Azerkail
Member
|
755 can be obtained from or falls in the categories of the following (Quotes box any formulae)...
Numbers of form n + (n+1)^2.
Can be obtained as sum of Smarandache mirror sequence terms.
Values of Fibbonacci polynomial n^2-n-1 for n=2,3,4,5,..
quote:
Not of form k + [ sqrt(k) ], k integer.
a(n)=sqrt( n(n+1)(n+2)(n+3) + 1 ).
Number of benzenoids with 22 hexagons, C_(3h) symmetry, and containing n carbon atoms
Numbers whose set of base 5 digits is {0,1}.
Number of fractions in Farey series of order n.
quote:
a(n) = 1+Sum_{i=1..n} phi(i).
a(n) = n(n+3)/2 - Sum(k = 2 to n, a([n/k])).
Expansion of x^2*(1+x-x^2)/((1-x^2)*(1-x)^2).
quote:
a(0)=0, a(n) = floor(a(n-1)+sqrt(a(n-1))+1) for n > 0
Primes with digits (0,1) taken as base 2 and converted to base 10.
a(1)=a(2)=1; for n >= 1, a(n+2)=a(n+1)+a(n)+(-1)^n.
quote:
for n>4 a(n-2) = floor( 2*PHI^n/sqrt(5)) +(1+(-1)^n)/2
a(n) = 2*Fibonacci(n-2)+(-1)^n.
Square-free numbers beginning with 7.
Sum of three distinct positive cubes.
Numbers n such that n^4 is an interprime = average of two successive primes.
Numbers n such that numerator of Bernoulli(2n) is divisible by 67, the third irregular prime.
Least non-trivial multiple of the n-th prime beginning with 7.
Sum of digits of n-th prime equals sum of digits of n.
Positive integers n such that n+d+1 is prime for all divisors d of n.
Numbers n such that 2*6^n -1 is prime.
Let S_n be the infinite sequence formed by starting with n and repeatedly reversing the digits and adding 16 to get the next term. Sequence gives number of steps for S_n to reach a cycle, or -1 if no cycle is ever reached.
Multiples of 5 having only odd digits.
Conjectured values for the minimal number a(n) so that the 'reverse and add!'-algorithm in base n does not terminate in a palindrome. If there is no such number in base n, then a(n) := -1.
n - 4^k is a prime or 1 for all k > 0 and n > 4^k.
Main diagonal of square maze arrangement of natural numbers
n such that prime(n)*prime(n+1) - 1 is semiprime.
Triangle, read by rows, where T(n,k) equals the least m>0 that produces the maximum number of partial quotients in the simple continued fraction expansion of (1/n + 1/k + 1/m).
Numbers n such that Sum_{k=1..n} Catalan(k) == 2 mod 3.
Numbers n such that GCD(n!+1, n^n+1) > 1.
|
Printable Version
Icon Legend
New Messages
No New Messages
Hot Topic w/ New Messages
Hot Topic w/o New Messages
Locked w/ New Messages
Locked w/o New Messages
Post New Thread
