A321214 a(n) = ((2 + sqrt(5))^p + (2 - sqrt(5))^p - 2^(p+1))/p where p = prime(n).

5, 20, 260, 3460, 716100, 10877380, 2678663940, 43007216580, 11439823225220, 52423583379994820, 880012516784503300, 4260164250933079388740, 1237929447780495036788100, 21180545285375859022020420, 6239638330555928133105753860

Offset: 1

Author

Jinyuan Wang, Oct 31 2018

Keywords

• nonn

Comments

This is an integer sequence. For odd primes p, (2 + sqrt(5))^p + (2 - sqrt(5))^p - 2^(p+1) = binomial(p, 2)*2^(p-1)*5 + binomial(p, 4)*2^(p-3)*5^2 + ... + binomial(p, p-1)*2^2*5^((p-1)/2), and p divides binomial(p, k) for 1 <= k <= p - 1.

For n > 1, a(n) is divisible by 20.

Links

• Jinyuan Wang, Table of n, a(n) for n = 1..250

Crossrefs

Cf. A000040, A014448, A064535, A000032.

Programs

Mathematica

Table[Floor[(2+Sqrt[5])^(Prime[n]) + (2-Sqrt[5])^(Prime[n]) - 2^(Prime[n]+1)]/Prime[n], {n, 1, 10}]

PARI

a(n) = my(p=prime(n)); (floor((2*quadgen(5)+1)^p+(-2*quadgen(5)+3)^p+.) - 2^(p+1))/p; \\ Michel Marcus, Nov 04 2018

PARI

a(n) = my(p=prime(n)); (([1,1;1,0]^(3*p)*[1;2])[2,1] - 2^(p+1))/p \\ Jianing Song, Dec 22 2018

Formula

a(n) = Sum_{k=1..(p-1)/2} (binomial(p, 2*k)/p)*2^(p-2*k+1)*5^k with p = A000040(n), for n > 1.

a(n) = (A014448(prime(n)) - 4)/prime(n) - 2*A064535(n).

a(n) = (A000032(3*prime(n)) - 4)/prime(n) - 2*A064535(n). - Jianing Song, Dec 22 2018