OFFSET
1,3
COMMENTS
For p > 3, if p is prime, then p^2 divides a(p). Conjecture: for n > 3, if n^2 divides a(n), then n is prime. Primes p such that p^3 diviedes a(p) are probably A088164. - Thomas Ordowski, Mar 30 2025
LINKS
Eric Weisstein's World of Mathematics, Harary Index
Eric Weisstein's World of Mathematics, Halved Cube Graph
FORMULA
a(n) = -2^(n-1)*HarmonicNumber(n)-2^(2*n-1)*Re(LerchPhi(2,1,n+1)).
EXAMPLE
First few terms are 0, 1, 6, 26, 100, 1096/3, 3920/3, 13936/3, 16544, 296256/5, ....
MATHEMATICA
Table[-2^(n - 1) HarmonicNumber[n] - 2^(2 n - 1) Re[LerchPhi[2, 1, n + 1]], {n, 20}] // Numerator
CROSSREFS
KEYWORD
nonn,frac
AUTHOR
Eric W. Weisstein, Jul 28 2017
STATUS
approved
