n\r\n!Kcul doog","description_html":"\u003cp\u003eThere are a lot of problems in Cody that deal with Fibonacci numbers (1, 1, 2, 3, 5, 8, 13, 21 etc...) so let's turn things around a bit.  Literally.  This problem deals with the Iccanobif sequence.  Instead of adding the previous two numbers in the sequence, reverse the digits of the previous two numbers and add those together to get the next number in the sequence.\u003c/p\u003e\u003cp\u003eIt starts out exactly the same as the Fibonacci (1, 1, 2, 3, 5, 8, 13) as all of the one digit numbers equal themselves when reversed.  However, instead of 8+13=21 for the next term, reverse the digits in 13 to get 31. The next term is now 8+31, or 39.  The next term after that is 31+93 (13 and 39 reversed) to get 124, and so on.  Unlike the starndard Fibonacci sequence, the Iccanobif sequence can actually decrease in value between terms.\u003c/p\u003e\u003cp\u003eYou will be given a number, and you will be asked to calculate that term in the Iccanobif sequence.\u003c/p\u003e\u003cp\u003e!Kcul doog\u003c/p\u003e","function_template":"function y = iccanobif(x)\r\n  y = x;\r\nend","test_suite":"%%\r\nx = 1;y_correct = 1;\r\nassert(isequal(iccanobif(x),y_correct))\r\n%%\r\nx = 9;y_correct = 124;\r\nassert(isequal(iccanobif(x),y_correct))\r\n%%\r\nx = 43;y_correct=36181429034;\r\nassert(isequal(iccanobif(x),y_correct))\r\n%%\r\nfor flag=1:50\r\n    y(flag)=iccanobif(flag);\r\nend\r\ndy=diff(y);\r\nassert(isequal(max(dy(1:25))+min(dy(1:25)),250750))\r\nassert(isequal(max(dy)+min(dy),19910139546138))\r\nsdy=sign(dy);\r\nassert(isequal(sum(sdy==-1),8))\r\n[m1,w1]=min(y(1:10));\r\n[m2,w2]=min(y(11:20));\r\n[m3,w3]=min(y(21:30));\r\n[m4,w4]=min(y(31:40));\r\n[m5,w5]=min(y(41:50));\r\nassert(isequal([m1 m2 m3 m4 m5],[1 836 113815 106496242 21807674140]))\r\nassert(isequal(w1*w2*w3*w4*w5,15))","published":true,"deleted":false,"likes_count":3,"comments_count":1,"created_by":1615,"edited_by":null,"edited_at":null,"deleted_by":null,"deleted_at":null,"solvers_count":66,"test_suite_updated_at":null,"rescore_all_solutions":false,"group_id":61,"created_at":"2019-01-18T13:40:37.000Z","updated_at":"2026-04-01T15:09:11.000Z","published_at":"2019-01-18T13:40:37.000Z","restored_at":null,"restored_by":null,"spam":false,"simulink":false,"admin_reviewed":false,"description_opc":"{\"relationships\":[{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/document\",\"relationshipId\":\"rId1\",\"target\":\"/matlab/document.xml\"},{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/output\",\"relationshipId\":\"rId2\",\"target\":\"/matlab/output.xml\"}],\"parts\":[{\"partUri\":\"/matlab/document.xml\",\"relationship\":[],\"contentType\":\"application/vnd.mathworks.matlab.code.document+xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\"?\u003e\u003cw:document xmlns:w=\\\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\\\"\u003e\u003cw:body\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eThere are a lot of problems in Cody that deal with Fibonacci numbers (1, 1, 2, 3, 5, 8, 13, 21 etc...) so let's turn things around a bit. Literally. This problem deals with the Iccanobif sequence. Instead of adding the previous two numbers in the sequence, reverse the digits of the previous two numbers and add those together to get the next number in the sequence.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eIt starts out exactly the same as the Fibonacci (1, 1, 2, 3, 5, 8, 13) as all of the one digit numbers equal themselves when reversed. However, instead of 8+13=21 for the next term, reverse the digits in 13 to get 31. The next term is now 8+31, or 39. The next term after that is 31+93 (13 and 39 reversed) to get 124, and so on. Unlike the starndard Fibonacci sequence, the Iccanobif sequence can actually decrease in value between terms.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eYou will be given a number, and you will be asked to calculate that term in the Iccanobif sequence.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003e!Kcul doog\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003c/w:body\u003e\u003c/w:document\u003e\"},{\"partUri\":\"/matlab/output.xml\",\"contentType\":\"text/xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\" standalone=\\\"no\\\" ?\u003e\u003cembeddedOutputs\u003e\u003cmetaData\u003e\u003cevaluationState\u003emanual\u003c/evaluationState\u003e\u003clayoutState\u003ecode\u003c/layoutState\u003e\u003coutputStatus\u003eready\u003c/outputStatus\u003e\u003c/metaData\u003e\u003coutputArray type=\\\"array\\\"/\u003e\u003cregionArray type=\\\"array\\\"/\u003e\u003c/embeddedOutputs\u003e\"}]}"}],"problem_search":{"errors":[],"problems":[{"id":44836,"title":"Iccanobif numbers 1","description":"There are a lot of problems in Cody that deal with Fibonacci numbers (1, 1, 2, 3, 5, 8, 13, 21 etc...) so let's turn things around a bit.  Literally.  This problem deals with the Iccanobif sequence.  Instead of adding the previous two numbers in the sequence, reverse the digits of the previous two numbers and add those together to get the next number in the sequence.\r\n\r\nIt starts out exactly the same as the Fibonacci (1, 1, 2, 3, 5, 8, 13) as all of the one digit numbers equal themselves when reversed.  However, instead of 8+13=21 for the next term, reverse the digits in 13 to get 31. The next term is now 8+31, or 39.  The next term after that is 31+93 (13 and 39 reversed) to get 124, and so on.  Unlike the starndard Fibonacci sequence, the Iccanobif sequence can actually decrease in value between terms.\r\n\r\nYou will be given a number, and you will be asked to calculate that term in the Iccanobif sequence.\r\n\r\n!Kcul doog","description_html":"\u003cp\u003eThere are a lot of problems in Cody that deal with Fibonacci numbers (1, 1, 2, 3, 5, 8, 13, 21 etc...) so let's turn things around a bit.  Literally.  This problem deals with the Iccanobif sequence.  Instead of adding the previous two numbers in the sequence, reverse the digits of the previous two numbers and add those together to get the next number in the sequence.\u003c/p\u003e\u003cp\u003eIt starts out exactly the same as the Fibonacci (1, 1, 2, 3, 5, 8, 13) as all of the one digit numbers equal themselves when reversed.  However, instead of 8+13=21 for the next term, reverse the digits in 13 to get 31. The next term is now 8+31, or 39.  The next term after that is 31+93 (13 and 39 reversed) to get 124, and so on.  Unlike the starndard Fibonacci sequence, the Iccanobif sequence can actually decrease in value between terms.\u003c/p\u003e\u003cp\u003eYou will be given a number, and you will be asked to calculate that term in the Iccanobif sequence.\u003c/p\u003e\u003cp\u003e!Kcul doog\u003c/p\u003e","function_template":"function y = iccanobif(x)\r\n  y = x;\r\nend","test_suite":"%%\r\nx = 1;y_correct = 1;\r\nassert(isequal(iccanobif(x),y_correct))\r\n%%\r\nx = 9;y_correct = 124;\r\nassert(isequal(iccanobif(x),y_correct))\r\n%%\r\nx = 43;y_correct=36181429034;\r\nassert(isequal(iccanobif(x),y_correct))\r\n%%\r\nfor flag=1:50\r\n    y(flag)=iccanobif(flag);\r\nend\r\ndy=diff(y);\r\nassert(isequal(max(dy(1:25))+min(dy(1:25)),250750))\r\nassert(isequal(max(dy)+min(dy),19910139546138))\r\nsdy=sign(dy);\r\nassert(isequal(sum(sdy==-1),8))\r\n[m1,w1]=min(y(1:10));\r\n[m2,w2]=min(y(11:20));\r\n[m3,w3]=min(y(21:30));\r\n[m4,w4]=min(y(31:40));\r\n[m5,w5]=min(y(41:50));\r\nassert(isequal([m1 m2 m3 m4 m5],[1 836 113815 106496242 21807674140]))\r\nassert(isequal(w1*w2*w3*w4*w5,15))","published":true,"deleted":false,"likes_count":3,"comments_count":1,"created_by":1615,"edited_by":null,"edited_at":null,"deleted_by":null,"deleted_at":null,"solvers_count":66,"test_suite_updated_at":null,"rescore_all_solutions":false,"group_id":61,"created_at":"2019-01-18T13:40:37.000Z","updated_at":"2026-04-01T15:09:11.000Z","published_at":"2019-01-18T13:40:37.000Z","restored_at":null,"restored_by":null,"spam":false,"simulink":false,"admin_reviewed":false,"description_opc":"{\"relationships\":[{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/document\",\"relationshipId\":\"rId1\",\"target\":\"/matlab/document.xml\"},{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/output\",\"relationshipId\":\"rId2\",\"target\":\"/matlab/output.xml\"}],\"parts\":[{\"partUri\":\"/matlab/document.xml\",\"relationship\":[],\"contentType\":\"application/vnd.mathworks.matlab.code.document+xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\"?\u003e\u003cw:document xmlns:w=\\\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\\\"\u003e\u003cw:body\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eThere are a lot of problems in Cody that deal with Fibonacci numbers (1, 1, 2, 3, 5, 8, 13, 21 etc...) so let's turn things around a bit. Literally. This problem deals with the Iccanobif sequence. Instead of adding the previous two numbers in the sequence, reverse the digits of the previous two numbers and add those together to get the next number in the sequence.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eIt starts out exactly the same as the Fibonacci (1, 1, 2, 3, 5, 8, 13) as all of the one digit numbers equal themselves when reversed. However, instead of 8+13=21 for the next term, reverse the digits in 13 to get 31. The next term is now 8+31, or 39. The next term after that is 31+93 (13 and 39 reversed) to get 124, and so on. Unlike the starndard Fibonacci sequence, the Iccanobif sequence can actually decrease in value between terms.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eYou will be given a number, and you will be asked to calculate that term in the Iccanobif sequence.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003cw:jc w:val=\\\"left\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003e!Kcul doog\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003c/w:body\u003e\u003c/w:document\u003e\"},{\"partUri\":\"/matlab/output.xml\",\"contentType\":\"text/xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\" standalone=\\\"no\\\" 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