David A. Madore - cp's OEIS frontend

A372325 Numbers whose binary expansion has an even number of 1's among positions listed in this sequence.

0, 2, 5, 7, 8, 10, 13, 15, 16, 18, 21, 23, 24, 26, 29, 31, 33, 35, 36, 38, 41, 43, 44, 46, 49, 51, 52, 54, 57, 59, 60, 62, 64, 66, 69, 71, 72, 74, 77, 79, 80, 82, 85, 87, 88, 90, 93, 95, 97, 99, 100, 102, 105, 107, 108, 110, 113, 115, 116, 118, 121, 123, 124

Offset: 1

David A. Madore, Apr 27 2024

			118 is in the sequence because 118 = 2^6 + 2^5 + 2^4 + 2^2 + 2^1, and an even number of the exponents 6,5,4,2,1 (namely 2,5) are in the sequence.
8192 is not in the sequence because 8192 = 2^13, and 13 is in the sequence.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A133457, A272011.

R:= 0: RL:= [1]: nextp:= 2: m:= 1: count:= 0:
for i from 1 while count < 100 do
  L:= convert(i,base,2);
  if i = nextp then
    nextp:= 2*nextp;
    if R[1+nops(RL)] = m then RL:= [op(RL),m+1] fi;
    m:= m+1;
  fi;
  if convert(L[RL],`+`)::even
  then R:= R,i; count:= count+1
  fi
od:
R; # Robert Israel, May 28 2024

from itertools import count, islice
def agen():  # generator of terms
    aset = 0 # stored as a bitmask
    for k in count(0):
        if (k&aset).bit_count()%2 == 0:
            yield k
            aset += (1<Michael S. Branicky, Apr 28 2024

A353173 Dimension of space of invariants of n-th tensor power of the 26-dimensional fundamental (or "standard") irreducible representation of F_4.

Original entry on oeis.org

1, 0, 1, 1, 5, 15, 70, 330, 1820, 10858, 70875, 497135, 3727955, 29658410, 248989676, 2194891440, 20231692430, 194286848280, 1937546532820, 20008993160460, 213436182918652, 2346406693816315, 26531060178217182, 307987244037724262, 3664579007885995952

Offset: 0

David A. Madore, Apr 28 2022

nonn

It is known that a(n) satisfies a linear recurrence relation with polynomial coefficients. The limit of a(n+1)/a(n) is 26.

			a(1)=0 because there is no F_4-invariant linear form on the 26-dimensional representation; a(2)=1 because there is, up to scalars, precisely one invariant quadratic form.

MathOverflow, Invariants for the exceptional complex simple Lie algebra F4

The analogous sequence for the (52-dimensional) adjoint representation of F_4 is: A179685.
A similar sequence for G_2 (for its 7-dimensional fundamental irreducible representation) is: A059710.
A similar sequence for B_2 (for its standard 5-dimensional irreducible representation) is: A095922.
For A_n the similar sequence (omitting some 0's) is given by the (n+1)-dimensional Catalan numbers, e.g., A005789 for A_2.

p_tensor(n,[0,0,0,1],F4)|[0,0,0,0] # Returns the value of a(n).

A342817 Power series expansion of AQM(1,1-8x) where AQM denotes the arithmetic-quadratic mean.

Original entry on oeis.org

1, -4, 4, 16, 52, 112, -48, -1984, -11212, -33360, 6224, 713536, 4441872, 13004480, -17374656, -432012032, -2525831628, -6454496208, 21147389392, 326358047552, 1794285832464, 4124461926592, -19727734694848, -263598020446976, -1416694290412784, -3151402998261312

Offset: 0

David A. Madore, Mar 22 2021

sign

Generating function: AQM(1,1-8x) where AQM(u,v) (arithmetic-quadratic mean of u and v) is the fixed point obtained by iterating ((u+v)/2, sqrt((u^2+v^2)/2)) (we choose 1-8x in order to avoid denominators, as in A060691).

			First steps of iteration of ((u+v)/2, sqrt((u^2+v^2)/2)) are (1, 1-8x), (1 - 4*x, 1 - 4*x + 8*x^2 + 32*x^3 + 96*x^4 + O(x^5)), then (1 - 4*x + 4*x^2 + 16*x^3 + 48*x^4 + O(x^5), 1 - 4*x + 4*x^2 + 16*x^3 + 56*x^4 + O(x^5)) and (1 - 4*x + 4*x^2 + 16*x^3 + 52*x^4 + O(x^5), 1 - 4*x + 4*x^2 + 16*x^3 + 52*x^4 + O(x^5)), so the first terms of this sequence are 1, -4, 4, 16, 52.

Compare A060691 for the arithmetic-geometric mean.

seq(n)={my(p=1, q=1-8*x+O(x*x^n)); while(p!=q, my(t=p+q); q = sqrt((p^2 + q^2)/2); p=t/2); Vec(p)} \\ Andrew Howroyd, Mar 22 2021

R. = PowerSeriesRing(QQ, default_prec=50)
(a,b) = (1,1-8*x)
for i in range(50):
    (a,b) = ((a+b)/2, sqrt((a^2+b^2)/2))
a.coefficients()

A331814 Fourier coefficients of the boundary of the Mandelbrot set (normalized by a power of two).

Original entry on oeis.org

-1, 1, -8, 15, 0, -94, -512, 987, 0, -7346, 65536, -122058, 0, -2757820, -22020096, 59250963, 0, -329425898, 2617245696, -4805611678, -34359738368, -19403249316, 498216206336, -36302282082, 0, 14136557849100, -71399536328704, -88183884706356

Offset: 0

David A. Madore, Jan 27 2020

sign

a(n) = 2^(2*n+1)*b(n) where b(n) is the unique sequence of complex numbers such that f(z) := z + Sum_{n>=0} (b(n)*z^-n) defines an analytic homeomorphism (biholomorphic bijection) between the complement of the unit disk and the complement of the Mandelbrot set, sometimes known as the "Jungreis function". (The b(n) are rationals, so we multiply them by the appropriate power of two to make them integers; this is equivalent to a simple rescaling of the complex plane.) It is conjectured that |b(n)| <= 1/n, so |a(n)| <= 2^(2*n+1)/n.

Note that the table given in Ewing and Schober (1992) gives the coefficients of the inverse series (contrary to what the text itself says): it's not wrong, it's just mislabeled.

			a(0)=-1 because B(1,1)=0 and B(0,1) = 8*B(1,1) - 2*B(0,0) = -1; then a(1)=1 because B(1,2)=0 and B(0,2) = 8*B(1,2) - B(0,1)^2 - 2*B(0,1) = 1.

John H. Ewing and Glenn Schober, On the Coefficients of the Mapping to the Exterior of the Mandelbrot set, Michigan Math. J. 37 (1990), 315-320.
John H. Ewing & Glenn Schober, The area of the Mandelbrot set, Numer. Math. 61 (1992) 59-72 (note that table 1 gives the coefficients of the INVERSE series).
Irwin Jungreis, The uniformization of the complement of the Mandelbrot set, Duke Math. J. 4 (1985), 935-938.
David A. Madore, Sage code

a(m)=B(0,m+1) where B(0,0)=1/2 and by downwards induction on k we have B(k-1,m) = 2^(2^(k+1)-1)*B(k,m) - 2^(2^(k+1)-4)*Sum_{j=2^k-1..m-2^k+1} (B(k-1,j)*B(k-1,m-j) - 2*B(0,m-2^k+1)) if m >= 2^k-1, 0 otherwise.

A218452 Number of ways to factor (1 + x + x^2+ ... + x^(n - 1))^2 as the product of two monic polynomials of degree n - 1 with positive coefficients (counting order).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 5, 7, 9, 9, 13, 11, 17, 19, 33

Offset: 1

David A. Madore, Oct 28 2012

nonn

a(n) is the number of ways one can divide the unit square in n possibly irregular lines * n possibly irregular columns (parallel to the sides) so that each of the diagonals of the n X n irregular checkerboard thus constructed has the same area as it would in a regular checkerboard. Alternatively, this is the number of ways to construct a pair of n-sided dice (probability distribution on the n sides, labeled 0 through n-1), no face having probability 0, so that the sum of the two dice follows the expected probability distribution for the sum of two fair n-sided dice. Note that a(n) is always odd because there is always the obvious factorization of (1+x+...+x^(n-1))^2 as 1+x+...+x^(n-1) times itself, and each other factorization counts twice.

			For n=12 we have a(n)=3 because apart from the obvious factorization of (1+x+...+x^11)^2 as (1+x+...+x^11) times itself, there exist the factorizations p*q and q*p where p = (1-sqrt(3)*x+x^2) * (1-x+x^2) * (1+x^2) * (1+x+x^2)^2 * (1+x) and q = (1-sqrt(3)*x+x^2) * (1-x+x^2) * (1+x^2) * (1+sqrt(3)*x+x^2)^2 * (1+x), both of which have positive coefficients, and those are the only two possible.

R. = AA['x']
def has_positive_coefficients(pol):
    return not any(c <= 0 for c in pol.coeffs())
def trydie(m):
    results = []
    tmp = list(factor(sum([x^i for i in range(m)])))
    facs = [f for (f,_) in tmp]
    n = len(facs)
    for i in range((3^n+1)//2):
        exps = [(i//(3^k))%3 for k in range(n)]
        coexps = [2-v for v in exps]
        pol = R(prod([facs[k]^exps[k] for k in range(n)]))
        copol = R(prod([facs[k]^coexps[k] for k in range(n)]))
        if pol.degree()

A181746 List of dimensions for which there exist several non-isomorphic irreducible representations of E8.

Original entry on oeis.org

8634368000, 175898504162692612600853299200000, 569584588357708246352221171200000000000, 30903356351561222538825891668691517440000, 75783579745006249731939558173442048000000, 221669803220344213547594484715842176665500, 11477043671509412495692698759458678235987968000

Offset: 1

David A. Madore, Nov 08 2010

nonn

Terms in this sequence are the terms which are repeated in A121732.

This sequence is infinitely long. For example, there exists two non-isomorphic irreducible E8 representations of dimension n^120 * 8634368000 for any n>0 (the dimension formula is homogeneous of degree 120 after a suitable change of variables). I have found no dimension with more than two non-isomorphic irreducible E8 representations. - Andy Huchala, Nov 22 2020

			With the fundamental weights numbered as in Bourbaki, the highest weights 10100000 and 10000011 both correspond to irreducible representations of dimension 8634368000. The highest weights 23000130 and 12000231 both correspond to irreducible representations of dimension 175898504162692612600853299200000.

N. Bourbaki, Lie Groups and Lie Algebras Chapters 4-6, Springer, 1968, 283-285.

Andy Huchala, Table of n, a(n) for n = 1..332
Andy Huchala, Java program
Andy Huchala, C++ program

Cf. A121732.

A178854 Asymptotic value of odd Catalan numbers mod 2^n.

Original entry on oeis.org

0, 1, 1, 5, 13, 29, 29, 93, 221, 221, 733, 1757, 3805, 7901, 7901, 24285, 57053, 122589, 122589, 384733, 384733, 384733, 2481885, 2481885, 10870493, 10870493, 10870493, 10870493, 145088221

Offset: 0

David A. Madore, Jun 18 2010

nonn

For every n, the odd Catalan numbers C(2^m-1) are eventually constant mod 2^n (namely for m >= n-1): then a(n) is the asymptotic value of the remainder.

			The odd Catalan numbers mod 2^6=64 are 1,5,45,61,29,29,29, so a(6)=29.

Shu-Chung Liu and Jean C.-C. Yeh, Catalan numbers modulo 2^k, J. Int. Seq. 13 (2010), article 10.5.4.

Cf. A038003 (odd Catalan numbers).

A000108 := proc(n) binomial(2*n,n)/(n+1) ; end proc:
A038003 := proc(n) A000108(2^n-1) ; end proc:
A178854 := proc(n) if n = 0 then 0; else modp(A038003(n-1),2^n) ; end if; end proc:
for n from 0 do printf("%d,\n",A178854(n)) ; end do: # R. J. Mathar, Jun 28 2010

(* first do *) Needs["DiscreteMath`CombinatorialFunctions`"] (* then *) f[n_] := Mod[ CatalanNumber[2^n - 1], 2^n]; Array[f, 25, 0] (* Robert G. Wilson v, Jun 28 2010 *)

a(n) = remainder(Catalan(2^m-1), 2^n) for any m >= n-1.

a(12)-a(24) from Robert G. Wilson v, Jun 28 2010

a(25)-a(28) from Robert G. Wilson v, Jul 23 2010

A174391 Triangle, read by row, of constant term of X^k modulo n-th cyclotomic polynomial.

Original entry on oeis.org

1, 1, -1, 1, 0, -1, 1, 0, -1, 0, 1, 0, 0, 0, -1, 1, 0, -1, -1, 0, 1, 1, 0, 0, 0, 0, 0, -1, 1, 0, 0, 0, -1, 0, 0, 0, 1, 0, 0, 0, 0, 0, -1, 0, 0, 1, 0, 0, 0, -1, -1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 1, 0, 0, 0, -1, 0, -1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 1, 0, 0, 0, 0, 0, -1

Offset: 1

David A. Madore, Mar 18 2010

a(n,k) (for n>=1 and 0<=k

			For n=6, the 6th cyclotomic polynomial is Phi_6(X) = X^2-X+1, and the remainders of 1, X, X^2,..., X^5 mod Phi_6 are 1, X, X-1, -1, -X, -X+1, so a(6,k)=1,0,-1,-1,0,1 for k from 0 to 5. (This gives the terms 16 to 22 of this sequence.)

R. = QQ['x']; [[((x^k)%(R.cyclotomic_polynomial(n))).subs(0) for k in range(n)] for n in range(1,31)]

A152049 Number of conjugacy classes of primitive elements in GF(2^n) which have trace 0.

Original entry on oeis.org

0, 0, 1, 1, 3, 2, 9, 9, 23, 29, 89, 72, 315, 375, 899, 1031, 3855, 3886, 13797, 12000, 42328, 59989, 178529, 138256, 647969, 859841, 2101143, 2370917, 9204061, 8911060, 34636833, 33556537, 105508927, 168423669, 464635937

Offset: 1

David A. Madore, Nov 21 2008

Also number of primitive polynomials of degree n over GF(2) whose second-highest coefficient is 0.

Always less than A011260 (and exactly one half of it when 2^n-1 is prime).

			a(3)=1 because of the two primitive degree 3 polynomials over GF(2), namely t^3+t+1 and t^3+t^2+1, only the former has a zero next-to-highest coefficient.
Similarly, a(13)=315, because of half (4096) of the 8192 elements of GF(2^13) have trace 0 and all except 0 (since 1 has trace 1) are primitive, so there are 4095/13=315 conjugacy classes of primitive elements of trace 0.

Cf. A192507 (GF(3^n)), A192508 (GF(5^n)), A192509 (GF(7^n)), A192510 (GF(11^n)), A192511 (GF(13^n)).

a := function(n)
    local q,k,cnt,x; q:=2^n; k:=GF(2,n); cnt:=0;
    for x in k do
        if Trace(k, GF(2), x)=0*Z(2) and Order(x)=q-1 then
            cnt := cnt+1;
        fi;
    od;
    return cnt/n;
end;
for n in [1..32] do  Print (a(n), ", ");  od;

a(n) = A192211(n)/n. [Joerg Arndt, Jul 03 2011]

More terms (13797...8911060) by Joerg Arndt, Jun 26 2011.

More terms (34636833...464635937) by Joerg Arndt, Jul 03 2011.

A130721 Sum of the cubes of the number of standard Young tableaux over all partitions of n.

Original entry on oeis.org

1, 1, 2, 10, 64, 596, 8056, 130432, 2534960, 59822884, 1718480368, 56754444440, 2110577206816, 87981286785328, 4129351961475872, 218382856010529472, 12813477368159567200, 822337333595479929044, 57213666993723455063392, 4305630141314873304140008

Offset: 0

David A. Madore, Jul 03 2007

nonn

The sum of the zeroth power of the number f(p) of standard Young tableaux gives the partition function (A000041), the sum of the first power of f(p) gives the involution function (A000085), the sum of the squares of f(p) gives the factorial function (A000142), so this sequence is the natural one after them.

			a(4) = 1^3 + 3^3 + 2^3 + 3^3 + 1^3 because the five partitions of 4 (namely 4, 3+1, 2+2, 2+1+1, 1+1+1+1) have respectively 1, 3, 2, 3, 1 standard Young tableaux.

Alois P. Heinz, Table of n, a(n) for n = 0..60

Cf. A000041, A000085, A000142.

Column k=3 of A208447.

h[l_] := With[{n=Length[l]}, Sum[i, {i, l}]!/Product[Product[1 + l[[i]] - j + Sum[If[l[[k]] >= j, 1, 0], {k, i+1, n}], {j, 1, l[[i]]}], {i, 1, n}]];
g[n_, i_, k_, l_] := g[n, i, l, k] = If[n == 0, h[l]^k, If[i < 1, 0, g[n, i - 1, k, l] + If[i > n, 0, g[n - i, i, k, Append[l, i]]]]];
a[n_] := If[n == 0, 1, g[n, n, 3, {}]];
Table[a[n], {n, 0, 20}] (* Jean-François Alcover, May 18 2017, after Alois P. Heinz *)

For p a partition of n, let f(p) be the number of standard Young tableaux with shape p. Then a(n) = sum(f(p)^3) where the sum ranges over all partitions p of n.

More terms from Alois P. Heinz, Feb 26 2012

cp's OEIS Frontend

User: David A. Madore

A372325 Numbers whose binary expansion has an even number of 1's among positions listed in this sequence.

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A353173 Dimension of space of invariants of n-th tensor power of the 26-dimensional fundamental (or "standard") irreducible representation of F_4.

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A342817 Power series expansion of AQM(1,1-8x) where AQM denotes the arithmetic-quadratic mean.

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PARI

Sage

A331814 Fourier coefficients of the boundary of the Mandelbrot set (normalized by a power of two).

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A218452 Number of ways to factor (1 + x + x^2+ ... + x^(n - 1))^2 as the product of two monic polynomials of degree n - 1 with positive coefficients (counting order).

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Sage

A181746 List of dimensions for which there exist several non-isomorphic irreducible representations of E8.

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A178854 Asymptotic value of odd Catalan numbers mod 2^n.

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A174391 Triangle, read by row, of constant term of X^k modulo n-th cyclotomic polynomial.

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Sage

A152049 Number of conjugacy classes of primitive elements in GF(2^n) which have trace 0.

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