A111775 -id:A111775 - cp's OEIS frontend

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

Offset: 1

David W. Wilson

nonn

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

			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

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.

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

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

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

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

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

[sloane.A109814(n) for n in range(1,20)]
# Jaap Spies, Aug 16 2005

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)

Edited by N. J. A. Sloane, Aug 23 2008 at the suggestion of R. J. Mathar

A111787 a(n) is the least k >= 3 such that n can be written as sum of k consecutive integers. a(n)=0 if such a k does not exist.

Original entry on oeis.org

0, 0, 0, 0, 0, 3, 0, 0, 3, 4, 0, 3, 0, 4, 3, 0, 0, 3, 0, 5, 3, 4, 0, 3, 5, 4, 3, 7, 0, 3, 0, 0, 3, 4, 5, 3, 0, 4, 3, 5, 0, 3, 0, 8, 3, 4, 0, 3, 7, 5, 3, 8, 0, 3, 5, 7, 3, 4, 0, 3, 0, 4, 3, 0, 5, 3, 0, 8, 3, 5, 0, 3, 0, 4, 3, 8, 7, 3, 0, 5, 3, 4, 0, 3, 5, 4, 3, 11, 0, 3, 7, 8, 3, 4, 5, 3, 0, 7, 3, 5, 0, 3, 0, 13

Offset: 1

Jaap Spies, Aug 16 2005

a(n)=0 if n is an odd prime or a power of 2. For numbers of the third kind we proceed as follows: suppose n is to be written as sum of k consecutive integers starting with m, then 2n = k(2m + k - 1). Let p be the smallest odd prime divisor of n then a(n) = min(p,2n/p).

			a(15)=3 because 15=4+5+6 (k=3) and 15=2+3+4+5 (k=4)

Nieuw Archief voor Wiskunde 5/6 nr. 2 Problems/UWC Problem C part 3, Jun 2005, pp. 181-182

Donald Alan Morrison, Table of n, a(n) for n=1..10000
Nieuw Archief voor Wiskunde 5/6 nr. 2 Problems/UWC, Problem C: solution
J. Spies, Sage program for computing A111787

Cf. A111774, A111775, A109814.

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; A111787:= proc(n) local d, k; k:=0; if isprime(n) or ispoweroftwo(n) then return(0); fi; for d from 3 by 2 to n do if n mod d = 0 then k:=min(d,2*n/d); break; fi; od; return(k); end proc; seq(A111787(i),i=1..150);

A202238 Characteristic function of positive integers not prime and not a power of 2.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 1

Offset: 1

Michael Somos, Dec 16 2011

Antti Karttunen, Table of n, a(n) for n = 1..65537
Index entries for characteristic functions

Cf. A010051, A111774, A111775, A111787, A209229.

{a(n) = n>0 && !isprime(n) && n != 2^valuation(n, 2)}

from sympy import isprime
def A202238(n): return int(not isprime(n) and bool((n&-n)^n)) # Chai Wah Wu, Mar 11 2025

A111775(n) = 0 if and only if a(n) = 0.
A111787(n) = 0 if and only if a(n) = 0.
a(n) = 1 for n in A111774. - Michel Marcus, Aug 28 2017

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A109814 a(n) is the largest k such that n can be written as sum of k consecutive positive integers.

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A111787 a(n) is the least k >= 3 such that n can be written as sum of k consecutive integers. a(n)=0 if such a k does not exist.

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A202238 Characteristic function of positive integers not prime and not a power of 2.

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