%I #49 Jul 20 2022 22:37:52

%S 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,

%T 8,33,16,43,25,6,35,15,46,25,3,36,13,48,24,61,36,10,49,22,63,35,6,49,

%U 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

%N 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).

%C 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

%C 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

%F From _Jon E. Schoenfield_, Jun 23 2022: (Start)

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

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

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

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

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

%e a(6) = -s(6) + t(6):

%e s(6) is the sum of the first 6 nonnegative integers = 6*5 / 2 = 15.

%e t(6) is the smallest sum k of consecutive integers starting from n = 6 such that k > s(6) = 15.

%e The first few sets of consecutive integers starting from 6 are

%e {6}, whose elements add up to 6,

%e {6, 7}, whose elements add up to 13,

%e {6, 7, 8}, whose elements add up to 21,

%e {6, 7, 8, 9}, whose elements add up to 30,

%e ...

%e the smallest sum that is > 15 is 21, so t(6) = 21, so a(6) = -15 + 21 = 6.

%o (JavaScript)

%o function a(n) {

%o var sum = 0;

%o for (var i = 0; i < n; i++)

%o sum -= i;

%o while (sum <= 0)

%o sum += i++;

%o return sum;

%o }

%o (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

%o (Python)

%o from math import isqrt

%o def A355182(n): return ((m:=(isqrt(((k:=n*(n-1))<<3)+1)+1)>>1)*(m+1)>>1)-k # _Chai Wah Wu_, Jul 14 2022

%Y Cf. A000217, A001477, A093001.

%Y Cf. A288998.

%K nonn

%O 1,3

%A _Andrea La Rosa_, Jun 23 2022