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#9 by Peter Woodward at Wed Feb 07 10:08:02 EST 2024
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Discussion
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Fri Mar 01
| 18:02
| Sean A. Irvine: Not convinced that these sequences with such a small number of displayable terms are useful. I find it hard to imagine anyone would search for this.
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Sat Mar 02
| 13:58
| N. J. A. Sloane: Not of general interest; recycled
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#8 by Peter Woodward at Wed Feb 07 10:07:44 EST 2024
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Discussion
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Wed Feb 07
| 10:07
| Peter Woodward: Yes, I added that, thank you.
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#3 by Peter Woodward at Thu Feb 01 10:28:18 EST 2024
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#2 by Peter Woodward at Thu Feb 01 10:26:16 EST 2024
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| NAME
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allocateda(1)=1, a(n) = H_(n-2)(a(n-2), a(n-1)) where H_n is forthe Petern-th Woodwardhyperoperator.
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| DATA
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1, 2, 3, 6, 216
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| OFFSET
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1,2
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| COMMENTS
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The sequence follows the hierarchy of arithmetic operations (successorship, addition, multiplication, exponentiation, tetration, ...) with the recurrence a(n-1)?a(n-2)=a(n), where "?" follows the sequence of S (successor), +, *, ^, ^^ (tetration), ...
a(6) (=216^^6) is too large to be represented.
Essentially a Fibonacci generalization: the Fibonacci hyperoperation sequence starting with 1.
Diverges from Cf. A367897 starting at a(5) because this is where commutativity is lost (exponentiation and tetration are not commutative).
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| LINKS
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Wikipedia, <a href="https://en.wikipedia.org/wiki/Binary_operation">Binary operation</a>
Wikipedia, <a href="https://en.wikipedia.org/wiki/Hyperoperation">Hyperoperation</a>
Wikipedia, <a href="https://en.wikipedia.org/wiki/Generalizations_of_Fibonacci_numbers">Generalizations of Fibonacci numbers</a>
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| EXAMPLE
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a(1)=1, a(2)=S(a(1)), a(3)=a(2)+a(1), a(4)=a(3)*a(2), a(5)=a(4)^a(3), a(6)=a(5)^^a(4), ...a(1) = 1
a(2) = H_0(a(1), a(1)) = 1 + 1 = 2 (successor of 1 = 2)
a(3) = H_1(a(2), a(1)) = 2 + 1 = 3
a(4) = H_2(a(3), a(2)) = 3 * 2 = 6
a(5) = H_3(a(4), a(3)) = 6^3 = 216
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| CROSSREFS
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Cf. A367897.
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| KEYWORD
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allocated
nonn,new
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| AUTHOR
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Peter Woodward, Feb 01 2024
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| STATUS
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approved
editing
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Discussion
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Thu Feb 01
| 10:28
| Peter Woodward: A054871 with recursion reversed on last two terms. Diverges starting on a(5) because 3^6 ≠ 6^3.
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#1 by Peter Woodward at Thu Feb 01 10:26:16 EST 2024
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| NAME
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allocated for Peter Woodward
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| KEYWORD
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allocated
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| STATUS
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approved
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#17 by Peter Woodward at Sun Jan 21 00:22:41 EST 2024
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#16 by Peter Woodward at Sun Jan 21 00:18:06 EST 2024
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| NAME
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a(1)=1, a(n) = H_(n-2)(a(n-12), a(n-21)) where H_n is the n-th hyperoperator.
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| STATUS
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approved
editing
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Discussion
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Sun Jan 21
| 00:22
| Peter Woodward: Switched (n-1) and (n-2) in the title.
a(1) = 1 (given)
a(2) = H_0(a(1), a(1)) = 1 + 1 = 2 (successor of 1 = 2)
a(3) = H_1(a(1), a(2)) = 1 + 2 = 3
a(4) = H_2(a(2), a(3)) = 2 * 3 = 6
a(5) = H_3(a(3), a(4)) = 3^6 = 729
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#14 by Peter Woodward at Tue Dec 26 14:29:27 EST 2023
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Discussion
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Tue Dec 26
| 14:58
| Michel Marcus: yes
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#13 by Peter Woodward at Tue Dec 26 14:19:31 EST 2023
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| LINKS
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Wikipedia, <a href="https://en.wikipedia.org/wiki/Generalizations_of_Fibonacci_numbers">Generalizations of Fibonacci numbers</a>
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| STATUS
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proposed
editing
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Discussion
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Tue Dec 26
| 14:29
| Peter Woodward: @Michel Marcus: I fixed wiki link, sorry. Yes A189896 has 4 terms: successor, addition, multiplication, exponentiation, and for the 5th states: The term is too big to be included (4^^4).
I meant "arbitrary to stop at a(5)" in the conceptual sense, I understand 6^^729 is an actual stumbling block for written depiction, which we cannot even estimate as far as I know.
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#12 by Peter Woodward at Tue Dec 26 11:40:37 EST 2023
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Discussion
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Tue Dec 26
| 12:00
| Michel Marcus: the new link is not ok
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| 12:03
| Michel Marcus: yes A189896 .... 4 terms ....
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| 12:08
| Michel Marcus: arbitrary to stop at a(5) ?? how big would it be ???
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