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.

Showing 1-5 of 5 results.

A357990 Square array T(n, k), n >= 0, k > 0, read by antidiagonals, where T(0, k) = 1 for k > 0 and where T(n, k) = R(n, k+1) - R(n, k) for n > 0, k > 0. Here R(n, k) = T(A053645(n), k)*k^(A290255(n) + 1).

Original entry on oeis.org

1, 1, 1, 3, 1, 1, 1, 5, 1, 1, 7, 1, 7, 1, 1, 3, 19, 1, 9, 1, 1, 7, 5, 37, 1, 11, 1, 1, 1, 11, 7, 61, 1, 13, 1, 1, 15, 1, 15, 9, 91, 1, 15, 1, 1, 7, 65, 1, 19, 11, 127, 1, 17, 1, 1, 17, 19, 175, 1, 23, 13, 169, 1, 19, 1, 1, 3, 43, 37, 369, 1, 27, 15, 217, 1, 21
Offset: 0

Views

Author

Mikhail Kurkov, Nov 20 2022

Keywords

Examples

			Square array begins:
   1,  1,   1,   1,   1,    1,    1,    1, ...
   1,  1,   1,   1,   1,    1,    1,    1, ...
   3,  5,   7,   9,  11,   13,   15,   17, ...
   1,  1,   1,   1,   1,    1,    1,    1, ...
   7, 19,  37,  61,  91,  127,  169,  217, ...
   3,  5,   7,   9,  11,   13,   15,   17, ...
   7, 11,  15,  19,  23,   27,   31,   35, ...
   1,  1,   1,   1,   1,    1,    1,    1, ...
  15, 65, 175, 369, 671, 1105, 1695, 2465, ...
		

Crossrefs

Programs

  • PARI
    R(n,k)=my(L=logint(n, 2), A=n - 2^L); T(A, k)*k^(L - if(A>0, logint(A, 2) + 1) + 1)
    T(n,k)=if(n==0, 1, R(n, k+1) - R(n, k))
    
  • PARI
    T(n, k) = my(A = 2*n+1, B, C, v1, v2); v1 = []; while(A > 0, B=valuation(A, 2); v1=concat(v1, B+1); A \= 2^(B+1)); v1 = Vecrev(v1); A = #v1; v2 = vector(A, i, 1); for(i=1, A-1, B = A-i; for(j=1, B, C = B-j+k+1; v2[j] = v2[j]*C^v1[B] - v2[j+1]*(C-1)^v1[B])); v2[1] \\ Mikhail Kurkov, Apr 30 2024

Formula

Conjecture: T(n, 1) = A329369(n).

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

A290256 a(n) is the number of parts equal to 1 in the integer partition having viabin number n.

Original entry on oeis.org

1, 2, 0, 3, 1, 0, 0, 4, 2, 1, 1, 0, 0, 0, 0, 5, 3, 2, 2, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 6, 4, 3, 3, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 7, 5, 4, 4, 3, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2
Offset: 1

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Author

Emeric Deutsch, Sep 03 2017

Keywords

Comments

The viabin number of an integer partition is defined in the following way. Consider the southeast border of the Ferrers board of the integer partition and consider the binary number obtained by replacing each east step with 1 and each north step, except the last one, with 0. The corresponding decimal form is, by definition, the viabin number of the given integer partition. "Viabin" is coined from "via binary". For example, consider the integer partition [2,2,2,1]. The southeast border of its Ferrers board yields 10100, leading to the viabin number 20.
a(n) is the number of 0's following directly the first 1 in the binary representation of 2n.

Examples

			a(9) = 2. Indeed, the binary form of 9 is 1001; with an additional 0 at the end, it leads to the path ENNEN, where E=(1,0), N=(0,1); this path is the southeast border of the Ferrers board of the integer partition [2,1,1], having 2 parts equal to 1.
		

Crossrefs

Cf. A290255.

Programs

  • Maple
    b := proc(n) if type(log[2](n), integer) then log[2](n) else b(floor((1/2)*n)) end if end proc: seq(b(2*n), n = 1 .. 78);

Formula

a(n) = A290255(2n).

A358631 Irregular table T(n, k), n >= 0, k > 0, read by rows of extended (due to binary expansion of n) Stirling numbers of the first kind.

Original entry on oeis.org

1, 1, 2, 3, 1, 4, 5, 1, 6, 11, 6, 1, 6, 7, 1, 12, 20, 9, 1, 18, 26, 9, 1, 24, 50, 35, 10, 1, 8, 9, 1, 18, 29, 12, 1, 30, 41, 12, 1, 48, 94, 59, 14, 1, 36, 47, 12, 1, 72, 130, 71, 14, 1, 96, 154, 71, 14, 1, 120, 274, 225, 85, 15, 1, 10, 11, 1, 24, 38, 15, 1, 42
Offset: 0

Views

Author

Mikhail Kurkov, Nov 24 2022

Keywords

Comments

Row n length is A000120(n) + 2.

Examples

			Irregular table begins:
    1,   1;
    2,   3,   1;
    4,   5,   1;
    6,  11,   6,  1;
    6,   7,   1;
   12,  20,   9,  1;
   18,  26,   9,  1;
   24,  50,  35, 10,  1;
    8,   9,   1;
   18,  29,  12,  1;
   30,  41,  12,  1;
   48,  94,  59, 14,  1;
   36,  47,  12,  1;
   72, 130,  71, 14,  1;
   96, 154,  71, 14,  1;
  120, 274, 225, 85, 15, 1;
		

Crossrefs

Programs

  • PARI
    b1(n)=if(n>0, my(A=n - 2^logint(n, 2)); if(A>0, logint(A, 2) + 1))
    b2(n)=if(n>0, my(A=b1(3*2^logint(n, 2) - n - 1)); n + if(A>0, 2^(A-1)))
    P(n,k)=if(n==0 || k==1, (n==0 && k<3) + (k==1 && n>0), my(L=logint(n, 2), A=n - 2^L); (hammingweight(A) + 2)*P(A, k-1)*(L - b1(n) + 1) + P(b2(A), k))
    T(n, k)=my(A=hammingweight(n)); if(k<=(A + 2), P(n, A - k + 3))

Formula

T(n, k) = P(n, wt(n) - k + 3) for n >= 0, 0 < k <= wt(n) + 2 where wt(n) = A000120(n).
P(n, 1) = 1 for n > 0 with P(0, 1) = P(0, 2) = 1.
P(n, k) = (A000120(q(n)) + 2)*P(q(n), k-1)*(A290255(n) + 1) + P(s(q(n)), k) for n > 0, k > 1 where q(n) = A053645(n) and where s(n) = n + [A063250(n) > 0]*2^(A063250(n) - 1).
T(2^n - 1, k) = abs(Stirling1(n+2, k)) for n >= 0, k > 0.
Conjectures: (Start)
T(n, 1) = (A000120(n) + 1)!*A347205(n) for n >= 0.
Sum_{k=1..A000120(n) + 2} T(n, k)*(-1)^k = 0 for n >= 0.
Sum_{k=0..2^n - 1} Sum_{j=1..A000120(k) + 2} T(k, j) = 2*A052852(n+1) for n >= 0.
Sum_{i=1..wt(k) + 2} m^(i-1)*T(k, i) = (wt(k) + 1)!*A347205(2^m*(2k+1)) for m >= 0, k >= 0 where wt(n) = A000120(n). (End)

Extensions

Offset corrected by Mikhail Kurkov, Nov 07 2024

A338971 Linear representation of the complete labeled binary trees of all integer heights, where the nodes at level i, 0 <= i <= n, of a binary tree with height n are labeled with the number n-i.

Original entry on oeis.org

0, 1, 0, 0, 2, 1, 1, 0, 0, 0, 0, 3, 2, 2, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 4, 3, 3, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 5, 4, 4, 3, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
Offset: 0

Views

Author

Marc van Dongen, Dec 18 2020

Keywords

Examples

			First few terms where each line represents a complete binary tree:
  n=0:  0
  n=1:  1 0 0
  n=2:  2 1 1 0 0 0 0
  n=3:  3 2 2 1 1 1 1 0 0 0 0 0 0 0 0
  n=4:  4 3 3 ...
Using this representation, the first row r(0) is given by [0]; row(n+1) is given by adding 1 to each member of r(n) and appending 2^(n+1) 0's: r(0) = [0], r(n+1) = [ i + 1 | i <- r(n) ] ++ [ 0 | i <- [1..2^(n+1)] ].
		

Crossrefs

Cf. A290255, A126646 (row lengths).

Programs

  • Haskell
    concat [ tree n | n <- [0..] ]
      where tree 0 = [0]
            tree n = [ i+1 | i <- tree (n-1) ] ++ [ 0 | i <- [1..2^n] ]
Showing 1-5 of 5 results.