"Fermat's little theorem" - cp's OEIS frontend

A062481 a(n) = n^prime(n).

1, 8, 243, 16384, 48828125, 13060694016, 232630513987207, 144115188075855872, 8862938119652501095929, 100000000000000000000000000000, 191943424957750480504146841291811, 8505622499821102144576131684114829934592, 4695452425098908797088971409337422035076128813

Offset: 1

Labos Elemer, Jul 09 2001

Harry J. Smith, Table of n, a(n) for n = 1..50

Cf. A000040, A000312, A051674, A096250.

Table[n^Prime[n],{n,20}] (* Harvey P. Dale, Jun 12 2014 *)

a(n)={n^prime(n)} \\ Harry J. Smith, Aug 08 2009

from sympy import prime
def a(n): return n**prime(n)
print([a(n) for n in range(1, 14)]) # Michael S. Branicky, Jun 15 2022

From Amiram Eldar, Nov 18 2020: (Start)
a(n) = n^A000040(n).
Sum_{n>=1} 1/a(n) = A096250. (End)
a(n) == n (mod prime(n)). [Fermat's little theorem] - Nicolas Bělohoubek, Jun 14 2022

More terms from Harvey P. Dale, Jun 12 2014

A083737 Pseudoprimes to bases 2, 3 and 5.

Original entry on oeis.org

1729, 2821, 6601, 8911, 15841, 29341, 41041, 46657, 52633, 63973, 75361, 101101, 115921, 126217, 162401, 172081, 188461, 252601, 294409, 314821, 334153, 340561, 399001, 410041, 488881, 512461, 530881, 552721, 658801, 670033, 721801, 748657

Offset: 1

Serhat Sevki Dincer (sevki(AT)ug.bilkent.edu.tr), May 05 2003

a(n) = n-th positive integer k(>1) such that 2^(k-1) == 1 (mod k), 3^(k-1) == 1 (mod k) and 5^(k-1) == 1 (mod k)
See A153580 for numbers k > 1 such that 2^k-2, 3^k-3 and 5^k-5 are all divisible by k but k is not a Carmichael number (A002997).
Note that a(1)=1729 is the Hardy-Ramanujan number. - Omar E. Pol, Jan 18 2009

			a(1)=1729 since it is the first number such that 2^(k-1) == 1 (mod k), 3^(k-1) == 1 (mod k) and 5^(k-1) == 1 (mod k).

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000 (first 102 from R. J. Mathar)
J. Bernheiden, Pseudoprimes (Text in German)
F. Richman, Primality testing with Fermat's little theorem
Index entries for sequences related to pseudoprimes

Proper subset of A052155. Superset of A230722. Cf. A153580, A002997, A001235, A011541.

Select[ Range[838200], !PrimeQ[ # ] && PowerMod[2, # - 1, # ] == 1 && PowerMod[3, 1 - 1, # ] == 1 && PowerMod[5, # - 1, # ] == 1 & ]

is(n)=!isprime(n)&&Mod(2,n)^(n-1)==1&&Mod(3,n)^(n-1)==1&&Mod(5,n)^(n-1)==1 \\ Charles R Greathouse IV, Apr 12 2012

Edited by Robert G. Wilson v, May 06 2003

Edited by N. J. A. Sloane, Jan 14 2009

A087265 Lucas numbers L(8*n).

Original entry on oeis.org

2, 47, 2207, 103682, 4870847, 228826127, 10749957122, 505019158607, 23725150497407, 1114577054219522, 52361396397820127, 2459871053643326447, 115561578124838522882, 5428934300813767249007, 255044350560122222180447

Offset: 0

Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Oct 19 2003

a(n+1)/a(n) converges to (47+sqrt(2205))/2 = 46.9787137... a(0)/a(1)=2/47; a(1)/a(2)=47/2207; a(2)/a(3)=2207/103682; a(3)/a(4)=103682/4870847; etc. Lim_{n->infinity} a(n)/a(n+1) = 0.02128623625... = 2/(47+sqrt(2205)) = (47-sqrt(2205))/2.
a(n) = a(-n). - Alois P. Heinz, Aug 07 2008
From Peter Bala, Oct 14 2019: (Start)
Let F(x) = Product_{n >= 0} (1 + x^(4*n+1))/(1 + x^(4*n+3)). Let Phi = 1/2*(sqrt(5) - 1). This sequence gives the partial denominators in the simple continued fraction expansion of the number F(Phi^8) = 1.0212763906... = 1 + 1/(47 + 1/(2207 + 1/(103682 + ...))).
Also F(-Phi^8) = 0.9787231991... has the continued fraction representation 1 - 1/(47 - 1/(2207 - 1/(103682 - ...))) and the simple continued fraction expansion 1/(1 + 1/((47 - 2) + 1/(1 + 1/((2207 - 2) + 1/(1 + 1/((103682 - 2) + 1/(1 + ...))))))).
F(Phi^8)*F(-Phi^8) = 0.9995468962... has the simple continued fraction expansion 1/(1 + 1/((47^2 - 4) + 1/(1 + 1/((2207^2 - 4) + 1/(1 + 1/((103682^2 - 4) + 1/(1 + ...))))))).
1/2 + 1/2*F(Phi^8)/F(-Phi^8) = 1.0217391349... has the simple continued fraction expansion 1 + 1/((47 - 2) + 1/(1 + 1/((103682 - 2) + 1/(1 + 1/(228826127 - 2) + 1/(1 + ...))))). (End)

			a(4) = 4870847 = 47*a(3) - a(2) = 47*103682 - 2207=((47+sqrt(2205))/2)^4 + ( (47-sqrt(2205))/2)^4 =4870846.999999794696 + 0.000000205303 = 4870847.

J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p. 91.
R. P. Stanley. Enumerative combinatorics. Vol. 2, volume 62 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge, 1999.

Indranil Ghosh, Table of n, a(n) for n = 0..596
Tanya Khovanova, Recursive Sequences
A. V. Zarelua, On Matrix Analogs of Fermat's Little Theorem, Mathematical Notes, vol. 79, no. 6, 2006, pp. 783-796. Translated from Matematicheskie Zametki, vol. 79, no. 6, 2006, pp. 840-855.
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)
Index entries for linear recurrences with constant coefficients, signature (47,-1).

Cf. A000032. Cf. Lucas(k*n): A005248 (k = 2), A014448 (k = 3), A056854 (k = 4), A001946 (k = 5), A087215 (k = 6), A087281 (k = 7), A087287 (k = 9), A065705 (k = 10), A089772 (k = 11), A089775 (k = 12).

a(n) = A000032(8n).

[ Lucas(8*n) : n in [0..100]]; // Vincenzo Librandi, Apr 14 2011

a:= n-> (Matrix([[2,47]]). Matrix([[47,1],[ -1,0]])^(n))[1,1]:
seq(a(n), n=0..14);  # Alois P. Heinz, Aug 07 2008

LucasL[8*Range[0,20]] (* or *) LinearRecurrence[{47,-1},{2,47},20] (* Harvey P. Dale, Oct 23 2017 *)

a(n) = 47*a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 47.
a(n) = ((47+sqrt(2205))/2)^n + ((47-sqrt(2205))/2)^n
(a(n))^2 = a(2n)+2.
G.f.: (2-47*x)/(1-47*x+x^2). - Alois P. Heinz, Aug 07 2008
From Peter Bala, Oct 14 2019: (Start)
a(n) = F(8*n+8)/F(8) - F(8*n-8)/F(8) = A049668(n+1) - A049668(n-1).
a(n) = trace(M^n), where M is the 2 X 2 matrix [0, 1; 1, 1]^8 = [13, 21; 21, 34].
Consequently the Gauss congruences hold: a(n*p^k) = a(n*p^(k-1)) ( mod p^k ) for all prime p and positive integers n and k. See Zarelua and also Stanley (Ch. 5, Ex. 5.2(a) and its solution).
45*Sum_{n >= 1} 1/(a(n) - 49/a(n)) = 1: (49 = Lucas(8) + 2 and 45 = Lucas(8) - 2)
49*Sum_{n >= 1} (-1)^(n+1)/(a(n) + 45/a(n)) = 1.
x*exp(Sum_{n >= 1} a(n)*x^/n) = x + 47*x^2 + 2208*x^3 + ... is the o.g.f. for A049668. (End)
E.g.f.: 2*exp(47*x/2)*cosh(21*sqrt(5)*x/2). - Stefano Spezia, Oct 18 2019
From Peter Bala, Apr 16 2025: (Start)
a(n) = Lucas(2*n)^4 - 4*Lucas(2*n)^2 + 2 = 2*T(4, (1/2)*Lucas(2*n)), where T(k, x) denotes the k-th Chebyshev polynomial of the first kind; more generally, for k >= 0, Lucas(2*k*n) = 2*T(k, Lucas(2*n)/2).
Sum_{n >= 1} 1/a(n) = (1/4) * (theta_3( (47 - sqrt(2205))/2 )^2 - 1) and
Sum_{n >= 1} (-1)^(n+1)/a(n) = (1/4) * (1 - theta_3( (sqrt(2205) - 47)/2 )^2),
where theta_3(x) = 1 + 2*Sum_{n >= 1} x^(n^2) (see A000122). See Borwein and Borwein, Proposition 3.5 (i), p. 91. Cf. A153415 and A003499. (End)

Terms a(22)-a(27) from John W. Layman, Jun 14 2004

A170915 Write 1 + sin x = Product_{n>=1} (1 + g_n * x^n); a(n) = denominator(g_n).

Original entry on oeis.org

1, 1, 6, 6, 120, 120, 5040, 280, 72576, 362880, 39916800, 11975040, 1245404160, 88957440, 1307674368000, 11675664000, 71137485619200, 1067062284288000, 121645100408832000, 101370917007360000, 10218188434341888000, 5109094217170944000, 25852016738884976640000

Offset: 1

N. J. A. Sloane, Jan 30 2010

From Petros Hadjicostas, Oct 06 2019: (Start)
The recurrence about (A(m,n): m,n >= 1) in the Formula section follows from Theorem 3 in Gingold et al. (1988); see also Gingold and Knopfmacher (1995, p. 1222). A(m=1,n) equals the n-th coefficient of the Taylor expansion of 1 + sin(x).
If 1 + sin(x) = 1/Product_{n>=1} (1 + f_n * x^n) (inverse power product expansion), then Gingold and Knopfmacher (1995) and Alkauskas (2008, 2009) proved that f_n = -g_n for n odd, and Sum_{s|n} (-g_{n/s})^s/s = -Sum_{s|n} (-f_{n/s})^s/s. [We caution that different authors may use -g_n for g_n, or -f_n for f_n, or both.] We have A328191(n) = numerator(f_n) and A328186(n) = denominator(f_n).
Wolfdieter Lang (see the link below) examined inverse power product expansions both for ordinary g.f.'s and for exponential g.f.'s.
In all cases, we assume the g.f.'s are unital, i.e., the g.f.'s start with a constant 1.
(End)

			g_n = 1, 0, -1/6, 1/6, -19/120, 19/120, -659/5040, 37/280, -7675/72576, ...

Giedrius Alkauskas, One curious proof of Fermat's little theorem, arXiv:0801.0805 [math.NT], 2008.
Giedrius Alkauskas, A curious proof of Fermat's little theorem, Amer. Math. Monthly 116(4) (2009), 362-364.
H. Gingold, H. W. Gould, and Michael E. Mays, Power Product Expansions, Utilitas Mathematica 34 (1988), 143-161.
H. Gingold and A. Knopfmacher, Analytic properties of power product expansions, Canad. J. Math. 47 (1995), 1219-1239.
W. Lang, Recurrences for the general problem, 2009.

Numerators are in A170914.

Cf. A170910, A170911, A170912, A170913, A328186, A328191.

# Calculates the fractions g_n (choose L much larger than M):
PPE_sin := proc(L, M)
local t1, t0, g, t2, n, t3;
if L < 2.5*M then print("Choose larger value for L");
else
t1 := 1 + sin(x);
t0 := series(t1, x, L);
g := []; t2 := t0;
for n to M do
t3 := coeff(t2, x, n);
t2 := series(t2/(1 + t3*x^n), x, L);
g := [op(g), t3];
end do;
end if;
[seq(g[n], n = 1 .. nops(g))];
end proc;
# Calculates the denominators of g_n:
h1 := map(denom, PPE_sin(100, 40)); # Petros Hadjicostas, Oct 06 2019 by modifying N. J. A. Sloane's program from A170912 and A170913.

A[m_, n_] :=
  A[m, n] =
   Which[m == 1, (1-(-1)^n)*(-1)^Floor[(n-1)/2]/(2*n!), m > n >= 1, 0, True,
    A[m - 1, n] - A[m - 1, m - 1]*A[m, n - m + 1]];
a[n_] := Denominator[A[n, n]];
a /@ Range[1, 55] (* Petros Hadjicostas, Oct 06 2019, courtesy of Jean-François Alcover *)

From Petros Hadjicostas, Oct 07 2019: (Start)
a(2*n+1) = A328186(2*n+1) for n >= 0.
Define (A(m,n): n,m >= 1) by A(m=1,2*n+1) = (-1)^n/(2*n+1)! for n >= 0, A(m=1,2*n) = 0 for n >= 1, A(m,n) = 0 for m > n >= 1 (upper triangular), and A(m,n) = A(m-1,n) - A(m-1,m-1) * A(m,n-m+1) for n >= m >= 2. Then g_n = A(n,n). (End)

A174275 a(n) = 2^(n-1) mod M(n) where M(n) = A014963(n) is the exponential of the Mangoldt function.

Original entry on oeis.org

0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0

Offset: 1

Mats Granvik, Mar 14 2010

Appears to be always either 0 or 1.
This follows from Fermat's Little Theorem. - Charles R Greathouse IV, Feb 13 2011
Characteristic function for odd prime powers (larger than one). - Antti Karttunen, Sep 14 2017, after Charles R Greathouse IV's Feb 13 2011 formula.

Antti Karttunen, Table of n, a(n) for n = 1..16385
Index entries for characteristic functions

Cf. A062173.

a[n_] := Mod[2^(n - 1), Exp[MangoldtLambda[n]]] (* Steven Foster Clark, Sep 04 2023 *)

vector(70, n, ispower(k=n, , &k); isprime(k)&k!=2) \\ Charles R Greathouse IV, Feb 13 2011

a(n) = A000079(n-1) mod A014963(n).

a(n) = 1 if n = p^k for k > 0 and p a prime not equal to 2, a(n) = 0 otherwise. - Charles R Greathouse IV, Feb 13 2011

More terms from Antti Karttunen, Sep 14 2017

Name corrected by Steven Foster Clark, Sep 05 2023

A170914 Write 1 + sin x = Product_{n>=1} (1 + g_n * x^n); a(n) = numerator(g_n).

Original entry on oeis.org

1, 0, -1, 1, -19, 19, -659, 37, -7675, 40043, -3578279, 1123009, -95259767, 7091713, -85215100151, 832857559, -4180679675171, 63804880881241, -6399968826052559, 5697831990097981, -478887035449041839, 252737248941887573, -1123931378903214542099, 35703551772944759

Offset: 1

N. J. A. Sloane, Jan 30 2010

From Petros Hadjicostas, Oct 06 2019: (Start)
The recurrence about (A(m,n): m,n >= 1) in the Formula section follows from Theorem 3 in Gingold et al. (1988); see also Gingold and Knopfmacher (1995, p. 1222). A(m=1,n) equals the n-th coefficient of the Taylor expansion of 1 + sin(x).
If 1 + sin(x) = 1/Product_{n>=1} (1 + f_n * x^n) (inverse power product expansion), then Gingold and Knopfmacher (1995) and Alkauskas (2008, 2009) proved that f_n = -g_n for n odd, and Sum_{s|n} (-g_{n/s})^s/s = -Sum_{s|n} (-f_{n/s})^s/s. [We caution that different authors may use -g_n for g_n, or -f_n for f_n, or both.] We have A328191(n) = numerator(f_n) and A328186(n) = denominator(f_n).
Wolfdieter Lang (see the link below) examined inverse power product expansions both for ordinary g.f.'s and for exponential g.f.'s.
In all cases, we assume the g.f.'s are unital, i.e., the g.f.'s start with a constant 1.
(End)

			g_n = 1, 0, -1/6, 1/6, -19/120, 19/120, -659/5040, 37/280, -7675/72576, ...

Giedrius Alkauskas, One curious proof of Fermat's little theorem, arXiv:0801.0805 [math.NT], 2008.
Giedrius Alkauskas, A curious proof of Fermat's little theorem, Amer. Math. Monthly 116(4) (2009), 362-364.
H. Gingold, H. W. Gould, and Michael E. Mays, Power Product Expansions, Utilitas Mathematica 34 (1988), 143-161.
H. Gingold and A. Knopfmacher, Analytic properties of power product expansions, Canad. J. Math. 47 (1995), 1219-1239.
W. Lang, Recurrences for the general problem, 2009.

Cf. Denominators are in A170915.

Cf. A170910, A170911, A170912, A170913, A328186, A328191.

# Calculates the fractions g_n (choose L much larger than M):
PPE_sin := proc(L, M)
local t1, t0, g, t2, n, t3;
if L < 2.5*M then print("Choose larger value for L");
else
t1 := 1 + sin(x);
t0 := series(t1, x, L);
g := []; t2 := t0;
for n to M do
t3 := coeff(t2, x, n);
t2 := series(t2/(1 + t3*x^n), x, L);
g := [op(g), t3];
end do;
end if;
[seq(g[n], n = 1 .. nops(g))];
end proc;
# Calculates the numerators of g_n:
h1 := map(numer, PPE_sin(100, 40)); # Petros Hadjicostas, Oct 06 2019 by modifying N. J. A. Sloane's program from A170912 and A170913.

A[m_, n_] :=
  A[m, n] =
   Which[m == 1, (1-(-1)^n)*(-1)^Floor[(n-1)/2]/(2*n!), m > n >= 1, 0, True,
    A[m - 1, n] - A[m - 1, m - 1]*A[m, n - m + 1]];
a[n_] := Numerator[A[n, n]];
a /@ Range[1, 55] (* Petros Hadjicostas, Oct 06 2019, courtesy of Jean-François Alcover *)

From Petros Hadjicostas, Oct 07 2019: (Start)
a(2*n+1) = -A328191(2*n+1) for n >= 0.
Define (A(m,n): n,m >= 1) by A(m=1,2*n+1) = (-1)^n/(2*n+1)! for n >= 0, A(m=1,2*n) = 0 for n >= 1, A(m,n) = 0 for m > n >= 1 (upper triangular), and A(m,n) = A(m-1,n) - A(m-1,m-1) * A(m,n-m+1) for n >= m >= 2. Then g_n = A(n,n). (End)

A204187 a(n) = Sum_{m=1..n-1} m^(n-1) modulo n.

Original entry on oeis.org

0, 1, 2, 0, 4, 3, 6, 0, 6, 5, 10, 0, 12, 7, 10, 0, 16, 9, 18, 0, 14, 11, 22, 0, 20, 13, 18, 0, 28, 15, 30, 0, 22, 17, 0, 0, 36, 19, 26, 0, 40, 21, 42, 0, 21, 23, 46, 0, 42, 25, 34, 0, 52, 27, 0, 0, 38, 29, 58, 0, 60, 31, 42, 0, 52, 33, 66, 0, 46, 35, 70, 0

Offset: 1

Jonathan Sondow, Jan 12 2012

a(n) = n - 1 if n is 1 or a prime, by Fermat's little theorem. It is conjectured that the converse is also true; see A055032 and A201560 and note that a(n) = n-1 <==> A055032(n) = 1 <==> A201560(n) = 0.

As of 1991, Giuga and Bedocchi had verified no composite n < 10^1700 satisfies a(n) = n - 1 (Ribemboim, 1991). - Alonso del Arte, May 10 2013

			Sum(m^3, m = 1 .. 3) = 1^3 + 2^3 + 3^3 = 36 == 0 (mod 4), so a(4) = 0.

Steve Dinh, The Hard Mathematical Olympiad Problems And Their Solutions, AuthorHouse, 2011, Problem 6 of Hong Kong Mathematical Olympiad 2007 (find a(7)), page 134.
Richard K. Guy, Unsolved Problems in Number Theory, A17.
Paulo Ribemboim, The Little Book of Big Primes. New York: Springer-Verlag (1991): 17.

Ivan Neretin, Table of n, a(n) for n = 1..10000
John Clark, On a conjecture involving Fermat's Little Theorem, Thesis, 2008, University of South Florida.
Hong Kong Mathematics Olympiad (2007-2008), Final Event 2 (Group), problem 2, p. 437.
Bernard Schott, Proof that a(n)=n/2 if n is of the form 4k+2.
Index to sequences related to Olympiads.

Cf. A055023, A055030, A055031, A055032, A201560.

Cf. A191677 (zeros).

Table[Mod[Sum[i^(n - 1), {i, n - 1}], n], {n, 75}] (* Alonso del Arte, May 10 2013 *)

a(n) = lift(sum(i=1, n, Mod(i, n)^(n-1))); \\ Michel Marcus, Feb 23 2020

def a(n): return sum(pow(m, n-1, n) for m in range(1, n))%n
print([a(n) for n in range(1, 73)]) # Michael S. Branicky, Jan 02 2022

a(p) = p - 1 if p is prime, and a(4n) = 0.
a(n) + 1 == A201560(n) (mod n).
a(n) = n/2 iff n is of the form 4k+2 (conjectured). - Ivan Neretin, Sep 23 2016
a(4*k+2) = 2*k+1; for a proof see corresponding link. - Bernard Schott, Dec 29 2021

A208536 Number of 5-bead necklaces of n colors not allowing reversal, with no adjacent beads having the same color.

Original entry on oeis.org

0, 0, 6, 48, 204, 624, 1554, 3360, 6552, 11808, 19998, 32208, 49764, 74256, 107562, 151872, 209712, 283968, 377910, 495216, 639996, 816816, 1030722, 1287264, 1592520, 1953120, 2376270, 2869776, 3442068, 4102224, 4859994, 5725824, 6710880

Offset: 1

R. H. Hardin, Feb 27 2012

This sequence would be better defined as a(n) = (n^5-n)/5 with offset 0, which is an integer by Fermat's little theorem. - N. J. A. Sloane, Nov 13 2023

			All solutions for n=3:
..1....1....1....1....1....1
..3....3....2....2....2....2
..1....2....1....3....3....1
..3....3....3....2....1....2
..2....2....2....3....3....3

R. H. Hardin, Table of n, a(n) for n = 1..210
Jack Jeffries, Differentiating by prime numbers, Notices Amer. Math. Soc., 70:11 (2023), 1772-1779.
Wikipedia, p-derivation.
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Row 5 of A208535.
Also, row 5 (with different offset) of A074650. - Eric M. Schmidt, Dec 08 2017
Cf. A208537.

A208536[n_]:=((n-1)^5-(n-1))/5;Array[A208536,50] (* Paolo Xausa, Nov 14 2023 *)

Vec(6*x^3*(1 + x)^2 / (1 - x)^6 + O(x^40)) \\ Colin Barker, Nov 11 2017

Empirical: a(n) = (1/5)*n^5 - 1*n^4 + 2*n^3 - 2*n^2 + (4/5)*n.
Equivalently: a(n) = ((n-1)^5 - (n-1))/5. - M. F. Hasler, Mar 05 2016
Empirical formula confirmed by Petros Hadjicostas, Nov 05 2017 (see A208535).
a(n+2) = delta(-n) = -delta(n) for n >= 0, where delta is the p-derivation over the integers with respect to prime p = 5. - Danny Rorabaugh, Nov 10 2017
From Colin Barker, Nov 11 2017: (Start)
G.f.: 6*x^3*(1 + x)^2 / (1 - x)^6.
a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6) for n>6.
(End)

A208537 Number of 7-bead necklaces of n colors not allowing reversal, with no adjacent beads having the same color.

Original entry on oeis.org

0, 0, 18, 312, 2340, 11160, 39990, 117648, 299592, 683280, 1428570, 2783880, 5118828, 8964072, 15059070, 24408480, 38347920, 58619808, 87460002, 127695960, 182857140, 257298360, 356336838, 486403632, 655210200, 871930800, 1147401450

Offset: 1

R. H. Hardin, Feb 27 2012

This sequence would be better defined as a(n) = (n^7-n)/7 with offset 0, which is an integer by Fermat's little theorem. - N. J. A. Sloane, Nov 13 2023
Row 7 of A208535.
Also, row 7 (with different offset) of A074650. - Eric M. Schmidt, Dec 08 2017

			All solutions for n=3:
..1...1...1...1...1...1...1...1...1...1...1...1...1...1...1...1...1...1
..2...2...2...3...2...2...2...2...2...3...3...3...2...2...2...2...2...2
..1...1...1...1...1...1...3...3...3...2...2...1...3...1...3...1...3...1
..2...3...2...3...3...3...2...1...2...3...1...3...2...2...1...3...1...2
..3...2...1...2...1...2...1...3...3...2...3...1...3...3...3...1...2...1
..1...3...3...3...2...1...3...1...2...3...2...3...1...2...2...3...3...2
..3...2...2...2...3...3...2...3...3...2...3...2...3...3...3...2...2...3

J. Jeffries, Differentiating by prime numbers, Notices Amer. Math. Soc., 70:11 (2023), 1772-1779.

R. H. Hardin, Table of n, a(n) for n = 1..210
Wikipedia, p-derivation.
Index entries for linear recurrences with constant coefficients, signature (8,-28,56,-70,56,-28,8,-1).

Cf. A208535.

A208537[n_]:=((n-1)^7-(n-1))/7;Array[A208537,50] (* Paolo Xausa, Nov 14 2023 *)

Vec(6*x^3*(3 + 28*x + 58*x^2 + 28*x^3 + 3*x^4) / (1 - x)^8 + O(x^40)) \\ Colin Barker, Nov 11 2017

Empirical: a(n) = (1/7)*n^7 - 1*n^6 + 3*n^5 - 5*n^4 + 5*n^3 - 3*n^2 + (6/7)*n.
Empirical formula confirmed by Petros Hadjicostas, Nov 05 2017 (see A208535).
a(n+2) = delta(-n) = -delta(n) for n >= 0, where delta is the p-derivation over the integers with respect to prime p = 7. - Danny Rorabaugh, Nov 10 2017
From Colin Barker, Nov 11 2017: (Start)
G.f.: 6*x^3*(3 + 28*x + 58*x^2 + 28*x^3 + 3*x^4) / (1 - x)^8.
a(n) = 8*a(n-1) - 28*a(n-2) + 56*a(n-3) - 70*a(n-4) + 56*a(n-5) - 28*a(n-6) + 8*a(n-7) - a(n-8) for n>8.
(End)
a(n) = ((n-1)^7 - (n-1))/7. (inspired by Hassler's formula in A208536) - Eric M. Schmidt, Dec 08 2017

A232210 Let b_k=3...3 consist of k>=1 3's. Then a(n) is the smallest k such that the concatenation prime(n)b_k is prime, or a(n)=0 if there is no such prime.

Original entry on oeis.org

1, 0, 1, 1, 1, 14, 1, 1, 1, 1, 1, 1, 2, 1, 2, 2, 1, 1, 1, 3, 1, 2, 6, 2, 2, 1, 1, 2, 1, 4, 4, 23, 1, 2, 1, 6, 2, 2, 5, 1, 10, 2, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 2, 4, 2, 1, 1, 1, 2, 4, 1, 2, 5, 4, 2, 3, 1, 1, 5, 4, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 6, 4, 2, 14, 2, 4, 1, 3

Offset: 1

Vladimir Shevelev, Sep 13 2014

Conjecture: for n>=3, a(n)>0.
Records are 1,14,23,50,252,4752,...
The corresponding primes are 2,13,131,653,883,1279,...
These primes beginning with the second one we call "stubborn primes".
Counter-conjecture: a(2889)=0. - Hans Havermann, Oct 15 2014
If a(n)=1, then the resulting primes are in A092993 and form A055782; if a(n)=2, then they form sequence 4133,4733,5333,7933,..., etc. - Vladimir Shevelev, Oct 16 2014
If a prime p divides Pb_k, then it also divides Pb_{k+m(p-1)} for all m>=0. This follows from Fermat's little theorem applied to b_x=(10^x-1)/3 with x=p-1. - M. F. Hasler, Oct 20 2014

			For n=1, start with prime(1)=2 and get already at the first step the prime 23. So a(1)=1.
For n=2, starting with prime(2)=3, one never gets a prime by appending further digits "3", therefore a(2)=0.
For n=3, n=4, n=5, one gets after the first step the primes 53, 73, 113, and therefore a(n)=1.
For n=6, start with prime(6)=13; one has to append 14 "3"s in order to get a new prime, so a(6)=14.
For n=2889, start with prime(2889) = 26293. (Do not mix up with prime(2899) = 26393...!) Appending 2k-1 or 6k-4 or 6k-2 or 18k-6 or 36k-18 or 180k-144 digits "3" yields a number divisible by 11 resp. 7 resp. 13 resp. 19 resp. 101 resp. 31. For 18k-12 and 36k (with k <> 1 (mod 5)) digits "3" there is no simple pattern and both yield sometimes large primes in the factorization, but (so far) always composite numbers 26293...3 (up to several thousand digits). - _M. F. Hasler_, Oct 16 2014

Hans Havermann, Table of n, a(n) for n = 1..2888 (first 200 terms from Michel Marcus)
Hans Havermann, "Google+ discussion relating to this sequence"
Vladimir Shevelev, "Stubborn primes"
Robert G. Wilson v, Table of n, a(n) for n = 1..10000 with -1 for those entries where a(n) has not yet been found

Cf. A055782, A092993, A242775, A247341, A247342, A248919.

f[n_] := Block[{k = 1, p = Prime@ n}, While[ !PrimeQ[p*10^k + (10^k - 1)/3], k++]; k]; f[2] = 0; Array[f, 100] (* Robert G. Wilson v, Apr 24 2015 *)
m3[n_]:=Module[{k=10n+3},While[!PrimeQ[k],k=10k+3];IntegerLength[k]-IntegerLength[ n]]; Join[{1,0},m3/@Prime[Range[3,90]]] (* Harvey P. Dale, Feb 11 2018 *)

a(n) = {if (n==2, return (0)); p = prime(n); k = 1; while (! isprime(p = p*10+3), k++); k;} \\ Michel Marcus, Sep 13 2014

More terms from Peter J. C. Moses, Sep 13 2014

cp's OEIS Frontend

A062481 a(n) = n^prime(n).

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A083737 Pseudoprimes to bases 2, 3 and 5.

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A087265 Lucas numbers L(8*n).

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A170915 Write 1 + sin x = Product_{n>=1} (1 + g_n * x^n); a(n) = denominator(g_n).

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A174275 a(n) = 2^(n-1) mod M(n) where M(n) = A014963(n) is the exponential of the Mangoldt function.

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A170914 Write 1 + sin x = Product_{n>=1} (1 + g_n * x^n); a(n) = numerator(g_n).

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A204187 a(n) = Sum_{m=1..n-1} m^(n-1) modulo n.

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