A338161 Successive sums of successive terms produce the successive natural numbers (see the Comments section).
1, -2, 4, -3, -5, 11, -6, -7, 8, 9, 10, 12, 13, -14, -16, -15, -17, 18, 19, -20, 21, 22, 23, 24, 25, -26, -27, -34, -28, -29, -30, 31, 32, 33, 35, -36, 37, 38, 39, 40, 41, -42, -43, -44, -57, -45, -46, -47, 48, 49, -50, 51, 52, 53, -55, 54, 56, 58, 59, 60, 61, -62, -63, -64, -65, -83, -66, -67, -68, -69, 70, 71
Offset: 1
Keywords
Examples
1 = 1 (1 term); 2 = - 2 + 4 (2 terms); 3 = - 3 - 5 + 11 (3 terms); 4 = - 6 - 7 + 8 + 9 (4 terms); 5 = 10 + 12 + 13 - 14 - 16 (5 terms); 6 = - 15 - 17 + 18 + 19 - 20 + 21 (6 terms); etc. How are the plus and minus signs split between the terms to get the above six equations? Here is the method -- with an example: 1) no absolute value of any term can be present twice or more in the sequence; 2) to start a new equation, always use the set of smallest absolute values not yet used; say, for the above 5-term equation, that they are a, b, c, d and e; 3) the set of unused values for a, b, c, d and e is here 10, 12, 13, 14, 15; 4) try all the possible mix of values and signs to find one or more solutions (the try 5 = 10 + 12 - 13 - 14 + 15, for instance, doesn't work as we would get 5 = 10); 5) if no such mix leads to a solution (which is the case here), add 1 to the biggest integer of the values' set and try again; 6) the above set would then become 10, 12, 13, 14, 16 -- and a quick computer search gives the solution 5 = 10 + 12 + 13 - 14 - 16; 7) had we found more than one solution, we would have kept the lexicographically earliest one (-10 comes before +10); 8) if a new mix doesn't lead to a solution, add again 1 to the biggest integer of the values' set and try again; etc.
Links
- Carole Dubois, Table of n, a(n) for n = 1..406
- Carole Dubois, Program (Python)
Crossrefs
Cf. A330903.
Programs
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Python
# see Link section
Comments