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 1-4 of 4 results.

A146533 Catalan transform of A135092.

Original entry on oeis.org

1, 7, 21, 70, 245, 882, 3234, 12012, 45045, 170170, 646646, 2469012, 9464546, 36402100, 140408100, 542911320, 2103781365, 8167621770, 31762973550, 123708423300, 482462850870, 1883902560540, 7364346373020, 28817007546600, 112866612890850, 442437122532132, 1735714865318364
Offset: 0

Views

Author

Philippe Deléham, Oct 31 2008

Keywords

Crossrefs

Cf. A088218, A000984, A029651, A100320, A029609, A144706.

Programs

  • Mathematica
    a[n_]:=(7*Binomial[2n,n]-5*KroneckerDelta[n,0])/2; Array[a,27,0] (* Stefano Spezia, Feb 14 2025 *)

Formula

a(n) = Sum_{k=0..n} A039599(n,k)*A010687(k) = Sum_{k=0..n} A106566(n,k)*A135092(k).
a(n) = (7*C(2n,n) - 5*0^n)/2.
From Stefano Spezia, Feb 14 2025: (Start)
G.f.: (7/sqrt(1 - 4*x) - 5)/2.
E.g.f.: (7*exp(2*x)*BesselI(0, 2*x) - 5)/2. (End)

Extensions

a(23)-a(26) from Stefano Spezia, Feb 14 2025

A047274 Numbers that are congruent to {0, 1} mod 7.

Original entry on oeis.org

0, 1, 7, 8, 14, 15, 21, 22, 28, 29, 35, 36, 42, 43, 49, 50, 56, 57, 63, 64, 70, 71, 77, 78, 84, 85, 91, 92, 98, 99, 105, 106, 112, 113, 119, 120, 126, 127, 133, 134, 140, 141, 147, 148, 154, 155, 161, 162, 168, 169
Offset: 1

Views

Author

N. J. A. Sloane

Keywords

Comments

Nonnegative k such that k or 6*k + 1 is divisible by 7. - Bruno Berselli, Feb 13 2018

Links

Crossrefs

Cf. A030308, A135092.

Programs

Formula

a(n) = 7*n - a(n-1) - 13 for n>1, a(1)=0. - Vincenzo Librandi, Aug 05 2010
From R. J. Mathar, Oct 08 2011: (Start)
a(n) = 7*n/2 - 19/4 - 5*(-1)^n/4.
G.f.: x^2*(1 + 6*x) / ((1 + x)*(x - 1)^2). (End)
a(n+1) = Sum_{k>=0} A030308(n,k)*A135092(k). - Philippe Deléham, Oct 17 2011
E.g.f.: 6 + ((14*x - 19)*exp(x) - 5*exp(-x))/4. - David Lovler, Aug 31 2022

A146541 Binomial transform of A010688.

Original entry on oeis.org

1, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 65536, 131072, 262144, 524288, 1048576, 2097152, 4194304, 8388608, 16777216, 33554432, 67108864, 134217728, 268435456, 536870912, 1073741824, 2147483648, 4294967296, 8589934592
Offset: 0

Views

Author

Philippe Deléham, Oct 31 2008

Keywords

Comments

Hankel transform is := 1,-48,0,0,0,0,0,0,0,...

Links

Crossrefs

Cf. A000007, A011782, A000079, A003945, A046055, A146523, A082505, A135092.
Essentially the same as A132479, A077552, A020707.

Programs

  • Mathematica
    Join[{1},2^Range[3,40]] (* Harvey P. Dale, Feb 28 2016 *)
  • PARI
    Vec((1+6*x)/(1-2*x) + O(x^50)) \\ Colin Barker, Mar 17 2016

Formula

a(0)=1, a(n) = 2^(n+2) for n>0.
a(n) = Sum_{k, 0..n} A109466(n,k)*A146534(k).
a(n) = A132479(n), n>1. - R. J. Mathar, Nov 02 2008
G.f.: (1+6*x) / (1-2*x). - Colin Barker, Mar 17 2016

Extensions

Corrected and extended by Harvey P. Dale, Feb 28 2016

A356120 Irregular triangle read by rows: the n-th row lists the values 0..2^n-1 representing all subsets of a set of n elements. When its elements are linearly ordered, the subsets are lexicographically ordered.

Original entry on oeis.org

0, 0, 1, 0, 2, 3, 1, 0, 4, 6, 7, 5, 2, 3, 1, 0, 8, 12, 14, 15, 13, 10, 11, 9, 4, 6, 7, 5, 2, 3, 1, 0, 16, 24, 28, 30, 31, 29, 26, 27, 25, 20, 22, 23, 21, 18, 19, 17, 8, 12, 14, 15, 13, 10, 11, 9, 4, 6, 7, 5, 2, 3, 1, 0, 32, 48, 56, 60, 62, 63, 61, 58, 59, 57, 52, 54, 55, 53, 50, 51, 49, 40, 44, 46, 47, 45, 42
Offset: 0

Views

Author

Valentin Bakoev, Jul 27 2022

Keywords

Comments

The sequence in the n-th row of the triangle is denoted by row(n). It contains the values in the range 0..2^n-1 corresponding to the binary vectors of length n. The integers in row (n), considered as characteristic vectors of the set B_n = {b_n, b_(n-1), ..., b_2, b_1}, whose elements are linearly ordered: b_n < b_(n-1) < ... < b_2 < b_1, define all subsets of B_n in lexicographic order. To obtain row(n) we reason inductively as follows. Obviously, row(0) = 0 = a(0) and corresponds to the empty set {}. Assume that the sequence row(n-1) = i_0, i_1, ..., i_(2^(n-1)-1) is obtained. It defines a lexicographic order of the subsets of B_(n-1) = {b_(n-1) , ..., b_2, b_1} - note that its elements linearly ordered. Then row (n) = i_0, 2^(n-1) + i_0, 2^(n-1) + i_1, ..., 2^(n-1) + i_(2^(n-1)-1), i_1, ..., i_(2^(n-1)-1) defines all subsets of B_n in lexicographic order. In other words, row(n) = 0, (row(n-1) + 2^(n-1)), (row(n-1) without the first term 0), i.e., row(n-1) is taken twice: first, all terms of row(n-1) are incremented by 2^(n-1) and second, the resulting sequence is inserted after the first term of row(n-1), which is always 0 and corresponds to {}.

Examples

			For n = 1, 2, 3, the sets B_n, their subsets (the column under B_n), binary characteristic words (column bin.) and corresponding integers (column dec.) are:
B_1 = {c}  bin.  dec. | B_2 = {b, c}  bin.   dec. | B_3 = {a, b, c}   bin.   dec.
      {}    0     0   |       {}       00     0   |       {}          000     0
      {c}   1     1   |       {b}      10     2   |       {a}         100     4
                      |       {b, c}   11     3   |       {a, b}      110     6
                      |          {c}   01     1   |       {a, b, c}   111     7
                                                  |       {a, c}      101     5
                                                  |          {b}      010     2
                                                  |          {b, c}   011     3
                                                  |             {c}   001     1
As seen, when B = {a, b, c}, its subsets {}, {a}, {a, b}, {a, b, c}, {a, c}, {b}, {b, c}, {c} are in lexicographic order, the corresponding binary words of length 3 are 000, 100, 110, 111, 101, 010, 011, 001, and so row(3) = 0, 4, 6, 7, 5, 2, 3, 1.
Triangle T(n,k) begins:
      k=0  1  2  3  4  5  6  7 ...
  n=0:  0;
  n=1:  0, 1;
  n=2:  0, 2, 3, 1;
  n=3:  0, 4, 6, 7, 5, 2, 3, 1;
  n=4:  0, 8, 12, 14, 15, 13, 10, 11, 9, 4, 6, 7, 5, 2, 3, 1;
  n=5:  0, 16, 24, 28, 30, 31, 29, 26, 27, 25, 20, 22, 23, 21, 18, 19, 17, 8, 12, 14, 15, 13, 10, 11, 9, 4, 6, 7, 5, 2, 3, 1,
  ...

References

  • Donald E. Knuth, The Art of Computer Programming, Volume 4A, Section 7.2.1.3 Exercise 19 (Binomial tree traversed in post-order).

Links

Crossrefs

Cf. A006516 (row sums), A000225 (main diagonal, the n-th term of row(n)).
row(n) without leading 0, when read in reverse order, gives the first 2^n-1 terms of A108918.
Starting with the n-th term of row(n), columns 0..4 are: A000004, A131577 without 0, A007283, A135092, A110286.

Programs

  • Mathematica
    (* computing row(n) *)
n = 5;
Array[row, 2^n];
row[0] = 0; row[1] = 1;
len = 2;
For[i = 2, i <= n, i++,
  For[j = 1, j < len, j++, row[j + len] = row[j]];
  For[j = len, j > 0, j--, row[j] = row[j - 1] + len];
  len = len*2;
];

Formula

row(n) is defined as:
row(0) = 0;
row(n) = 0, (2^(n-1)+row(n-1)), (row(n-1)\{0}),
where (2^(n-1) + row(n-1)) means a subsequence obtained by adding 2^(n-1) to every term of row(n-1), and (row(n-1)\{0}) means a subsequence row(n-1) without its first term 0, for n = 0, 1, 2, ...).
Then a(n) is defined as:
a(2^m - 1) = 0, for m = 0, 1, 2, ..., and
a(2^m + k) = 2^(m-1) + a(2^(m-1) + k - 1), for k = 0, 1, ..., 2^(m-1) - 1, and
a(2^m + 2^(m-1) + k) = a(2^(m-1)+k), for k = 0, 1, ..., 2^(m-1) - 2.
Showing 1-4 of 4 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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