A108618 -id:A108618 - cp's OEIS frontend

A110613 a(n+3) = 5a(n+2) - 2a(n+1) - 8*a(n), a(0) = 1, a(1) = 0, a(2) = 3.

1, 0, 3, 7, 29, 107, 421, 1659, 6597, 26299, 105029, 419771, 1678405, 6712251, 26846277, 107379643, 429507653, 1718008763, 6871991365, 27487878075, 109951337541, 439805000635, 1759219303493, 7036875815867, 28147500467269

Offset: 0

Creighton Dement, Jul 31 2005

A Jacobsthal related sequence (A001045). This sequence was calculated using the same rules given for A108618; the "initial seed" is the floretion given in the program code, below.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,-2,-8).

Cf. A001045, A007582, A007581, A049775, A087440, A108618, A110614.

seriestolist(series((1-5*x+5*x^2)/((4*x-1)*(2*x-1)*(x+1)), x=0,25)); -or- Floretion Algebra Multiplication Program, FAMP Code: 2tessumseq[(.5'i - .5'k - .5i' + .5k' - .5'ij' - .5'ji' - .5'jk' - .5'kj')('i + j' + 'ij' + 'ji')] Sumtype is set to:sum[Y[15]] = sum(*) (from 3rd term, disregarding signs)

LinearRecurrence[{5, -2, -8}, {1, 0, 3}, 50] (* G. C. Greubel, Sep 01 2017 *)

x='x+O('x^50); Vec((1-5*x+5*x^2)/((4*x-1)*(2*x-1)*(x+1))) \\ G. C. Greubel, Sep 01 2017

G.f.: (1-5*x+5*x^2)/((4*x-1)*(2*x-1)*(x+1)).
Program "Superseeker" finds:
a(n) + a(n+1) = A007582(n) = A007581(n+1) - A007581(n).
a(n+2) - a(n) = A049775(n).
a(n) + 2*a(n+1) + a(n+2) = A087440(n+1).

A119953 Y = X = 'i + .25(ii + jj + kk + e); Z = 'i - i' + .5(jj + kk - jk + kj) + e. See pdf-file for an exact definition (this sequence gives an initial term 1); Version "jes".

Original entry on oeis.org

1, 1, 4, 2, 1, 6, 7, 4, 2, 6, 6, 1, 2, 4, 7, 1, 2, 4, -1, -3, 0, 4, -1, -11, -6, -1, -5, -13, -10, -4, -10, -14, -10, -9, -13, -17, -12, -11, -19, -18, -13, -15, -19, -18, -15, -18, -23, -19, -15, -18, -25, -23, -18, -22, -30, -25, -20, -27, -34, -30, -24, -30, -39, -35, -26, -33, -44, -39, -31, -35, -46, -42, -34, -39, -47

Offset: 0

Creighton Dement, May 30 2006

Compare with A108618. Beginning at a(16350) = 560, the sequence apparently enters a loop and repeats the 60 terms: 560, 382, 327, 503, 558, 383, 328, 503, 557, 383, 327, 504, 558, 382, 327, 504, 560, 381, 327, 506, 559, 380, 327, 507, 559, 377, 326, 508, 558, 377, 328, 508, 559, 377, 326, 509, 560, 377, 325, 509, 559, 378, 326, 508, 559, 378, 328, 507, 559, 380, 327, 506, 559, 381, 327, 503, 558, 382, 326, 503.
Let f(n) give the frequency of occurrence of the number n in the above 60 term set. Then f(560) = 3, f(382) = 3, f(327) = 7, f(503) = 4, f(558) = 4, f(383) = 2, f(328) = 3, f(557) = 1, f(504) = 2, f(381) = 2, f(506) = 2, f(559) = 7, f(380) = 2, f(507) = 2, f(377) = 4, f(326) = 4, f(508) = 3, f(509) = 2, f(325) = 1, f(378) = 2
Example MUSICALGORITHMS settings (link): Pitch: Scale values 11-66, Duration: Scaling 0-2 (perform division operation).

Creighton Dement, Table of n, a(n) for n = 0..30000
Creighton Dement, Construction of an integer sequence with musical properties
J. Middleton, Musicalgorithms.

Cf. A119954, A108618.

A119954 Y = X = 'i + .25(ii + jj + kk + e); Z = 'i - i' + .5(jj + kk - jk + kj) + e. See pdf-file and comment for an exact definition (this sequence gives an initial term 3); Version "les".

Original entry on oeis.org

3, 2, -2, 9, 24, 33, 21, -2, -6, 18, 47, 30, -13, -20, 4, -7, -32, -42, -59, -80, -77, -66, -74, -107, -128, -98, -67, -81, -127, -151, -142, -119, -107, -117, -151, -190, -176, -136, -123, -158, -193, -202, -173, -140, -133, -165, -204, -188, -140, -113, -151, -205, -195, -127, -82, -88, -120

Offset: 0

Creighton Dement, Jun 09 2006

sign

To obtain this sequence, follow the same instructions given for A119953. A119953(n) was obtained by adding the coefficients of 'i and i' at the end of the n-th iteration. a(n) is obtained by adding the coefficients of the basis vectors ij, ik, ji, jk, ki, kj at the end of the n-th iteration. Note: Some of these coefficients are always 0. "Version les" refers to the 6 basis vectors mentioned above.

Creighton Dement, Table of n, a(n) for n = 0..30000
Creighton Dement, Construction of an integer sequence with musical properties.

Cf. A119953, A108618.

A126626 A floretion-generated sequence based on the iterative procedure defined in the link given.

Original entry on oeis.org

1, 0, -1, 2, -3, 2, -1, 2, 1, 0, 3, -2, 5, -4, 3, -4, 1, -4, -1, -2, -3, 0, -5, 2, -7, 4, -9, 4, -7, 4, -5, 4, -3, 4, -1, 4, 1, 2, 3, 0, 5, -2, 7, -4, 9, -6, 11, -8, 13, -8, 11, -8, 9, -8, 7, -8, 5, -8, 3, -8, 1, -8, -1, -6, -3, -4, -5, -2, -7, 0, -9, 2, -11, 4, -13, 6, -15, 8, -17, 10, -19, 12, -17, 12, -15, 12, -13, 12, -11, 12, -9, 12, -7, 12

Offset: 0

Creighton Dement, Nov 21 2006

Changing Step 4 in this sequence results in A117154 (see link). Plotting against A125141 returns an "elliptic spiral". As music (example settings): Compute pitch by scaling: 11-66, perform division operation. Compute duration by scaling: 0-2, perform division operation

Creighton Dement, Table of n, a(n) for n = 0..5000
Creighton Dement, Notes on A126626 and A117154
Creighton Dement, Floretions 2009, DRAFT. See section 4.1 Algorithms.

Cf. A125141, A117154, A108618.

This sequence is calculated by noting the coefficient of the unit basis vector of the floretion Y after each iteration (see link for further details). Note: this basis vector may also be represented as the unit 4 X 4 matrix.

A108985 Expansion of (x+1)(x^3-x^2-x-1)/((1-x)(x^2+2x-1)(x^2+x+1)).

Original entry on oeis.org

1, 4, 11, 27, 66, 161, 389, 940, 2271, 5483, 13238, 31961, 77161, 186284, 449731, 1085747, 2621226, 6328201, 15277629, 36883460, 89044551, 214972563, 518989678, 1252951921, 3024893521, 7302738964, 17630371451, 42563481867, 102757335186

Offset: 0

Creighton Dement, Jun 15 2005

A floretion-generated sequence calculated using the same rules given for A108618 with initial seed x = - .5'ii' - .5'jj' + .5'kk' - .5'jk' - .5'kj' + .5e; version: "tes". Note: replacing "tessum(*)seq" with "tesseq" in the program code gives A001333.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,1,1,-2,-1).

Cf. A108618, A108986, A108987, A108988.

Vec((1 + x)*(1 + x + x^2 - x^3) / ((1 - x)*(1 + x + x^2)*(1 - 2*x - x^2)) + O(x^30)) \\ Colin Barker, May 06 2019

a(n) = 2*a(n-1) + a(n-2) + a(n-3) - 2*a(n-4) - a(n-5) for n>4. - Colin Barker, May 06 2019

A117153 The floretion + .25'i + .25'j + .25'k + .25i' + .25j' + .25k' + .25'ii' + .25'jj' + .25'kk' + .25'ij' + .25'ik' + .25'ji' + .25'jk' + .25'ki' + .25'kj' + .25e.

Original entry on oeis.org

3, -3, 9, -12, -15, 0, 33, -42, -36, 30, 15, -54, -6, 63, -63, -15, 54, -36, -42, 24, -12, -15, 33, -36, 12, 39, -30, 3, 15, 27, 39, 9, 54, 12, 33, 51, 18, 99, 60, 0, 87, 84, 36, 60, 96, 45, 102, 93, 42, 93, 111, 57, 114, 129, 57, 132, 138, 78, 81, 165, 81

Offset: 1

Creighton Dement, Apr 21 2006

The precise definition of this sequence seems to have been lost. The sequence stays in the OEIS, however, for historical reasons, because it is mentioned in the "Music" web page, and because we hope that one day the definition will be recovered. - N. J. A. Sloane, Jan 09 2011.

This is one of several sequence that were listed on my website at the time. These were in large part only posted as examples of simple floretion algorithms with various parameters set and not yet actually meant to be published. The definition would have been very similar to that of A108618. See especially Step 2, Loop 1: Add the fractional parts of the real coefficient basis vectors of x*y. - Creighton Dement, Jan 03 2018

Creighton Dement, First 1001 terms
Creighton Dement, Floretion Online Multiplier [broken link].

Cf. A108618.

A272694 a(n) = (c(n)+f(c(n)))/2, where c() = A272693() and f(k) = sign(k)*(k mod 2).

Original entry on oeis.org

0, -1, -1, -3, -2, -2, -2, 0, 3, 1, -2, -3, -4, -3, 1, 5, 2, -3, -5, -5, -2, 3, 6, 3, -5, -11, -6, 3, 7, 5, -3, -10, -9, -2, 8, 8, 1, -8, -11, -6, 6, 10, 5, -6, -13, -9, 1, 11, 8, -3, -11, -11, -2, 9, 12, 3, -11, -17, -8, 9, 18, 9, -11, -23, -12, 9, 19, 11, -9, -22, -15, 4, 20, 14, -6, -20, -17, 2, 19, 17

Offset: 0

N. J. A. Sloane, Jun 08 2016

sign

Created in an attempt to simplify the definition of A108618.

N. J. A. Sloane, Table of n, a(n) for n = 0..10000

Cf. A108618, A108619, A272693.

M:=1000;
a:=Array(0..M,0); # A108618 (with different offset)
b:=Array(0..M,0); # A108619 (with different offset)
c:=Array(0..M,0); # A272693
f:=n->sign(n)*(n mod 2);
a[0]:=0; b[0]:=0; c[0]:=0;
for n from 1 to M do
b[n]:=1+(a[n-1]+b[n-1])/2;
a[n]:=1+c[n-1]+f(c[n-1])+3*f(b[n]-1);
c[n]:=(a[n]-3*b[n])/2;
od:
[seq(a[n],n=0..M)];
[seq(b[n],n=0..M)];
[seq(c[n],n=0..M)];
[seq((c[n]+f(c[n]))/2,n=0..M)]; # A272694

A308496 Numbers with digits 1,2,4,7 when written in base 8.

Original entry on oeis.org

1, 2, 4, 7, 9, 10, 12, 15, 17, 18, 20, 23, 33, 34, 36, 39, 57, 58, 60, 63, 73, 74, 76, 79, 81, 82, 84, 87, 97, 98, 100, 103, 121, 122, 124, 127, 137, 138, 140, 143, 145, 146, 148, 151, 161, 162, 164, 167, 185, 186, 188, 191, 265, 266, 268, 271, 273

Offset: 1

Creighton Dement, Jun 01 2019

Floretions of all orders. - Creighton Dement, Oct 28 2022
For any natural number n, the set of terms of this sequence between indices (4^n-1)/3 and (4^(n+1)-4)/3 is "isomorphic" to the group of n-th order floretions. In this case, group multiplication is given by bitwise logical operations (see EXAMPLE). Note that the case of n = 1 is simply the quaternions.
In the table below, the left column is the binary representation, the middle column the terms of ((a(n)) and the right column the conventional notation. Multiply x*y (disregarding signs) using the bitwise XNOR operation, where x and y are any floretions of the same order. The XNOR operation returns a 1 if the number of 1's in its inputs is even, and a 0 if the number of 1's is odd. This operation is used to calculate the base vector of the result.
**** 1st-order floretions (= quaternions) ****
         | binary  | decimal | floretion
  1      | 001     |    1    |   i
  2      | 010     |    2    |   j
  4      | 100     |    4    |   k
  7      | 111     |    7    |   e (unit)
**** 2nd-order floretions ****
  1_1    | 001_001 |    9    |   ii
  1_2    | 001_010 |   10    |   ij
  1_4    | 001_100 |   12    |   ik
  1_7    | 001_111 |   15    |   ie
  2_1    | 010_001 |   17    |   ji
  2_2    | 010_010 |   18    |   jj
  2_4    | 010_100 |   20    |   jk
  2_7    | 010_111 |   23    |   je
  4_1    | 100_001 |   33    |   ki
  4_2    | 100_010 |   34    |   kj
  4_4    | 100_100 |   36    |   kk
  4_7    | 100_111 |   39    |   ke
  7_1    | 111_001 |   57    |   ei
  7_2    | 111_010 |   58    |   ej
  7_4    | 111_100 |   60    |   ek
  7_7    | 111_111 |   63    |   ee
**** 3rd-order floretions ****
  1_1_1
  1_1_2
  ...
Note that for a floretion of order n, two digits from any one of its "binary triplets" abc determine the other since XOR(a,b,c) = 1.
When working with a floretion algebra over the reals, i.e., elements of the form x = q_1*f_1 + ... q_m*f_m where q_1,...,q_m are real numbers and f_1,...,f_m are any floretions of the same order, then x may also be referred to as a "floretion". In this case f_1,...,f_m (i.e., terms of this sequence) may be referred to as "floretion base vectors" to avoid confusion.
Taking signs into account:
Given two binary representations (ab) and (cd) for quaternion elements, define multiplication as:
    Compute (XNOR(a,c))(XNOR(b,d)) to get the base vector of the result.
    Compute AND(b,c), AND(XNOR(a,b),d), and AND(a,XNOR(c,d)). These are all bitwise AND operations.
    The sign is negative if and only if the total number of 1's in the results is even.
For example, with k*j = (10)*(01) = -i, compute:
    The base vector as (XNOR(1,0) XNOR(0,1)) = (0)(0) = i.
    The signs as AND(0,1), AND(XNOR(1,0),1), AND(1, XNOR(0,1)) = 0, 0, 0. There are zero 1's in total, which is an even number, so the result is negative.
An example of image processing: take for example a quaternion x = .2i + .5j + .3k + e. Assume we have a square monitor (aspect ratio). Furthermore, assume the screen is divided into 4 squares- one for i (bottom left), one for j (top left), one for k (top right) one for e (bottom right) and that the coefficient is the amount the pixels are lit up on the screen (1 being full brightness, 0 being off- this could be modified later to accomodate negative numbers).  Now imagine we have square monitor of resolution 2^n x 2^n. Then we can represent any black and white image with that resolution with an n-th order floretion. This means we can multiply images together, with some parallels to Fourier analysis.
Multiplying an image by an idempotent floretion would allow one to repeatedly apply a specific transformation (e.g., a rotation, scaling, or some other operation) to an image, and then undo all of those transformations by continuing to apply the same operation a certain number of times. It could be used in applications such as data encryption, where an image could be "scrambled" using a specific floretion and then "unscrambled" by continuing to apply the same floretion.
A compact definition of multiplication is x*y = (ab)(cd) = (-1)^{m+1} (aqc)(bqd) where m = b&c + (aqb)&d + a&(cqd) and "q", "&" are the bitwise XNOR and AND operators respectively. - Creighton Dement, Jul 09 2023

Creighton Dement, Table of n, a(n) for n = 1..10000
Creighton Dement, Floretions and Actions of D3 (DRAFT)
Stefan Rakonjac, The Sounds of Sequences: Floretions

Cf. A007094, A108618, A047541.

A308496Q[n_]:=ContainsOnly[IntegerDigits[n,8],{1,2,4,7}];
Select[Range[1000],A308496Q] (* Paolo Xausa, Dec 31 2023 *)

is(n)=!#setminus(Set(digits(n,8)),[1,2,4,7]);
a(n) =
{
  local(total_count, index);
  until(total_count == n+1, if(is(index)==1, total_count++); index++);
  index-1;
}

a(n,b=8,d=[1,2,4,7]) = { for (w=1, oo, if (n>#d^w, n-=#d^w, return (fromdigits(apply(x -> d[1+x], digits(#d^w+n-1, #d))[2..-1],b)))) } \\ Rémy Sigrist, Jun 01 2019

A109620 a(n) = (1/3)n^3 - n^2 - (1/3)n - 1.

Original entry on oeis.org

-1, -2, -3, -2, 3, 14, 33, 62, 103, 158, 229, 318, 427, 558, 713, 894, 1103, 1342, 1613, 1918, 2259, 2638, 3057, 3518, 4023, 4574, 5173, 5822, 6523, 7278, 8089, 8958, 9887, 10878, 11933, 13054, 14243, 15502, 16833, 18238, 19719, 21278, 22917, 24638, 26443, 28334, 30313, 32382, 34543, 36798

Offset: 0

Creighton Dement, Aug 01 2005

It is interesting that this sequence is generated using the same rules as those given for A108618 (version "jes"; the initial seed is the floretion given in program code, below). In reference to those rules, we have: **Loop 0** + .5'i + .5i' + .5'ik' + .5'ji' + e **Loop 1** + 1.5'i - .5'j + 1.5i' - .5k' + .5'ii' + 1.5'ik' + 1.5'ji' + .5'kj' + 3e **Loop 2** + 3.5'i - 2'j + 3.5i' - 2k' + 2'ii' + 3.5'ik' + 3.5'ji' + 2'kj' + 5e **Loop 3** + 6.5'i - 5.5'j + 6.5i' - 5.5k' + 5.5'ii' + 6.5'ik' + 6.5'ji' + 5.5'kj' + 7e **Loop 4** + 10.5'i - 12'j + 10.5i' - 12k' + 12'ii' + 10.5'ik' + 10.5'ji' + 12'kj' + 9e **Loop 5** + 15.5'i - 22.5'j + 15.5i' - 22.5k' + 22.5'ii' + 15.5'ik' + 15.5'ji' + 22.5'kj' + 11e **Loop 6** + 21.5'i - 38'j + 21.5i' - 38k' + 38'ii' + 21.5'ik' + 21.5'ji' + 38'kj' + 13e **Loop 7** + 28.5'i - 59.5'j + 28.5i' - 59.5k' + 59.5'ii' + 28.5'ik' + 28.5'ji' + 59.5'kj' + 15e **Loop 8** + 36.5'i - 88'j + 36.5i' - 88k' + 88'ii' + 36.5'ik' + 36.5'ji' + 88'kj' + 17e **Loop 9** + 45.5'i - 124.5'j + 45.5i' - 124.5k' + 124.5'ii' + 45.5'ik' + 45.5'ji' + 124.5'kj' + 19e. a(n) is calculated by adding the real number coefficients of 'i, 'j and 'k (which is always 0 here) from the n-th loop and multiplying the result by -2.

Index entries for linear recurrences with constant coefficients, signature (4, -6, 4, -1).

Cf. A006527, A002061.

seriestolist(series((2*x-1)*(x^2+1)/(x-1)^4, x=0,50)); -or- Floretion Algebra Multiplication Program, FAMP Code: -2jessumseq[ + .5'i + .5i' + .5'ik' + .5'ji' + e], Sumtype: sum[Y[15]] = sum[ * ]. Note: 2ibasesumseq = A002061, apart from initial term, -2jbasesumseq = A006527.

Table[n^3/3-n^2-n/3-1,{n,0,50}] (* or *) LinearRecurrence[{4,-6,4,-1},{-1,-2,-3,-2},60] (* Harvey P. Dale, Jul 21 2013 *)

a(n) = A006527(n) - A002061(n+1), g.f. (2*x-1)*(x^2+1)/(x-1)^4

a(0)=-1, a(1)=-2, a(2)=-3, a(3)=-2, a(n)=4*a(n-1)-6*a(n-2)+4*a(n-3)- a(n-4). - Harvey P. Dale, Jul 21 2013

cp's OEIS Frontend

A110613 a(n+3) = 5*a(n+2) - 2*a(n+1) - 8*a(n), a(0) = 1, a(1) = 0, a(2) = 3.

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A119953 Y = X = 'i + .25(ii + jj + kk + e); Z = 'i - i' + .5(jj + kk - jk + kj) + e. See pdf-file for an exact definition (this sequence gives an initial term 1); Version "jes".

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A119954 Y = X = 'i + .25(ii + jj + kk + e); Z = 'i - i' + .5(jj + kk - jk + kj) + e. See pdf-file and comment for an exact definition (this sequence gives an initial term 3); Version "les".

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A126626 A floretion-generated sequence based on the iterative procedure defined in the link given.

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A108985 Expansion of (x+1)*(x^3-x^2-x-1)/((1-x)*(x^2+2*x-1)*(x^2+x+1)).

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A117153 The floretion + .25'i + .25'j + .25'k + .25i' + .25j' + .25k' + .25'ii' + .25'jj' + .25'kk' + .25'ij' + .25'ik' + .25'ji' + .25'jk' + .25'ki' + .25'kj' + .25e.

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A272694 a(n) = (c(n)+f(c(n)))/2, where c() = A272693() and f(k) = sign(k)*(k mod 2).

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Maple

A308496 Numbers with digits 1,2,4,7 when written in base 8.

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Mathematica

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PARI

A109620 a(n) = (1/3)*n^3 - n^2 - (1/3)*n - 1.

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A110613 a(n+3) = 5a(n+2) - 2a(n+1) - 8*a(n), a(0) = 1, a(1) = 0, a(2) = 3.

A108985 Expansion of (x+1)(x^3-x^2-x-1)/((1-x)(x^2+2x-1)(x^2+x+1)).

A109620 a(n) = (1/3)n^3 - n^2 - (1/3)n - 1.