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A215099
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a(0)=0, a(1)=1, a(n) = least k>a(n-1) such that k+a(n-2) is prime.
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2
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0, 1, 2, 4, 5, 7, 8, 10, 11, 13, 18, 24, 25, 29, 34, 38, 39, 41, 44, 48, 53, 55, 56, 58, 71, 73, 78, 84, 85, 89, 94, 102, 103, 109, 120, 124, 131, 133, 138, 144, 145, 149, 162, 164, 169, 173, 178, 180, 181, 187, 192, 196, 197, 201
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OFFSET
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0,3
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COMMENTS
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For n>0 and (n mod 4)<2, a(n) is odd.
Same definition, but k+a(n-2) is a
Fibonacci number: A006498 except first two terms,
Lucas number: A000045 except first two terms,
Example of a related sequence definition: a(0)=0, a(1)=1, a(n) = least k>a(n-1) such that k+a(n-2) is a cube.
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LINKS
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PROG
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(Python)
# primes = [2, ..., 401]
prpr = 0
prev = 1
for n in range(77):
print prpr,
b = c = 0
while c<=prev:
c = primes[b] - prpr
b+=1
prpr = prev
prev = c
(PARI) first(n) = my(res = vector(n, i, i-1), k); for(x=3, n, k=res[x-1]+1; while(!isprime(k+res[x-2]), k++); res[x]=k); res \\ Iain Fox, Apr 22 2019 (corrected by Iain Fox, Apr 25 2019)
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CROSSREFS
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Cf. A062042: a(1) = 2, a(n) = least k>a(n-1) such that k+a(n-1) is a prime.
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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