A062965 Positive numbers which are one less than a perfect square that is also another power.

15, 63, 80, 255, 624, 728, 1023, 1295, 2400, 4095, 6560, 9999, 14640, 15624, 16383, 20735, 28560, 38415, 46655, 50624, 59048, 65535, 83520, 104975, 117648, 130320, 159999, 194480, 234255, 262143, 279840, 331775, 390624, 456975, 531440, 614655

Offset: 1

Author

Jason Earls, Jul 16 2001

Keywords

• nonn

Examples

			a(2) = 63 because the perfect square 64 = 8^2 = 4^3.
		

References

• William Dunham, Euler: The Master of Us All, The Mathematical Association of America, Washington D.C., 1999, p. 65.
• Leonhard Euler, "Variae observationes circa series infinitas," Opera Omnia, Ser. 1, Vol. 14, pp. 216-244.
• Nicolao Fvss, "Demonstratio Theorematvm Qvorvndam Analyticorvm," Nova Acta Academiae Scientiarum Imperialis Petropolitanae, 8 (1794) 223-226.

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000
• Leonhard Euler, Variae observationes circa series infinitas.
• Eric Weisstein's World of Mathematics, Perfect Power.

Crossrefs

Cf. A037450, A062834, A062757, A001597.

Cf. A131605.

Programs

Mathematica

Take[ Select[ Range[ 2, 150 ], GCD@@(Last/@FactorInteger[ # ])>1& ]^2-1] (* corrected by Jon Maiga, Sep 28 2019 *)

Python

from sympy import mobius, integer_nthroot
def A062965(n):
    def f(x): return int(n-1+x+sum(mobius(k)*(integer_nthroot(x,k)[0]-1) for k in range(2,x.bit_length())))
    kmin, kmax = 1,2
    while f(kmax) >= kmax:
        kmax <<= 1
    while True:
        kmid = kmax+kmin>>1
        if f(kmid) < kmid:
            kmax = kmid
        else:
            kmin = kmid
        if kmax-kmin <= 1:
            break
    return kmax**2-1 # Chai Wah Wu, Aug 14 2024

Formula

From Terry D. Grant, Oct 25 2020: (Start)

a(n) = A001597(n+1)^2 - 1.

Sum_{k>=1} 1/a(k) = 7/4 - Pi^2/6 = 7/4 - zeta(2).

Sum_{k>=1} 1/(a(k)+1) = Sum_{k>=2} mu(k)*(1-zeta(2*k)).

(End)

Extensions

More terms from Dean Hickerson, Jul 24 2001