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a(n) is the largest k such that n can be written as sum of k consecutive positive integers.
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%I #64 Dec 23 2022 17:25:42

%S 1,1,2,1,2,3,2,1,3,4,2,3,2,4,5,1,2,4,2,5,6,4,2,3,5,4,6,7,2,5,2,1,6,4,

%T 7,8,2,4,6,5,2,7,2,8,9,4,2,3,7,5,6,8,2,9,10,7,6,4,2,8,2,4,9,1,10,11,2,

%U 8,6,7,2,9,2,4,10,8,11,12,2,5,9,4,2,8,10,4,6,11,2,12,13,8,6,4,10,3,2,7,11,8,2,12

%N a(n) is the largest k such that n can be written as sum of k consecutive positive integers.

%C n is the sum of at most a(n) consecutive positive integers. As suggested by _David W. Wilson_, Aug 15 2005: Suppose n is to be written as sum of k consecutive integers starting with m, then 2n = k(2m + k - 1). Only one of the factors is odd. For each odd divisor d of n there is a unique corresponding k = min(d,2n/d). a(n) is the largest among those k. - _Jaap Spies_, Aug 16 2005

%C The numbers that can be written as a sum of k consecutive positive integers are those in column k of A141419 (as a triangle). - _Peter Munn_, Mar 01 2019

%C The numbers that cannot be written as a sum of two or more consecutive positive integers are the powers of 2. So a(n) = 1 iff n = 2^k for k >= 0. - _Bernard Schott_, Mar 03 2019

%H Donovan Johnson, <a href="/A109814/b109814.txt">Table of n, a(n) for n = 1..10000</a>

%H K. S. Brown's Mathpages, <a href="http://www.mathpages.com/home/kmath107.htm">Partitions into Consecutive Integers</a>

%H A. Heiligenbrunner, <a href="http://ah9.at/ahsummen.htm">Sum of adjacent numbers (in German)</a>.

%H Jaap Spies, <a href="http://www.jaapspies.nl/mathfiles/problem2005-2C.pdf">Problem C NAW 5/6 nr. 2 June 2005</a>, July 2005 (solution to problem below).

%H Jaap Spies, <a href="http://www.jaapspies.nl/oeis/a111776.sage">Sage program for computing A109814</a>

%H Universitaire Wiskunde Competitie, <a href="http://www.nieuwarchief.nl/serie5/pdf/naw5-2005-06-2-181.pdf">Problem C</a>, Nieuw Archief voor Wiskunde, 5/6, no. 2, Problems/UWC, Jun 2005, pp. 181-182.

%F From _Reinhard Zumkeller_, Apr 18 2006: (Start)

%F a(n)*(a(n)+2*A118235(n)-1)/2 = n;

%F a(A000079(n)) = 1;

%F a(A000217(n)) = n. (End)

%e Examples provided by _Rainer Rosenthal_, Apr 01 2008:

%e 1 = 1 ---> a(1) = 1

%e 2 = 2 ---> a(2) = 1

%e 3 = 1+2 ---> a(3) = 2

%e 4 = 4 ---> a(4) = 1

%e 5 = 2+3 ---> a(5) = 2

%e 6 = 1+2+3 ---> a(6) = 3

%e a(15) = 5: 15 = 15 (k=1), 15 = 7+8 (k=2), 15 = 4+5+6 (k=3) and 15 = 1+2+3+4+5 (k=5). - _Jaap Spies_, Aug 16 2005

%p A109814:= proc(n) local m, k, d; m := 0; for d from 1 by 2 to n do if n mod d = 0 then k := min(d, 2*n/d): fi; if k > m then m := k fi: od; return(m); end proc; seq(A109814(i),i=1..150); # _Jaap Spies_, Aug 16 2005

%t a[n_] := Reap[Do[If[OddQ[d], Sow[Min[d, 2n/d]]], {d, Divisors[n]}]][[2, 1]] // Max; Table[a[n], {n, 1, 102}]

%o (Sage)

%o [sloane.A109814(n) for n in range(1,20)]

%o # _Jaap Spies_, Aug 16 2005

%o (Python)

%o from sympy import divisors

%o def a(n): return max(min(d, 2*n//d) for d in divisors(n) if d&1)

%o print([a(n) for n in range(1, 103)]) # _Michael S. Branicky_, Dec 23 2022

%Y Cf. A001227, A111774, A111775, A141419, A138591.

%Y Cf. A000079 (powers of 2), A000217 (triangular numbers).

%K nonn

%O 1,3

%A _David W. Wilson_

%E Edited by _N. J. A. Sloane_, Aug 23 2008 at the suggestion of _R. J. Mathar_