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

A003500 a(n) = 4*a(n-1) - a(n-2) with a(0) = 2, a(1) = 4.

2, 4, 14, 52, 194, 724, 2702, 10084, 37634, 140452, 524174, 1956244, 7300802, 27246964, 101687054, 379501252, 1416317954, 5285770564, 19726764302, 73621286644, 274758382274, 1025412242452, 3826890587534, 14282150107684, 53301709843202, 198924689265124

Offset: 0

N. J. A. Sloane

a(n) gives values of x satisfying x^2 - 3*y^2 = 4; corresponding y values are given by 2*A001353(n).
If M is any given term of the sequence, then the next one is 2*M + sqrt(3*M^2 - 12). - Lekraj Beedassy, Feb 18 2002
For n > 0, the three numbers a(n) - 1, a(n), and a(n) + 1 form a Fleenor-Heronian triangle, i.e., a Heronian triangle with consecutive sides, whose area A(n) may be obtained from the relation [4*A(n)]^2 = 3([a(2n)]^2 - 4); or A(n) = 3*A001353(2*n)/2 and whose semiperimeter is 3*a[n]/2. The sequence is symmetrical about a[0], i.e., a[-n] = a[n].
For n > 0, a(n) + 2 is the number of dimer tilings of a 2*n X 2 Klein bottle (cf. A103999).
Tsumura shows that, for prime p, a(p) is composite (contrary to a conjecture of Juricevic). - Charles R Greathouse IV, Apr 13 2010
Except for the first term, positive values of x (or y) satisfying x^2 - 4*x*y + y^2 + 12 = 0. - Colin Barker, Feb 04 2014
Except for the first term, positive values of x (or y) satisfying x^2 - 14*x*y + y^2 + 192 = 0. - Colin Barker, Feb 16 2014
A268281(n) - 1 is a member of this sequence iff A268281(n) is prime. - Frank M Jackson, Feb 27 2016
a(n) gives values of x satisfying 3*x^2 - 4*y^2 = 12; corresponding y values are given by A005320. - Sture Sjöstedt, Dec 19 2017
Middle side lengths of almost-equilateral Heronian triangles. - Wesley Ivan Hurt, May 20 2020
For all elements k of the sequence, 3*(k-2)*(k+2) is a square. - Davide Rotondo, Oct 25 2020

B. C. Berndt, Ramanujan's Notebooks Part IV, Springer-Verlag, see p. 82.
J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p.91.
Michael P. Cohen, Generating Heronian Triangles With Consecutive Integer Sides. Journal of Recreational Mathematics, vol. 30 no. 2 1999-2000 p. 123.
L. E. Dickson, History of The Theory of Numbers, Vol. 2 pp. 197;198;200;201. Chelsea NY.
Charles R. Fleenor, Heronian Triangles with Consecutive Integer Sides, Journal of Recreational Mathematics, Volume 28, no. 2 (1996-7) 113-115.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
R. P. Stanley. Enumerative combinatorics. Vol. 2, volume 62 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge, 1999.
V. D. To, "Finding All Fleenor-Heronian Triangles", Journal of Recreational Mathematics vol. 32 no.4 2003-4 pp. 298-301 Baywood NY.

T. D. Noe, Table of n, a(n) for n=0..200
P. Bala, Some simple continued fraction expansions for an infinite product, Part 1
R. A. Beauregard and E. R. Suryanarayan, The Brahmagupta Triangles, The College Mathematics Journal 29(1) 13-7 1998 MAA.
Hacène Belbachir, Soumeya Merwa Tebtoub and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
Daniel Birmajer, Juan B. Gil and Michael D. Weiner, Linear recurrence sequences with indices in arithmetic progression and their sums, arXiv preprint arXiv:1505.06339 [math.NT], 2015.
K. S. Brown's Mathpages, Some Properties of the Lucas Sequence(2, 4, 14, 52, 194, ...)
H. W. Gould, A triangle with integral sides and area, Fib. Quart., 11 (1973), 27-39.
Tanya Khovanova, Recursive Sequences
E. Keith Lloyd, The Standard Deviation of 1, 2, ..., n: Pell's Equation and Rational Triangles, Math. Gaz. vol 81 (1997), 231-243.
S. Northshield, An Analogue of Stern's Sequence for Z[sqrt(2)], Journal of Integer Sequences, 18 (2015), #15.11.6.
Hideyuki Ohtskua, proposer, Problem B-1351, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 62, No. 3 (2024), p. 258.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Jeffrey Shallit, An interesting continued fraction, Math. Mag., 48 (1975), 207-211.
Jeffrey Shallit, An interesting continued fraction, Math. Mag., 48 (1975), 207-211. [Annotated scanned copy]
Yu Tsumura, On compositeness of special types of integers, arXiv:1004.1244 [math.NT], 2010.
Eric Weisstein's World of Mathematics, Heronian Triangle
Wikipedia, Heronian triangle
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 (4,-1).

Cf. A001075, A001353, A001835.
Cf. A011945 (areas), A334277 (perimeters).
Cf. this sequence (middle side lengths), A016064 (smallest side lengths), A335025 (largest side lengths).
Cf. A001570, A002530, A005320, A006051, A048788, A174500, A268281.
Cf. A011943, A102344, A019673, A105531.

a003500 n = a003500_list !! n
a003500_list = 2 : 4 : zipWith (-)
   (map (* 4) $ tail a003500_list) a003500_list
-- Reinhard Zumkeller, Dec 17 2011

I:=[2,4]; [n le 2 select I[n] else 4*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Nov 14 2018

A003500 := proc(n) option remember; if n <= 1 then 2*n+2 else 4*procname(n-1)-procname(n-2); fi;
end proc;

a[0]=2; a[1]=4; a[n_]:= a[n]= 4a[n-1] -a[n-2]; Table[a[n], {n, 0, 23}]
LinearRecurrence[{4,-1},{2,4},30] (* Harvey P. Dale, Aug 20 2011 *)
Table[Round@LucasL[2n, Sqrt[2]], {n, 0, 20}] (* Vladimir Reshetnikov, Sep 15 2016 *)

x='x+O('x^99); Vec(-2*(-1+2*x)/(1-4*x+x^2)) \\ Altug Alkan, Apr 04 2016

[lucas_number2(n,4,1) for n in range(0, 24)] # Zerinvary Lajos, May 14 2009

a(n) = ( 2 + sqrt(3) )^n + ( 2 - sqrt(3) )^n.
a(n) = 2*A001075(n).
G.f.: 2*(1 - 2*x)/(1 - 4*x + x^2). Simon Plouffe in his 1992 dissertation.
a(n) = A001835(n) + A001835(n+1).
a(n) = trace of n-th power of the 2 X 2 matrix [1 2 / 1 3]. - Gary W. Adamson, Jun 30 2003 [corrected by Joerg Arndt, Jun 18 2020]
From the addition formula, a(n+m) = a(n)*a(m) - a(m-n), it is easy to derive multiplication formulas, such as: a(2*n) = (a(n))^2 - 2, a(3*n) = (a(n))^3 - 3*(a(n)), a(4*n) = (a(n))^4 - 4*(a(n))^2 + 2, a(5*n) = (a(n))^5 - 5*(a(n))^3 + 5*(a(n)), a(6*n) = (a(n))^6 - 6*(a(n))^4 + 9*(a(n))^2 - 2, etc. The absolute values of the coefficients in the expansions are given by the triangle A034807. - John Blythe Dobson, Nov 04 2007
a(n) = 2*A001353(n+1) - 4*A001353(n). - R. J. Mathar, Nov 16 2007
From Peter Bala, Jan 06 2013: (Start)
Let F(x) = Product_{n=0..infinity} (1 + x^(4*n + 1))/(1 + x^(4*n + 3)). Let alpha = 2 - sqrt(3). This sequence gives the simple continued fraction expansion of 1 + F(alpha) = 2.24561 99455 06551 88869 ... = 2 + 1/(4 + 1/(14 + 1/(52 + ...))). Cf. A174500.
Also F(-alpha) = 0.74544 81786 39692 68884 ... has the continued fraction representation 1 - 1/(4 - 1/(14 - 1/(52 - ...))) and the simple continued fraction expansion 1/(1 + 1/((4 - 2) + 1/(1 + 1/((14 - 2) + 1/(1 + 1/((52 - 2) + 1/(1 + ...))))))).
F(alpha)*F(-alpha) has the simple continued fraction expansion 1/(1 + 1/((4^2 - 4) + 1/(1 + 1/((14^2 - 4) + 1/(1 + 1/((52^2 - 4) + 1/(1 + ...))))))).
(End)
a(2^n) = A003010(n). - John Blythe Dobson, Mar 10 2014
a(n) = [x^n] ( (1 + 4*x + sqrt(1 + 8*x + 12*x^2))/2 )^n for n >= 1. - Peter Bala, Jun 23 2015
E.g.f.: 2*exp(2*x)*cosh(sqrt(3)*x). - Ilya Gutkovskiy, Apr 27 2016
a(n) = Sum_{k=0..floor(n/2)} (-1)^k*n*(n - k - 1)!/(k!*(n - 2*k)!)*4^(n - 2*k) for n >= 1. - Peter Luschny, May 10 2016
From Peter Bala, Oct 15 2019: (Start)
a(n) = trace(M^n), where M is the 2 X 2 matrix [0, 1; -1, 4].
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).
2*Sum_{n >= 1} 1/( a(n) - 6/a(n) ) = 1.
6*Sum_{n >= 1} (-1)^(n+1)/( a(n) + 2/a(n) ) = 1.
8*Sum_{n >= 1} 1/( a(n) + 24/(a(n) - 12/(a(n))) ) = 1.
8*Sum_{n >= 1} (-1)^(n+1)/( a(n) + 8/(a(n) + 4/(a(n))) ) = 1.
Series acceleration formulas for sums of reciprocals:
Sum_{n >= 1} 1/a(n) = 1/2 - 6*Sum_{n >= 1} 1/(a(n)*(a(n)^2 - 6)),
Sum_{n >= 1} 1/a(n) = 1/8 + 24*Sum_{n >= 1} 1/(a(n)*(a(n)^2 + 12)),
Sum_{n >= 1} (-1)^(n+1)/a(n) = 1/6 + 2*Sum_{n >= 1} (-1)^(n+1)/(a(n)*(a(n)^2 + 2)) and
Sum_{n >= 1} (-1)^(n+1)/a(n) = 1/8 + 8*Sum_{n >= 1} (-1)^(n+1)/(a(n)*(a(n)^2 + 12)).
Sum_{n >= 1} 1/a(n) = ( theta_3(2-sqrt(3))^2 - 1 )/4 = 0.34770 07561 66992 06261 .... See Borwein and Borwein, Proposition 3.5 (i), p.91.
Sum_{n >= 1} (-1)^(n+1)/a(n) = ( 1 - theta_3(sqrt(3)-2)^2 )/4. Cf. A003499 and A153415. (End)
a(n) = tan(Pi/12)^n + tan(5*Pi/12)^n. - Greg Dresden, Oct 01 2020
From Wolfdieter Lang, Sep 06 2021: (Start)
a(n) = S(n, 4) - S(n-2, 4) = 2*T(n, 2), for n >= 0, with S and T Chebyshev polynomials, with S(-1, x) = 0 and S(-2, x) = -1. S(n, 4) = A001353(n+1), for n >= -1, and T(n, 2) = A001075(n).
a(2*k) = A067902(k), a(2*k+1) = 4*A001570(k+1), for k >= 0. (End)
a(n) = sqrt(2 + 2*A011943(n+1)) = sqrt(2 + 2*A102344(n+1)), n>0. - Ralf Steiner, Sep 23 2021
Sum_{n>=1} arctan(3/a(n)^2) = Pi/6 - arctan(1/3) = A019673 - A105531 (Ohtskua, 2024). - Amiram Eldar, Aug 29 2024

More terms from James Sellers, May 03 2000

Additional comments from Lekraj Beedassy, Feb 14 2002

A051783 Numbers k such that 3^k + 2 is prime.

Original entry on oeis.org

0, 1, 2, 3, 4, 8, 10, 14, 15, 24, 26, 36, 63, 98, 110, 123, 126, 139, 235, 243, 315, 363, 386, 391, 494, 1131, 1220, 1503, 1858, 4346, 6958, 7203, 10988, 22316, 33508, 43791, 45535, 61840, 95504, 101404, 106143, 107450, 136244, 178428, 361608, 504206, 1753088

Offset: 1

Jud McCranie, Dec 09 1999

From Farideh Firoozbakht and M. F. Hasler, Dec 06 2009: (Start)
If Q is a perfect number such that gcd(Q, 3(3^a(n) + 2)) = 1, then x = 3^(a(n) - 1)*(3^a(n) + 2)*Q is a solution of the equation sigma(x) = 3(x - Q). This is a result of the following theorem:
Theorem: If Q is a (q-1)-perfect number for some prime q, then for all integers t, the equation sigma(x) = q*x - (2t+1)*Q has the solution x = q^(k-1)*p*Q whenever k is a positive integer such that p = q^k + 2t is prime, gcd(q^(k-1), p) = 1 and gcd(q^(k-1)*p,Q) = 1.
Note that by taking t = -1/2(m*q+1), this theorem gives us some solutions of the equation sigma(x) = q *(x + m*Q). See comment lines of the sequence A058959. (End)
No further terms < 200000. - Ray Chandler, Jul 31 2011
A090649 implies that 361608 is a member of this sequence. - Robert Price, Aug 18 2014
No further terms < 320000. - Luke W. Richards, Mar 04 2018
a(45) and a(46) are probable primes because a primality certificate has not yet been found. They have been verified PRP with mprime. - Luke W. Richards, May 04 2018
No further terms < 1300000. - Luke W. Richards, May 17 2018
No further terms < 1400000. - Luke W. Richards, Jul 28 2020
Conjecture: The number n = 3^k + 2 is prime if and only if 2^((n-1)/2) == -1 (mod n). - Maheswara Rao Valluri, Jun 01 2020. [Note that this is an if and only if assertion, so it does not follow from Fermat's Little Theorem. - N. J. A. Sloane, Sep 07 2020]

			3^8 + 2 = 6563 is prime, so 8 is in the sequence.
3^26 + 2 = 2541865828331, a prime number, so 26 is in the sequence.

F. Firoozbakht and M. F. Hasler, Variations on Euclid's formula for Perfect Numbers, JIS 13 (2010) #10.3.1.
Henri and Renaud Lifchitz, PRP Records.

Cf. A057735, A087885, A014224, A058959.

A051783 = Select[Range[0, 20000], PrimeQ[3^# + 2] &]

is(n)=ispseudoprime(3^n+2) \\ Charles R Greathouse IV, Mar 21 2013

{4346, 6958, 7203} from David J. Rusin, Sep 29 2000
10988 from Ray Chandler, Nov 21 2004
{22316, 33508} found by Henri Lifchitz, Sep-Oct 2002
{43791, 45535, 61840} found by Henri Lifchitz, Oct-Nov 2004
95504 found by Wojciech Florek Dec 15 2005. - Alexander Adamchuk, Mar 02 2008
Edited by N. J. A. Sloane, Dec 19 2009
{101404, 106143, 107450, 136244} from Mike Oakes, Nov 2009
178428 from Ray Chandler, Jul 29 2011
a(45)-a(46) from Luke W. Richards, May 04 2018
a(47) from Paul Bourdelais, Mar 29 2022

A005935 Pseudoprimes to base 3.

Original entry on oeis.org

91, 121, 286, 671, 703, 949, 1105, 1541, 1729, 1891, 2465, 2665, 2701, 2821, 3281, 3367, 3751, 4961, 5551, 6601, 7381, 8401, 8911, 10585, 11011, 12403, 14383, 15203, 15457, 15841, 16471, 16531, 18721, 19345, 23521, 24046, 24661, 24727, 28009, 29161

Offset: 1

N. J. A. Sloane

nonn

Theorem: If q>3 and both numbers q and (2q-1) are primes then n=q*(2q-1) is a pseudoprime to base 3 (i.e. n is in the sequence). So for n>2, A005382(n)*(2*A005382(n)-1) is in the sequence (see Comments lines for the sequence A122780). 91,703,1891,2701,12403,18721,38503,49141... are such terms. This sequence is a subsequence of A122780. - Farideh Firoozbakht, Sep 13 2006
Composite numbers n such that 3^(n-1) == 1 (mod n).
Theorem (R. Steuerwald, 1948): if n is a pseudoprime to base b and gcd(n,b-1)=1, then (b^n-1)/(b-1) is a pseudoprime to base b. In particular, if n is an odd pseudoprime to base 3, then (3^n-1)/2 is a pseudoprime to base 3. - Thomas Ordowski, Apr 06 2016
Steuerwald's theorem can be strengthened by weakening his assumption as follows: if n is a weak pseudoprime to base b and gcd(n,b-1)=1, then ... - Thomas Ordowski, Feb 23 2021

J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 91, p. 33, Ellipses, Paris 2008.
R. K. Guy, Unsolved Problems in Number Theory, A12.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

R. J. Mathar, T. D. Noe and Hiroaki Yamanouchi, Table of n, a(n) for n = 1..102839 (terms a(1)-a(798) from R. J. Mathar, a(799)-a(1000) from T. D. Noe)
J. Bernheiden, Pseudoprimes (Text in German)
C. Pomerance & N. J. A. Sloane, Correspondence, 1991
F. Richman, Primality testing with Fermat's little theorem
Carlos Rivera, Conjecture 105: conjecture about twin primes and pseudo twin primes to base 3, The Prime Puzzles & Problems Connection.
Rudolf Steuerwald, Über die Kongruenz a^(n-1) == 1 (mod n), Sitzungsber. math.-naturw. Kl. Bayer. Akad. Wiss. München, 1948, pp. 69-70.
Eric Weisstein's World of Mathematics, Fermat Pseudoprime
Index entries for sequences related to pseudoprimes

Pseudoprimes to other bases: A001567 (2), A005936 (5), A005937 (6), A005938 (7), A005939 (10).
Subsequence of A122780.
Cf. A005382.

base = 3; t = {}; n = 1; While[Length[t] < 100, n++; If[! PrimeQ[n] && PowerMod[base, n-1, n] == 1, AppendTo[t, n]]]; t (* T. D. Noe, Feb 21 2012 *)

is_A005935(n)={Mod(3,n)^(n-1)==1 & !ispseudoprime(n) & n>1}  \\ M. F. Hasler, Jul 19 2012

More terms from David W. Wilson, Aug 15 1996

A014117 Numbers n such that m^(n+1) == m (mod n) holds for all m.

Original entry on oeis.org

1, 2, 6, 42, 1806

Offset: 1

David Broadhurst

"Somebody incorrectly remembered Fermat's little theorem as saying that the congruence a^{n+1} = a (mod n) holds for all a if n is prime" (Zagier). The sequence gives the set of integers n for which this property is in fact true.
If i == j (mod n), then m^i == m^j (mod n) for all m. The latter congruence generally holds for any (m, n)=1 with i == j (mod k), k being the order of m modulo n, i.e., the least power k for which m^k == 1 (mod n). - Lekraj Beedassy, Jul 04 2002
Also, numbers n such that n divides denominator of the n-th Bernoulli number B(n) (cf. A106741). Also, numbers n such that 1^n + 2^n + 3^n + ... + n^n == 1 (mod n). Equivalently, numbers n such that B(n)*n == 1 (mod n). Equivalently, Sum_{prime p, (p-1) divides n} n/p == -1 (mod n). It is easy to see that for n > 1, n must be an even squarefree number. Moreover, the set P of prime divisors of all such n satisfies the property: if p is in P, then p-1 is the product of distinct elements of P. This set is P = {2, 3, 7, 43}, implying that the sequence is finite and complete. - Max Alekseyev, Aug 25 2013
In 2005, B. C. Kellner proved E. W. Weisstein's conjecture that denom(B_n) = n only if n = 1806. - Jonathan Sondow, Oct 14 2013
Squarefree numbers n such that b^n == 1 (mod n^2) for every b coprime to n. Squarefree terms of A341858. - Thomas Ordowski, Aug 05 2024
Conjecture: Numbers n such that gcd(d+1, n) > 1 for every proper divisor d of n. Verified up to 10^696. - David Radcliffe, May 29 2025

M. A. Alekseyev, J. M. Grau, and A. M. Oller-Marcen. Computing solutions to the congruence 1^n + 2^n + ... + n^n == p (mod n). Discrete Applied Mathematics, 2018. doi:10.1016/j.dam.2018.05.022 arXiv:1602.02407 [math.NT], 2016-2018.
John H. Castillo and Jhony Fernando Caranguay Mainguez, The set of k-units modulo n, arXiv:1708.06812 [math.NT], 2017.
Yongyi Chen and Tae Kyu Kim, On Generalized Carmichael Numbers, arXiv:2103.04883 [math.NT], 2021.
J. Dyer-Bennet, A Theorem in Partitions of the Set of Positive Integers, Amer. Math. Monthly, 47(1940) pp. 152-4.
L. Halbeisen and N. Hungerbühler, On generalised Carmichael numbers, Hardy-Ramanujan Society, 1999, 22 (2), pp.8-22. (hal-01109575)
J. M. Grau, A. M. Oller-Marcen, and Jonathan Sondow, On the congruence 1^m + 2^m + ... + m^m == n (mod m) with n|m, arXiv 1309.7941 [math.NT], 2013; Monatsh. Math., 177 (2015), 421-436.
B. C. Kellner, The equation denom(B_n) = n has only one solution, preprint 2005.
Don Reble, A014117 and related OEIS sequences (shows there are no other terms)
Jonathan Sondow and K. MacMillan, Reducing the Erdos-Moser equation 1^n + 2^n + … + k^n = (k+1)^n modulo k and k^2, arXiv:1011.2154 [math.NT], 2010; see Prop. 2; Integers, 11 (2011), article A34.
Eric Weisstein's World of Mathematics, Bernoulli Number
D. Zagier, Problems posed at the St Andrews Colloquium, 1996

Squarefree terms of A124240. - Robert Israel and Thomas Ordowski, Jun 23 2017

Cf. A007018, A031971, A054377, A100016, A242927, A341858.

r[n_] := Reduce[ Mod[m^(n+1) - m, n] == 0, m, Integers]; ok[n_] := Range[n]-1 === Simplify[ Mod[ Flatten[ m /. {ToRules[ r[n][[2]] ]}], n], Element[C[1], Integers]]; ok[1] = True; A014117 = {}; Do[ If[ok[n], Print[n]; AppendTo[ A014117, n] ], {n, 1, 2000}] (* Jean-François Alcover, Dec 21 2011 *)
Select[Range@ 2000, Function[n, Times @@ Boole@ Map[Function[m, PowerMod[m, n + 1, n] == Mod[m, n]], Range@ n] > 0]] (* Michael De Vlieger, Dec 30 2016 *)

[n for n in range(1, 2000) if all(pow(m, n+1, n) == m for m in range(n))] # David Radcliffe, May 29 2025

For n <= 5, a(n) = a(n-1)^2 + a(n-1) with a(0) = 1. - Raphie Frank, Nov 12 2012

a(n+1) = A007018(n) = A054377(n) = A100016(n) for n = 1, 2, 3, 4. - Jonathan Sondow, Oct 01 2013

A073817 Tetranacci numbers with different initial conditions: a(n) = a(n-1) + a(n-2) + a(n-3) + a(n-4) starting with a(0)=4, a(1)=1, a(2)=3, a(3)=7.

Original entry on oeis.org

4, 1, 3, 7, 15, 26, 51, 99, 191, 367, 708, 1365, 2631, 5071, 9775, 18842, 36319, 70007, 134943, 260111, 501380, 966441, 1862875, 3590807, 6921503, 13341626, 25716811, 49570747, 95550687, 184179871, 355018116, 684319421, 1319068095, 2542585503

Offset: 0

Mario Catalani (mario.catalani(AT)unito.it), Aug 12 2002

These tetranacci numbers follow the same pattern as Lucas and generalized tribonacci(A001644) numbers: Binet's formula is a(n) = r1^n + r2^n + r3^n + r4^n, with r1, r2, r3, r4 roots of the characteristic polynomial.

For n >= 4, a(n) is the number of cyclic sequences consisting of n zeros and ones that do not contain four consecutive ones provided the positions of the zeros and ones are fixed on a circle. This is proved in Charalambides (1991) and Zhang and Hadjicostas (2015). For example, a(4)=15 because only the sequences 1110, 1101, 1011, 0111, 0011, 0101, 1001, 1010, 0110, 1100, 0001, 0010, 0100, 1000, 0000 avoid four consecutive ones on a circle. (For n=1,2,3 the statement is still true provided we allow the sequence to wrap around itself on a circle. For example, a(2)=3 because only the sequences 00, 01, 10 avoid four consecutive ones when wrapped around on a circle.) - Petros Hadjicostas, Dec 18 2016

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Martin Burtscher, Igor Szczyrba, and Rafał Szczyrba, Analytic Representations of the n-anacci Constants and Generalizations Thereof, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.5.
C. A. Charalambides, Lucas numbers and polynomials of order k and the length of the longest circular success run, The Fibonacci Quarterly, 29 (1991), 290-297.
Spiros D. Dafnis, Andreas N. Philippou, and Ioannis E. Livieris, An Alternating Sum of Fibonacci and Lucas Numbers of Order k, Mathematics (2020) Vol. 9, 1487.
Petros Hadjicostas, Cyclic Compositions of a Positive Integer with Parts Avoiding an Arithmetic Sequence, Journal of Integer Sequences, 19 (2016), #16.8.2.
Tony D. Noe and Jonathan Vos Post, Primes in Fibonacci n-step and Lucas n-step Sequences, J. of Integer Sequences, Vol. 8 (2005), Article 05.4.4
J. L. Ramírez and V. F. Sirvent, A Generalization of the k-Bonacci Sequence from Riordan Arrays, The Electronic Journal of Combinatorics, 22(1) (2015), #P1.38.
Yüksel Soykan, Gaussian Generalized Tetranacci Numbers, arXiv:1902.03936 [math.NT], 2019.
Yüksel Soykan, Tetranacci and Tetranacci-Lucas Quaternions, arXiv:1902.05868 [math.RA], 2019.
Yüksel Soykan, Summation Formulas for Generalized Tetranacci Numbers, Asian Journal of Advanced Research and Reports (2019) Vol. 7, No. 2, Article No. AJARR.52434, 1-12.
Yüksel Soykan, Properties of Generalized (r, s, t, u)-Numbers, Earthline J. of Math. Sci. (2021) Vol. 5, No. 2, 297-327.
Kai Wang, Identities, generating functions and Binet formula for generalized k-nacci sequences, 2020.
Kai Wang, Identities for generalized enneanacci numbers, Generalized Fibonacci Sequences (2020).
Eric Weisstein's World of Mathematics, Fibonacci n-Step Number.
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.
L. Zhang and P. Hadjicostas, On sequences of independent Bernoulli trials avoiding the pattern '11..1', Math. Scientist, 40 (2015), 89-96.
Index entries for linear recurrences with constant coefficients, signature (1,1,1,1).

Cf. A000078, A001630, A001644, A000032, A106295 (Pisano periods). Two other versions: A001648, A074081.

a:=[4,1,3,7];; for n in [5..40] do a[n]:=a[n-1]+a[n-2]+a[n-3] +a[n-4]; od; a; # G. C. Greubel, Feb 19 2019

I:=[4,1,3,7]; [n le 4 select I[n] else Self(n-1) +Self(n-2) +Self(n-3) +Self(n-4): n in [1..40]]; // G. C. Greubel, Feb 19 2019

a[0]=4; a[1]=1; a[2]=3; a[3]=7; a[4]=15; a[n_]:= 2*a[n-1] -a[n-5]; Array[a, 34, 0]
CoefficientList[Series[(4-3x-2x^2-x^3)/(1-x-x^2-x^3-x^4), {x, 0, 40}], x]
LinearRecurrence[{1,1,1,1},{4,1,3,7},40] (* Harvey P. Dale, Jun 01 2015 *)

Vec((4-3*x-2*x^2-x^3)/(1-x-x^2-x^3-x^4) + O(x^40)) \\ Michel Marcus, Jan 29 2016

((4-3*x-2*x^2-x^3)/(1-x-x^2-x^3-x^4)).series(x, 40).coefficients(x, sparse=False) # G. C. Greubel, Feb 19 2019

G.f.: (4 - 3*x - 2*x^2 - x^3)/(1 - x - x^2 - x^3 - x^4).
a(n) = 2*a(n-1) - a(n-5), with a(0)=4, a(1)=1, a(2)=3, a(3)=7, a(4)=15. - Vincenzo Librandi, Dec 20 2010
a(n) = A000078(n+2) + 2*A000078(n+1) + 3*A000078(n) + 4*A000078(n-1). - Advika Srivastava, Aug 22 2019
a(n) = 8*a(n-3) - a(n-5) - 2*a(n-6) - 4*a(n-7). - Advika Srivastava, Aug 25 2019
a(n) = Trace(M^n), where M is the 4 X 4 matrix [0, 0, 0, 1; 1, 0, 0, 1; 0, 1, 0, 1; 0, 0, 1, 1], the companion matrix to the monic polynomial x^4 - x^3 - x^2 - x - 1. It follows that the sequence satisfies the Gauss congruences: a(n*p^r) == a(n*p^(r-1)) (mod p^r) for positive integers n and r and all primes p. See Zarelua. - Peter Bala, Dec 31 2022
a(n) = Sum_{r root of x^4-x^3-x^2-x-1} r^n. - Fabian Pereyra, Feb 28 2025

Typo in definition corrected by Vincenzo Librandi, Dec 20 2010

A056854 a(n) = Lucas(4*n).

Original entry on oeis.org

2, 7, 47, 322, 2207, 15127, 103682, 710647, 4870847, 33385282, 228826127, 1568397607, 10749957122, 73681302247, 505019158607, 3461452808002, 23725150497407, 162614600673847, 1114577054219522, 7639424778862807, 52361396397820127, 358890350005878082, 2459871053643326447

Offset: 0

Barry E. Williams, Aug 29 2000

a(n) and b(n) := A004187(n) are the nonnegative proper and improper solutions of the Pell equation a(n)^2 - 5*(3*b(n))^2 = +4. See the cross-reference to A004187 below. - Wolfdieter Lang, Jun 26 2013

Lucas numbers of the form n^2-2. - Michel Lagneau, Aug 11 2014

			Pell equation: n = 0, 2^2 - 45*0^2 = +4 (improper);  n = 1, 7^2 - 5*(3*1)^2 = +4; n=2, 47^2 - 5*(3*7)^2 = +4. - _Wolfdieter Lang_, Jun 26 2013

R. P. Stanley. Enumerative combinatorics. Vol. 2, volume 62 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge, 1999.

G. C. Greubel, Table of n, a(n) for n = 0..1190
Peter Bala, Some simple continued fraction expansions for an infinite product, Part 1
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242.
Dale Gerdemann, Combinatorial proofs of Zeckendorf family identities, Fib. Quart. 46/47 (2009) 249. [_N. J. A. Sloane_, Dec 05 2009]
A. F. Horadam, Special Properties of the Sequence W(n){a,b; p,q}, Fib. Quart., Vol. 5, No. 5 (1967), pp. 424-434.
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 sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (7,-1).

Cf. A004187, A005248.
Cf. quadrisection of A000032: this sequence (first), A056914 (second), A246453 (third, without 11), A288913 (fourth).
Cf. Lucas(k*n): A000032 (k = 1), A005248 (k = 2), A014448 (k = 3), A001946 (k = 5), A087215 (k = 6), A087281 (k = 7), A087265 (k = 8), A087287 (k = 9), A065705 (k = 10), A089772 (k = 11), A089775 (k = 12).

[Lucas(4*n): n in [0..100]]; // Vincenzo Librandi, Apr 14 2011

a[0] = 2; a[1] = 7; a[n_] := 7a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 19}] (* Robert G. Wilson v, Jan 30 2004 *)
LinearRecurrence[{7,-1},{2,7},25] (* or *) LucasL[4*Range[0,25]] (* Harvey P. Dale, Aug 08 2011 *)

a(n)=if(n<0,0,polsym(1-7*x+x^2,n)[n+1])

a(n)=if(n<0,0,2*subst(poltchebi(n),x,7/2))

[lucas_number2(n,7,1) for n in range(27)] #Zerinvary Lajos, Jun 25 2008

a(n) = 7*a(n-1) - a(n-2) with a(0)=2, a(1)=7.
a(n) = A000032(4*n), where A000032 = Lucas numbers.
a(n) = 7*S(n-1, 7) - 2*S(n-2, 7) = S(n, 7) - S(n-2, 7) = 2*T(n, 7/2), with S(n, x) := U(n, x/2), S(-1, x) := 0, S(-2, x) := -1. U(n, x), resp. T(n, x), are Chebyshev's polynomials of the second, resp. first, kind. S(n-1, 7) = A004187(n), n>=0. See A049310 and A053120.
a(n) = ((7+sqrt(45))/2)^n + ((7-sqrt(45))/2)^n.
G.f.: (2-7x)/(1-7x+x^2).
a(n) = A005248(2*n); bisection of A005248.
a(n) = Fibonacci(8*n)/Fibonacci(4*n), n>0. - Gary Detlefs, Dec 26 2010
a(n) = 2 + 5*Fibonacci(2*n)^2 = 2 + 5*A049684(n), n >= 0. This is in Koshy's book (reference under A065563) 15. on p. 88. Compare with the above Chebyshev T formula. - Wolfdieter Lang, Aug 27 2012
From Peter Bala, Jan 06 2013: (Start)
Let F(x) = Product_{n = 0..inf} (1 + x^(4*n+1))/(1 + x^(4*n+3)). Let alpha = 1/2*(7 - 3*sqrt(5)). This sequence gives the simple continued fraction expansion of 1 + F(alpha) = 2.14242 42709 40138 85949 ... = 2 + 1/(7 + 1/(47 + 1/(322 + ...))).
Also F(-alpha) = 0.85670 72882 04563 14901 ... has the continued fraction representation 1 - 1/(7 - 1/(47 - 1/(322 - ...))) and the simple continued fraction expansion 1/(1 + 1/((7-2) + 1/(1 + 1/((47-2) + 1/(1 + 1/((322-2) + 1/(1 + ...))))))). Cf. A005248.
F(alpha)*F(-alpha) has the simple continued fraction expansion 1/(1 + 1/((7^2-4) + 1/(1 + 1/((47^2-4) + 1/(1 + 1/((322^2-4) + 1/(1 + ...))))))).
Added Oct 13 2019: 1/2 + (1/2)*F(alpha)/F(-alpha) = 1.16675297774947414828... has the simple continued fraction expansion 1 + 1/((7 - 2) + 1/(1 + 1/((322 - 2) + 1/(1 + 1/(15127 - 2) + 1/(1 + ...))))). (End)
a(n) = Fibonacci(4*n+2) - Fibonacci(4*n-2), where Fibonacci(-2) = -1. - Bruno Berselli, May 25 2015
a(n) = sqrt(45*(A004187(n))^2+4).
From Peter Bala, Oct 13 2019: (Start)
a(n) = F(4*n+4)/F(4) - F(4*n-4)/F(4) = A004187(n+1) - A004187(n-1).
a(n) = trace(M^n), where M is the 2 X 2 matrix [0, 1; 1, 1]^4 = [2, 3; 3, 5].
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).
5*Sum_{n >= 1} 1/(a(n) - 9/a(n)) = 1: (9 = Lucas(4)+2 and 5 = Lucas(4)-2)
9*Sum_{n >= 1} (-1)^(n+1)/(a(n) + 5/a(n)) = 1.
Sum_{n >= 1} 1/a(n) = (1/4)*( theta_3((7-3*sqrt(5))/2)^2 - 1 ), where theta_3(q) = 1 + 2*Sum_{n >= 1} q^n^2. Cf. A153415.
Sum_{n >= 1} (-1)^(n+1)/a(n) = (1/4)*( 1 - theta_3((3*sqrt(5)-7)/2)^2 ).
x*exp(Sum_{n >= 1} a(n)*x^/n) = x + 7*x^2 + 48*x^3 + 329*x^4 + ... is the o.g.f. for A004187. (End)
E.g.f.: 2*exp(7*x/2)*cosh(3*sqrt(5)*x/2). - Stefano Spezia, Oct 18 2019
a(2k+1)/7 is the numerator of the continued fraction [3*sqrt(5), 3*sqrt(5), ..., 3*sqrt(5)] with 2k copies of 3*sqrt(5), for k>0. - Greg Dresden and Tracy Z. Wu, Sep 10 2020
a(n) = Sum_{k>=1} Lucas(2*n*k)/(Lucas(2*n)^k). - Diego Rattaggi, Jan 20 2025

More terms from James Sellers, Aug 31 2000

Chebyshev comments from Wolfdieter Lang, Oct 31 2002

A039678 Smallest number m > 1 such that m^(p-1)-1 is divisible by p^2, where p = n-th prime.

Original entry on oeis.org

5, 8, 7, 18, 3, 19, 38, 28, 28, 14, 115, 18, 51, 19, 53, 338, 53, 264, 143, 11, 306, 31, 99, 184, 53, 181, 43, 164, 96, 68, 38, 58, 19, 328, 313, 78, 226, 65, 253, 259, 532, 78, 176, 276, 143, 174, 165, 69, 330, 44, 33, 332, 94, 263, 48, 79, 171, 747, 731, 20, 147, 91, 40

Offset: 1

Felice Russo

Using Fermat's little theorem twice, it is easy to see that m=p^2-1 solves this problem for all odd primes p. In fact, there appear to be exactly p-1 values of m with 1 <= m <= p^2 for which m^(p-1) == 1 (mod p^2). See A096082 for the related open problem. - T. D. Noe, Aug 24 2008
That there are exactly p-1 values of 1 <= m <= p^2 for which m^(p-1) == 1 (mod p^2) follows immediately from Hensel's lifting lemma and Fermat's little theorem - every solution mod p corresponds to a unique solution mod p^2. - Phil Carmody, Jan 10 2011
For n > 2, prime(n) does not divide a(n)^2 - 1, so a(n) is the smallest m > 1 such that (m^(prime(n)-1) - 1)/(m^2 - 1) == 0 (mod prime(n)^2). - Thomas Ordowski, Nov 24 2018

			For n=3, p=5 is the third prime and 5^2 = 25 divides 7^4 - 1 = 2400.

P. Ribenboim, The New Book of Prime Number Records, Springer, 1996, 345-349.

T. D. Noe, Table of n, a(n) for n = 1..10000

Cf. A185103.

dpa[n_]:=Module[{p=Prime[n],a=2},While[PowerMod[a,p-1,p^2]!=1,a++];a]; Array[dpa,70] (* Harvey P. Dale, Sep 05 2012 *)

a(n) = my(p=prime(n)); for(a=2, oo, if(Mod(a, p^2)^(p-1)==1, return(a))) \\ Felix Fröhlich, Nov 24 2018

from sympy import prime
from sympy.ntheory.residue_ntheory import nthroot_mod
def A039678(n): return 2**2+1 if n == 1 else int(nthroot_mod(1,(p:= prime(n))-1,p**2,True)[1]) # Chai Wah Wu, May 18 2022

a(n) = A185103(A000040(n)).

More terms from David W. Wilson
Definition adjusted by Felix Fröhlich, Jun 24 2014
Edited by Felix Fröhlich, Nov 24 2018

A007663 Fermat quotients: (2^(p-1)-1)/p, where p=prime(n).

Original entry on oeis.org

1, 3, 9, 93, 315, 3855, 13797, 182361, 9256395, 34636833, 1857283155, 26817356775, 102280151421, 1497207322929, 84973577874915, 4885260612740877, 18900352534538475, 1101298153654301589, 16628050996019877513, 64689951820132126215, 3825714619033636628817

Offset: 2

N. J. A. Sloane, Sep 19 1994

The only terms that are squares are a(2) = 1 and a(4) = 9. - Nick Hobson, May 20 2007
From Jonathan Sondow, Jul 19 2010: (Start)
a(n) == 0 (mod 3) if n > 2, since p = prime(n) > 3
and 0 = (-1)^(p-1)-1 == 2^(p-1)-1 (mod 3). (End)
p is in A001220 if and only if p | (2^(p-1)-1)/p, i.e., a(n) is divisible by prime(n). - Felix Fröhlich, Jun 20 2014
In general, every prime p that is 1 mod q-1 will create a numerator that is 0 mod q via Fermat's Little Theorem, meaning every p with this property (except q) will have a Fermat quotient divisible by q. - Roderick MacPhee, May 12 2017

Albert H. Beiler, Recreations in the theory of numbers, New York, Dover, (2nd ed.) 1966. See pp. 47, 308.
L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 1, p. 105.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
D. Wells, The Penguin Dictionary of Curious and Interesting Numbers, Penguin Books, NY, 1986, p. 70.

T. D. Noe, Table of n, a(n) for n=2..100
Nick Hobson, Solution to puzzle 158: Fermat squares.
Jonathan Sondow, Lerch Quotients, Lerch Primes, Fermat-Wilson Quotients, and the Wieferich-non-Wilson Primes 2, 3, 14771, in Proceedings of CANT 2011, arXiv:1110.3113 [math.NT], 2011-2012.
Jonathan Sondow, Lerch Quotients, Lerch Primes, Fermat-Wilson Quotients, and the Wieferich-non-Wilson Primes 2, 3, 14771, Combinatorial and Additive Number Theory, CANT 2011 and 2012, Springer Proc. in Math. & Stat., vol. 101 (2014), pp. 243-255.
Andrei K. Svinin, On some class of sums, arXiv:1610.05387 [math.CO], 2016-2017. See p. 11.
H. S. Vandiver, Fermat's Quotients And Related Arithmetic Functions, PNAS 1945 31 (1) pp. 55-60.
H. S. Vandiver, New Types Of Congruences Involving Bernoulli Numbers and Fermat's Quotient, PNAS 1948 34 (3) pp. 103-110.
H. S. Vandiver, On Congruences Which Relate The Fermat And Wilson Quotients To The Bernoulli Numbers, PNAS 1949 35 (6) pp. 332-337.

Cf. A002322, A001917, A096060, A001045.

A007663:= n-> map (p-> (2^(p-1)-1)/p, ithprime(n)):
seq (A007663(n), n=2..20); # Jani Melik, Jan 24 2011

A007663[n_Integer?Positive]:=(-1+2^(Prime[n]-1))/Prime[n]/;(n>1) (* Enrique Pérez Herrero, Sep 08 2010 *)
Table[(2^(n-1)-1)/n,{n,Prime[Range[2,20]]}] (* Harvey P. Dale, Nov 07 2016 *)

forprime(p=3, 100, print1((2^(p-1)-1)/p ", ")) \\ Satish Bysany, Mar 11 2017

From Alexander Adamchuk, Oct 01 2006: (Start)
a(n) = 3*A096060(n) for n > 2.
a(n) = 3*A001045(prime(n)-1)/prime(n) for n > 1. (End)
a(n) = Sum_{i=0..(p-3)/2} 2^i*(p-i-2)!/((i+1)!*(p-2*(i+1))!) where p = prime(n), for n >= 2. - Vladimir Pletser, Jan 26 2023

A061286 Smallest integer for which the number of divisors is the n-th prime.

Original entry on oeis.org

2, 4, 16, 64, 1024, 4096, 65536, 262144, 4194304, 268435456, 1073741824, 68719476736, 1099511627776, 4398046511104, 70368744177664, 4503599627370496, 288230376151711744, 1152921504606846976

Offset: 1

Labos Elemer, May 22 2001

Seems to be the same as "Even numbers with prime number of divisors" - Jason Earls, Jul 04 2001

Except for the first term, smallest number == 1 (mod prime(n)) having n divisors (by Fermat's little theorem). - Amarnath Murthy and Meenakshi Srikanth (menakan_s(AT)yahoo.com), Jun 20 2003

Karl V. Keller, Jr., Table of n, a(n) for n = 1..460
Index to divisibility sequences

Cf. A000005, A005179, A003680, A061283, A061286, A006093, A005097, A006254, A119523, A196202.

Table[2^(p-1),{p,Table[Prime[n],{n,1,18}]}] (* Geoffrey Critzer, May 26 2013 *)

forstep(n=2,100000000,2,x=numdiv(n); if(isprime(x),print(n)))

a(n)=2^(prime(n)-1) \\ Charles R Greathouse IV, Apr 08 2012

from sympy import isprime, divisor_count as tau
[2] + [2**(2*n) for n in range(1, 33) if isprime(tau(2**(2*n)))] # Karl V. Keller, Jr., Jul 10 2020

a(n) = 2^(prime(n)-1) = 2^A006093(n).
a(n) = A005179(prime(n)). - R. J. Mathar, Aug 09 2019
Sum_{n>=1} 1/a(n) = A119523. - Amiram Eldar, Aug 11 2020

A290261 Write 1 - x/(1-x) as an inverse power product 1/(1 + a(1)x) 1/(1 + a(2)x^2) 1/(1 + a(3)x^3) 1/(1 + a(4)x^4) ...

Original entry on oeis.org

1, 2, 2, 6, 6, 10, 18, 54, 54, 114, 186, 334, 630, 1314, 2106, 5910, 7710, 15642, 27594, 57798, 97902, 207762, 364722, 712990, 1340622, 2778930, 4918482, 10437702, 18512790, 37500858, 69273666, 154021590, 258155910, 535004610, 981288906

Offset: 1

Gus Wiseman, Jul 24 2017

nonn

The initial terms are given on page 1234 of Gingold, Knopfmacher, 1995.
From Petros Hadjicostas, Oct 04 2019: (Start)
In Section 3 of Gingold and Knopfmacher (1995), it is proved that, if f(z) = Product_{n >= 1} (1 + g(n))*z^n = 1/(Product_{n >= 1} (1 - h(n))*z^n), then g(2*n - 1) = h(2*n - 1) and Sum_{d|n} (1/d)*h(n/d)^d = -Sum_{d|n} (1/d)*(-g(n/d))^d. The same results were proved more than ten years later by Alkauskas (2008, 2009). [If we let a(n) = -g(n), then Alkauskas works with f(z) = Product_{n >= 1} (1 - a(n))*z^n; i.e., a(2*n - 1) = -h(2*n - 1) etc.]
The PPE of 1 - x/(1-x) is given in A220418, which is also studied in Gingold and Knopfmacher (1995) starting at p. 1223.
(End)

Seiichi Manyama, Table of n, a(n) for n = 1..3333
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.

Cf. A220418, A273866, A289501, A290262.

nn=20;Solve[Table[Expand[SeriesCoefficient[Product[1/(1+a[k]x^k),{k,n}],{x,0,n}]]==-1,{n,nn}],Table[a[n],{n,nn}]][[1,All,2]]
(* Second program: *)
A[m_, n_] := A[m, n] = Which[m == 1, 2^(n-1), m > n >= 1, 0, True, A[m - 1, n] - A[m - 1, m - 1]*A[m, n - m + 1] ];
a[n_] := A[n, n];
a /@ Range[1, 55] (* Petros Hadjicostas, Oct 04 2019, courtesy of Jean-François Alcover *)

a(n) = Sum_t (-1)^v(t) where the sum is over all enriched p-trees of weight n (see A289501 for definition) and v(t) is the number of nodes (branchings and leaves) in t.
From Petros Hadjicostas, Oct 04 2019: (Start)
a(n) satisfies Sum_{d|n} (1/d)*(-a(n/d))^d = -(2^n - 1)/n. Thus, a(n) = Sum_{d|n, d>1} (1/d)*(-a(n/d))^d + (2^n - 1)/n.
a(2*n - 1) = A220418(2*n - 1) for n >= 1 because A220418 gives the PPE of 1 - x/(1-x).
Define (A(m,n): n,m >= 1) by A(m=1,n) = 2^(n-1) 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 a(n) = A(n,n). [Theorem 3 in Gingold et al. (1988).]
(End)

cp's OEIS Frontend

A003500 a(n) = 4*a(n-1) - a(n-2) with a(0) = 2, a(1) = 4.

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A051783 Numbers k such that 3^k + 2 is prime.

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A005935 Pseudoprimes to base 3.

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A014117 Numbers n such that m^(n+1) == m (mod n) holds for all m.

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A073817 Tetranacci numbers with different initial conditions: a(n) = a(n-1) + a(n-2) + a(n-3) + a(n-4) starting with a(0)=4, a(1)=1, a(2)=3, a(3)=7.

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A056854 a(n) = Lucas(4*n).

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A290261 Write 1 - x/(1-x) as an inverse power product 1/(1 + a(1)x) 1/(1 + a(2)x^2) 1/(1 + a(3)x^3) 1/(1 + a(4)x^4) ...