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[ the sum of consecutive positive integers in seven ways - notation (#,F) Let # = number of consecutive integers Let F = first integer of the sequence then 2002 equals (4,499) = (7,283) = (11,177) = (13,148) = (28,58) = (44,24) = (52,13) I asked Carlos Rivera and Terry Trotter if there exist a mathematical procedure to find all combinations. Is there a way to get them all straightaway, not only for 2002 but for any arbitrary integer ? Carlos replied with No more integer solutions other than the seven found, for b in S=(a+b)(b-a+1)/2 = 2002, for 1 < = a <= 1001 ps. a is the first member of the sequence and b is the last member of the sequence. Terry used his Excel spreadsheet to solve the problem in a more elaborate way. The basic equation is nx + n(n – 1)/2 = 2002 and he set it up this way
For any arbitrary integer, replace 2002 with another number 3 terms, starting with 666 now, for next year, 2003 gives 2 terms, starting with 1001 Terry excelled further and got more data regarding 2000 has 3 ways (5,398) (25,68) (32,47)
2001 has 6 ways (3,666) (6,331) (23,76) (29,55) (46,21) (58,6) 2002 has 7 ways (4,499) (7,283) (11,177) (13,148) (28,58) (44,24) (52,13) 2003 has 1 way (2,1001) 2004 has 3 ways (3,667) (8,247) (24,72) 2005 has 3 ways (2,1002) (5,399) (10,196) 2006 has 3 ways (4,500) (17,110) (59,5) 2007 has 5 ways (2,1003) (3,668) (6,332) (9,219) (18,103) 2008 has 1 way (16,118) 2009 has 5 ways (2,1004) (7,284) (14,137) (41,29) (49,17) 2010 has 7 ways (3,669) (4,501) (5,400) (12,162) (15,127) (20,91) (60,4) Terry worked a little more and allowed negatives as well as positives. The following new items emerged 2002 has 3 more ways (#77,–12, up to 64) (#91,–23, up to 67) (#143,–57, up to 85) All can be verified by students as they use the famous textbook formula S = (n/2)[2a + (n–1)d] The analogue case for numbers as sums of (at least two) consecutive positive primes seems much more difficult ! Some examples for 2002 Carlos Rivera formulated a hard but nice challenge The earliest primes expressible as...
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