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A002048
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Segmented numbers, or prime numbers of measurement.
(Formerly M0972 N0363)
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12
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1, 2, 4, 5, 8, 10, 14, 15, 16, 21, 22, 25, 26, 28, 33, 34, 35, 36, 38, 40, 42, 46, 48, 49, 50, 53, 57, 60, 62, 64, 65, 70, 77, 80, 81, 83, 85, 86, 90, 91, 92, 100, 104, 107, 108, 116, 119, 124, 127, 132, 133, 137, 141, 144, 145, 148, 150, 151, 154, 158, 159, 163, 165
(list;
graph;
refs;
listen;
history;
text;
internal format)
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OFFSET
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1,2
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COMMENTS
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The segmented numbers are the positive integers excluding those equal to the sum of two or more consecutive smaller terms. The prime numbers of measurement are their partial sums, cf. A002049. - M. F. Hasler, Jun 26 2019
Without the requirement that the smaller terms be consecutive, the sequence becomes the sequence of powers of 2 (A000079). - Alonso del Arte, Jan 25 2020
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REFERENCES
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R. K. Guy, Unsolved Problems in Number Theory, E30.
Š. Porubský, On MacMahon's segmented numbers and related sequences. Nieuw Arch. Wisk. (3) 25 (1977), no. 3, 403--408. MR0485763 (58 #5575)
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
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R. L. Graham and C. B. A. Peck, Problem E1910, Amer. Math. Monthly, 75 (1968), 80-81.
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FORMULA
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Andrews conjectures that lim_{n -> oo} n log n / (a(n) loglog n) = 1. - N. J. A. Sloane, Dec 01 2013
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EXAMPLE
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Although 5 is the sum of the terms 1 and 4, those prior terms are not consecutive, and therefore 5 is in the sequence.
6 is not in the sequence because it is the sum of consecutive prior terms 2 and 4.
7 is not in the sequence either because it is also the sum of consecutive prior terms, in this case 1, 2, 4.
8 is in the sequence because no sum whatsoever of distinct prior terms adds up to 8.
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MAPLE
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A002048 := proc(anmax::integer, printlist::boolean)
local a, asum, su, i, piv, j;
a := [];
for i from 1 to anmax do
a := [op(a), i];
od:
if printlist then
printf("%d %d\n", 1, a[1]);
printf("%d %d\n", 2, a[2]);
fi;
asum := [a[1]+a[2], a[2]];
for i from 3 to anmax do
asum := [op(asum), 0];
od:
piv := 3;
while piv <= nops(a) do
for i from 1 to piv-2 do
a := remove(has, a, asum[i]);
od:
if printlist then
printf("%a %a\n", piv, a[piv]);
fi;
for i from 1 to piv do
asum := subsop(i=asum[i]+a[piv], asum);
od:
piv := piv+1;
od;
RETURN(a);
end:
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MATHEMATICA
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A002048[anmax_] := (a = {}; Do[AppendTo[a, i], {i, anmax}]; asum = {a[[1]] + a[[2]], a[[2]]}; Do[AppendTo[asum, 0], {i, 3, anmax}]; piv = 3; While[piv <= Length[a], Do[a = DeleteCases[a, asum[[i]]], {i, 1, piv - 2}]; Do[asum[[i]] += a[[piv]], {i, piv}]; piv = piv + 1; ]; a); A002048[63] (* Jean-François Alcover, Jul 28 2011, converted from R. J. Mathar's Maple prog. *)
searchMax = 200; segmNums = {1}; curr = 2; While[curr < searchMax, If[Not[MemberQ[Apply[Plus, Subsequences[segmNums], 1], curr]], AppendTo[segmNums, curr], ]; curr = curr + 1]; segmNums (* Alonso del Arte, Jan 25 2020 *)
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PROG
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(C++)
#include <iostream>
#include <vector>
#include <algorithm>
#define NMAX 400
using namespace std;
int main(int argc, char *argv[])
{ vector<int> a; for(int i = 0; i < NMAX; i++) a.push_back(i+1); for(int piv = 2; piv < a.size(); piv++) for(int i = 0; i < piv - 1 && i < a.size() - 1; i++) { int su = a[i] + a[i + 1]; remove(a.begin(), a.end(), su); for(int j = i + 2; j < piv && j < a.size(); j++) { su += a[j]; remove(a.begin(), a.end(), su); if(su > NMAX) break; } } for(int i = 0; i < a.size() && a[i] < NMAX; i++) cout << a[i] << ", "; return 0;
(Haskell)
import Data.List ((\\))
a002048 n = a002048_list !! (n-1)
a002048_list = f [1..] [] where
f (x:xs) ys = x : f (xs \\ scanl (+) x ys) (x : ys)
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CROSSREFS
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KEYWORD
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nonn,nice
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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