A093001 -id:A093001 - cp's OEIS frontend

A093000 Least k such that Sum_{r=n+1..k} r >= n!.

2, 3, 5, 8, 16, 38, 101, 284, 852, 2694, 8935, 30952, 111598, 417560, 1617204, 6468816, 26671611, 113158064, 493244565, 2205856753, 10108505545, 47413093714, 227385209453, 1113955476429, 5569777382146, 28400403557929

Offset: 1

Amarnath Murthy, Mar 29 2004

Equivalently, least k such that the product of the first n positive integers is less than the sum of the integers from n+1 through k.

a(n) = floor(sqrt(2*n! + n^2)) for most values of n; the exceptions are 1,2,3,7,..., in which case a(n) = floor(sqrt(2*n! + n^2)) + 1.

			a(4) = 8 because 4! = 24 and 5+6+7+8 = 26 > 24, but 5+6+7 = 18.
a(5) = 16 because 5! = 120 and 6+7+8+...+15+16 = 121 > 120.

Cf. A093001.

{ for(n=1,20, s=0; found=0; for(k=n+1,10000000, if( k*(k+1)-n*(n+1)>= 2*n!, print1(k,","); found=1; break; ); ); if(found==0, print(0); ); ); } \\ R. J. Mathar, Apr 21 2006

Least k such that {k(k+1)/2 - n(n+1)/2} >= n!.

a(n) = ceiling((-1 + sqrt(1 + 8n! + 4n^2 + 4n))/2) and ignoring the -1 outside the sqrt and the 1 inside gives the approximate formula in the comment. - Joshua Zucker, May 08 2006

More terms from R. J. Mathar, Apr 21 2006
More terms from Joshua Zucker, May 08 2006
Name simplified by Jon E. Schoenfield, Jun 15 2019

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).

Original entry on oeis.org

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

Andrea La Rosa, Jun 23 2022

nonn

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

			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.

Cf. A000217, A001477, A093001.

Cf. A288998.

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

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

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

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)

cp's OEIS Frontend

A093000 Least k such that Sum_{r=n+1..k} r >= n!.

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