Patrick McKinley - cp's OEIS frontend

A369346 Continued fraction expansion of the real root of x^3 - x^2 - 1 = 0.

1, 2, 6, 1, 3, 5, 4, 22, 1, 1, 4, 1, 2, 84, 1, 3, 1, 6, 1, 3, 1, 9, 1, 1, 1, 1, 19, 3, 1, 2, 1, 5, 1, 5, 2, 2, 1, 1, 1, 1, 76, 6, 8, 1, 1, 5, 1, 5, 1, 1, 25, 1, 2, 1, 116, 2, 1, 8, 1, 1, 3, 1, 53, 5, 276, 2, 1, 1, 1, 3, 3, 2, 1, 1, 4, 13, 1, 1, 1, 4, 1, 1, 1, 9, 9, 1, 1, 9, 6, 1, 2, 32

Offset: 0

Patrick McKinley, Jan 20 2024

Patrick McKinley, Table of n, a(n) for n = 0..12174
Eric Weisstein's World of Mathematics, Supergolden Ratio.

Cf. A092526 (decimal expansion), A381124, A381125 (convergents).

ContinuedFraction[x/.First[Solve[x^3-x^2-1==0,x]],92] (* Stefano Spezia, Jan 21 2024 *)

\p100 \\ realprecision
contfrac(solve(x = 1, 2, x^3 - x^2 - 1),, 80) \\ Hugo Pfoertner, Jan 21 2024

/* The "test" calculation evaluates the cubic to confirm the calculation of the root. */
define iter(frac)
{j = 0
 while(frac > 1){
   frac -= 1;
   j+=1}
 j
 return 1/frac}
scale=12578
f=(1+(e(l(((29+3*sqrt(93))/2))/3))+(e(l(((29-3*sqrt(93))/2))/3)))/3
psi=f
test=(psi-1)*psi*psi-1
for(i=0;i<12175;i++)f=iter(f)

Offset changed by Andrew Howroyd, Feb 14 2025

A257092 Square array read by antidiagonals: Nimsum function for "Take-or-Break" Nim where a legal move is defined as: 1) Remove a nonzero number of counters from any pile up to the size of the selected pile OR 2) Split any pile of size greater than one into two nonzero piles (removing no counters from the board).

Original entry on oeis.org

0, 1, 1, 2, 0, 2, 3, 4, 4, 3, 4, 5, 0, 5, 4, 5, 2, 6, 6, 2, 5, 6, 3, 1, 0, 1, 3, 6, 7, 8, 8, 8, 8, 8, 8, 7, 8, 9, 3, 1, 0, 1, 3, 9, 8, 9, 6, 10, 2, 6, 6, 2, 10, 6, 9, 10, 7, 5, 11, 5, 0, 5, 11, 5, 7, 10, 11, 12, 12, 4, 12, 4, 4, 12, 4, 12, 12, 11, 12, 13, 7, 13, 3, 13, 0, 13, 3, 13, 7, 13, 12, 13, 10, 14, 14, 10, 2, 14, 14, 2, 10, 14, 14, 10, 13, 14, 11, 9, 7, 9, 11, 1, 0, 1, 11, 9, 7, 9, 11, 14

Offset: 0

Patrick McKinley, Apr 19 2015

Observe that many "safe" three-pile positions in Nim (1-2-3, 1-4-5, 2-4-6, etc.) consist of one pile whose size is the sum of the sizes of the other two. The "Break" move is designed to trivially defeat these positions by breaking the large pile into copies of the other two. This leaves a clear winning position where every pile has a "twin".

Like the standard Nimsum function defined in A003987, this relation constitutes an Abelian group over nonnegative integers where every element is its own inverse.

			The square table defining the relation begins:
0  1  2  3  4  5  6  7  8  9   ...
1  0  4  5  2  3  8  9  6  7   ...
2  4  0  6  1  8  3 10  5 12   ...
3  5  6  0  8  1  2 11  4 13   ...
4  2  1  8  0  6  5 12  3 10   ...
5  3  8  1  6  0  4 13  2 11   ...
6  8  3  2  5  4  0 14  1 16   ...
7  9 10 11 12 13 14  0 16  1   ...
8  6  5  4  3  2  1 16  0 14   ...
9  7 12 13 10 11 16  1 14  0   ...
.  .  .  .  .  .  .  .  .  .
Reading from the table, 1-2-4, 1-3-5 and 2-3-6 are safe positions in Take-or-Break Nim.

Cf. A003987.

flip(x) = if (x==0, 0, if (x % 2, x+1, x-1));
tabl(nn) = {for (n=0, nn, for (k=0, nn, print1(flip(bitxor(flip(n), flip(k))), ", ");); print(););} \\ Michel Marcus, Apr 23 2015

NSum(x,y) = Flip(Flip(x) XOR Flip(y))
Where XOR is the bitwise exclusive OR characteristic of A003987.
Flip(n) = 0    if n == 0.
        = n+1  if n is odd.
        = n-1  if n is even.

A215508 Smallest m such that the period of the continued fraction of sqrt(m) is A215485(n); records of A013646.

Original entry on oeis.org

1, 2, 3, 41, 58, 106, 193, 337, 586, 949, 1061, 1117, 1153, 1249, 1669, 2381, 3733, 5857, 6577, 6781, 8389, 11173, 14293, 15817, 17137, 17209, 23017, 37921, 38377, 46261, 47293, 56929, 82561, 90121, 113173, 122401, 148957, 151057, 161149, 163729, 193873, 206209, 225769, 322513, 497473, 576529, 676129, 686893, 706621, 862921, 946489, 992281, 1032649, 1198081, 1597033, 1655677, 1779409, 1930021, 2299489, 2367481, 2584081, 3209281, 3528409, 3933073, 4068241, 4160521, 4283689, 4726009, 4833901

Offset: 0

Patrick McKinley, Aug 13 2012

nonn

The continued fractions of these numbers have the "hard to get" lengths listed in sequence A215485. They fill the last gaps in the table when computing A013646.

			The lengths of the continued fractions of sqrt(1), sqrt(2), sqrt(3) and sqrt(41) are 0, 1, 2 and 3 respectively. The rest of the sequence follows A215485 similarly.

Patrick McKinley, Table of n, a(n) for n = 0..253

Cf. A013646, A215485.

a(n) = A013646(A215485(n)). - Pontus von Brömssen, Nov 24 2024

A215485 Periods of square root continued fractions at which A013646 sets a new record.

Original entry on oeis.org

0, 1, 2, 3, 7, 9, 13, 19, 23, 27, 35, 41, 43, 45, 53, 55, 71, 77, 101, 127, 129, 135, 147, 163, 169, 189, 199, 201, 247, 283, 335, 353, 367, 459, 465, 503, 537, 587, 625, 637, 643, 739, 767, 827, 1009, 1135, 1325, 1423, 1433, 1543, 1561, 1775, 1781, 1951, 2011

Offset: 0

Patrick McKinley, Aug 12 2012

nonn

Each term of this sequence takes a turn at being the smallest unknown period for a square root continued fraction. Periods 1 and 2 are seen as the periods of sqrt(2) and sqrt(3) respectively, but a period of 3 is not seen until sqrt(41).
By convention, the period for perfect squares (e.g., 1) is 0.
Open question: Are there any more even terms after the 2?

			When a square root continued fraction with a period of 3 is first seen (at sqrt(41)), the lowest period not yet seen is 7, which first occurs as the period of sqrt(58).

Patrick McKinley, Table of n, a(n) for n = 0..254

Cf. A013646.

A215160 Odd numbers n with the property that the binary representation of n is the same as the decimal representation of the smallest multiple of n that can be represented with only 1's and 0's.

Original entry on oeis.org

1, 21, 2231, 28261, 611123, 1200341, 3427673, 2202416417, 11102657671

Offset: 1

Patrick McKinley, Aug 05 2012

All numbers that are a power of 2 times a member of the sequence share the property that the binary representation is the same as the decimal representation of the first 1's and 0's multiple.

Of the values listed, only 1200341 and 3427673 are primes. - Jonathan Vos Post, Aug 09 2012

			For example 21*481=10101 (the first multiple of 21 containing only 1's and 0's) and the binary representation of 21 is 10101.

Cf. A079339.

rebase := proc(n,bin,bout)
    local a,c,i;
    a := 0 ;
    c := convert(n,base,bin) ;
    add( op(i,c)*bout^(i-1),i=1..nops(c)) ;
end proc:
isA079339 := proc(n,c)
    local c2,b;
    if modp(c,n) > 0 then
        return false;
    end if;
    c2 := rebase(c,10,2) ;
    for b from 1 to c2-1 do
        if modp( rebase(b,2,10),n) = 0 then
            return false;
        end if;
    end do:
    return true ;
end proc:
for n from 1 by 2 do
    sb := rebase(n,2,10) ;
    if isA079339(n,sb) then
        print(n);
    end if;
end do: # R. J. Mathar, Aug 09 2012

{odd n: n*A079339(n) = A007088(n)} . - R. J. Mathar, Aug 09 2012

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A369346 Continued fraction expansion of the real root of x^3 - x^2 - 1 = 0.

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A215508 Smallest m such that the period of the continued fraction of sqrt(m) is A215485(n); records of A013646.

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A215485 Periods of square root continued fractions at which A013646 sets a new record.

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A215160 Odd numbers n with the property that the binary representation of n is the same as the decimal representation of the smallest multiple of n that can be represented with only 1's and 0's.

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