[ *January 9, 2002* ]

Numbers equal to the sum of (at least two) consecutive integers

& numbers equal to the sum of (at least two) consecutive primes.

As a consequence of finding equal to

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

cell A1 = 1 | B1 is empty | C1 is empty |

cell A2 = A1 + 1 | B2 = A2*(A2–1)/2 | C2 = (2002-B2)/A2 |

cell A3 = "copy" of A2 | B3 = "copy" of B2 | C3 = "copy" of C2 |

... | ... | ... |

cell A63 = "copy" of A2 | B63 = "copy" of B2 | C63 = "copy" of C2 |

For any arbitrary integer, replace 2002 with another number

E.g. 2001 gives integer solutions for

3 terms, starting with 666

6 terms, starting with 331

23 terms, starting with 76

29 terms, starting with 56

46 terms, starting with 21

58 terms, starting with 6

now, for next year, 2003 gives __only__

2 terms, starting with 1001

Terry excelled further and got more data regarding

sums of consecutive positive integers

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

491 + 499 + 503 + 509 (#4)

107 up to 179 (#14)

Carlos Rivera formulated a hard but nice challenge

Who can provide the following sequence ?

The earliest numbers that are expressible as a sum of consecutive primes

in K ways, for K = 1, 2, 3, ... 20.

The earliest primes expressible as...

were already covered by Carlos (cr) in his own primepuzzle website

yet he gave also some solutions for the smallest nonprimes.

See cases K = 6 and K = 7.

K | Earliest number | Sums of (at least two) consecutive primes | Who | Earliest prime |

1 | 5 | (#2,2) 2 + 3 | pdg | 5 |

2 | 36 | (#2,17) 17 + 19
(#4,5) 5 + 7 + 11 + 13 | pdg | 41 |

3 | 240 | (#2,113) (#4,53) (#8,17) | pdg | 311 |

4 | 311 | (#3,101) (#5,53) (#7,31) (#11,11) | pdg | 311 |

5 | 16277 | (#7,2297) (#11,1451) (#13,1213) (#35,359) (#37,331) | pdg | 34421 |

6 | 130638 | (#8,16273) (#10,13009) (#12,10847) (#56,2113) (#140,461) (#206,29) | cr | 442019 |

7 | 218918 | (#12,18199) (#16,13619) (#22,9851) (#28,7691) (#38,5623) (#46,4561) (#62,3301) | cr | 3634531 |

8 | ? | ? | ? | 48205429 |

9 | ? |

10 | ? |

11 | ? |

12 | ? |

13 | ? |

14 | ? |

15 | ? |

16 | ? |

17 | ? |

18 | ? |

19 | ? |

20 | ? |