STATUS
proposed
approved
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Revision History for A004253
(Bold, blue-underlined text is an addition; faded, red-underlined text is aShowing entries 1-10 | older changes
STATUS
editing
proposed
COMMENTS
(sqrt(21)+5))/2 = 4.7912878... = exp(arccosh(5/2)) = 4 + 3/4 + 3/(4*19) + 3/(19*91) + 3/(91*436) + ... - Gary W. Adamson, Dec 18 2007
STATUS
approved
editing
STATUS
reviewed
approved
STATUS
proposed
reviewed
STATUS
editing
proposed
FORMULA
For n, j, k in Z, a(n+2)*a(n+j+k) - a(n+j)*a(n+k) = 3*A004254(j)*A004254(k). The case j = 1)^, k = 2 = 3is given above.
More generally, for n, k in Z, a(n+2*k)*a(n) - a(n+k)^2 = 3*A004254(k)^2 = A003690(k+1).
a(n)*a(n+k+1) - a(n+1)*a(n+k) = 3*A004254(k). The case k = 2 is noted above.
More generally, for n, j, k in Z, a(n)*a(n+j+k) - a(n+j)*a(n+k) = 3*A004254(j)* A004254(k ).
FORMULA
a(n) = a(1-n).
a(n) = A004254(n) + A004254(1-n).
For n >= 1, a(n+2)*a(n) - a(n+1)^2 = 3.
More generally, for n, k >= 0, in Z, a(n+2*k)*a(n) - a(n+k)^2 = 3*A004254(k)^2 = A003690(k+1).
For k >= 0, a(n)*a(n+k+1) - a(n+1)*a(n+k) = 3*A004254(k). The case k = 2 is noted above.
More generally, for n, j, k, r >= 0, in Z, a(n)*a(n+j+k+r) - a(n+kj)*a(n+rk) = 3*A004254(rj)* A004254(kj).
a(n)^2 + a(n-+1)^2 - 5*a(n)*a(n+1) = - 3. (End)
More generally, a(n)^2 + a(n+k)^2 - (A004254(k+1) - A004254(k-1))*a(n)*a(n+k) = -3*A004254(k)^2. (End)
FORMULA
For n >= 1, a(n+2)*a(n-2) - a(n-+1)^2 = 3.
More generally, for k >= 0, a(n)^+2 + *k)*a(n-1)^2 - 5*a(n+k)^2 = 3*aA004254(k)^2 = A003690(nk+1) = -3. - _Peter Bala_, Feb 10 2024
For k >= 0, a(n)*a(n+k+1) - a(n+1)*a(n+k) = 3*A004254(k). The case k = 2 is noted above.
More generally, for k, r >= 0, a(n)*a(n+k+r) - a(n+k)*a(n+r) = A004254(r)* A004254(k).
a(n)^2 + a(n-1)^2 - 5*a(n)*a(n+1) = -3. (End)