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 1-7 of 7 results.

A001901 Successive numerators of Wallis's approximation to Pi/2 (reduced).

Original entry on oeis.org

1, 2, 4, 16, 64, 128, 256, 2048, 16384, 32768, 65536, 262144, 1048576, 2097152, 4194304, 67108864, 1073741824, 2147483648, 4294967296, 17179869184, 68719476736, 137438953472, 274877906944, 2199023255552
Offset: 0

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Author

Keywords

Comments

If p is prime, then a(p-2) == - A001902(p-2) (mod p). Cf. A064169 (third comment) and my formula here. Such pseudoprimes are 1467, 7831, ... Primes p such that a(p-2) == - A001902(p-2) (mod p^2) are 5, 45827, ... Cf. A355959, see also A330719 (third comment). - Thomas Ordowski, Oct 19 2024

Examples

			From _Wolfdieter Lang_, Dec 07 2017: (Start)
The Wallis numerators (N) and denominators (D) with partial products A(n) = A001900(n) and B(n) = A000246(n+1) in unreduced form, and a(n) and b(n) = A001902(n) in reduced form.
n, k:     0  1  2  3  4   5    6     7      8        9       10 ...
N(k):     1  2  2  4  4   6    6     8      8       10       10 ...
D(k):     1  1  3  3  5   5    7     7      9        9        9 ...
A(n):     1  2  4 16 64 384 2304 18432 147456  1474560 14745600 ...
B(n):     1  1  3  9 45 225 1575 11025  99225   893025  9823275 ...
a(n):     1  2  4 16 64 128  256  2048  16384    32768    65536 ...
b(n):     1  1  3  9 45  75  175  1225  11025    19845    43659 ...
n = 5: numerator(1*2*2*4*4*6/(1*1*3*3*5*5)) = numerator(384/225) = numerator(128/75) = 128. (End)
		

References

  • H.-D. Ebbinghaus et al., Numbers, Springer, 1990, p. 146.

Crossrefs

Denominators are A001902. Subsequence of A000079.

Programs

  • Mathematica
    a[n_?EvenQ] := n!!^2/((n - 1)!!^2*(n + 1)); a[n_?OddQ] := ((n - 1)!!^2*(n + 1))/n!!^2; Table[a[n] // Numerator, {n, 0, 23}] (* Jean-François Alcover, Jun 19 2013 *)

Formula

(2*2*4*4*6*6*8*8*...*2n*2n*...)/(1*3*3*5*5*7*7*9*...*(2n-1)*(2n+1)*...) for n >= 1.
From Wolfdieter Lang, Dec 07 2017: (Start)
1/1 * 2/1 * 2/3 * 4/3 * 4/5 * 6/5 * 6/7 * ...; partial products (reduced). Here the numerators with offset 0.
a(n) = numerator(W(n)), for n >= 0, with W(n) = Product_{k=0..n} N(k)/D(k) (reduced), with N(k) = 2*floor((k+1)/2) for k >= 1 and N(0) = 1, and D(k) = 2*floor(k/2) + 1, for k >= 0. (End)
a(n) is the numerator of the continued fraction [1;1,1/2,1/3,...,1/n]. - Thomas Ordowski, Oct 19 2024

A355960 Primes p such that (p+5)^(p-1) == 1 (mod p^2).

Original entry on oeis.org

3, 23, 1574773
Offset: 1

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Author

Felix Fröhlich, Jul 21 2022

Keywords

Comments

Equivalently, primes p such that 5^p == p+5 (mod p^2), or Fermat quotient q_p(5) == 1/5 (mod p). - Max Alekseyev, Sep 16 2024

Crossrefs

(p+k)^(p-1) == 1 (mod p^2): A355959 (k=2), A355961 (k=6), A355962 (k=7), A355963 (k=8), A355964 (k=9), A355965 (k=10).

Programs

  • PARI
    forprime(p=1, , if(Mod(p+5, p^2)^(p-1)==1, print1(p, ", ")))

A355961 Primes p such that (p+6)^(p-1) == 1 (mod p^2).

Original entry on oeis.org

13, 47, 3803, 151051, 240727, 259933847
Offset: 1

Views

Author

Felix Fröhlich, Jul 21 2022

Keywords

Crossrefs

(p+k)^(p-1) == 1 (mod p^2): A355959 (k=2), A355960 (k=5), A355962 (k=7), A355963 (k=8), A355964 (k=9), A355965 (k=10).

Programs

  • PARI
    forprime(p=1, , if(Mod(p+6, p^2)^(p-1)==1, print1(p, ", ")))

A355962 Primes p such that (p+7)^(p-1) == 1 (mod p^2).

Original entry on oeis.org

2, 3, 229, 701, 31446553, 1016476523, 8918351831
Offset: 1

Views

Author

Felix Fröhlich, Jul 21 2022

Keywords

Comments

a(8) > 10^13 if it exists. - Jason Yuen, May 12 2024
Equivalently, primes p such that 7^p == p+7 (mod p^2), or Fermat quotient q_p(7) == 1/7 (mod p). - Max Alekseyev, Sep 16 2024

Crossrefs

(p+k)^(p-1) == 1 (mod p^2): A355959 (k=2), A355960 (k=5), A355961 (k=6), A355963 (k=8), A355964 (k=9), A355965 (k=10).

Programs

  • PARI
    forprime(p=1, , if(Mod(p+7, p^2)^(p-1)==1, print1(p, ", ")))

Extensions

a(7) from Jason Yuen, May 12 2024

A355963 Primes p such that (p+8)^(p-1) == 1 (mod p^2).

Original entry on oeis.org

1531, 7445287
Offset: 1

Views

Author

Felix Fröhlich, Jul 21 2022

Keywords

Comments

a(3) > 34294200797 if it exists.

Crossrefs

(p+k)^(p-1) == 1 (mod p^2): A355959 (k=2), A355960 (k=5), A355961 (k=6), A355962 (k=7), A355964 (k=9), A355965 (k=10).

Programs

  • PARI
    forprime(p=1, , if(Mod(p+8, p^2)^(p-1)==1, print1(p, ", ")))

A355964 Primes p such that (p+9)^(p-1) == 1 (mod p^2).

Original entry on oeis.org

13, 19, 2207, 26041, 332698495781
Offset: 1

Views

Author

Felix Fröhlich, Jul 21 2022

Keywords

Comments

a(6) > 10^13 if it exists. - Jason Yuen, May 12 2024

Crossrefs

(p+k)^(p-1) == 1 (mod p^2): A355959 (k=2), A355960 (k=5), A355961 (k=6), A355962 (k=7), A355963 (k=8), A355965 (k=10).

Programs

  • PARI
    forprime(p=1, , if(Mod(p+9, p^2)^(p-1)==1, print1(p, ", ")))

Extensions

a(5) from Jason Yuen, May 12 2024

A355965 Primes p such that (p+10)^(p-1) == 1 (mod p^2).

Original entry on oeis.org

13, 41, 97, 809, 1151, 1657, 800011
Offset: 1

Views

Author

Felix Fröhlich, Jul 21 2022

Keywords

Comments

A computer search taking less than 3 seconds shows there are no further terms below the one millionth prime. - Harvey P. Dale, Mar 04 2024
I ran the PARI program below for 8.5 hours and it did not find any further terms. (I do not know how far it searched.) - N. J. A. Sloane, Mar 05 2024

Crossrefs

(p+k)^(p-1) == 1 (mod p^2): A355959 (k=2), A355960 (k=5), A355961 (k=6), A355962 (k=7), A355963 (k=8), A355964 (k=9).

Programs

  • Mathematica
    Select[Prime[Range[70000]],PowerMod[#+10,#-1,#^2]==1&] (* Harvey P. Dale, Mar 04 2024 *)
  • PARI
    forprime(p=1, , if(Mod(p+10, p^2)^(p-1)==1, print1(p, ", ")))
Showing 1-7 of 7 results.