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A349644
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Array read by antidiagonals, n >= 2, m >= 0: T(n,m) is the smallest prime p = prime(k) such that all n-th differences of (prime(k), ..., prime(k+n+m)) are zero.
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5
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OFFSET
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2,1
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COMMENTS
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T(n,m) = prime(k), where k is the smallest positive integer such that A095195(j,n) = 0 for k+n <= j <= k+n+m.
Equivalently, T(n,m) is the smallest prime p = prime(k) such that there is a polynomial f of degree at most n-1 such that f(j) = prime(j) for k <= j <= k+n+m.
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LINKS
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FORMULA
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T(n,m) <= T(n-1,m+1).
T(n,m) <= T(n, m+1).
Sum_{j=0..n} (-1)^j*binomial(n,j)*prime(k+i+j) = 0 for 0 <= i <= m, where prime(k) = T(n,m).
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EXAMPLE
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Array begins:
n\m| 0 1 2 3 4
---+------------------------------------------------
2 | 3 251 9843019 121174811 ?
3 | 17 347 2903 15373 128981
4 | 347 2903 15373 128981 19641263
5 | 41 8081 128981 19641263 245333213
6 | 211 128981 19641263 245333213 245333213
7 | 271 386471 81028373 245333213 27797667517
8 | 23 2022971 245333213 27797667517 ?
9 | 191 7564091 10246420463 ? ?
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PROG
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(Python)
from sympy import nextprime
d = [float('inf')]*(n-1)
p = [0]*(n+m)+[2]
c = 0
while 1:
del p[0]
p.append(nextprime(p[-1]))
d.insert(0, p[-1]-p[-2])
for i in range(1, n):
d[i] = d[i-1]-d[i]
if d.pop() == 0:
if c == m: return p[0]
c += 1
else:
c = 0
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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