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%I #30 Dec 09 2022 07:07:12
%S 15,63,80,255,624,728,1023,1295,2400,4095,6560,9999,14640,15624,16383,
%T 20735,28560,38415,46655,50624,59048,65535,83520,104975,117648,130320,
%U 159999,194480,234255,262143,279840,331775,390624,456975,531440,614655
%N Positive numbers which are one less than a perfect square that is also another power.
%D William Dunham, Euler: The Master of Us All, The Mathematical Association of America, Washington D.C., 1999, p. 65.
%D Leonhard Euler, "Variae observationes circa series infinitas," Opera Omnia, Ser. 1, Vol. 14, pp. 216-244.
%D Nicolao Fvss, "Demonstratio Theorematvm Qvorvndam Analyticorvm," Nova Acta Academiae Scientiarum Imperialis Petropolitanae, 8 (1794) 223-226.
%H Amiram Eldar, <a href="/A062965/b062965.txt">Table of n, a(n) for n = 1..10000</a>
%H Leonard Euler, <a href="https://scholarlycommons.pacific.edu/euler-works/72/">Variae observationes circa series infinitas</a>.
%H Leonard Euler, <a href="http://math.dartmouth.edu/~euler/pages/E072.html">Variae observationes circa series infinitas</a>.
%H Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/PerfectPower.html">Perfect Power</a>.
%F From _Terry D. Grant_, Oct 25 2020: (Start)
%F a(n) = A001597(n+1)^2 - 1.
%F Sum_{k>=1} 1/a(k) = 7/4 - Pi^2/6 = 7/4 - zeta(2).
%F Sum_{k>=1} 1/(a(k)+1) = Sum_{k>=2} mu(k)*(1-zeta(2*k)).
%F (End)
%e a(2) = 63 because the perfect square 64 = 8^2 = 4^3.
%t Take[ Select[ Range[ 2, 150 ], GCD@@(Last/@FactorInteger[ # ])>1& ]^2-1] (* corrected by _Jon Maiga_, Sep 28 2019 *)
%Y Cf. A037450, A062834, A062757, A001597.
%Y Cf. A131605.
%K nonn
%O 1,1
%A _Jason Earls_, Jul 16 2001
%E More terms from _Dean Hickerson_, Jul 24 2001
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