OFFSET
1,2
COMMENTS
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
REFERENCES
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).
LINKS
R. J. Mathar, Table of n, a(n) for n = 1..7836
G. E. Andrews, MacMahon's prime numbers of measurement, Amer. Math. Monthly, 82 (1975), 922-923.
Thomas Bloom, Problem 474, Erdős Problems.
R. L. Graham and C. B. A. Peck, Problem E1910, Amer. Math. Monthly, 75 (1968), 80-81.
R. K. Guy, Letter to G. E. Andrews, Apr 14 1975
P. A. MacMahon, The prime numbers of measurement on a scale, Proc. Camb. Phil. Soc. 21 (1923), 651-654; reprinted in Coll. Papers I, pp. 797-800.
Samuel B. Reid, C program for A002048
Eric Weisstein's World of Mathematics, Prime Number of Measurement.
FORMULA
Andrews conjectures that lim_{n -> oo} n log n / (a(n) loglog n) = 1. - N. J. A. Sloane, Dec 01 2013
EXAMPLE
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.
MAPLE
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:
A002048(40000, true);
# R. J. Mathar, Jun 04 2006
MATHEMATICA
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 *)
PROG
(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;
} /* R. J. Mathar, May 31 2006 */
(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)
-- Reinhard Zumkeller, May 23 2013
(C) // See Links section for C program by Samuel B. Reid, Jan 26 2020
CROSSREFS
KEYWORD
nonn,nice
AUTHOR
EXTENSIONS
More terms from R. J. Mathar, May 31 2006
STATUS
approved