A277730 Irregular triangle read by rows: T(n,k) = number of times a gap of 2k occurs between the first n successive odd primes.

1, 2, 2, 1, 3, 1, 3, 2, 4, 2, 4, 3, 4, 3, 1, 5, 3, 1, 5, 3, 2, 5, 4, 2, 6, 4, 2, 6, 5, 2, 6, 5, 3, 6, 5, 4, 7, 5, 4, 7, 5, 5, 7, 6, 5, 8, 6, 5, 8, 6, 6, 8, 7, 6, 8, 7, 7, 8, 7, 7, 1, 8, 8, 7, 1, 9, 8, 7, 1, 9, 9, 7, 1, 10, 9, 7, 1, 10, 10, 7, 1, 10, 10, 7, 1, 0, 0, 1, 10, 11, 7, 1, 0, 0, 1

Offset: 2

Author

N. J. A. Sloane, Nov 06 2016

Keywords

• nonn
• look
• tabf

Comments

T(m, 2) = max{T(m, k): k >= 1} for m >= 100. - Ya-Ping Lu, Dec 25 2024

Examples

			Triangle begins:
  1,
  2,
  2, 1,
  3, 1, <- gaps in 3,5,7,11,13 are 2 (3 times), 4 (once)
  3, 2,
  4, 2,
  4, 3,
  4, 3, 1,
  5, 3, 1,
  5, 3, 2,
  5, 4, 2,
  6, 4, 2,
  6, 5, 2,
  6, 5, 3,
  6, 5, 4,
  7, 5, 4,
  7, 5, 5,
  7, 6, 5,
  8, 6, 5,
  ...
		

Links

• Lars Blomberg, Table of n, a(n) for n = 2..9997 (the first 703 rows)

Crossrefs

Cf. A277729, A213930.

Programs

Python

from sympy import nextprime; p = 3; L = []
for n in range(2, 32):
    np = nextprime(p); k = (np - p)//2
    if len(L) < k: {L.append(0) for i in range(len(L), k-1)}; L.append(1)
    else: L[k-1] += 1
print(*L, sep =", ", end = ", "); p = np  # Ya-Ping Lu, Dec 25 2024