A332115
a(n) = (10^(2n+1)-1)/9 + 4*10^n.
Original entry on oeis.org
5, 151, 11511, 1115111, 111151111, 11111511111, 1111115111111, 111111151111111, 11111111511111111, 1111111115111111111, 111111111151111111111, 11111111111511111111111, 1111111111115111111111111, 111111111111151111111111111, 11111111111111511111111111111, 1111111111111115111111111111111
Offset: 0
- Brady Haran and Simon Pampena, Glitch Primes and Cyclops Numbers, Numberphile video (2015).
- Patrick De Geest, Palindromic Wing Primes: (1)5(1), updated: June 25, 2017.
- Makoto Kamada, Factorization of 11...11511...11, updated Dec 11 2018.
- Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).
Cf.
A138148 (cyclops numbers with binary digits),
A002113 (palindromes).
Cf.
A332125 ..
A332195 (variants with different repeated digit 2, ..., 9).
Cf.
A332112 ..
A332119 (variants with different middle digit 2, ..., 9).
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A332115 := n -> (10^(2*n+1)-1)/9+4*10^n;
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Array[(10^(2 # + 1)-1)/9 + 4*10^# &, 15, 0]
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apply( {A332115(n)=10^(n*2+1)\9+4*10^n}, [0..15])
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def A332115(n): return 10**(n*2+1)//9+4*10**n
A331869
Numbers n for which R(n) + 4*10^floor(n/2) is prime, where R(n) = (10^n-1)/9 (repunit: A002275).
Original entry on oeis.org
1, 3, 4, 15, 76, 91, 231, 1363, 1714, 1942, 2497, 4963, 5379, 12397, 23224, 26395
Offset: 1
For n = 1, R(1) + 4*10^floor(1/2) = 5 is prime.
For n = 3, R(3) + 4*10^floor(3/2) = 151 is prime.
For n = 4, R(4) + 4*10^floor(4/2) = 1511 is prime.
For n = 15, R(15) + 4*10^floor(15/2) = 111111151111111 is prime.
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Select[Range[0, 2500], PrimeQ[(10^# - 1)/9 + 4*10^Floor[#/2]] &]
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for(n=0,9999,ispseudoprime(p=10^n\9+4*10^(n\2))&&print1(n","))
Showing 1-2 of 2 results.
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