A176594 -id:A176594 - cp's OEIS frontend

A011764 a(n) = 3^(2^n) (or: write in base 3, read in base 9).

3, 9, 81, 6561, 43046721, 1853020188851841, 3433683820292512484657849089281, 11790184577738583171520872861412518665678211592275841109096961

Offset: 0

Stephan Y Solomon (ilans(AT)way.com)

a(n) is the second-highest value k such that A173419(k) = n+2. - Charles R Greathouse IV, Oct 03 2012
Let b(0) = 6; b(n+1) = smallest number such that b(n+1) + Product_{i=0..n} b(i) divides b(n+1)*Product_{i=0..n} b(i). Then b(n+1) = a(n) for n >= 0. - Derek Orr, Dec 13 2014
Changing "+" to "-": Let b(0) = 6; b(n+1) = smallest number such that b(n+1) - Product_{i=0..n} b(i) divides b(n+1)*Product_{i=0..n} b(i). Then b(n+2) = a(n) for n >= 0. - Derek Orr, Jan 04 2015
With offset = 1, a(n) is the number of collections C of subsets of {1,2,...,n} such that if S is in C then the complement of S is not in C. - Geoffrey Critzer, Feb 06 2017

Vincenzo Librandi, Table of n, a(n) for n = 0..11

Cf. A001146, A078885, A176594.

Subsequence of A000244 (powers of 3).

[3^(2^n): n in [0..8]]; // Vincenzo Librandi, Sep 15 2011

3^(2^Range[0,10]) (* Harvey P. Dale, Oct 14 2012 *)

a(n)=3^2^n \\ Charles R Greathouse IV, Oct 03 2012

def A011764(n): return 3**(1<Chai Wah Wu, Oct 09 2024

a(0) = 3 and a(n+1) = a(n)^2. - Benoit Jubin, Jun 27 2009
Sum_{n>=0} 1/a(n) = A078885. - Amiram Eldar, Nov 09 2020
Product_{n>=0} (1 + 1/a(n)) = 3/2. - Amiram Eldar, Jan 29 2021
a(n) = A000244(A000079(n)), or A011764 = A000244 o A000079. - M. F. Hasler, Jul 20 2023

A329050 Square array A(n,k) = prime(n+1)^(2^k), read by descending antidiagonals (0,0), (0,1), (1,0), (0,2), (1,1), (2,0), ...; Fermi-Dirac primes (A050376) in matrix form, sorted into rows by their prime divisor.

Original entry on oeis.org

2, 4, 3, 16, 9, 5, 256, 81, 25, 7, 65536, 6561, 625, 49, 11, 4294967296, 43046721, 390625, 2401, 121, 13, 18446744073709551616, 1853020188851841, 152587890625, 5764801, 14641, 169, 17, 340282366920938463463374607431768211456, 3433683820292512484657849089281, 23283064365386962890625, 33232930569601, 214358881, 28561, 289, 19

Offset: 0

Antti Karttunen and Peter Munn, Nov 02 2019

This sequence is a permutation of A050376, so every positive integer is the product of a unique subset, S_factors, of its terms. If we restrict S_factors to be chosen from a subset, S_0, consisting of numbers from specified rows and/or columns of this array, there are notable sequences among those that may be generated. See the examples. Other notable sequences can be generated if we restrict the intersection of S_factors with specific rows/columns to have even cardinality. In any of the foregoing cases, the numbers in the resulting sequence form a group under the binary operation A059897(.,.).
Shares with array A246278 the property that columns grow downward by iterating A003961, and indeed, this array can be obtained from A246278 by selecting its columns 1, 2, 8, 128, ..., 2^((2^k)-1), for k >= 0.
A(n,k) is the image of the lattice point with coordinates X=n and Y=k under the inverse of the bijection f defined in the first comment of A306697. This geometric relationship can be used to construct an isomorphism from the polynomial ring GF(2)[x,y] to a ring over the positive integers, using methods similar to those for constructing A297845 and A306697. See A329329, the ring's multiplicative operator, for details.

			The top left 5 X 5 corner of the array:
  n\k |   0     1       2           3                   4
  ----+-------------------------------------------------------
   0  |   2,    4,     16,        256,              65536, ...
   1  |   3,    9,     81,       6561,           43046721, ...
   2  |   5,   25,    625,     390625,       152587890625, ...
   3  |   7,   49,   2401,    5764801,     33232930569601, ...
   4  |  11,  121,  14641,  214358881,  45949729863572161, ...
Column 0 continues as a list of primes, column 1 as a list of their squares, column 2 as a list of their 4th powers, and so on.
Every nonnegative power of 2 (A000079) is a product of a unique subset of numbers from row 0; every squarefree number (A005117) is a product of a unique subset of numbers from column 0. Likewise other rows and columns generate the sets of numbers from sequences:
Row 1:                 A000244 Powers of 3.
Column 1:              A062503 Squares of squarefree numbers.
Row 2:                 A000351 Powers of 5.
Column 2:              A113849 4th powers of squarefree numbers.
Union of rows 0 and 1:     A003586 3-smooth numbers.
Union of columns 0 and 1:  A046100 Biquadratefree numbers.
Union of row 0 / column 0: A122132 Oddly squarefree numbers.
Row 0 excluding column 0:  A000302 Powers of 4.
Column 0 excluding row 0:  A056911 Squarefree odd numbers.
All rows except 0:         A005408 Odd numbers.
All columns except 0:      A000290\{0} Positive squares.
All rows except 1:         A001651 Numbers not divisible by 3.
All columns except 1:      A252895 (have odd number of square divisors).
If, instead of restrictions on choosing individual factors of the product, we restrict the product to be of an even number of terms from each row of the array, we get A262675. The equivalent restriction applied to columns gives us A268390; applied only to column 0, we get A028260 (product of an even number of primes).

Antti Karttunen, Table of n, a(n) for n = 0..77; the first 12 antidiagonals of array
Wikipedia, Polynomial ring

Transpose: A329049.
Permutation of A050376.
Rows 1-4: A001146, A011764, A176594, A165425 (after the two initial terms).
Columns 1-5: A000040, A001248, A030514, A179645, A030635.
Antidiagonal products: A191555.
Subtable of A182944, A242378, A246278, A329332.
A000290, A003961, A225546 are used to express relationship between terms of this sequence.
Related binary operations: A059897, A306697, A329329.
See also the table in the example section.

Table[Prime[#]^(2^k) &[m - k + 1], {m, 0, 7}, {k, m, 0, -1}] // Flatten (* Michael De Vlieger, Dec 28 2019 *)

up_to = 105;
A329050sq(n,k) = (prime(1+n)^(2^k));
A329050list(up_to) = { my(v = vector(up_to), i=0); for(a=0, oo, for(col=0, a, i++; if(i > up_to, return(v)); v[i] = A329050sq(col, a-col))); (v); };
v329050 = A329050list(up_to);
A329050(n) = v329050[1+n];
for(n=0,up_to-1,print1(A329050(n),", ")); \\ Antti Karttunen, Nov 06 2019

A(0,k) = 2^(2^k), and for n > 0, A(n,k) = A003961(A(n-1,k)).
A(n,k) = A182944(n+1,2^k).
A(n,k) = A329332(2^n,2^k).
A(k,n) = A225546(A(n,k)).
A(n,k+1) = A000290(A(n,k)) = A(n,k)^2.

Example annotated for clarity by Peter Munn, Feb 12 2020

A078886 Decimal expansion of Sum {n>=0} 1/5^(2^n).

Original entry on oeis.org

2, 4, 1, 6, 0, 2, 5, 6, 0, 0, 0, 6, 5, 5, 3, 6, 0, 0, 0, 0, 0, 0, 4, 2, 9, 4, 9, 6, 7, 2, 9, 6, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 8, 4, 4, 6, 7, 4, 4, 0, 7, 3, 7, 0, 9, 5, 5, 1, 6, 1, 6, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 4, 0, 2, 8, 2, 3, 6, 6, 9, 2, 0, 9, 3, 8, 4

Offset: 0

Robert G. Wilson v, Dec 11 2002

Decimal expansion has increasingly large gaps of zeros, the digits delimited by these zeros are equal to 2^(2^m) as m=0,1,2,3,... The continued fraction expansion (A122165) and consists entirely of 3's, 5's and 7's, after an initial partial quotient of 4. - Paul D. Hanna, Aug 22 2006

			0.241602560006553600000...
From _Paul D. Hanna_, Aug 22 2006: (Start)
Decimal expansion consists of large gaps of zeros between strings of digits that form powers of 2; this can be seen by grouping the digits as follows:
x = .2 4 16 0 256 000 65536 000000 4294967296 000000000000 ...= 0.24160256000655360000004294...
and then recognizing the substrings as powers of 2:
2 = 2^(2^0), 4 = 2^(2^1), 16 = 2^(2^2), 65536 = 2^(2^4), 4294967296 = 2^(2^5), 18446744073709551616 = 2^(2^6), ... (End)

Aubrey J. Kempner, On Transcendental Numbers, Transactions of the American Mathematical Society, volume 17, number 4, October 1916, pages 476-482.
Index entries for transcendental numbers

Cf. A122165 (continued fraction), A176594.

Similar sums: A007404, A078885, A078585, A078887, A078888, A078889, A078890, A036987.

RealDigits[ N[ Sum[1/5^(2^n), {n, 0, Infinity}], 110]][[1]]

{a(n)=local(x=sum(k=0,ceil(3+log(n+1)),1/5^(2^k)));(floor(10^n*x))%10} \\ Paul D. Hanna, Aug 22 2006

Equals -Sum_{k>=1} mu(2*k)/(5^k - 1), where mu is the Möbius function (A008683). - Amiram Eldar, Jul 12 2020

Edited by R. J. Mathar, Aug 02 2008

A185457 a(n) = abs( Im((2+i)^(2^n)) ).

Original entry on oeis.org

1, 4, 24, 336, 354144, 116749235904, 22940770664883067253376, 182503181432559739767250904458105698387204864

Offset: 0

Carmine Suriano, Feb 04 2011

The next term is too large to be displayed here.

Old name was: Leg of primitive Pythagorean triangle generated by repeated application of the basic formula y(n) = 2*x(n-1)*y(n-1), x(1)=2, y(1)=1.

			y(2)=24 since x(1)=3, y(1)=4 are the two legs of Pythagorean triangle obtained by p=2, q=1; second iteration p=3, q=4 gives 2*3*4=24.

Alois P. Heinz, Table of n, a(n) for n = 0..11

Cf. A176594, A120905.

Cf. A099456 ( imaginary part of (2+i)^n ).

a:= n-> abs(Im((2+I)^(2^n))):
seq(a(n), n=0..8);  # Alois P. Heinz, Apr 25 2013

Table[Abs[Im[(2 + I)^(2^n)]], {n, 0, 10}] (* G. C. Greubel, Jul 07 2017 *)

a(n) = abs(imag((2+I)^(2^n))); \\ Joerg Arndt, Apr 25 2013

from sympy import im, I
def a(n): return abs(im((2 + I)**(2**n)))
print([a(n) for n in range(11)]) # Indranil Ghosh, Jul 08 2017

a(n) = abs( Im((2+i)^(2^n)) ).

Better name from Joerg Arndt, Apr 25 2013

A225157 Denominators of the sequence of fractions f(n) defined recursively by f(1) = 5/1; f(n+1) is chosen so that the sum and the product of the first n terms of the sequence are equal.

Original entry on oeis.org

1, 4, 21, 541, 345181, 136901485261, 21135572172649245550621, 496712610012943408146407697714437299262548141, 271328559212953102170688304392824035451911661168940831351173011072850527195615099225368381

Offset: 1

Martin Renner, Apr 30 2013

Numerators of the sequence of fractions f(n) is A165423(n+1), hence sum(A165423(i+1)/a(i),i=1..n) = product(A165423(i+1)/a(i),i=1..n) = A165423(n+2)/A225164(n) = A176594(n-1)/A225164(n).

			f(n) = 5, 5/4, 25/21, 625/541, ...
5 + 5/4 = 5 * 5/4 = 25/4; 5 + 5/4 + 25/21 = 5 * 5/4 * 25/21 = 625/84; ...

Cf. A100441, A165423, A176594, A225164.

b:=n->5^(2^(n-2)); # n > 1
b(1):=5;
p:=proc(n) option remember; p(n-1)*a(n-1); end;
p(1):=1;
a:=proc(n) option remember; b(n)-p(n); end;
a(1):=1;
seq(a(i),i=1..9);

a(n) = 5^(2^(n-2)) - product(a(i),i=1..n-1), n > 1 and a(1) = 1.

a(n) = 5^(2^(n-2)) - p(n) with a(1) = 1 and p(n) = p(n-1)*a(n-1) with p(1) = 1.

A241241 If x is in the sequence then so are x^2 and x(x+1)/2.

Original entry on oeis.org

0, 1, 2, 3, 4, 6, 9, 10, 16, 21, 36, 45, 55, 81, 100, 136, 231, 256, 441, 666, 1035, 1296, 1540, 2025, 3025, 3321, 5050, 6561, 9316, 10000, 18496, 26796, 32896, 53361, 65536, 97461, 194481, 222111, 443556, 536130, 840456, 1071225, 1186570, 1679616, 2051325

Offset: 1

Reinhard Zumkeller, Apr 17 2014

nonn

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000

Cf. A000290, A000217.

Subsequence of A005214; some subsequences: A001146, A007501, A011764, A176594, A173501, A050909.

import Data.Set (singleton, deleteFindMin, insert)
a241241 n = a241241_list !! (n-1)
a241241_list = 0 : 1 : f (singleton 2) where
   f s = m : f (insert (a000290 m) $ insert (a000217 m) s')
         where (m, s') = deleteFindMin s

Nest[Flatten[{#,#^2,(#(#+1))/2}]&,{0,1,2},5]//Union (* Harvey P. Dale, Aug 12 2016 *)

Initial 0 and 1 prepended by Jon Perry, Apr 17 2014

A225164 Denominators of the sequence s(n) of the sum resp. product of fractions f(n) defined recursively by f(1) = 5/1; f(n+1) is chosen so that the sum and the product of the first n terms of the sequence are equal.

Original entry on oeis.org

1, 4, 84, 45444, 15686405364, 2147492192737717340004, 45388476229808808857318702720533556450342484

Offset: 1

Martin Renner, Apr 30 2013

nonn

Numerators of the sequence s(n) of the sum resp. product of fractions f(n) is A165423(n+2), hence s(n) = sum(A165423(i+1)/A225157(i),i=1..n) = product(A165423(i+1)/A225157(i),i=1..n) = A165423(n+2)/a(n) = A176594(n-1)/a(n).

			f(n) = 5, 5/4, 25/21, 625/541, ...
5 + 5/4 = 5 * 5/4 = 25/4; 5 + 5/4 + 25/21 = 5 * 5/4 * 25/21 = 625/84; ...
s(n) = 1/b(n) = 5, 25/4, 625/84, ...

Cf. A076628, A165423, A176594, A225157.

b:=proc(n) option remember; b(n-1)-b(n-1)^2; end:
b(1):=1/5;
a:=n->5^(2^(n-1))*b(n);
seq(a(i),i=1..8);

a(n) = 5^(2^(n-1))*b(n) where b(n)=b(n-1)-b(n-1)^2 with b(1)=1/5.

A361253 If n = m^2 for some m > 1 then a(n) = a(m), otherwise a(n) = n.

Original entry on oeis.org

0, 1, 2, 3, 2, 5, 6, 7, 8, 3, 10, 11, 12, 13, 14, 15, 2, 17, 18, 19, 20, 21, 22, 23, 24, 5, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 6, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 7, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 8, 65, 66, 67, 68

Offset: 0

Rémy Sigrist, Mar 06 2023

All terms belong to A000037 U { 0, 1 }.
All terms of A000037 appear infinitely many times.
This sequence can be seen as the limit of the k-th iterate of A097448 as k tends to infinity.

			a(9) = a(3^2) = a(3) = 3 (as 3 is not a square).

Michael De Vlieger, Table of n, a(n) for n = 0..10000

Cf. A000037, A001146, A011764, A176594, A097448.

nn = 120; Array[Set[a[#], #] &, 2, 0]; Do[If[IntegerQ[#], Set[k, a[#]], Set[k, n]] &[Sqrt[n]]; Set[a[n], k], {n, nn}]; Array[a, nn] (* Michael De Vlieger, Mar 06 2023 *)

a(n) = my (m); { while (n > 1 && issquare(n, &m), n = m); return (n) }

from sympy import integer_nthroot
def A361253(n):
    if n <= 1:
        return n
    a, b = integer_nthroot(c:=n,2)
    while b:
        a, b = integer_nthroot(c:=a,2)
    return c # Chai Wah Wu, Mar 17 2023

a(a(n)) = a(n).
a(n) <= A097448(n).
a(n) = 2 iff n belongs to A001146.
a(n) = 3 iff n belongs to A011764.
a(n) = 5 iff n belongs to A176594.

cp's OEIS Frontend

A011764 a(n) = 3^(2^n) (or: write in base 3, read in base 9).

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A329050 Square array A(n,k) = prime(n+1)^(2^k), read by descending antidiagonals (0,0), (0,1), (1,0), (0,2), (1,1), (2,0), ...; Fermi-Dirac primes (A050376) in matrix form, sorted into rows by their prime divisor.

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A078886 Decimal expansion of Sum {n>=0} 1/5^(2^n).

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A185457 a(n) = abs( Im((2+i)^(2^n)) ).

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A225157 Denominators of the sequence of fractions f(n) defined recursively by f(1) = 5/1; f(n+1) is chosen so that the sum and the product of the first n terms of the sequence are equal.

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A241241 If x is in the sequence then so are x^2 and x(x+1)/2.

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A225164 Denominators of the sequence s(n) of the sum resp. product of fractions f(n) defined recursively by f(1) = 5/1; f(n+1) is chosen so that the sum and the product of the first n terms of the sequence are equal.

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