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
A255770
Number of distinct prime factors of A220161(n).
Original entry on oeis.org
1, 2, 3, 4, 6, 8, 9, 11, 13, 17, 19, 21
Offset: 0
A220161(0) = 7 so a(0) = 1.
A220161(1) = 3*7 so a(1) = 2.
A220161(2) = 3*7*13 so a(2) = 3.
A220161(3) = 3*7*13*241 so a(3) = 4.
A220161(4) = 3*7*13*97*241*673 so a(4) = 6.
- Arthur Engel, Problem-Solving Strategies, Springer, 1998, pages 121-122 (E3, said to be a "recent competition problem from the former USSR").
Showing 1-4 of 4 results.
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