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

1,2

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

Links

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

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

Example

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

Maple

a:= proc(n) local j, k, s; j, k, s:= 1$3;
      while s<>n do
         if s<n then k:= k+1; s:= s+k
                else s:= s-j; j:= j+1 fi
      od: j
    end:
seq(a(n), n=1..100);  # Alois 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 *)

Prog

(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 */

Crossrefs

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

Sequence in context: A243060 A286321 A349192 * A346702 A304624 A083653

Adjacent sequences: A118232 A118233 A118234 * A118236 A118237 A118238

Keyword

nonn,look

Author

Reinhard Zumkeller, Apr 18 2006

Status

approved