cp's OEIS Frontend

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.

Previous Showing 11-18 of 18 results.

A092920 Number of strongly monotone partitions of [n].

Original entry on oeis.org

1, 1, 2, 4, 9, 22, 58, 164, 496, 1601, 5502, 20075, 77531, 315947, 1354279, 6087421, 28611385, 140239297, 715116827, 3785445032, 20760746393, 117759236340, 689745339984, 4165874930885, 25911148634728, 165775085602106, 1089773992530717, 7353740136527305
Offset: 0

Views

Author

Ralf Stephan, Apr 17 2004

Keywords

Comments

A partition is strongly monotone if its blocks can be written in increasing order of their least element and increasing order of their greatest element, simultaneously.
a(n) is the number of strongly nonoverlapping partitions of [n] where "strongly nonoverlapping" means nonoverlapping (see A006789 for definition) and, in addition, no singleton block is a subset of the span (interval from minimum to maximum) of another block. For example, 13-24 is nonnesting and 14-23 is strongly nonoverlapping but neither has the other property. The Motzkin number M_n (A001006) counts strongly noncrossing partitions of [n]. - David Callan, Sep 20 2007
Strongly monotone partitions can also be described as partitions in which no block is contained in the span of another, where span denotes the interval from smallest to largest entries. For example, 134/25/6 is strongly monotone but 135/24/6 is not because the block 24 is contained in the interval [1,5]. - David Callan, Aug 27 2014

Crossrefs

Similar recurrences: A284005, A329369, A341392.

Programs

  • Maple
    G:=1/(1-x-x^2/(1-x-x^2/(1-2*x-x^2/(1-3*x-x^2/(1-4*x-x^2/(1-5*x-x^2/(1-6*x-x^2/(1-7*x-x^2/(1-8*x-x^2/(1-9*x-x^2/(1-10*x-x^2/(1-11*x-x^2/(1-12*x-x^2/(1-13*x-x^2/(1-14*x-x^2/(1-15*x-x^2/(1-16*x-x^2/(1-17*x-x^2)))))))))))))))))): Gser:=series(G,x=0,32): seq(coeff(Gser, x, n), n=0..28);  # Emeric Deutsch, Apr 13 2005
  • Mathematica
    terms = 26;
    f[1] = 1; f[k_ /; k>1] = -x^2;
    g[1] = 1-x; g[k_ /; k>1] := 1-(k-1)x;
    A[x_] = ContinuedFractionK[f[k], g[k], {k, 1, Ceiling[terms/2]}];
    CoefficientList[A[x] + O[x]^terms, x] (* Jean-François Alcover, Aug 07 2018 *)

Formula

G.f.: Sum_{n>=0} a(n)*x^n = 1/(1-x-x^2/(1-x-x^2/(1-2*x-x^2/(1-3*x-x^2/...)))) = 1/(1-x-x^2*B(x)) where B(x) is g.f. for the Bessel numbers A006789.
a(n) = leftmost column terms of M^n*V, where M = an infinite tridiagonal matrix with all 1's in the super and subdiagonals and (1,1,2,3,4,5,...) as the main diagonal; and the rest zeros. V = vector [1,0,0,0,...]. - Gary W. Adamson, Jun 16 2011
G.f.: 1/Q(0) where Q(k) = 1-x*(k+2)+x/(1+x/Q(k+1)); (continued fraction). - Sergei N. Gladkovskii, Apr 17 2013
Conjecture: a(n) = Sum_{k=0..2^(n-1) - 1} b(k) for n > 0 with a(0) = 1 where b(2^m*(2n+1)) = Sum_{k=0..[m > 0]*(m-1)} binomial(m-1, k)*b(2^k*n) for m >= 0, n >= 0 with b(0) = 1. - Mikhail Kurkov, Apr 24 2023

Extensions

More terms from Emeric Deutsch, Apr 13 2005

A347204 a(n) = a(f(n)/2) + a(floor((n+f(n))/2)) for n > 0 with a(0) = 1 where f(n) = A129760(n).

Original entry on oeis.org

1, 2, 3, 5, 4, 7, 10, 15, 5, 9, 13, 20, 17, 27, 37, 52, 6, 11, 16, 25, 21, 34, 47, 67, 26, 43, 60, 87, 77, 114, 151, 203, 7, 13, 19, 30, 25, 41, 57, 82, 31, 52, 73, 107, 94, 141, 188, 255, 37, 63, 89, 132, 115, 175, 235, 322, 141, 218, 295, 409, 372, 523, 674
Offset: 0

Views

Author

Mikhail Kurkov, Aug 23 2021 [verification needed]

Keywords

Comments

Modulo 2 binomial transform of A243499(n).

Crossrefs

Programs

  • MATLAB
    function a = A347204(max_n)
        a(1) = 1;
        a(2) = 2;
        for nloop = 3:max_n
            n = nloop-1;
            s = 0;
            for k = 0:floor(log2(n))-1
                s = s + a(1+A053645(n)-2^k*(mod(floor(n/(2^k)),2)));
            end
            a(nloop) = 2*a(A053645(n)+1) + s;
        end
    end
    function a_n = A053645(n)
        a_n = n - 2^floor(log2(n));
    end % Thomas Scheuerle, Oct 25 2021
  • Mathematica
    f[n_] := BitAnd[n, n - 1]; a[0] = 1; a[n_] := a[n] = a[f[n]/2] + a[Floor[(n + f[n])/2]]; Array[a, 100, 0] (* Amiram Eldar, Nov 19 2021 *)
  • PARI
    f(n) = bitand(n, n-1); \\ A129760
    a(n) = if (n<=1, n+1, if (n%2, a(n\2)+a(n-1), a(f(n/2)) + a(n/2+f(n/2)))); \\ Michel Marcus, Oct 25 2021
    
  • PARI
    \\ Also see links.
    
  • PARI
    A129760(n) = bitand(n, n-1);
    memoA347204 = Map();
    A347204(n) = if (n<=1, n+1, my(v); if(mapisdefined(memoA347204,n,&v), v, v = if(n%2, A347204(n\2)+A347204(n-1), A347204(A129760(n/2)) + A347204(n/2+A129760(n/2))); mapput(memoA347204,n,v); (v))); \\ (Memoized version of Michel Marcus's program given above) - Antti Karttunen, Nov 20 2021
    

Formula

a(n) = a(n - 2^f(n)) + (1 + f(n))*a((n - 2^f(n))/2) for n > 0 with a(0) = 1 where f(n) = A007814(n).
a(2n+1) = a(n) + a(2n) for n >= 0.
a(2n) = a(n - 2^f(n)) + a(2n - 2^f(n)) for n > 0 with a(0) = 1 where f(n) = A007814(n).
a(n) = 2*a(f(n)) + Sum_{k=0..floor(log_2(n))-1} a(f(n) - 2^k*T(n,k)) for n > 1 with a(0) = 1, a(1) = 2, and where f(n) = A053645(n), T(n,k) = floor(n/2^k) mod 2.
Sum_{k=0..2^n - 1} a(k) = A035009(n+1) for n >= 0.
a((4^n - 1)/3) = A002720(n) for n >= 0.
a(2^n - 1) = A000110(n+1),
a(2*(2^n - 1)) = A005493(n),
a(2^2*(2^n - 1)) = A005494(n),
a(2^3*(2^n - 1)) = A045379(n),
a(2^4*(2^n - 1)) = A196834(n),
a(2^m*(2^n-1)) = T(n,m+1) is the n-th (m+1)-Bell number for n >= 0, m >= 0 where T(n,m) = m*T(n-1,m) + Sum_{k=0..n-1} binomial(n-1,k)*T(k,m) with T(0,m) = 1.
a(n) = Sum_{j=0..2^A000120(n)-1} A243499(A295989(n,j)) for n >= 0. Also A243499(n) = Sum_{j=0..2^f(n)-1} (-1)^(f(n)-f(j)) a(A295989(n,j)) for n >= 0 where f(n) = A000120(n). In other words, a(n) = Sum_{j=0..n} (binomial(n,j) mod 2)*A243499(j) and A243499(n) = Sum_{j=0..n} (-1)^(f(n)-f(j))*(binomial(n,j) mod 2)*a(j) for n >= 0 where f(n) = A000120(n).
Generalization:
b(n, x) = (1/x)*b((n - 2^f(n))/2, x) + (-1)^n*b(floor((2n - 2^f(n))/2), x) for n > 0 with b(0, x) = 1 where f(n) = A007814(n).
Sum_{k=0..2^n - 1} b(k, x) = (1/x)^n for n >= 0.
b((4^n - 1)/3, x) = (1/x)^n*n!*L_{n}(x) for n >= 0 where L_{n}(x) is the n-th Laguerre polynomial.
b((8^n - 1)/7, x) = (1/x)^n*Sum_{k=0..n} (-x)^k*A265649(n, k) for n >= 0.
b(2^n - 1, x) = (1/x)^n*Sum_{k=0..n} (-x)^k*A008277(n+1, k+1),
b(2*(2^n - 1), x) = (1/x)^n*Sum_{k=0..n} (-x)^k*A143494(n+2, k+2),
b(2^2*(2^n - 1), x) = (1/x)^n*Sum_{k=0..n} (-x)^k*A143495(n+3, k+3),
b(2^m*(2^n - 1), x) = (1/x)^n*Sum_{k=0..n} (-x)^k*T(n+m+1, k+m+1, m+1) for n >= 0, m >= 0 where T(n,k,m) is m-Stirling numbers of the second kind.

A358612 Irregular table T(n, k), n >= 0, k > 0, read by rows of extended (due to binary expansion of n) Stirling numbers of the second kind.

Original entry on oeis.org

1, 1, 1, 3, 1, 1, 5, 2, 1, 7, 6, 1, 1, 9, 4, 1, 11, 11, 2, 1, 13, 15, 3, 1, 15, 25, 10, 1, 1, 17, 8, 1, 19, 21, 4, 1, 21, 28, 6, 1, 23, 44, 19, 2, 1, 25, 39, 9, 1, 27, 58, 27, 3, 1, 29, 68, 34, 4, 1, 31, 90, 65, 15, 1, 1, 33, 16, 1, 35, 41, 8, 1, 37, 54, 12, 1
Offset: 0

Views

Author

Mikhail Kurkov, Nov 23 2022

Keywords

Comments

Row n length is A000120(n) + 2.

Examples

			Irregular table begins:
  1,  1;
  1,  3,  1;
  1,  5,  2;
  1,  7,  6,  1;
  1,  9,  4;
  1, 11, 11,  2;
  1, 13, 15,  3;
  1, 15, 25, 10,  1;
  1, 17,  8;
  1, 19, 21,  4;
  1, 21, 28,  6;
  1, 23, 44, 19,  2;
  1, 25, 39,  9;
  1, 27, 58, 27,  3;
  1, 29, 68, 34,  4;
  1, 31, 90, 65, 15, 1;
		

Crossrefs

Programs

  • PARI
    T(n, k)=if(n==0 || k==1, (n==0 && k<3) + (k==1 && n>0), k*T(n\2, k) + T(n\2, k-1) - if(n%2==0, (T(n, k-1) + T(n\2,k-1))/(k-1)))
    
  • PARI
    row(n) = my(A, v1, v2); v1 = [1, 1]; if(n == 0, v1, forstep(i=logint(n, 2), 0, -1, A = bittest(n, i); v2 = vector(#v1+A, i, 0); v2[1] = 1; for(j=2, #v2, v2[j] = j*if(j==#v1+1, 0, v1[j]) + v1[j-1] - if(A, 0, (v2[j-1] + v1[j-1])/(j-1))); v1 = v2); v1) \\ Mikhail Kurkov, Apr 30 2024

Formula

T(n, 1) = 1 for n > 0 with T(0, 1) = T(0, 2) = 1.
T(2n+1, k) = k*T(n, k) + T(n, k-1) for n >= 0, k > 1.
T(2n, k) = k*T(n, k) + T(n, k-1) - (T(2n, k-1) + T(n, k-1))/(k-1) for n > 0, k > 1.
T(2^n - 1, k) = Stirling2(n+2, k) for n >= 0, k > 0.
T(n, 2) = 2n+1 for n >= 0.
Conjectured formulas: (Start)
T(n, A000120(n) + 2) = A341392(n) for n >= 0.
Sum_{i=1..wt(k) + 2} i!*i^m*T(k, i)*(-1)^(wt(k) - i + 2) = A329369(2^m*(2k+1)) for m >= 0, k >= 0 where wt(n) = A000120(n). (End)
Conjecture: T(n, k) = (k-1)^g(n)*T(h(n), k-1) + k^(g(n)+1)*T(h(n), k) for n > 0, k > 1 with T(n, 1) = T(0, 2) = 1 where g(n) = A007814(n) and where h(n) = A025480(n-1). - Mikhail Kurkov, Jun 21 2024

A373183 Irregular table T(n, k), n >= 0, k > 0, read by rows with row polynomials R(n, x) such that R(2n+1, x) = x*R(n, x) for n >= 0, R(2n, x) = x*(R(n, x+1) - R(n, x)) for n > 0 with R(0, x) = x.

Original entry on oeis.org

1, 0, 1, 1, 2, 0, 0, 1, 3, 4, 0, 1, 2, 1, 3, 3, 0, 0, 0, 1, 7, 8, 0, 3, 4, 3, 8, 6, 0, 0, 1, 2, 7, 15, 9, 0, 1, 3, 3, 1, 4, 6, 4, 0, 0, 0, 0, 1, 15, 16, 0, 7, 8, 7, 18, 12, 0, 0, 3, 4, 17, 34, 18, 0, 3, 8, 6, 3, 11, 15, 8, 0, 0, 0, 1, 2, 31, 57, 27, 0, 7, 15
Offset: 0

Views

Author

Mikhail Kurkov, May 27 2024

Keywords

Comments

Row n length is A000120(n) + 1.

Examples

			Irregular table begins:
  1;
  0,  1;
  1,  2;
  0,  0, 1;
  3,  4;
  0,  1, 2;
  1,  3, 3;
  0,  0, 0, 1;
  7,  8;
  0,  3, 4;
  3,  8, 6
  0,  0, 1, 2
  7, 15, 9;
  0,  1, 3, 3;
  1,  4, 6, 4;
  0,  0, 0, 0, 1;
		

Crossrefs

Programs

  • PARI
    row(n) = my(x = 'x, A = x); forstep(i=if(n == 0, -1, logint(n, 2)), 0, -1, A = if(bittest(n, i), x*A, x*(subst(A, x, x+1) - A))); Vecrev(A/x)

Formula

Conjectured formulas: (Start)
R(2n, x) = R(n, x) + R(n - 2^f(n), x) + R(2n - 2^f(n), x) where f(n) = A007814(n) (see A329369).
b(2^m*n + q) = Sum_{i=A001511(n+1)..A000120(n)+1} T(n, i)*b(2^m*(2^(i-1)-1) + q) for n >= 0, m >= 0, q >= 0 where b(n) = A329369(n). Note that this formula is recursive for n != 2^k - 1.
R(n, x) = c(n, x)
where c(2^k - 1, x) = x^(k+1) for k >= 0,
c(n, x) = Sum_{i=0..s(n)} p(n, s(n)-i)*Sum_{j=0..i} (s(n)-j+1)^A279209(n)*binomial(i, j)*(-1)^j,
p(n, k) = Sum_{i=0..k} c(t(n) + (2^i - 1)*A062383(t(n)), x)*L(s(n), k, i) for 0 <= k < s(n) with p(n, s(n)) = c(t(n) + (2^s(n) - 1)*A062383(t(n)), x),
s(n) = A090996(n), t(n) = A087734(n),
L(n, k, m) are some integer coefficients defined for n > 0, 0 <= k < n, 0 <= m <= k that can be represented as W(n-m, k-m, m+1)
and where W(n, k, m) = (k+m)*W(n-1, k, m) + (n-k)*W(n-1, k-1, m) + [m > 1]*W(n, k, m-1) for 0 <= k < n, m > 0 with W(0, 0, m) = 1, W(n, k, m) = 0 for n < 0 or k < 0.
In particular, W(n, k, 1) = A173018(n, k), W(n, k, 2) = A062253(n, k), W(n, k, 3) = A062254(n, k) and W(n, k, 4) = A062255(n, k).
Here s(n), t(n) and A279209(n) are unique integer sequences such that n can be represented as t(n) + (2^s(n) - 1)*A062383(t(n))*2^A279209(n) where t(n) is minimal. (End)
Conjectures from Mikhail Kurkov, Jun 19 2024: (Start)
T(n, k) = d(n, 1, A000120(n) - k + 2) where d(n, m, k) = (m+1)^g(n)*d(h(n), m+1, k) - m^(g(n)+1)*d(h(n), m, k-1) for n > 0, m > 0, k > 0 with d(n, m, 0) = 0 for n >= 0, m > 0, d(0, m, k) = [k <= m]*abs(Stirling1(m, m-k+1)) for m > 0, k > 0, g(n) = A290255(n) and where h(n) = A053645(n). In particular, d(n, 1, 1) = A341392(n).
Sum_{i=A001511(n+1)..wt(n)+k} d(n, k, wt(n)-i+k+1)*A329369(2^m*(2^(i-1)-1) + q) = k!*A357990(2^m*n + q, k) for n >= 0, k > 0, m >= 0, q >= 0 where wt(n) = A000120(n).
If we change R(0, x) to Product_{i=0..m-1} (x+i), then for resulting irregular table U(n, k, m) we have U(n, k, m) = d(n, m, A000120(n) - k + m + 1).
T(n, k) = (-1)^(wt(n)-k+1)*Sum_{i=1..wt(n)-k+3} Stirling1(wt(n)-i+3, k+1)*A358612(n, wt(n)-i+3) for n >= 0, k > 0 where wt(n) = A000120(n). (End)
Conjecture: T(2^m*(2k+1), q) = (-1)^(wt(k)-q)*Sum_{i=q..wt(k)+2} Stirling1(i,q)*A358612(k,i)*i^m for m >= 0, k >= 0, q > 0 where wt(n) = A000120(n). - Mikhail Kurkov, Jan 17 2025

A379817 Irregular table T(n, k), n >= 0, k >= 0, read by rows such that T(n,k) = f(n,k)/f(2^k-1,k) where f(n,k) is defined in Comments.

Original entry on oeis.org

1, 1, 1, 3, 2, 1, 3, 1, 7, 4, 3, 7, 2, 7, 12, 3, 1, 7, 6, 1, 15, 8, 7, 15, 4, 17, 26, 6, 3, 17, 13, 2, 31, 42, 9, 7, 31, 21, 3, 15, 50, 30, 4, 1, 15, 25, 10, 1, 31, 16, 15, 31, 8, 37, 54, 12, 7, 37, 27, 4, 69, 88, 18, 17, 69, 44, 6, 37, 112, 63, 8, 3, 37, 56
Offset: 0

Views

Author

Mikhail Kurkov, Jan 03 2025

Keywords

Comments

Here f(n,k) = b(2^k*(2n+1)) - Sum_{j=1..k} b(2^(j-1)*(2n+1))*R(k,j) for n >= 0, k >= 0 where b(n) = A329369(n) and where R(k,j) is the unique solution to b(2^k*(2^i-1)) = Sum_{j=1..k} b(2^(j-1)*(2^i-1))*R(k,j) for k > 0, 1 <= i <= k.
Row n length is A000120(n) + 1.

Examples

			Irregular table begins:
   1;
   1,  1;
   3,  2;
   1,  3,  1;
   7,  4;
   3,  7,  2;
   7, 12,  3;
   1,  7,  6,  1;
  15,  8;
   7, 15,  4;
  17, 26,  6;
   3, 17, 13,  2;
  31, 42,  9;
   7, 31, 21,  3;
  15, 50, 30,  4;
   1, 15, 25, 10, 1;
		

Crossrefs

Programs

  • PARI
    upto(n) = my(A, v1); v1 = vector(n+1, i, 0); v1[1] = 1; for(i=1, n, v1[i+1] = v1[i\2+1] + if(i%2, 0, A = 1 << valuation(i/2, 2); v1[i/2-A+1] + v1[i-A+1])); v1 \\ from A329369
    R(k) = my(v1, M1, M2); v1 = upto(2^k*(2^k-1)); M1 = matrix(k, k, i, j, v1[2^(j-1)*(2^i-1)+1]); M2 = matrix(k, 1, i, j, v1[2^k*(2^i-1)+1]); M1 = matsolve(M1, M2)
    row(n) = my(A = hammingweight(n), v1, v2, v3); v1 = upto(2^A*(2*n+1)); v2 = vector(A, i, R(i)); v3 = vector(A, i, (v1[2^i*(2*n+1)+1] - sum(j=1, i, v1[2^(j-1)*(2*n+1)+1]*v2[i][j,1]))/(v1[2^i*(2*(2^i-1)+1)+1] - sum(j=1, i, v1[2^(j-1)*(2*(2^i-1)+1)+1]*v2[i][j,1]))); concat(v1[n+1], v3)

Formula

Conjectures: (Start)
f(2^k-1,k) = ((k+1)!)^2 for k >= 0.
R(k,j) = -Stirling1(k+2, j+1) for k > 0, 1 <= j <= k.
T(2^n-1, k) = Stirling2(n+1, k+1) for n >= 0, 0 <= k <= n.
T(n,k) = c(n,wt(n)-k) for n >= 0, 0 <= k <= wt(n) where c(2n+1,k) = c(n,k) + (wt(n)-k+2)*c(n,k-1), c(2n,k) = (wt(n)-k+1)*c(2n+1,k) for n > 0, k > 0 with c(n,0) = A341392(n) for n >= 0, c(0,k) = 0 for k > 0 and where wt(n) = A000120(n). (End)

A379818 a(2n+1) = a(n) for n >= 0, a(2n) = a(n) + a(n - 2^f(n)) + a(2n - 2^f(n)) + a(A025480(n-1)) for n > 0 with a(0) = 1 where f(n) = A007814(n).

Original entry on oeis.org

1, 1, 4, 1, 10, 4, 10, 1, 22, 10, 28, 4, 49, 10, 22, 1, 46, 22, 64, 10, 118, 28, 64, 4, 190, 49, 118, 10, 190, 22, 46, 1, 94, 46, 136, 22, 256, 64, 148, 10, 424, 118, 292, 28, 478, 64, 136, 4, 661, 190, 478, 49, 796, 118, 256, 10, 1177, 190, 424, 22, 661, 46
Offset: 0

Views

Author

Mikhail Kurkov, Jan 03 2025

Keywords

Crossrefs

Programs

  • PARI
    upto(n) = my(A, v1); v1 = vector(n+1, i, 0); v1[1] = 1; for(i=1, n, v1[i+1] = v1[i\2+1] + if(i%2, 0, A = 1 << valuation(i/2, 2); v1[i/2-A+1] + v1[i-A+1] + v1[i\(4*A)+1])); v1

Formula

Conjecture: a(2^m*(2k+1)) = Sum_{j=0..m} (binomial(m+2, j+1) - binomial(m, j))*a(2^j*k) for m >= 0, k >= 0 with a(0) = 1.

A379819 Irregular table T(n, k), n >= 0, k >= 0, read by rows such that T(n,k) = f(n,k)/f(2^k-1,k) where f(n,k) is defined in Comments.

Original entry on oeis.org

1, 1, 1, 4, 2, 1, 3, 1, 10, 4, 4, 8, 2, 10, 13, 3, 1, 7, 6, 1, 22, 8, 10, 18, 4, 28, 30, 6, 4, 20, 14, 2, 49, 47, 9, 10, 36, 22, 3, 22, 56, 31, 4, 1, 15, 25, 10, 1, 46, 16, 22, 38, 8, 64, 64, 12, 10, 46, 30, 4, 118, 102, 18, 28, 88, 48, 6, 64, 138, 68, 8, 4
Offset: 0

Views

Author

Mikhail Kurkov, Jan 03 2025

Keywords

Comments

Here f(n,k) = b(2^k*(2n+1)) - Sum_{j=1..k} b(2^(j-1)*(2n+1))*R(k,j) for n >= 0, k >= 0 where b(n) = A379818(n) and where R(k,j) is the unique solution to b(2^k*(2^i-1)) = Sum_{j=1..k} b(2^(j-1)*(2^i-1))*R(k,j) for k > 0, 1 <= i <= k.
Row n length is A000120(n) + 1.

Examples

			Irregular table begins:
   1;
   1,  1;
   4,  2;
   1,  3,  1;
  10,  4;
   4,  8,  2;
  10, 13,  3;
   1,  7,  6,  1;
  22,  8;
  10, 18,  4;
  28, 30,  6;
   4, 20, 14,  2;
  49, 47,  9;
  10, 36, 22,  3;
  22, 56, 31,  4;
   1, 15, 25, 10, 1;
		

Crossrefs

Programs

  • PARI
    upto(n) = my(A, v1); v1 = vector(n+1, i, 0); v1[1] = 1; for(i=1, n, v1[i+1] = v1[i\2+1] + if(i%2, 0, A = 1 << valuation(i/2, 2); v1[i/2-A+1] + v1[i-A+1] + v1[i\(4*A)+1])); v1 \\ from A379818
    R(k) = my(v1, M1, M2); v1 = upto(2^k*(2^k-1)); M1 = matrix(k, k, i, j, v1[2^(j-1)*(2^i-1)+1]); M2 = matrix(k, 1, i, j, v1[2^k*(2^i-1)+1]); M1 = matsolve(M1, M2)
    row(n) = my(A = hammingweight(n), v1, v2, v3); v1 = upto(2^A*(2*n+1)); v2 = vector(A, i, R(i)); v3 = vector(A, i, (v1[2^i*(2*n+1)+1] - sum(j=1, i, v1[2^(j-1)*(2*n+1)+1]*v2[i][j,1]))/(v1[2^i*(2*(2^i-1)+1)+1] - sum(j=1, i, v1[2^(j-1)*(2*(2^i-1)+1)+1]*v2[i][j,1]))); concat(v1[n+1], v3)

Formula

Conjectures: (Start)
f(2^k-1,k) = (k+1)*A130032(k+1) for k >= 0.
T(2^n-1, k) = Stirling2(n+1, k+1) for n >= 0, 0 <= k <= n.
T(n,k) = c(n,wt(n)-k) for n >= 0, 0 <= k <= wt(n) where c(2n+1,k) = c(n,k) + (wt(n)-k+2)*c(n,k-1), c(2n,k) = (wt(n)-k+1)*c(2n+1,k) + [(wt(n)-k+1) > 0]*c(n,k-1) for n > 0, k > 0 with c(n,0) = A341392(n) for n >= 0, c(0,k) = 0 for k > 0 and where wt(n) = A000120(n). (End)

A363417 a(n) = Sum_{j=0..2^n - 1} b(j) for n >= 0 where b(n) = (A023416(n) + 1)*b(A053645(n)) + [A036987(n) = 0]*b(A266341(n)) for n > 0 with b(0) = 1.

Original entry on oeis.org

1, 2, 6, 23, 106, 566, 3415, 22872, 167796, 1334596, 11414192, 104270906, 1011793389, 10379989930, 112134625986, 1271209859403, 15077083642150, 186588381229340, 2403775013224000, 32168379148440968, 446341838086450308, 6410107231501731012, 95136428354649665256
Offset: 0

Views

Author

Mikhail Kurkov, Jun 11 2023 [verification needed]

Keywords

Comments

Note that [A036987(n) = 0]*b(A266341(n)) is the same as max((1 - T(n, j))*b(A053645(n) + 2^j*(1 - T(n, j))) | 0 <= j <= A000523(n)) where T(n, k) = floor(n/2^k) mod 2.
In fact b(n) is a generalization of A347205 just as A329369 is a generalization of A341392.

Crossrefs

Similar recurrences: A284005, A329369, A341392, A347205.

Programs

  • PARI
    A063250(n)=my(L=logint(n, 2), A=0); for(i=0, L, my(B=n\2^(L-i)+1); A++; A-=logint(B, 2)==valuation(B, 2)); A
    upto(n)=my(v, v1); v=vector(2^n, i, 0); v[1]=1; v1=vector(n+1, i, 0); v1[1]=1; for(i=1, #v-1, my(L=logint(i, 2), A=i - 2^L, B=A063250(i)); v[i+1]=(L - hammingweight(i) + 2)*v[A+1] + if(B>0, v[A + 2^(B-1) + 1])); for(i=1, n, v1[i+1]=v1[i] + sum(j=2^(i-1)+1, 2^i, v[j])); v1
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