This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
a = {}; Do[Do[w = 4x^3y + 4x y^3; If[w < 10000, AppendTo[a, w]], {x, y, 1000}], {y, 1, 1000}]; Union[a] (*Artur Jasinski*)
M(0)= 1
M(1)= 2, 1 - given by definition. For n > 1, we use an inductive approach (i.e., iteration) instead of recursion since a straightforward recursive function for computing the terms of M(n) is not efficient.
To obtain M(2) we apply the recursive formula for n=2:
#m(2,0) = 2^(2^(2-1))*#m(1,0) = 4*2 = 8;
#m(2,1) = 2^(2^(2-1))*#m(1,1) + #m(1,0) = 4*1 + 2 = 6;
#m(2,2) = 1.
Thus M(2) = 8, 6, 1.
To obtain M(3) we compute:
#m(3,0) = 2^(2^(3-1))*#m(2,0) = 16*8 = 128;
#m(3,1) = 2^(2^(3-1))*#m(2,1) + #m(2,0) = 16*6 + 8 = 104;
#m(3,2) = 2^(2^(3-1))*#m(2,2) + #m(2,1) = 16*1 + 6 = 22;
#m(3,3) = 1.
Therefore M(3) = 128, 104, 22, 1.
Let's clarify the meaning of its terms. The lexicographic order of {0,1}^3 is (0,0,0), (0,0,1), (0,1,0), (0,1,1), (1,0,0), (1,0,1), (1,1,0), (1,1,1), and the corresponding sequence of (serial) numbers is 0, 1, 2, ..., 7. We have:
m(3,0) = (1,0,0,0,0,0,0,0) is the characteristic vector of the unique vector of weight 0 and hereby #m(3,0) = 2^7 = 128;
m(3,1) = (0,1,1,0,1,0,0,0) is the characteristic vector of all vectors of weight 1 and #m(3,1)=104;
m(3,2) = (0,0,0,1,0,1,1,0) is the characteristic vector of all vectors of weight 2 and then #m(3,2)=22;
m(3,3) = (0,0,0,0,0,0,0,1) the characteristic vector of the unique vector of weight 3 and #m(3,3)=1.
This sequence is bijectively related to the sequence A294648, representing the Weight-Lexicographic Order of the vectors of {0,1}^n. The row M(n) bijectively corresponds to the row L(n) of A294648, for n= 1,2, ... For example, if n=3, the third row of A294648 is
L(3) = l(3,0),...,l(3,3) = (0), (1, 2, 4), (3, 5, 6), (7) = 0, 1, 2, 4, 3, 5, 6, 7.
We recall that the coordinates of m(n,k) are numbered from left to right by 0, 1, ..., 2^n-1, and L(n) contains all these integers, partitioned in subsequences.
For the subsequences of L(3) we note:
l(3,0) = 0, which is the serial number of the unique vector (0,0,0) of {0,1}^3 of weight 0. It corresponds to m(3,0) since m(3,0) contains a unit in its leftmost coordinate only.
l(3,1) = 1, 2, 4, which are the serial numbers of all 3 vectors of {0,1}^3 of weight 1. So l(3,1) corresponds to m(3,1) -- note that 1, 2 and 4 are the coordinates in which m(3,1) contains units.
l(3,2) = 3, 5, 6, which are the serial numbers of all 3 vectors of {0,1}^3 of weight 2. It corresponds to m(3,1) -- note that 3, 5 and 6 are the coordinates in which m(3,2) contains units.
l(3,3) = 7, which is the serial number of the unique vector (1,1,1) of {0,1}^3 of weight 3. It corresponds to m(3,3).
From _Michel Marcus_, Jul 10 2018: (Start)
Triangle begins:
1,
2, 1,
8, 6, 1,
128, 104, 22, 1,
32768, 26752, 5736, 278, 1,
2147483648, 1753251840, 375941248, 18224744, 65814, 1,
... (End)
T[n_, k_] := If[n == 1, If[k == 0, 2, 1], If[k == 0, 2^(2^n - 1), If[k == n, 1, 2^(2^(n-1)) T[n-1, k] + T[n-1, k-1]]]];
Table[T[n, k], {n, 0, 6}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 25 2018, after Michel Marcus *)
T(n, k) = if (n==1, if (k==0, 2, 1), if (k==0, 2^(2^n-1), if (k==n, 1, 2^(2^(n-1))*T(n-1,k) + T(n-1,k-1)))); tabf(nn) = for (n=1, nn, for (k=0, n, print1(T(n, k), ", ")); print); \\ Michel Marcus, Jul 10 2018
a(0) = 2^1 = 2. a(1) = 2^2 - 2^1 = 2. a(2) = 2^4 - 2^2 + 2^1 = 14. a(3) = 2^8 - 2^4 + 2^2 - 2^1 = 242. a(4) = 2^16 - 2^8 + 2^4 - 2^2 + 2^1 = 65294.
a[n_] := (-1)^n * Sum[(-1)^k * 2^(2^k), {k, 0, n}]; Array[a, 9, 0] (* Amiram Eldar, May 07 2021 *)
{a(n) = (-1)^n*sum(k=0,n,(-1)^k*2^2^k)}
Let n = 25 = 1 + 8 + 16 = 2^0 + 2^3 + 2^4.
Then a(n) = 65794 = 2 + 256 + 65536 = 2^(2^0) + 2^(2^3) + 2^(2^4).
The binary indices of n are {0, 3, 4}. Those of a(n) are {1, 8, 16}.
a := proc(n) select(d -> d[2] <> 0, ListTools:-Enumerate(convert(n,base,2))): add(2^(2^(%[j][1] - 1)), j = 1..nops(%)) end: seq(a(n), n = 0..34); # Peter Luschny, Oct 31 2022
a[n_] := Total[2^(2^Range[If[n == 0, 1, IntegerLength[n,2]] - 1, 0, -1]) * IntegerDigits[n, 2]]; Array[a, 35, 0] (* Amiram Eldar, Oct 31 2022 *)
a(n) = my(d=Vecrev(digits(n,2))); for (k=1, #d, d[k] *= 2^(2^(k-1))); vecsum(d); \\ Michel Marcus, Oct 31 2022
def a(n):
binary_string = "{0:b}".format(n)[::-1] # little-endian
result = 0
for i, binary_digit in enumerate(binary_string):
if binary_digit == '1':
result += 1 << (1 << i) # 2 ** (2 ** i)
return result
Triangle begins:
{1, 2},
{1, 2, 2, 4},
{1, 2, 2, 4, 2, 4, 4, 8, 2, 4, 4, 8, 4, 8, 8, 16}
...
a = {}; Do[Do[w = x^4 + 6x^2 y^2 + y^4; If[w < 10000, AppendTo[a, w]], {x, y, 1000}], {y, 1, 1000}]; Union[a]
(* alternate program *)
Union[Select[#[[1]]^4+6#[[1]]^2 #[[2]]^2+#[[2]]^4&/@Tuples[Range[ 1000],2], #<10000&]] (* Harvey P. Dale, Oct 07 2012 *)
a = {}; Do[Do[w = x^5 + 10x^3 y^2 + 5x y^4; If[w < 100000, AppendTo[a, w]], {x, 1, 1000}], {y, 1, 1000}]; Union[a]
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