login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

A321214
a(n) = ((2 + sqrt(5))^p + (2 - sqrt(5))^p - 2^(p+1))/p where p = prime(n).
1
5, 20, 260, 3460, 716100, 10877380, 2678663940, 43007216580, 11439823225220, 52423583379994820, 880012516784503300, 4260164250933079388740, 1237929447780495036788100, 21180545285375859022020420, 6239638330555928133105753860
OFFSET
1,1
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
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
MATHEMATICA
Table[Floor[(2+Sqrt[5])^(Prime[n]) + (2-Sqrt[5])^(Prime[n]) - 2^(Prime[n]+1)]/Prime[n], {n, 1, 10}]
PROG
(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
CROSSREFS
KEYWORD
nonn
AUTHOR
Jinyuan Wang, Oct 31 2018
STATUS
approved