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 1-6 of 6 results.

A288676 Numbers k such that A108394(k) is not a prime.

Original entry on oeis.org

5, 22, 25, 26, 31, 36, 39
Offset: 1

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Author

Seiichi Manyama, Jun 13 2017

Keywords

Comments

A288641(a(n)) > A108394(a(n)).
a(8) > 61.

Examples

			a(1) = 5. So A288641(5) > A108394(5), that is 251 > 214.
		

Crossrefs

A003504 a(0)=a(1)=1; thereafter a(n+1) = (1/n)*Sum_{k=0..n} a(k)^2 (a(n) is not always integral!).

Original entry on oeis.org

1, 1, 2, 3, 5, 10, 28, 154, 3520, 1551880, 267593772160, 7160642690122633501504, 4661345794146064133843098964919305264116096, 1810678717716933442325741630275004084414865420898591223522682022447438928019172629856
Offset: 0

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Keywords

Comments

The sequence appears with a different offset in some other sources. - Michael Somos, Apr 02 2006
Also known as Göbel's (or Goebel's) Sequence. Asymptotically, a(n) ~ n*C^(2^n) where C=1.0478... (A115632). A more precise asymptotic formula is given in A116603. - M. F. Hasler, Dec 12 2007
Let s(n) = (n-1)*a(n). By considering the p-adic representation of s(n) for primes p=2,3,...,43, one finds that a(44) is the first nonintegral value in this sequence. Furthermore, for n>44, the valuation of s(n) w.r.t. 43 is -2^(n-44), implying that both s(n) and a(n) are nonintegral. - M. F. Hasler and Max Alekseyev, Mar 03 2009
a(44) is approximately 5.4093*10^178485291567. - Hans Havermann, Nov 14 2017.
The fractional part is simply 24/43 (see page 709 of Guy (1988)).
The more precise asymptotic formula is a(n+1) ~ C^(2^n) * (n + 2 - 1/n + 4/n^2 - 21/n^3 + 138/n^4 - 1091/n^5 + ...). - Michael Somos, Mar 17 2012

Examples

			a(3) = (1 * 2 + 2^2) / 2 = 3 given a(2) = 2.
		

References

  • R. K. Guy, Unsolved Problems in Number Theory, 3rd edition, Sect. E15.
  • Clifford Pickover, A Passion for Mathematics, 2005.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A005166, A005167, A097398, A108394, A115632, A116603 (asymptotic formula).

Programs

  • Maple
    a:=2: L:=1,1,a: n:=15: for k to n-2 do a:=a*(a+k)/(k+1): L:=L,a od:L; # Robert FERREOL, Nov 07 2015
  • Mathematica
    a[n_] := a[n] = Sum[a[k]^2, {k, 0, n-1}]/(n-1); a[0] = a[1] = 1; Table[a[n], {n, 0, 13}] (* Jean-François Alcover, Feb 06 2013 *)
    With[{n = 14}, Nest[Append[#, (#.#)/(Length[#] - 1)] &, {1, 1}, n - 2]] (* Jan Mangaldan, Mar 21 2013 *)
  • PARI
    A003504(n,s=2)=if(n-->0,for(k=1,n-1,s+=(s/k)^2);s/n,1) \\ M. F. Hasler, Dec 12 2007
    
  • Python
    a=2; L=[1,1,a]; n=15
    for k in range(1,n-1):
        a=a*(a+k)//(k+1)
        L.append(a)
    print(L) # Robert FERREOL, Nov 07 2015

Formula

a(n+1) = ((n-1) * a(n) + a(n)^2) / n if n > 1. - Michael Somos, Apr 02 2006
0 = a(n)*(+a(n)*(a(n+1) - a(n+2)) - a(n+1) - a(n+1)^2) +a(n+1)*(a(n+1)^2 - a(n+2)) if n>1. - Michael Somos, Jul 25 2016

Extensions

a(0)..a(43) are integral, but from a(44) onwards every term is nonintegral - H. W. Lenstra, Jr.
Corrected and extended by M. F. Hasler, Dec 12 2007
Further corrections from Max Alekseyev, Mar 04 2009

A005166 a(0) = 1; a(n) = (1 + a(0)^3 + ... + a(n-1)^3)/n (not always integral!).

Original entry on oeis.org

1, 2, 5, 45, 22815, 2375152056927, 2233176271342403475345148513527359103
Offset: 0

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Comments

Terms are integers until n=A097398(2,2)=89.
Guy states that by computing the sequence modulo 89 it is easy to show that a(89) is not integral. - T. D. Noe, Sep 17 2007

References

  • R. K. Guy, Unsolved Problems in Number Theory, Springer, 1st edition, 1981. See section E15.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Mathematica
    a[0]=1; a[n_]:=(1 + Sum[a[k]^3, {k,0,n-1}])/n; Array[a,7,0] (* Stefano Spezia, Oct 13 2024 *)

A005167 a(n+1) = (1 + a(0)^4 + ... + a(n)^4 )/(n+1) (not always integral!).

Original entry on oeis.org

1, 2, 9, 2193, 5782218987645, 223567225753623833253893162919867828939456664850241
Offset: 0

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Comments

Terms are integer until n=A097398(3,2)=97.
Guy states that by computing the sequence modulo 97 it is easy to show that a(97) is not integral. - T. D. Noe, Sep 17 2007
The next term -- a(6) -- has 201 digits. - Harvey P. Dale, Nov 20 2018

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Mathematica
    nxt[{n_,a_,t_}]:={n+1,(1+t)/(n+1),t+((1+t)/(n+1))^4}; NestList[nxt,{0,1,1},5][[All,2]] (* Harvey P. Dale, Nov 20 2018 *)

A288641 Define the sequence {b_n(k)} as the solutions of the recursion (k+1) * b_n(k+1) = b_n(k) * (b_n(k)^(n-1) + k) with b_n(0) = 1. a(n) is the least prime p where p * b_n(p) is not 0 mod p.

Original entry on oeis.org

43, 89, 97, 251, 19, 239, 37, 79, 83, 239, 31, 431, 19, 79, 23, 827, 43, 173, 31, 103, 179, 73, 19, 431, 193, 101, 53, 811, 47, 1427, 19, 251, 29, 311, 137, 71, 23, 499, 43, 47, 19, 419, 31, 191, 83, 337, 59, 1559, 19, 127, 109, 163, 67, 353, 83, 191, 83, 107
Offset: 2

Views

Author

Seiichi Manyama, Jun 13 2017

Keywords

Comments

If A108394(n) is a prime, a(n) = A108394(n).

Examples

			(k+1) * b_2(k+1) = b_2(k) * (b_2(k) + k) with b_2(0) = 1.
b_2(1) == 2, b_2(2) == 3, b_2(3) == 5, ... , b_2(42) == 33 mod 43.
So 43 * b_2(43) == b_2(42) * (b_2(42) + 42) == 24 (> 0) mod 43.
		

Crossrefs

Cf. A003504 ({b_2(n+1)}), A005166 ({b_3(n)}), A005167 ({b_4(n)}), A108394, A288676.

A378851 Values of p for which x_p = s_p/p in the modified Göbel sequence is nonintegral.

Original entry on oeis.org

43, 61, 67, 83, 103, 107, 109, 157, 163, 167, 311, 317, 349, 353, 367, 389, 409, 457, 521, 523, 593, 661, 683, 739, 821, 827, 877, 881, 887, 967, 1051, 1249, 1277, 1319, 1321, 1433, 1439, 1447, 1481, 1483, 1499, 1667, 1693, 1709, 1747, 1867, 1877, 1931, 1933, 2063
Offset: 1

Views

Author

Eric W. Weisstein, Dec 09 2024

Keywords

Crossrefs

Showing 1-6 of 6 results.