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.

A055670 a(n) = prime(n) - (-1)^prime(n).

Original entry on oeis.org

1, 4, 6, 8, 12, 14, 18, 20, 24, 30, 32, 38, 42, 44, 48, 54, 60, 62, 68, 72, 74, 80, 84, 90, 98, 102, 104, 108, 110, 114, 128, 132, 138, 140, 150, 152, 158, 164, 168, 174, 180, 182, 192, 194, 198, 200, 212, 224, 228, 230, 234, 240, 242, 252, 258, 264, 270, 272, 278, 282, 284
Offset: 1

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Author

N. J. A. Sloane, Jun 09 2000

Keywords

Comments

Number of right-inequivalent prime Hurwitz quaternions of norm p, where p = n-th rational prime (indexed by A000040).
Two primes are considered right-equivalent if they differ by right multiplication by one of the 24 units. - N. J. A. Sloane
Start of n-th run of consecutive nonprime numbers. Since 2 is the only even prime, for all other prime numbers the expression "- (-1)^(n-th prime)" works out to "+ 1." - Alonso del Arte, Oct 18 2011

Examples

			a(1) = 2 - (-1)^2 = 1, a(2) = 3 - (-1)^3 = 4.
		

References

  • L. E. Dickson, Algebras and Their Arithmetics, Dover, 1960, Section 91.
  • Lynn Arthur Steen and J. Arthur Seebach, Jr., Counterexamples in Topology, Dover, New York, 1978, page 134.

Crossrefs

a(n) = A083503(p) for n>1.

Programs

  • Mathematica
    Join[{1},Prime[Range[2,70]]+1] (* Harvey P. Dale, Oct 29 2013 *)

Formula

a(n) = prime(n)+1 = A008864(n) for n >= 2. a(n) = A055669(n)/24.

Extensions

More terms from David W. Wilson, May 02 2001
I would also like to get the sequences of inequivalent prime Hurwitz quaternions, where two primes are considered equivalent if they differ by left or right multiplication by one of the 24 units. This will give two more sequences, analogs of A055670 and A055672.
Edited by N. J. A. Sloane, Aug 16 2009

A110738 a(n) = common ratio for row n in A112668.

Original entry on oeis.org

1, 3, 4, 3, 6, 5, 8, 3, 4, 9, 12, 5, 14, 13, 16, 3, 18, 5, 20, 3, 4, 21, 24, 5, 6, 25, 4, 13, 30, 11, 32, 3, 34, 33, 36, 5, 38, 37, 16, 3, 42, 5, 44, 21, 16, 45, 48, 5, 8, 9, 52, 5, 54, 5, 16, 13, 7, 57, 60, 7, 62, 61, 4, 3, 66, 23, 68, 13, 70, 29, 72, 5, 74, 73, 16, 37, 78, 17, 80, 3
Offset: 1

Views

Author

Amarnath Murthy, Aug 10 2005

Keywords

Crossrefs

Programs

  • Maple
    A110738 := proc(n) local a ; if n = 1 then RETURN(1) ; else a := 2 ; while (1-a^n)/(1-a) mod n <> 0 do a := a+1 ; od ; RETURN(a) ; fi ; end: for n from 1 to 80 do printf("%d, ",A110738(n)) : od : # R. J. Mathar, Mar 13 2007

Extensions

More terms from R. J. Mathar, Mar 13 2007

A083502 Smallest k such that n*(n+k) + 1 is an n-th power.

Original entry on oeis.org

1, 2, 18, 16, 1550, 2598, 299586, 812, 29118, 348678430, 67546215506, 20345040, 61054982557998, 281241170407078, 76861433640456450, 2690404, 128583032925805678334, 211927625850, 275941052631578947368402, 174339200
Offset: 1

Views

Author

Amarnath Murthy and Meenakshi Srikanth (menakan_s(AT)yahoo.com), May 03 2003

Keywords

Comments

Sequence is obviously infinite.
If the sequence is restricted to only prime n's, the sequence increases absolutely. See comment in A083503.
[Since there is actually no comment in A083503: this probably means to say that (conjectural!) A083503(prime(n)) = A008864(n) which leads to a(p) = Sum_{s=2..p} binomial(p,s)*p^(s-1) for primes p, an increasing subsequence. - R. J. Mathar, Aug 01 2025]
a(n) = (x^n-1)/n - n, where x is the least integer > 1 with x^n == 1 (mod n). - Robert Israel, Aug 01 2025

Crossrefs

The i's in the above Mathematica coding, except for a(1), give A055670.

Programs

  • Maple
    A083502 := proc(n)
        local a,b ;
        if n = 1 then
            1 ;
        else
            for b from 2 do
                a := (b^n-1)/n-n ;
                if type( a,'integer') then
                    return  a;
                end if;
            end do:
        end if;
    end proc:
    seq(A083502(n),n=1..20) ; # R. J. Mathar, Aug 01 2025
    # alternative
    f:= proc(n) local X,S;
      S:= min(map(t -> subs(t,X), {msolve(X^n = 1, n)} minus {{X=1}}));
      if S = infinity then ((n+1)^n - 1)/n - n else (S^n-1)/n - n fi
    end proc:
    f(1):= 1:
    map(f, [$1..50]); # Robert Israel, Aug 01 2025
  • Mathematica
    Do[i = 2; While[k = (i^n - 1)/n - n; !IntegerQ[k], i++ ]; Print[k], {n, 1, 20}]

Extensions

Edited and extended by Robert G. Wilson v, May 11 2003
Showing 1-3 of 3 results.