A355182 a(n) = t(n) - s(n) where s(n) = n*(n-1)/2 is the sum of the first n nonnegative integers and t(n) is the smallest sum of consecutive integers starting from n such that t(n) > s(n).

1, 1, 4, 3, 1, 6, 3, 10, 6, 1, 10, 4, 15, 8, 21, 13, 4, 19, 9, 26, 15, 3, 22, 9, 30, 16, 1, 24, 8, 33, 16, 43, 25, 6, 35, 15, 46, 25, 3, 36, 13, 48, 24, 61, 36, 10, 49, 22, 63, 35, 6, 49, 19, 64, 33, 1, 48, 15, 64, 30, 81, 46, 10, 63, 26, 81, 43, 4, 61, 21, 80, 39, 100, 58, 15, 78, 34, 99

Offset: 1

Author

Andrea La Rosa, Jun 23 2022

Keywords

• nonn

Comments

Record high values of a(n)/n approach sqrt(2) and occur at values of n that are terms of A011900; a(A011900(k)) = A046090(k). - Jon E. Schoenfield, Jun 23 2022

It appears that the sequence 1,2,4,5,6,8,... (the largest integer in the t(n) sum) is A288998. - Michel Marcus, Jun 24 2022

Examples

			a(6) = -s(6) + t(6):
s(6) is the sum of the first 6 nonnegative integers = 6*5 / 2 = 15.
t(6) is the smallest sum k of consecutive integers starting from n = 6 such that k > s(6) = 15.
The first few sets of consecutive integers starting from 6 are
  {6}, whose elements add up to 6,
  {6, 7}, whose elements add up to 13,
  {6, 7, 8}, whose elements add up to 21,
  {6, 7, 8, 9}, whose elements add up to 30,
  ...
the smallest sum that is > 15 is 21, so t(6) = 21, so a(6) = -15 + 21 = 6.
		

Crossrefs

Cf. A000217, A001477, A093001.

Cf. A288998.

Programs

JavaScript

function a(n) {
    var sum = 0;
    for (var i = 0; i < n; i++)
        sum -= i;
    while (sum <= 0)
        sum += i++;
    return sum;
}

PARI

a(n) = my(t=0, s=n*(n-1)/2, k=n); until (t > s, t += k; k ++); t-s; \\ Michel Marcus, Jun 24 2022

Python

from math import isqrt
def A355182(n): return ((m:=(isqrt(((k:=n*(n-1))<<3)+1)+1)>>1)*(m+1)>>1)-k # Chai Wah Wu, Jul 14 2022

Formula

From Jon E. Schoenfield, Jun 23 2022: (Start)

a(n) = t(n) - s(n) where

s(n) = n*(n-1)/2,

j = floor(sqrt(8*n^2 - 8*n + 1)),

m = ceiling(j/2) - n + 1, and

t(n) = (m*(m + 2*n - 1))/2. (End)