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 1-8 of 8 results.

A342603 a(0) = 0, a(1) = 1; a(2*n) = a(n), a(2*n+1) = 6*a(n) + a(n+1).

Original entry on oeis.org

0, 1, 1, 7, 1, 13, 7, 43, 1, 19, 13, 85, 7, 85, 43, 259, 1, 25, 19, 127, 13, 163, 85, 517, 7, 127, 85, 553, 43, 517, 259, 1555, 1, 31, 25, 169, 19, 241, 127, 775, 13, 241, 163, 1063, 85, 1027, 517, 3109, 7, 169, 127, 847, 85, 1063, 553, 3361, 43, 775, 517, 3361, 259, 3109, 1555, 9331, 1
Offset: 0

Views

Author

Ilya Gutkovskiy, Mar 17 2021

Keywords

Crossrefs

Programs

  • Maple
    a:= proc(n) option remember; `if`(n<2, n, (q->
         `if`(d=1, 6*a(q)+a(q+1), a(q)))(iquo(n, 2, 'd')))
        end:
    seq(a(n), n=0..70);  # Alois P. Heinz, Mar 17 2021
  • Mathematica
    a[0] = 0; a[1] = 1; a[n_] := If[EvenQ[n], a[n/2], 6 a[(n - 1)/2] + a[(n + 1)/2]]; Table[a[n], {n, 0, 64}]
    nmax = 64; CoefficientList[Series[x Product[(1 + x^(2^k) + 6 x^(2^(k + 1))), {k, 0, Floor[Log[2, nmax]] + 1}], {x, 0, nmax}], x]

Formula

G.f.: x * Product_{k>=0} (1 + x^(2^k) + 6*x^(2^(k+1))).
a(2^n-1) = (6^n - 1)/5 = A003464(n); a(2^n) = 1; a(2^n+1) = 6*n + 1 = A016921(n). - Alois P. Heinz, Mar 17 2021

A342633 a(0) = 0, a(1) = 1; a(2*n) = a(n), a(2*n+1) = 3*a(n) + a(n+1).

Original entry on oeis.org

0, 1, 1, 4, 1, 7, 4, 13, 1, 10, 7, 25, 4, 25, 13, 40, 1, 13, 10, 37, 7, 46, 25, 79, 4, 37, 25, 88, 13, 79, 40, 121, 1, 16, 13, 49, 10, 67, 37, 118, 7, 67, 46, 163, 25, 154, 79, 241, 4, 49, 37, 136, 25, 163, 88, 277, 13, 118, 79, 277, 40, 241, 121, 364, 1, 19, 16, 61, 13, 88, 49, 157
Offset: 0

Views

Author

Ilya Gutkovskiy, Mar 17 2021

Keywords

Crossrefs

Programs

  • Maple
    a:= proc(n) option remember; `if`(n<2, n, (q->
         `if`(d=1, 3*a(q)+a(q+1), a(q)))(iquo(n, 2, 'd')))
        end:
    seq(a(n), n=0..71);  # Alois P. Heinz, Mar 17 2021
  • Mathematica
    a[0] = 0; a[1] = 1; a[n_] := If[EvenQ[n], a[n/2], 3 a[(n - 1)/2] + a[(n + 1)/2]]; Table[a[n], {n, 0, 71}]
    nmax = 71; CoefficientList[Series[x Product[(1 + x^(2^k) + 3 x^(2^(k + 1))), {k, 0, Floor[Log[2, nmax]] + 1}], {x, 0, nmax}], x]

Formula

G.f.: x * Product_{k>=0} (1 + x^(2^k) + 3*x^(2^(k+1))).

A342634 a(0) = 0, a(1) = 1; a(2*n) = a(n), a(2*n+1) = 4*a(n) + a(n+1).

Original entry on oeis.org

0, 1, 1, 5, 1, 9, 5, 21, 1, 13, 9, 41, 5, 41, 21, 85, 1, 17, 13, 61, 9, 77, 41, 169, 5, 61, 41, 185, 21, 169, 85, 341, 1, 21, 17, 81, 13, 113, 61, 253, 9, 113, 77, 349, 41, 333, 169, 681, 5, 81, 61, 285, 41, 349, 185, 761, 21, 253, 169, 761, 85, 681, 341, 1365, 1, 25, 21, 101, 17
Offset: 0

Views

Author

Ilya Gutkovskiy, Mar 17 2021

Keywords

Crossrefs

Programs

  • Maple
    a:= proc(n) option remember; `if`(n<2, n, (q->
         `if`(d=1, 4*a(q)+a(q+1), a(q)))(iquo(n, 2, 'd')))
        end:
    seq(a(n), n=0..68);  # Alois P. Heinz, Mar 17 2021
  • Mathematica
    a[0] = 0; a[1] = 1; a[n_] := If[EvenQ[n], a[n/2], 4 a[(n - 1)/2] + a[(n + 1)/2]]; Table[a[n], {n, 0, 68}]
    nmax = 68; CoefficientList[Series[x Product[(1 + x^(2^k) + 4 x^(2^(k + 1))), {k, 0, Floor[Log[2, nmax]] + 1}], {x, 0, nmax}], x]

Formula

G.f.: x * Product_{k>=0} (1 + x^(2^k) + 4*x^(2^(k+1))).
a(n) == 1 (mod 4) for n >= 1. - Hugo Pfoertner, Mar 17 2021

A342635 a(0) = 0, a(1) = 1; a(2*n) = a(n), a(2*n+1) = 5*a(n) + a(n+1).

Original entry on oeis.org

0, 1, 1, 6, 1, 11, 6, 31, 1, 16, 11, 61, 6, 61, 31, 156, 1, 21, 16, 91, 11, 116, 61, 311, 6, 91, 61, 336, 31, 311, 156, 781, 1, 26, 21, 121, 16, 171, 91, 466, 11, 171, 116, 641, 61, 616, 311, 1561, 6, 121, 91, 516, 61, 641, 336, 1711, 31, 466, 311, 1711, 156, 1561, 781, 3906, 1, 31, 26
Offset: 0

Views

Author

Ilya Gutkovskiy, Mar 17 2021

Keywords

Crossrefs

Programs

  • Maple
    a:= proc(n) option remember; `if`(n<2, n, (q->
         `if`(d=1, 5*a(q)+a(q+1), a(q)))(iquo(n, 2, 'd')))
        end:
    seq(a(n), n=0..66);  # Alois P. Heinz, Mar 17 2021
  • Mathematica
    a[0] = 0; a[1] = 1; a[n_] := If[EvenQ[n], a[n/2], 5 a[(n - 1)/2] + a[(n + 1)/2]]; Table[a[n], {n, 0, 66}]
    nmax = 66; CoefficientList[Series[x Product[(1 + x^(2^k) + 5 x^(2^(k + 1))), {k, 0, Floor[Log[2, nmax]] + 1}], {x, 0, nmax}], x]

Formula

G.f.: x * Product_{k>=0} (1 + x^(2^k) + 5*x^(2^(k+1))).
a(n) == 1 (mod 5) for n >= 1. - Hugo Pfoertner, Mar 17 2021

A342636 a(0) = 0, a(1) = 1; a(2*n) = a(n), a(2*n+1) = 7*a(n) + a(n+1).

Original entry on oeis.org

0, 1, 1, 8, 1, 15, 8, 57, 1, 22, 15, 113, 8, 113, 57, 400, 1, 29, 22, 169, 15, 218, 113, 799, 8, 169, 113, 848, 57, 799, 400, 2801, 1, 36, 29, 225, 22, 323, 169, 1198, 15, 323, 218, 1639, 113, 1590, 799, 5601, 8, 225, 169, 1296, 113, 1639, 848, 5993, 57, 1198, 799, 5993, 400, 5601, 2801
Offset: 0

Views

Author

Ilya Gutkovskiy, Mar 17 2021

Keywords

Crossrefs

Programs

  • Maple
    a:= proc(n) option remember; `if`(n<2, n, (q->
         `if`(d=1, 7*a(q)+a(q+1), a(q)))(iquo(n, 2, 'd')))
        end:
    seq(a(n), n=0..62);  # Alois P. Heinz, Mar 17 2021
  • Mathematica
    a[0] = 0; a[1] = 1; a[n_] := If[EvenQ[n], a[n/2], 7 a[(n - 1)/2] + a[(n + 1)/2]]; Table[a[n], {n, 0, 62}]
    nmax = 62; CoefficientList[Series[x Product[(1 + x^(2^k) + 7 x^(2^(k + 1))), {k, 0, Floor[Log[2, nmax]] + 1}], {x, 0, nmax}], x]

Formula

G.f.: x * Product_{k>=0} (1 + x^(2^k) + 7*x^(2^(k+1))).

A342638 a(0) = 0, a(1) = 1; a(2*n) = a(n), a(2*n+1) = 9*a(n) + a(n+1).

Original entry on oeis.org

0, 1, 1, 10, 1, 19, 10, 91, 1, 28, 19, 181, 10, 181, 91, 820, 1, 37, 28, 271, 19, 352, 181, 1639, 10, 271, 181, 1720, 91, 1639, 820, 7381, 1, 46, 37, 361, 28, 523, 271, 2458, 19, 523, 352, 3349, 181, 3268, 1639, 14761, 10, 361, 271, 2620, 181, 3349, 1720, 15571, 91, 2458, 1639, 15571, 820
Offset: 0

Views

Author

Ilya Gutkovskiy, Mar 17 2021

Keywords

Crossrefs

Programs

  • Maple
    a:= proc(n) option remember; `if`(n<2, n, (q->
         `if`(d=1, 9*a(q)+a(q+1), a(q)))(iquo(n, 2, 'd')))
        end:
    seq(a(n), n=0..60);  # Alois P. Heinz, Mar 17 2021
  • Mathematica
    a[0] = 0; a[1] = 1; a[n_] := If[EvenQ[n], a[n/2], 9 a[(n - 1)/2] + a[(n + 1)/2]]; Table[a[n], {n, 0, 60}]
    nmax = 60; CoefficientList[Series[x Product[(1 + x^(2^k) + 9 x^(2^(k + 1))), {k, 0, Floor[Log[2, nmax]] + 1}], {x, 0, nmax}], x]

Formula

G.f.: x * Product_{k>=0} (1 + x^(2^k) + 9*x^(2^(k+1))).

A178239 Triangle read by rows, antidiagonals of an array generated from a(n) = a(2n), a(2n+1) = r*a(n) + a(n+1).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 1, 1, 1, 1, 4, 1, 3, 1, 1, 1, 5, 1, 5, 2, 1, 1, 1, 6, 1, 7, 3, 3, 1, 1, 1, 7, 1, 9, 4, 7, 1, 1, 1, 1, 8, 1, 11, 5, 13, 1, 4, 1, 1, 1, 9, 1, 13, 6, 21, 1, 7, 3, 1, 1, 1, 10, 1, 15, 7, 31, 1, 10, 5, 5, 1
Offset: 0

Views

Author

Gary W. Adamson, May 23 2010

Keywords

Comments

Partial sums of array terms in groups of 1, next 2, next 4, ... 8 = powers of (r+2).
Row sums = A178240: (1, 2, 3, 5, 7, 11, 16, 23, ...).
Row 1 of the array = A002487.
Row 2 = .............A116528.
Row 3 = .............A342633.
Row 4 = .............A342634.
...
Row 10 = ............A178243.
Polcoeff row r of the array as f(x) satisfies f(x)/f(x^2) = (1 + x + r*x^2).
Let q(x) = (1 + x + r*x^2). Then polcoeff row 4 = q(x) * q(x^2) * q(x^4) * q(x^8) * ...

Examples

			First few rows of the array =
      n=1  n=2  n=3  n=4  n=5  n=6  n=7  n=8  n=9 n=10 n=11 n=12 n=13 n=14 n=15
  r=0:  1,   1,   1,   1,   1,   1,   1,   1,   1,   1,   1,   1,   1,   1,   1, ...
  r=1:  1,   1,   2,   1,   3,   2,   3,   1,   4,   3,   5,   2,   5,   3,   4, ...
  r=2:  1,   1,   3,   1,   5,   3,   7,   1,   7,   5,  13,   3,  13,   7,  15, ...
  r=3:  1,   1,   4,   1,   7,   4,  13,   1,  10,   7,  25,   4,  25,  13,  40, ...
  r=4:  1,   1,   5,   1,   9,   5,  21,   1,  13,   9,  41,   5,  41,  21,  85, ...
  r=5:  1,   1,   6,   1,  11,   6,  31,   1,  16,  11,  61,   6,  61,  31, 156, ...
  ...
Example: In row 3: (1, 1, 4, 1, 7, 4, 13, ...) = A342633, r = 3.
A342633(7) = 13 = 3*4 + 1. In blocks of 1, 2, 4, 8, ... terms, partial sums are powers of (r+2) = 5: (1, 5, 25, ...).
First few rows of the triangle =
  1;
  1, 1;
  1, 1,  1;
  1, 1,  2, 1;
  1, 1,  3, 1,  1;
  1, 1,  4, 1,  3,  1;
  1, 1,  5, 1,  5,  2,  1;
  1, 1,  6, 1,  7,  3,  3, 1;
  1, 1,  7, 1,  9,  4,  7, 1,  1;
  1, 1,  8, 1, 11,  5, 13, 1,  4,  1;
  1, 1,  9, 1, 13,  6, 21, 1,  7,  3,  1;
  1, 1, 10, 1, 15,  7, 31, 1, 10,  5,  5, 1;
  1, 1, 11, 1, 17,  8, 43, 1, 13,  7, 13, 2,  1;
  1, 1, 12, 1, 19,  9, 57, 1, 16,  9, 21, 3,  5, 1;
  1, 1, 13, 1, 21, 11, 73, 1, 19, 11, 31, 4, 13, 2, 1;
  ...
		

Crossrefs

Cf. A178240, A359250 (column polynomials).

Formula

Antidiagonals of an array generated from a(n) = a(2n); a(2n+1) = r*a(n) + a(n+1).
Given a triangle M with columns stepped down twice from the previous column, for columns > 0, with (1, 1, r, 0, 0, 0, ...) in each column, r-th row of the array = lim_{n->oo} M^n.

A359250 Irregular triangle read by rows where T(n,k) is the coefficient of y^k in polynomial P(n) defined by P(2n) = P(n) and P(2n+1) = y*P(n) + P(n+1) starting P(0) = 0, P(1) = 1.

Original entry on oeis.org

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

Views

Author

Kevin Ryde, Dec 28 2022

Keywords

Comments

Row n length is wt(n) = A000120(n), the binary weight of n, so that k ranges 0 <= k < wt(n).
Evaluated at y=1, the recurrence for P is per Stern's diatomic sequence so that row n has sum A002487(n).
Evaluated at y=2, the recurrence for P is per A116528 so that Sum T(n,k)*2^k = A116528(n), and similarly variations such as y=10 for A178243.
Array A178239(r,n) is P(n) evaluated at y=r, and in particular P(n) is the polynomial for the values in column n there.
Column k=1, when that value exists, is T(n,1) = A080791(n-1), the number of 0 bits in n-1, since expressing the recurrence in Q(m) = P(m+1) adds y*Q(m-1) at each 0 bit in m.
Reversing the bits of n is no change to P(n), so that P(n) = P(A030101(n)).

Examples

			Triangle begins:
      k=0  1  2
  n=0:  (empty)
  n=1:  1,
  n=2:  1,
  n=3:  1, 1,
  n=4:  1,
  n=5:  1, 2,
  n=6:  1, 1,
  n=7:  1, 1, 1,
  n=8:  1,
  n=9:  1, 3,
		

Crossrefs

Cf. A000120 (row lengths), A002487 (row sums).
Cf. A178239 (array), A030101 (bit reversal).
Cf. A125184.

Programs

  • MATLAB
    function a = A359250( max_n )
        ac = cell(1,1); ac{1} = [1];
        for n = 2:max_n
            m = floor(n/2);
            if 2*m == n
                ac{n} = ac{m};
            else
                ac{n} = [ac{m+1} zeros(1,(length(ac{m})+1)-length(ac{m+1}))] ...
                + [0 ac{m}];
            end
        end
        a = cell2mat(ac);
    end % Thomas Scheuerle, Dec 28 2022
    
  • PARI
    \\ See links.

Formula

G.f.: Sum_{n>=0} P(n)*x^n = x * Product_{e>=0} 1 + x^(2^e) + y*x^(2^(e+1)).
Showing 1-8 of 8 results.