A111775 Number of ways n can be written as a sum of at least three consecutive integers.

0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 2, 0, 0, 2, 0, 1, 2, 1, 0, 1, 1, 1, 2, 1, 0, 3, 0, 0, 2, 1, 2, 2, 0, 1, 2, 1, 0, 3, 0, 1, 4, 1, 0, 1, 1, 2, 2, 1, 0, 3, 2, 1, 2, 1, 0, 3, 0, 1, 4, 0, 2, 3, 0, 1, 2, 3, 0, 2, 0, 1, 4, 1, 2, 3, 0, 1, 3, 1, 0, 3, 2, 1, 2, 1, 0, 5, 2, 1, 2, 1, 2, 1, 0, 2, 4, 2, 0, 3, 0, 1

Offset

1,15

Comments

Powers of 2 and (odd) primes cannot be written as a sum of at least three consecutive integers. a(n) strongly depends on the number of odd divisors of n (A001227): 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 of n there is a unique corresponding k, k=1 and k=2 must be excluded.

When the initial 0 term is a(1), a(n) is the number of times n occurs after the second column in the square array of A049777. - Bob Selcoe, Feb 14 2014

For nonnegative integers x,y where x-y>=3: a(n) equals the number of ways n can be expressed as a function of (x*(x+1)-y*(y+1))/2 when the initial 0 term is a(1). - Bob Selcoe, Feb 14 2014

References

Nieuw Archief voor Wiskunde 5/6 nr. 2 Problems/UWC Problem C part 4, Jun 2005, p. 181-182

Links

Alois P. Heinz, 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)

Nieuw Archief voor Wiskunde 5/6 nr. 2 Problems/UWC, Problem C: solution of this Problem

J. Spies, Sage program for computing A111775

Formula

If n is even then a(n) = A001227(n)-1 = A069283(n) otherwise a(n) = A001227(n)-2, for n > 1.

G.f.: Sum_{n >= 2} x^(3*n)/(1 - x^(2*n)). - Peter Bala, Jan 12 2021

Example

a(15) = 2 because 15 = 4+5+6 and 15 = 1+2+3+4+5. The number of odd divisors of 15 is 4.
G.f. = x^6 + x^9 + x^10 + x^12 + x^14 + 2*x^15 + 2*x^18 + x^20 + 2*x^21 + x^22 + ...
a(30) = 3 because there are 3 ways to satisfy (x*(x+1)-y*(y+1))/2 = 30 when x-y>=3: x=8, y=3; x=9, y=5; and x=11, y=8. - Bob Selcoe, Feb 14 2014

Maple

A001227:= proc(n) local d, s; s := 0: for d from 1 by 2 to n do if n mod d = 0 then s:=s+1 fi: end do: return(s); end proc; A111775:= proc(n) local k; if n=1 then return(0) fi: k := A001227(n): if type(n, even) then k:=k-1 else k:=k-2 fi: return k; end proc; seq(A111775(i), i=1..150);

Mathematica

a[n_] := If[n == 1, 0, Total[Mod[Divisors[n], 2]] - Mod[n, 2] - 1];
a /@ Range[1, 100] (* Jean-François Alcover, Oct 14 2019 *)

Prog

(PARI) {a(n) = local(m); if( n<1, 0, sum( i=0, n, m=0; if( issquare( 1 + 8*(n + i * (i + 1)/2), &m), m\2 > i+2)))}; /* Michael Somos, Aug 27 2012 */

Crossrefs

Cf. A111774, A001227 (number of odd divisors), A069283.

Sequence in context: A161116 A262726 A112605 * A325166 A025844 A035461

Adjacent sequences: A111772 A111773 A111774 * A111776 A111777 A111778

Keyword

easy,nonn

Author

Jaap Spies, Aug 16 2005

Status

approved