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A111774
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Numbers that can be written as a sum of at least three consecutive positive integers.
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18
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6, 9, 10, 12, 14, 15, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30, 33, 34, 35, 36, 38, 39, 40, 42, 44, 45, 46, 48, 49, 50, 51, 52, 54, 55, 56, 57, 58, 60, 62, 63, 65, 66, 68, 69, 70, 72, 74, 75, 76, 77, 78, 80, 81, 82, 84, 85, 86, 87, 88, 90, 91, 92, 93, 94, 95, 96, 98, 99, 100, 102
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OFFSET
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1,1
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COMMENTS
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In this sequence there are no (odd) primes and there are no powers of 2.
So we have only three kinds of natural numbers: the odd primes, the powers of 2 and the numbers that can be represented as a sum of at least three consecutive integers.
Odd primes can only be written as a sum of two consecutive integers. Powers of 2 do not have a representation as a sum of k consecutive integers (other than the trivial n=n, for k=1).
Numbers of the form (x*(x+1)-y*(y+1))/2 for nonnegative integers x,y with x-y >= 3. - Bob Selcoe, Feb 21 2014
Numbers of the form (x + 1)*(x + 2*y)/2 for integers x,y with x >= 2 and y >= 1. For y = 1 only triangular numbers (A000217) >= 6 occur. - Ralf Steiner, Jun 27 2019
If k >= 1 sequences are c_k(n) = c_k(n - 1) + n + k - 1, c_k(0) = 0, means c_k(n) = n*(n + 2*k - 1)/2: A000217, A000096, A055998, A055999, A056000, ... then this sequence is the union of c_k(n), n >= 3. (End)
This sequence gives all positive integers that have at least one odd prime as proper divisor. The proof follows from the first two comments.
The set {a(n)}_{n>=1} equals the set {k positive integer : floor(k/2) - delta(k) >= 1}, where delta(k) = A055034(k). Proof: floor(k/2) gives the number of positive odd numbers < k, and delta(k), gives the number of positive odd numbers coprime to k. delta(1) = 1 but 1 is not < 1, therefore k = 1 is not a member of this set. Hence a member >= 2 of this set has at least one odd number > 1 and < k missing in the set of odd numbers relative prime to k. Therefore there exists at least one odd prime < k dividing k. (End)
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REFERENCES
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Paul Halmos, "Problems for Mathematicians, Young and Old", Dolciani Mathematical Expositions, 1991, Solution to problem 3G p. 179.
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LINKS
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Nieuw Archief voor Wiskunde 5/6 nr. 2 Problems/UWC, Problem C, Jun 2005, p. 181-182.
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EXAMPLE
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a(1)=6 because 6 is the first number that can be written as a sum of three consecutive positive integers: 6 = 1+2+3.
6 10 15 21 28 36 45 55 66
9 14 20 27 35 44 54 65
12 18 25 33 42 52 63
15 22 30 39 49 60
18 26 35 45 56
21 30 40 51
24 34 45
27 38
30
This is (x*(x+1)-y*(y+1))/2 for nonnegative integers x,y with x-y >= 3, because it is equivalent to 1+2+3/+4/+5/...+x/-0/-1/-2/-3/-4/-5/...-(x+3)/ for all possible strings of consecutive integers, which represents every possible way to sum three or more consecutive positive integers. So for example, 4+5+6+7 = 1+2+3+4+5+6+7-1-2-3 = 22, which is (x*(x+1)-y*(y+1))/2 when x=7, y=3. Notice that values can appear more than once in the array because some numbers can be represented as sums of more than one string of three or more consecutive positive integers. For example, 30 = (x*(x+1)-y*(y+1))/2 when (a) x=11, y=8: 9+10+11; (b) x=9, y=5: 6+7+8+9; and (c) x=8, y=3: 4+5+6+7+8. By definition, x-y is the number of integers in the string. (End)
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MAPLE
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ispoweroftwo := proc(n) local a, t; t := 1; while (n > t) do t := 2*t end do; if (n = t) then a := true else a := false end if; return a; end proc; f:= proc(n) if (not isprime(n)) and (not ispoweroftwo(n)) then return n end if; end proc; seq(f(i), i = 1..150);
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MATHEMATICA
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max=6!; lst={}; Do[z=n+(n+1); Do[z+=(n+x); If[z>max, Break[]]; AppendTo[lst, z], {x, 2, max}], {n, max}]; Union[lst] (* Vladimir Joseph Stephan Orlovsky, Mar 06 2010 *)
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PROG
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(PARI) isok(n) = !(n == 1) && !isprime(n) && !(isprimepower(n, &p) && (p == 2)); \\ Michel Marcus, Jul 02 2019
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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