A374578 Pierpont primes are primes of the form 2^t*3^u + 1; this sequence gives the u's in order.
0, 0, 0, 1, 1, 0, 2, 2, 2, 1, 3, 4, 1, 0, 3, 5, 2, 1, 2, 4, 6, 4, 6, 3, 5, 4, 1, 7, 2, 9, 8, 0, 7, 2, 8, 4, 10, 9, 6, 1, 8, 5, 2, 6, 3, 5, 4, 9, 4, 12, 11, 3, 14, 3, 15, 5, 7, 16, 13, 3, 10, 4, 17, 10, 11, 12, 3, 4, 1, 8, 5, 8, 4, 11, 7, 15, 12, 2, 10, 1, 22, 4
Offset: 1
Keywords
Examples
a(1) = 0, because the first Pierpont prime is 2 = 2^0 * 3^0 + 1. a(6) = 0, because the sixth Pierpont prime is 17 = 2^4 * 3^0 + 1. a(7) = 2, because the seventh Pierpont prime is 19 = 2^1 * 3^2 + 1.
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000
Programs
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Mathematica
With[{lim = 10^12}, IntegerExponent[Select[Sort@ Flatten@Table[2^i*3^j + 1, {i, 0, Log2[lim]}, {j, 0, Log[3, lim/2^i]}], PrimeQ] - 1, 3]] (* Amiram Eldar, Sep 02 2024 *)
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PARI
lista(lim) = {my(s = List()); for(i = 0, logint(lim, 2), for(j = 0, logint(lim >> i, 3), listput(s, 2^i * 3^j + 1))); s = Set(s); for(i = 1, #s, if(isprime(s[i]), print1(valuation(s[i] - 1, 3), ", ")));} \\ Amiram Eldar, Sep 02 2024
Extensions
More terms from Stefano Spezia, Jul 12 2024
Comments