A114409 Length of all-prime chain of prime[n] + successive even pentagonal numbers.

1, 2, 3, 2, 1, 2, 4, 1, 2, 4, 1, 2, 1, 2, 1, 2, 4, 4, 2, 1, 1, 1, 2, 2, 2, 1, 1, 1, 1, 4, 1, 2, 2, 1, 3, 1, 1, 2, 1, 2, 2, 1, 1, 1, 2, 4, 1, 2, 3, 1, 2, 1, 2, 2, 1, 2, 3, 1, 2, 1, 1, 1, 1, 1, 1, 1, 3, 2, 1, 1, 1, 3, 1, 1, 1, 2, 4, 1, 4, 2, 4, 2, 1, 1, 1, 2, 1

Offset: 2

Author

Jonathan Vos Post, Feb 11 2006

Keywords

• easy
• nonn

Comments

a(1) is undefined, as prime(1) is the only even prime, for which the length-5 chain is of 2 + successive odd pentagonal numbers A014632: 2 prime, 2+1 = 3 prime, 2+5 = 7 prime, 2+35 = 37 prime, 2+51 = 53 prime, but 2+117 = 119 = 7 * 17 nonprime. Pentagonal numbers A000326 = n*(3*n-1)/2 = 0, 1, 5, 12, 22, 35, 51, 70, 92, 117, 145, ... Even pentagonal numbers A014633 = 12, 22, 70, 92, 176, 210, 330, 376, 532, 590, 782, 852, 1080, ...

Examples

			a(2) = 1 because prime(2) = 3 is prime, but prime(2) + EvenPent(1) = 3 + 12 = 15 = 3 * 5 is nonprime, giving a chain of just 1 successive prime.
a(3) = 2 because 5 is prime, prime(3) + EvenPent(1) = 5 + 12 = 17 is prime, but prime(3) + EvenPent(2) = 5 + 22 = 27 = 3^3 is nonprime, giving a chain of 2 successive primes.
a(4) = 3 because 7 is prime, 7+12 = 19 is prime, 7+22 = 29 is prime, but 7+70 = 77 = 7*11 is nonprime, for a chain of 3 successive primes.
		

Crossrefs

Cf. A000040, A000326, A014633.

Programs

Mathematica

evp = Select[#*(3*# - 1)/2 &@ Range[200], EvenQ]; a[n_] := Block[{s = Prime@n, c = 1}, While[PrimeQ[s + evp[[c]]], c++]; c]; a /@ Range[2, 90] (* Giovanni Resta, Jun 14 2016 *)

Formula

a(n) = k = length of chain prime[n] + A014633(1) + ... + A014633(k) such that each term in the chain is prime.

Extensions

Corrected and extended by Giovanni Resta, Jun 14 2016