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.

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A066220 Least k > 0 such that t^k = 1 mod (prime(n) - t) for 0 < t < prime(n).

Original entry on oeis.org

1, 1, 2, 4, 6, 60, 60, 120, 144, 7920, 55440, 18480, 7920, 27720, 2520, 637560, 8288280, 480720240, 480720240, 480720240, 480720240, 480720240, 1442160720, 9854764920, 59128589520, 59128589520, 147821473800, 670124014560
Offset: 1

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Author

Michael Ulm (taga(AT)hades.math.uni-rostock.de), Dec 18 2001

Keywords

Comments

This sequence gives the period length of the base-p representation of HarmonicNumber[p-1]/p^2 (whose numerator is A061002).

Examples

			a(5) = 6 because 2^6 = 1 mod 9, 3^6 = 1 mod 8, 4^6 = 1 mod 7, 5^6 = 1 mod 6, 6^6 = 1 mod 5, 7^6 = 1 mod 4, 8^6 = 1 mod 3, 9^6 = 1 mod 2 and 6 is the minimal exponent that satisfies this.
		

Programs

  • Mathematica
    a[p_?PrimeQ] := Module[{e = 1}, While[! And @@ Table[Mod[PowerMod[i, e, p - i] - 1, p - i] == 0, {i, p - 1}], e++]; e]; a /@ Prime[Range[10]]

A120284 Numerator of absolute value of Sum_{k=1..n} (-1)^(k+1)*(2*k+1)*(Sum_{i=1..k} 1/i).

Original entry on oeis.org

3, 9, 25, 125, 147, 343, 761, 6849, 7381, 81191, 86021, 1118273, 1171733, 1171733, 2436559, 41421503, 42822903, 271211719, 279175675, 55835135, 19093197, 439143531, 1347822955, 33695573875, 34395742267, 309561680403, 315404588903
Offset: 1

Views

Author

Alexander Adamchuk, Jul 07 2006

Keywords

Comments

p^3 divides a(p-1) for prime p>3, a(p-1)/p^3=A061002[n], a(p-1)/p=A001008(p-1) for p>2. p^2 divides a(p-2) for prime p>3. p^3 divides a(p^2-1) for prime p>3. p divides a(p^2-2) for prime p>3. p^3 divides a(p^3-1) for prime p>3. p^3 divides a(p^4-1) for prime p>3.

Crossrefs

Programs

  • Mathematica
    Numerator[Abs[Table[Sum[(-1)^(k+1)*(2k+1)*Sum[1/i,{i,1,k}],{k,1,n}],{n,1,30}]]]

Formula

a(n) = numerator(abs(Sum_{k=1..n} (-1)^(k+1)*(2*k+1)*(Sum_{i=1..k} 1/i))).

A127063 Primes p such that denominator of Sum_{k=1..p-1} 1/k^2 is a square and denominator Sum_{k=1..p-1} 1/k^3 is a cube and denominator Sum_{k=1..p-1} 1/k^4 is a fourth power and denominator Sum_{k=1..p-1} 1/k^5 is a fifth power.

Original entry on oeis.org

2, 3, 5, 17, 439, 443, 16400183, 16400191, 16400201, 16400203, 16400221, 16400231, 16400233, 16400269, 16400273, 16400299, 16400309, 16400317, 16400347, 16400383, 16400387, 16400389, 16400411, 16400413, 16400429, 16400431
Offset: 1

Views

Author

Artur Jasinski, Jan 04 2007

Keywords

Comments

Subsequence of A127062 and of A127061. - Max Alekseyev, Feb 08 2007

Crossrefs

Extensions

More terms from Max Alekseyev, Feb 08 2007

A330014 When prime(n) is an odd prime (n >= 2) and N(n) / D(n) = Sum_{k=1..prime(n)-1} 1/k^3, then prime(n) divides N(n) and a(n) = N(n) / prime(n).

Original entry on oeis.org

3, 407, 4081, 1742192177, 1964289620189, 26430927136768997, 12913609418092462447, 14639800647032731764901, 21461951639001843544904995612963, 489697309796854053100609288112563213, 97796057728171000155497946604711651753457
Offset: 2

Views

Author

Bernard Schott, Nov 27 2019

Keywords

Comments

The idea of this sequence comes from the 1st exercise of "sélection de la délégation française" in 2005 for IMO 2006 where it was asked to prove that prime(n) divides N(n) [See reference].
The first fractions N(n)/D(n) are 9/8, 2035/1728, 28567/24000, 19164113947/16003008000, 25535765062457/21300003648000, ...

Examples

			For prime(4) = 7 then 1 + 1/2^3 + 1/3^3 + 1/4^3 + 1/5^3 + 1/6^3 = 28567/24000 and 28567/7 = 4081, a(4) = 4081.
		

References

  • Guy Alarcon and Yves Duval, TS: Préparation au Concours Général, RMS, Collection Excellence, Paris, 2010, chapitre 10, Exercices de sélection de la délégation française en Octobre 2005 pour OIM 2006, Exercice 1, p. 169, p. 179.

Crossrefs

Cf. A076637, A061002, A076637 (Wolstenholme's Theorem).

Programs

  • Magma
    [(Numerator(&+ [1/(k-1)^3:k in [2..NthPrime(n)]])) / NthPrime(n):n in [2..12]]; // Marius A. Burtea, Nov 27 2019
  • Mathematica
    a[n_] := Numerator[Sum[1/(i- 1)^3, {i, 2,(p = Prime[n])}]]/p; Array[a, 11, 2] (* Amiram Eldar, Nov 27 2019 *)
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