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.

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A238398 Numerators of inverse binomial transform of A198631(n)/A006519(n+1) with -1 instead of A198631(1)=1.

Original entry on oeis.org

1, -3, 2, -11, 4, -11, 6, -39, 8, -49, 10, 647, 12, -5487, 14, 929329, 16, -3202325, 18, 221930505, 20, -4722116563, 22, 968383680643, 24, -14717667114201, 26, 2093660879252563, 28, -86125672563201239, 30, 129848163681107300961, 32
Offset: 0

Views

Author

Paul Curtz, Feb 26 2014

Keywords

Comments

From modified fractional Euler numbers.
Inverse binomial transform:
1, -3/2, 2, -11/4, 4, -11/2, 6, -39/8, 8, -49/2, 10, 647/4, 12, -5487/2,... = a(n)/b(n). b(2n) = A004277(n).
Difference table of c(n) = 1, -1/2, 0, -1/4,... :
1, -1/2, 0, -1/4, 0, 1/2, 0,...
-3/2, 1/2, -1/4, 1/4, 1/2, -1/2, -17/8,...
2, -3/4, 1/2, 1/4, -1, -13/8, 17/4,...
-11/4, 5/4, -1/4, -5/4, -5/8, 47/8, 73/8,...
4, -3/2, -1, 5/8, 13/2, 13/4, -107/2,...
-11/2, 1/2, 13/8, 47/8, -13/4, -227/4, -227/8,
6, 9/8, 17/4, -73/8, -107/2, 227/8, 2957/4,...
etc.
c(n) + a(n)/b(n) = 2, -2, 2, -3, 4, -5, 6, -7, 8, -9,... = A233583(n+1) signed. (a(n) discovered in 2013)

Crossrefs

Cf. A235774.

Programs

  • Mathematica
    max = 40;(* b = A198631 *) b[0] = 1; b[1] = -1; b[n_] := Numerator[EulerE[n, 1]/(2^n-1)]; bb = Table[b[n]/2^IntegerExponent[n+1, 2], {n, 0, max}]; a[n_] := Differences[bb, n] // First // Numerator ; Table[a[n], {n, 0, max}]

A238800 Unreduced numerators in triangle that leads to the Euler numbers A198631(n)/A006519(n+1).

Original entry on oeis.org

1, 1, 1, -2, 1, -3, 1, -4, 2, 1, -5, 5, 1, -6, 9, -10, 1, -7, 14, -35, 1, -8, 20, -80, 26, 1, -9, 27, -150, 117, 1, -10, 35, -250, 325, -454, 1, -11, 44, -385, 715, -2497, 1, -12, 54, -560, 1365, -8172, 5914, 1, -13
Offset: 0

Views

Author

Paul Curtz, Mar 05 2014

Keywords

Comments

We use the array ASPEC mentioned in A191302:
2, 1, 1, 1, 1, 1, 1, 1,...
2, 3, 4, 5, 6, 7, 8, 9,...
2, 5, 9, 14, 20, 27, 35, 44,...
2, 7, 16, 30, 50, 77, 112, 156,...
with the first upper diagonal of the difference table of the autosequence A198631(n)/A006519(n+1), i.e., 1/2, -1/4, 1/4, -5/8, 13/4, -227/8, 2957/8,...
written by columns:
1/2
1/2,
1/2, -1/4,
1/2, -1/4,
1/2, -1/4, 1/4,
1/2, -1/4, 1/4,
1/2, -1/4, 1/4, -5/8,
1/2, -1/4, 1/4, -5/8,
etc.
Hence, by multiplication of this double triangle by ASPEC, the beginning of the double triangle ESPEC is obtained:
E(0) = 1 = 1
E(1) = 1/2 = 1/2
E(2) = 0 = 1/2 -2/4
E(3) = -1/4 = 1/2 -3/4
E(4) = 0 = 1/2 -4/4 +2/4
E(5) = 1/2 = 1/2 -5/4 +5/4
E(6) = 0 = 1/2 -6/4 +9/4 -10/8
E(7) = -17/8 = 1/2 -7/4 +14/4 -35/8
E(8) = 0 = 1/2 -8/4 +20/4 -80/8 +26/4.
The terms of the sequence are the reduced numerators. Like A192456(n) for Bernoulli numbers A164555(n)/A027642(n).

Examples

			a(n) by triangle
1,
1,
1, -2,
1, -3,
1, -4,  2,
1, -5,  5,
1, -6,  9, -10,
1, -7, 14, -35,
1, -8, 20, -80, 26,
etc.

A240980 Numerators of f(n) with 2*f(n+1) = f(n) + A198631(n)/A006519(n+1), f(0)=0.

Original entry on oeis.org

0, 1, 1, 1, 0, 0, 1, 1, -1, -1, 15, 15, -169, -169, 10753, 10753, -28713, -28713, 1586789, 1586789, -27542974, -13771487, 4694573547, 4694573547, -60230569205, -60230569205, 7328718272473, 7328718272473, -1043166080490099, -1043166080490099, 343459524172314625, 343459524172314625
Offset: 0

Views

Author

Paul Curtz, Aug 06 2014

Keywords

Comments

An autosequence is a sequence which has its inverse binomial transform equal to the signed sequence. (Examples: 1) A000045(n) is of the first kind. 2) 1/(n+1) is of the second kind).
f(n), companion to A198631(n)/A006519(n+1), is an autosequence of the first kind.
The difference table of f(n) is:
0, 1/2, 1/2, 1/4, 0, 0, ...
1/2, 0, -1/4, -1/4, 0, 1/4, ...
-1/2, -1/4, 0, 1/4, 1/4, -3/8, ...
1/4, 1/4, 1/4, 0, -5/8, -5/8, ...
etc.
The main diagonal is 0's=A000004. The first two upper diagonal are equal.
a(n) are the numerators of f(n).
f(n) is the first sequence of the family of alternated autosequences of the first and of the second kind
0, 1/2, 1/2, 1/4, 0, 0, ...
1, 1/2, 0, -1/4, 0, 1/2, ... = A198631(n)/A006519(n+1),
0, -1/2, -1/2, 1/4, 1, -1/2, ...
-1, -1/2, 1, 7/4, -2, -8, ...
etc.
Like A164555(n)/A027642(n), A198631(n)/A006519(n+1) is an autosequence which has its main diagonal equal to the first upper diagonal multiplied by 2. See A190339(n).
The first column is 0 followed by A122045(n).
For the numerators of the second column see A241209(n).

Examples

			2*f(1) = 0 + 1, f(1) = 1/2;
2*f(2) = 1/2 + 1/2, f(2) = 1/2;
2*f(3) = 1/2 + 0, f(3) = 1/4.

Links

Crossrefs

Cf. A122045, A190339, A233808.

Programs

  • Mathematica
    Clear[f]; f[0] = 0; f[1] = 1/2; f[n_] := f[n] = (1/2)*(EulerE[n-1, 1]/2^IntegerExponent[n-1, 2] + f[n-1]); Table[f[n] // Numerator, {n, 0, 31}] (* Jean-François Alcover, Aug 06 2014 *)

A329121 Smallest k such that m//A002275(n)//m is prime, where // denotes concatenation, A002275(n) is the n-th repunit (R_n) and m = k+10^(A006519(n)-1).

Original entry on oeis.org

0, 2, 3, 16, 73, 18, 7, 28, 117, 72, 7, 2, 7, 2, 33, 10, 29, 0, 21, 10, 191, 0, 1133, 30, 79, 20, 81, 58, 19, 2, 57, 116, 77, 108, 79, 48, 97, 40, 1043, 42, 47, 86, 13, 52, 149, 82, 47, 56, 9, 80, 41, 42, 61, 86, 51, 68, 311, 262, 1001, 222, 7, 496, 1367, 12
Offset: 0

Views

Author

Chai Wah Wu, Nov 05 2019

Keywords

Comments

a(n) = A263182(n) - 10^(A006519(n)-1). We use the convention that A007814(0) = 0 and A006519(0) = 1.
a(A004023(n)-2) = 0.

Links

Crossrefs

Cf. A002275, A004023, A006519, A007814, A263182.

A329486 a(n) = 3*A006519(n)/2 + n/2 where A006519(n) is the highest power of 2 dividing n.

Original entry on oeis.org

2, 4, 3, 8, 4, 6, 5, 16, 6, 8, 7, 12, 8, 10, 9, 32, 10, 12, 11, 16, 12, 14, 13, 24, 14, 16, 15, 20, 16, 18, 17, 64, 18, 20, 19, 24, 20, 22, 21, 32, 22, 24, 23, 28, 24, 26, 25, 48, 26, 28, 27, 32, 28, 30, 29, 40, 30, 32, 31, 36, 32, 34, 33, 128, 34, 36, 35, 40
Offset: 1

Views

Author

Markus Rissanen, Nov 14 2019

Keywords

Comments

A combination of sequences A006519 (highest power of 2 dividing n) and A003602 (Kimberling's paraphrases).

Links

Crossrefs

Cf. A006519, A003602.

Programs

  • Mathematica
    Array[3*2^(IntegerExponent[#, 2] - 1) + #/2 &, 68] (* Michael De Vlieger, Jul 10 2022 *)
  • PARI
    a(n) = (3*2^valuation(n, 2) + n)/2; \\ Michel Marcus, Mar 03 2020
  • Python
    def A329486(n): return (3*(n&-n)+n)>>1 # Chai Wah Wu, Jul 10 2022

Extensions

Edited and more terms from Michel Marcus, Mar 03 2020

A340619 n appears A006519(n) times.

Original entry on oeis.org

1, 2, 2, 3, 4, 4, 4, 4, 5, 6, 6, 7, 8, 8, 8, 8, 8, 8, 8, 8, 9, 10, 10, 11, 12, 12, 12, 12, 13, 14, 14, 15, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 17, 18, 18, 19, 20, 20, 20, 20, 21, 22, 22, 23, 24, 24, 24, 24, 24, 24, 24, 24, 25, 26, 26
Offset: 1

Views

Author

Rémy Sigrist, Jan 13 2021

Keywords

Comments

This sequence has similarities with the Cantor staircase function.
This sequence can be seen as an irregular table where the n-th row contains A006519(n) times the value n.
For any k > 1, the set of points { (n, a(n)), n = 1..A006520(2^k-1) } is symmetric; for example, for k = 3, we have the following configuration:
a(n)
^
| *
| **
| *
| ****
| *
| **
|*
+-------------> n

Examples

			The first rows, alongside A006519(n), are:
    n | n-th row               | A006519(n)
   ---+------------------------+-----------
    1 | 1                      |          1
    2 | 2, 2                   |          2
    3 | 3                      |          1
    4 | 4, 4, 4, 4             |          4
    5 | 5                      |          1
    6 | 6, 6                   |          2
    7 | 7                      |          1
    8 | 8, 8, 8, 8, 8, 8, 8, 8 |          8
    9 | 9                      |          1
   10 | 10, 10                 |          2

Links

Crossrefs

See A061392 and A340500 for similar sequences.
Cf. A006519, A006520, A046699.

Programs

  • Mathematica
    A340619[n_] := Array[n &, Table[BitAnd[BitNot[i - 1], i], {i, 1, n}][[n]]];
Table[A340619[n], {n, 1, 26}] // Flatten (* Robert P. P. McKone, Jan 19 2021 *)
  • PARI
    concat(apply(v -> vector(2^valuation(v,2), k, v), [1..26]))
  • PARI
    a(n) = my(ret=0); forstep(k=logint(n,2),0,-1, if(n > k<<(k-1), ret+=1<Kevin Ryde, Jan 18 2021

Formula

a(A006520(n)) = n.
a(A006520(n)+1) = n+1.
a(n) + a(A006520(2^k-1) + 1 - n) = 2^k for any k > 0 and n = 1..A006520(2^k-1).
a(n) = 2^k + (a(r) if r>0), where k such that k*2^(k-1) < n <= (k+1)*2^k and r = n - (k+2)*2^(k-1). - Kevin Ryde, Jan 18 2021

A343934 Irregular triangle read by rows: row n gives the sequence of iterations of k - A006519(k), starting with k=n, until 0 is reached.

Original entry on oeis.org

1, 2, 3, 2, 4, 5, 4, 6, 4, 7, 6, 4, 8, 9, 8, 10, 8, 11, 10, 8, 12, 8, 13, 12, 8, 14, 12, 8, 15, 14, 12, 8, 16, 17, 16, 18, 16, 19, 18, 16, 20, 16, 21, 20, 16, 22, 20, 16, 23, 22, 20, 16, 24, 16, 25, 24, 16, 26, 24, 16, 27, 26, 24, 16, 28, 24, 16
Offset: 1

Views

Author

Christian Perfect, May 04 2021

Keywords

Comments

Row n starts with n, then the highest power of 2 dividing n is subtracted to produce the next entry in the row.
n first appears at position A000788(n)+1.

Examples

			The triangle begins
1
2
3 2
4
5 4
6 4
7 6 4

Links

Crossrefs

Cf. A000120 (row widths), A000788, A006519, A129760, A298011 (row sums).

Programs

  • Mathematica
    Table[Most @ NestWhileList[# - 2^IntegerExponent[#, 2] &, n, # > 0 &], {n, 1, 30}] // Flatten (* Amiram Eldar, May 05 2021 *)
  • Python
    def gen_a():
    for n in range(1,100):
        k = n
        while k>0:
            yield k
            k = k & (k-1)
a = gen_a()

A368595 Alternating sum of A006519.

Original entry on oeis.org

-1, 1, 0, 4, 3, 5, 4, 12, 11, 13, 12, 16, 15, 17, 16, 32, 31, 33, 32, 36, 35, 37, 36, 44, 43, 45, 44, 48, 47, 49, 48, 80, 79, 81, 80, 84, 83, 85, 84, 92, 91, 93, 92, 96, 95, 97, 96, 112, 111, 113, 112, 116, 115, 117, 116, 124, 123, 125, 124, 128, 127, 129, 128
Offset: 1

Views

Author

Jeffrey Shallit, Dec 31 2023

Keywords

Comments

a(n) <= (n/2)*log_2 n, with equality at powers of 2.

Crossrefs

Cf. A006519. A006520 (all positive signs), A136013.
Cf. A093347 (with powers of 3).

Programs

  • Mathematica
    a[1]=-1;a[n_]:=If[OddQ[n],a[n-1]-2^IntegerExponent[n,2],a[n-1]+2^IntegerExponent[n,2]];Table[a[n],{n,63}] (* James C. McMahon, Dec 31 2023 *)
  • PARI
    a(n) = fromdigits(Vec(Pol(binary(n))'),2) - bitand(n,1); \\ Kevin Ryde, Jan 01 2024
  • Python
    def A368595(n): return sum(map(lambda x:(x[0]+1)*(1<Chai Wah Wu, Jan 01 2024

Formula

a(n) = Sum_{i=1..n} (-1)^i*A006519(i).
a(n) = A136013(n) - (n mod 2). - Kevin Ryde, Jan 01 2024

A377556 E.g.f.: exp(Sum_{n>=1} A006519(n) * x^n).

Original entry on oeis.org

1, 1, 5, 19, 193, 1181, 13021, 117895, 1868609, 20980153, 348219541, 4940639771, 98898110785, 1632238421269, 34910480911853, 672959412044431, 16733065940227201, 359936040496423025, 9469928134781142949, 229631546862609396643, 6716832478519734558401, 178344294076141938008461
Offset: 0

Views

Author

Vaclav Kotesovec, Nov 01 2024

Keywords

Crossrefs

Cf. A000123, A006519, A038500, A161809, A377555.

Programs

  • Mathematica
    nmax = 25; CoefficientList[Series[Exp[Sum[2^IntegerExponent[k, 2]*x^k, {k, 1, nmax}]], {x, 0, nmax}], x] * Range[0,nmax]!

A161803 G.f.: A(x) = exp( Sum_{n>=1} A162552(n) * 2*A006519(n) * x^n/n ).

Original entry on oeis.org

1, 2, 0, -2, 6, 12, 0, -8, 24, 44, 0, -30, 54, 104, 0, -60, 238, 466, 0, -402, 924, 1892, 0, -1228, 3264, 6006, 0, -4052, 6688, 13052, 0, -7452, 16536, 32140, 0, -24828, 39660, 85744, 0, -53592, 114336, 212406, 0, -141090, 190754, 386956, 0, -216572, 136078
Offset: 0

Views

Author

Paul D. Hanna, Jul 19 2009

Keywords

Comments

A162552 forms the l.g.f. of log[ Sum_{n>=0} x^(n^2) ], while
2*A006519 forms the l.g.f. of binary partitions (A000123) and
A006519(n) is the highest power of 2 dividing n.

Examples

			G.f.: 1 + 2*x - 2*x^3 + 6*x^4 + 12*x^5 - 8*x^7 + 24*x^8 + 44*x^9 +...

Crossrefs

Cf. A161800, A162552.

Programs

  • PARI
    {a(n)=local(SQ=sum(m=0, sqrtint(n+1), x^(m^2))+x*O(x^n), L=sum(m=1,n,2*2^valuation(m,2)*polcoeff(log(SQ),m)*x^m)+x*O(x^n)); polcoeff(exp(L),n)}
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