A118235 Smallest positive number starting an interval of consecutive integers with element sum n.

1, 2, 1, 4, 2, 1, 3, 8, 2, 1, 5, 3, 6, 2, 1, 16, 8, 3, 9, 2, 1, 4, 11, 7, 3, 5, 2, 1, 14, 4, 15, 32, 3, 7, 2, 1, 18, 8, 4, 6, 20, 3, 21, 2, 1, 10, 23, 15, 4, 8, 6, 3, 26, 2, 1, 5, 7, 13, 29, 4, 30, 14, 3, 64, 2, 1, 33, 5, 9, 7, 35, 4, 36, 17, 3, 6, 2, 1, 39, 14, 5, 19, 41, 7, 4, 20, 12, 3, 44, 2, 1, 8

Offset: 1

Author

Reinhard Zumkeller, Apr 18 2006

Keywords

• nonn
• look

Comments

Right border of A299765. - Omar E. Pol, Jul 24 2018

In other words: a(n) is smallest part of the partitions of n into consecutive parts. - Omar E. Pol, Mar 12 2019

Examples

			a(3)=1 since 3 = 1+2; a(5)=2 since 5 = 2+3; a(6)=1 since 6 = 1+2+3; etc.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 1..20000 (first 1000 terms from Paul D. Hanna)

Crossrefs

Cf. A000079, A000217, A001227, A002817, A104512, A109814, A212652, A299765.

Programs

Maple

a:= proc(n) local j, k, s; j, k, s:= 1$3;
      while s<>n do
         if sAlois P. Heinz, Aug 05 2018

Mathematica

a[n_] := Module[{j = 1, k = 1, s = 1}, While[True, If[s == n, Break[]]; If[s < n, k = k+1; s = s+k, s = s-j; j = j+1]]; j];
Array[a, 100] (* Jean-François Alcover, Mar 12 2019, after Alois P. Heinz *)

PARI

{a(n)=local(A=n);for(j=1,n,for(k=j,n+1,if(n==k*(k-1)/2-j*(j-1)/2,A=j;k=j=2*n+1)));A} /* Paul D. Hanna, Oct 28 2011 */

Formula

A109814(n) * (A109814(n) + 2*a(n) - 1) / 2 = n.

a(m) = n iff m = 2^k: a(A000079(n)) = A000079(n);

a(m) = 1 iff m = k*(k+1)/2: a(A000217(n)) = 1.

a(A002817(n-1)+1) = n; i.e., a(m) = n if m = k*(k-1)/2 + 1 and k = n*(n-1)/2 + 1. - Paul D. Hanna, Oct 28 2011

a(m) = 2 iff m = k*(k+3)/2: a(A000096(n)) = 2. - Bernard Schott, Mar 12 2019