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.

A110567 a(n) = n^(n+1) + 1.

Original entry on oeis.org

1, 2, 9, 82, 1025, 15626, 279937, 5764802, 134217729, 3486784402, 100000000001, 3138428376722, 106993205379073, 3937376385699290, 155568095557812225, 6568408355712890626, 295147905179352825857, 14063084452067724991010
Offset: 0

Views

Author

Jonathan Vos Post, Sep 12 2005

Keywords

Comments

For n >= 2, a(n) = the n-th positive integer such that a(n) (base n) has a block of exactly n consecutive zeros.
Comments from Alexander Adamchuk, Nov 12 2006 (Start)
(2n+1)^2 divides a(2n). a(2n)/(2n+1)^2 = {1,1,41,5713,1657009,826446281,633095889817,691413758034721,...} = A081215(2n).
p divides a(p-1) for prime p. a(p-1)/p = {1,3,205,39991,9090909091,8230246567621,...} = A081209(p-1) = A076951(p-1).
p^2 divides a(p-1) for an odd prime p. a(p-1)/p^2 = {1,41,5713,826446281,633095889817,1021273028302258913,1961870762757168078553, 14199269001914612973017444081,...} = A081215(p-1).
Prime p divides a((p-3)/2) for p = {13,17,19,23,37,41,43,47,61,67,71,89, 109,113,137,139,157,163,167,181,191,...}.
Prime p divides a((p-5)/4) for p = {29,41,61,89,229,241,281,349,421,509,601,641,661,701,709,769,809,821,881,...} = A107218(n) Primes of the form 4x^2+25y^2.
Prime p divides a((p-7)/6) for p = {79,109,127,151,313,421,541,601,613,751,757,787,...}.
Prime p divides a((p-9)/8) for p = {41,337,401,521,569,577,601,857,929,937,953,977,...} A subset of A007519(n) Primes of form 8n+1.
Prime p divides a((p-11)/10) for p = {41,181,331,601,761,1021,1151,1231,1801,...}.
Prime p divides a((p-13)/12) for p = {313,337,433,1621,1873,1993,2161,2677,2833,...}. (End)

Examples

			Examples illustrating the Comment:
a(2) = 9 because the first positive integer (base 2) with a block of 2 consecutive zeros is 100 (base 2) = 4, and the 2nd is 1001 (base 2) = 9 = 1 + 2^3.
a(3) = 82 because the first positive integer (base 3) with a block of 3 consecutive zeros is 1000 (base 3) = 81, the 2nd is 2000 (base 3) = 54 and the 3rd is 10001 (base 3) = 82 = 1 + 3^4.
a(4) = 1025 because the first positive integer (base 4) with a block of 4 consecutive zeros is 10000 (base 4) = 256, the 2nd is 20000 (base 4) = 512, the 3rd is 30000 (base 4) = 768 and the 4th 100001 (base 4) = 1025 = 1 + 4^5. and the 2nd is 1001 (base 2) = 9 = 1 + 2^3.

Links

Crossrefs

Cf. A007778: n^(n+1); A000312: n^n; A014566: Sierpinski numbers of the first kind: n^n + 1.
Cf also A081209, A076951, A081215. A110567.

Programs

  • Magma
    [n^(n+1) + 1: n in [0..25]]; // G. C. Greubel, Oct 16 2017
  • Mathematica
    Table[n^(n+1)+1,{n,0,30}] (* Harvey P. Dale, Oct 30 2015 *)
  • PARI
    for(n=0,25, print1(1 + n^(n+1), ", ")) \\ G. C. Greubel, Aug 31 2017

Formula

a(n) = A007778(n) + 1.
a(n) = A110567(n) for n > 1. - Georg Fischer, Oct 20 2018

Extensions

Entry revised by N. J. A. Sloane, Oct 20 2018 at the suggestion of Georg Fischer.

A076951 Smallest k > 0 such that nk-1 is an n-th power, or 0 if no such number exists.

Original entry on oeis.org

1, 1, 3, 0, 205, 0, 39991, 0, 57, 5905, 9090909091, 0, 8230246567621, 0, 10371206370520815, 0, 17361641481138401521, 0, 37275544492386193492507, 0, 22706531343006, 0, 326583187044036098379401213863, 0, 45035996273705
Offset: 1

Views

Author

Amarnath Murthy, Oct 20 2002

Keywords

Comments

p divides a(p) for odd primes p. a(p) = A081209(p-1) = p*A081215(p-1) for odd primes p. a(p)/p = A081215(p-1) = {1,41,5713,826446281,633095889817,1021273028302258913,...} - Alexander Adamchuk, Nov 12 2006

Crossrefs

Cf. A076952.
Cf. A081209, A081215.

Extensions

More terms from Sascha Kurz, Jan 21 2003

A273319 a(n) = ((2*n+1)^(n+1) + (-1)^n)/(n+1)^2.

Original entry on oeis.org

2, 2, 14, 150, 2362, 49210, 1280582, 40045166, 1464047858, 61310662578, 2894855376382, 152184891889030, 8817255144288554, 558260148630165098, 38351949989325264182, 2841496569324942436830
Offset: 0

Views

Author

Gionata Neri, May 19 2016

Keywords

Comments

When searching for the smallest k such that n^k + 1 is not squarefree, I noticed that if n is even then n^(n+1) + 1 is not squarefree, and if n is of the form 4*j + 1 (j>0) then n^((n+1)/2) + 1 is not squarefree.

Crossrefs

Cf. A081215.

Programs

  • Mathematica
    Table[((2n + 1)^(n + 1) + (-1)^n)/(n + 1)^2, {n, 0, 15}] (* Alonso del Arte, May 19 2016 *)

Formula

a(n) = Sum_{k = 0..n} (-1)^(n - k)*2^(k+1)*(n+1)^(k-1)*C(n+1, n - k).
Showing 1-3 of 3 results.

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

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