A127921 -id:A127921 - cp's OEIS frontend

A127922 1/24 of product of three numbers: n-th prime, previous and following number.

1, 5, 14, 55, 91, 204, 285, 506, 1015, 1240, 2109, 2870, 3311, 4324, 6201, 8555, 9455, 12529, 14910, 16206, 20540, 23821, 29370, 38024, 42925, 45526, 51039, 53955, 60116, 85344, 93665, 107134, 111895, 137825, 143450, 161239, 180441, 194054

Offset: 2

Artur Jasinski, Feb 06 2007

nonn

The product of (n-1), n, and (n+1) = n^3 - n. - Harvey P. Dale, Jan 17 2011
For n > 2, a(n) = A001318(n-2) * A007310(n-1), if A007310(n-1) is prime. Also a(n) is a subsequence of A000330. - Richard R. Forberg, Dec 25 2013
If p is an odd prime it can always be the side length of a leg of a primitive Pythagorean triangle. However it constrains the other leg to have a side length of (p^2-1)/2 and the hypotenuse to have a side length of (p^2+1)/2. The resulting triangle has an area equal to (p-1)*p*(p+1)/4. a(n) is 1/6 the area of such triangles. - Frank M Jackson, Dec 06 2017

G. C. Greubel, Table of n, a(n) for n = 2..1000
César Aguilera, Two Prime Number Objects and The Velucchi Numbers, hal-02909691 [math.NT], 2020.

Cf. A000040, A000330, A011842.

Cf. A036689, A034953, A127917, A127918, A127919, A127920, A127921, A011842.

Table[(Prime[n] + 1) Prime[n](Prime[n] - 1)/24, {n, 1, 100}] (#^3-#)/ 24&/@ Prime[Range[2,40]] (* Harvey P. Dale, Jan 17 2011 *)
((#-1)#(#+1))/24&/@Prime[Range[2,40]] (* Harvey P. Dale, Jan 20 2023 *)

for(n=2,25, print1((prime(n)+1)*prime(n)*(prime(n)-1)/24, ", ")) \\ G. C. Greubel, Jun 19 2017

a(n) = A011842(A000040(n) + 1) = A000330((A000040(n) - 1) / 2).

A380790 Length of the n-th Golomb ruler constructed by the Paul Erdős and Pál Turán formula.

Original entry on oeis.org

20, 110, 308, 1254, 2106, 4760, 6650, 11822, 23954, 29202, 49950, 68060, 78518, 102460, 147446, 203432, 225090, 298418, 354858, 386316, 489484, 568052, 700964, 907920, 1025150, 1086856, 1218944, 1289034, 1436456, 2039620, 2238790, 2561900, 2675472, 3296774, 3430418

Offset: 2

Darío Clavijo, Feb 03 2025

nonn

In October of 1941 Paul Erdős and Pál Turán found that a Golomb ruler could be constructed for every odd prime p.
Such a ruler has the property that the mark or notches are defined by: notch(k) = 2pk + (k^2 mod p) for k in {0..p-1}, with p=A000040(n).
Empirical observation: a(n) satisfies p^3-p^2 <= a(n)/p^3 <= 0.9999.
Except for n=2, a(n) is divisible by p.
Also partial sums of A217793.

			 n | p  | Golomb ruler notches                             | a(n)
---+----+--------------------------------------------------+-------
 2 | 3  | 0, 7,  13                                        | 20
 3 | 5  | 0, 11, 24, 34, 41                                | 110
 4 | 7  | 0, 15, 32, 44, 58, 74,  85                       | 308
 5 | 11 | 0, 23, 48, 75, 93, 113, 135, 159, 185, 202, 221  | 1254

P. Erdös and P. Turán, On a Problem of Sidon in Additive Number Theory, and on some Related Problems, Journal of the London Mathematical Society, Volume s1-16, Issue 4, October 1941, Pages 212-215.
Wikipedia, Golomb ruler
Index entries for sequences related to Golomb rulers

Cf. A000010, A000040, A000330, A048153, A076409, A100104, A127921, A135177, A160378, A217793.

a(n)= if(n==2, return(20));  my(p=prime(n)); if(bitand(p, 3)==1, return((p*(p-1)*(2*p+1))/2)); if(bitand(p, 3)==3, return((p*(p-1)*(2*p+1))/2 - p * qfbclassno(-p)));

from sympy import prime
from math import isqrt
def a(n):
  p = prime(n)
  if p & 3 == 1: return (p*(p-1)*(2*p+1))//2
  m = isqrt(p-1)
  return (p-1) * p**2 + (m*(m+1)*(2*m+1))//6 + sum(pow(k,2,p) for k in range(m+1,p))
print([a(n) for n in range(2, 37) ])

a(n) = Sum_{k=0..p-1} (2*k*p + k^2 mod p), where p is the n-th prime.
a(n) = (p-1)*p^2 + 1 + Sum_{k=2..p-1} (k^2 mod p), where p is the n-th prime.
a(n) = (p-1)*p^2 + A000330(m) + Sum_{k=m+1..p-1} (k^2 mod p), where m = floor(sqrt(p-1)) and p is the n-th prime.
a(n) = (p-1)*p^2 + p*(p-1)*(p+1)/12 - 2*p*(Sum_{k=1..(p-1)/2} floor(k^2/p)), where p is the n-th prime.
a(n) = A100104(A000040(n)) + A048153(A000040(n)) - 1.
a(n) = A100104(A000040(n)) + A076409(n).
a(n) = A160378(A000040(n)), iif A000040(n) = 1 (mod 4).
a(n) = A160378(A000040(n)) - A000040(n)*A355879(n), iif A000040(n) = 3 (mod 4).
a(n) < A000040(n)^3.
a(n) > A000040(n)^3 - A000040(n)^2.
a(n) = 0 mod A000040(n) for n >= 3.
a(n) = Sum_{k=0..A000040(n)-1} A217793(n - 1, k).
a(n) = A135177(n) + A127921(n) - 2*p*(Sum_{k=1..(p-1)/2} floor(k^2/p)), where p = A000040(n).

cp's OEIS Frontend

A127922 1/24 of product of three numbers: n-th prime, previous and following number.

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