A109814 - OEIS

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A109814

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

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, 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, 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

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,3

COMMENTS

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

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

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

LINKS

Donovan Johnson, Table of n, a(n) for n = 1..10000

K. S. Brown's Mathpages, Partitions into Consecutive Integers

A. Heiligenbrunner, Sum of adjacent numbers (in German).

Jaap Spies, Problem C NAW 5/6 nr. 2 June 2005, July 2005 (solution to problem below).

Jaap Spies, Sage program for computing A109814

Universitaire Wiskunde Competitie, Problem C, Nieuw Archief voor Wiskunde, 5/6, no. 2, Problems/UWC, Jun 2005, pp. 181-182.

FORMULA

From Reinhard Zumkeller, Apr 18 2006: (Start)

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

a(A000079(n)) = 1;

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

EXAMPLE

Examples provided by Rainer Rosenthal, Apr 01 2008:

1 = 1 ---> a(1) = 1

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

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

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

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

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

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

MAPLE

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

MATHEMATICA

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

PROG

(Sage)

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

# Jaap Spies, Aug 16 2005

(Python)

from sympy import divisors

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

print([a(n) for n in range(1, 103)]) # Michael S. Branicky, Dec 23 2022

CROSSREFS

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