A220294
a(n) = 1 - 2^(2^n) + 2^(2^(n+1)).
Original entry on oeis.org
3, 13, 241, 65281, 4294901761, 18446744069414584321, 340282366920938463444927863358058659841, 115792089237316195423570985008687907852929702298719625575994209400481361428481
Offset: 0
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[1 - 2^(2^n) + 2^(2^(n+1)): n in [0..10]]; // G. C. Greubel, Aug 10 2018
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Table[4^(2^m) - 2^(2^m) + 1, {m, 0, 7}] (* Michael De Vlieger, Aug 02 2016 *)
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A220294(n):=1 - 2^(2^n) + 2^(2^(n+1))$ makelist(A220294(n),n,0,10); /* Martin Ettl, Dec 10 2012 */
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{a(n) = if( n<0, 0, 1 - 2^(2^n) + 2^(2^(n+1)))};
A255772
Start with 7; thereafter, in order of appearance, the prime factors of A220294.
Original entry on oeis.org
7, 3, 13, 241, 97, 673, 193, 22253377, 18446744069414584321, 769, 442499826945303593556473164314770689, 349621839326921795694385454593, 331192380488114152600457428497953408512758882817, 212780015855109121
Offset: 1
- Arthur Engel, Problem-Solving Strategies, Springer, 1998, pages 121-122 (E3, said to be a "recent competition problem from the former USSR").
A220161
a(n) = 1 + 2^(2^n) + 2^(2^(n+1)).
Original entry on oeis.org
7, 21, 273, 65793, 4295032833, 18446744078004518913, 340282366920938463481821351505477763073, 115792089237316195423570985008687907853610267032561502502920958615344897851393
Offset: 0
- 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.
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[1 + 2^(2^n) + 2^(2^(n+1)): n in [0..10]]; // G. C. Greubel, Aug 10 2018
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Table[1+2^(2^n)+2^(2^(n+1)),{n,0,7}] (* Harvey P. Dale, Dec 16 2015 *)
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A220161(n):=1 + 2^(2^n) + 2^(2^(n+1))$ makelist(A220161(n),n,0,10); /* Martin Ettl, Dec 10 2012 */
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vector(10, n, n--; 1 + 2^(2^n) + 2^(2^(n+1))) \\ G. C. Greubel, Aug 10 2018
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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
A255771
Number of distinct prime factors of A220294(n).
Original entry on oeis.org
1, 1, 1, 2, 2, 1, 2, 2, 4, 2, 2
Offset: 0
A220294(0) = 3 so a(0) = 1.
A220294(1) = 13 so a(1) = 1.
A220294(2) = 241 so a(2) = 1.
A220294(3) = 97*673 so a(3) = 2.
A220294(4) = 193*22253377 so a(4) = 2.
- Arthur Engel, Problem-Solving Strategies, Springer, 1998, pages 121-122 (E3, said to be a "recent competition problem from the former USSR").
a(9) was found in 2008 by Geoffrey Reynolds. a(10) was found by Anders Björn and Hans Riesel. -
Arkadiusz Wesolowski, Aug 02 2016
Showing 1-4 of 4 results.
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