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 A220161 a(n) = 1 + 2^(2^n) + 2^(2^(n+1)). 5
 7, 21, 273, 65793, 4295032833, 18446744078004518913, 340282366920938463481821351505477763073, 115792089237316195423570985008687907853610267032561502502920958615344897851393 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,1 COMMENTS For n >= 1, W. Sierpiński proves that a(n) is divisible by 21. For n >= 1, A. Engel shows that a(n) = a(n-1) * A220294(n-1). - Hans Havermann, Mar 07 2015 REFERENCES Arthur Engel, Problem-Solving Strategies, Springer, 1998, pages 121-122 (E3, said to be a "recent competition problem from the former USSR"). W. Sierpiński, 250 Problems in Elementary Number Theory. New York: American Elsevier, 1970. Problem #123. LINKS G. C. Greubel, Table of n, a(n) for n = 0..10 K. Zelator, On a theorem on sums of the form 1+2^(2^n)+2^(2^n+1)+...+2^(2^n+m) and a result linking Fermat with Mersenne numbers, arXiv:0806.1514 [math.GM], 2008. FORMULA a(n) = A000215(n+1) + A000215(n) - 1. A070969(n) = sqrt(4*a(n) - 3). a(n+1) = a(n) * (1 + a(n) - A070969(n)) = a(n) * (1 + A087046(n+2)) hence a(n) divides a(n+1). - Michael Somos, Dec 10 2012 MATHEMATICA Table[1+2^(2^n)+2^(2^(n+1)), {n, 0, 7}] (* Harvey P. Dale, Dec 16 2015 *) PROG (Maxima) A220161(n):=1 + 2^(2^n) + 2^(2^(n+1))\$ makelist(A220161(n), n, 0, 10); /* Martin Ettl, Dec 10 2012 */ (PARI) vector(10, n, n--; 1 + 2^(2^n) + 2^(2^(n+1))) \\ G. C. Greubel, Aug 10 2018 (MAGMA) [1 + 2^(2^n) + 2^(2^(n+1)): n in [0..10]]; // G. C. Greubel, Aug 10 2018 (Python) def a(n): return 1 + 2**(2**n) + 2**(2**(n+1)) print([a(n) for n in range(8)]) # Michael S. Branicky, Jul 21 2021 CROSSREFS Cf. A000215, A070969, A087046, A220294, A255770, A255771, A255772. Sequence in context: A111878 A133279 A192734 * A213152 A082826 A130740 Adjacent sequences:  A220158 A220159 A220160 * A220162 A220163 A220164 KEYWORD nonn,changed AUTHOR Michel Marcus, Dec 06 2012 STATUS approved

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Last modified July 29 17:05 EDT 2021. Contains 346346 sequences. (Running on oeis4.)