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 1-3 of 3 results.

A047893 Number of decimal digits of Euler (Zig) numbers A000364.

Original entry on oeis.org

1, 1, 2, 4, 5, 7, 9, 11, 13, 15, 17, 20, 22, 25, 27, 30, 32, 35, 38, 41, 44, 46, 49, 52, 55, 58, 61, 64, 68, 71, 74, 77, 80, 84, 87, 90, 94, 97, 100, 104, 107, 111, 114, 118, 121, 125, 128, 132, 135, 139, 143, 146, 150, 154, 157, 161, 165, 168, 172, 176, 180, 183
Offset: 1

Views

Author

Labos Elemer

Keywords

Examples

			a(4) = floor(log_10(1385)) + 1 = 4, E(4) = 1385, the 4th Euler number.

References

  • J. Peters and J. Stein, Mathematische Tafeln, Revised Russian Edition, Moscow, 1968.

Links

Crossrefs

Cf. A000364, A034971, A034972, A000182 (tangent numbers), A047894.

Programs

  • Mathematica
    a[n_] := IntegerLength[EulerE[2*n]]; Array[a, 100] (* Amiram Eldar, Mar 20 2025 *)
  • PARI
    a(n) = #Str(subst(bernpol(2*n+1), 'x, 1/4)*4^(2*n+1)*(-1)^(n+1)/(2*n+1));

Formula

a(n) = A055642(A000364(n)) = floor(log_10(|E(2n)|)) + 1 for n >=2, where E is Euler's E function.

Extensions

a(1) corrected by Amiram Eldar, Mar 20 2025

A047894 Number of digits of A000182(n).

Original entry on oeis.org

1, 1, 2, 3, 4, 6, 8, 10, 12, 14, 16, 19, 21, 23, 26, 28, 31, 34, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 76, 79, 82, 85, 89, 92, 95, 99, 102, 105, 109, 112, 116, 119, 123, 126, 130, 134, 137, 141, 144, 148, 152, 155, 159, 163, 167, 170, 174, 178, 182, 185
Offset: 1

Views

Author

Labos Elemer

Keywords

Examples

			a(7) = 8 since the 7th tangential number is 22368256, which has 8 decimal digits.

References

  • J. Peters and J. Stein, Mathematische Tafeln. Revised Russian Edition, 1968, Moscow.

Links

Crossrefs

Cf. A000182, A055642, A000384, A034972, A034971, A047893.

Programs

  • Mathematica
    a[n_] := IntegerLength[(2^(2*n)*(2^(2*n) - 1)*Abs[BernoulliB[2*n]])/(2*n)]; Array[a, 100] (* Amiram Eldar, Mar 17 2025 after Stan Wagon at A000182 *)
  • Python
    from sympy import bernoulli
def A047894(n): return len(str(abs(((2-(2<<(m:=n<<1)))*bernoulli(m)<Chai Wah Wu, Apr 14 2023

Formula

a(n) = floor(log_10(A000182(n))) + 1. - Franklin T. Adams-Watters, Mar 29 2014
a(n) = A055642(A000182(n)). - Michel Marcus, Apr 15 2023

Extensions

a(1) corrected by Chai Wah Wu, Apr 14 2023

A047872 a(n) = floor(abs(B(2*n + 2)/B(2*n))) where B(n) is the n-th Bernoulli number.

Original entry on oeis.org

0, 0, 0, 1, 2, 3, 4, 6, 7, 9, 11, 13, 16, 19, 22, 25, 28, 31, 35, 39, 43, 47, 52, 57, 62, 67, 72, 78, 83, 89, 95, 102, 108, 115, 122, 129, 136, 144, 152, 160, 168, 176, 185, 193, 202, 212, 221, 231, 240, 250, 260, 271, 281, 292, 303, 314, 326, 337, 349, 361, 373
Offset: 0

Views

Author

Labos Elemer

Keywords

Examples

			a(3) = floor(abs(B(4)/B(3))) = floor((1/30)/(1/42)) = floor(7/5) = floor(1.4) = 1.
a(249) = floor(abs(B(250)/B(249))) = 6319.

References

  • Glaisher, J. W. L.; Tables of the first 250 Bernoulli numbers. Trans. Cambridge Phil. Soc. 12 (1873), 384-391.
  • Peters, J. and Stein, J., Matematische Tafeln. Revised Russian Edition, 1968, Moscow.

Links

Crossrefs

Cf. A034972, A034971, A002939.

Programs

  • Maple
    seq(floor(abs(bernoulli(2*n+2)/bernoulli(2*n))),n=0..200); # Robert Israel, Jun 27 2018
  • Mathematica
    Table[Floor[Abs[BernoulliB[2*n + 2]/BernoulliB[2*n]]], {n, 0, 60}] (* T. D. Noe, Jun 27 2013 *)
  • PARI
    a(n) = floor(abs(bernfrac(2*n+2)/bernfrac(2*n))) \\ Michel Marcus, Jun 27 2013

Formula

a(n) = floor( (n+1)*(2*n+1)/(2*Pi^2)) (conjectured). - Bill McEachen, Dec 08 2021
A002939(n+1)*B(2*n)/B(2*(n+1)) = -(2*Pi)^2*(1 + O(1/4^n)). See the StackExchange link. - Peter Luschny, Dec 08 2021
Showing 1-3 of 3 results.

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

Content is available under The OEIS End-User License Agreement.