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

A065705 a(n) = Lucas(10*n).

2, 123, 15127, 1860498, 228826127, 28143753123, 3461452808002, 425730551631123, 52361396397820127, 6440026026380244498, 792070839848372253127, 97418273275323406890123, 11981655542024930675232002, 1473646213395791149646646123, 181246502592140286475862241127

Offset: 0

Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Oct 25 2003

Lim_{n->infinity} a(n+1)/a(n) = (123 + sqrt(15125))/2 = 122.9918693812...
Lim_{n->infinity} a(n)/a(n+1) = (123 - sqrt(15125))/2 = 0.00813061875578...
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^10) = 1.0081300769... = 1 + 1/(123 + 1/(15127 + 1/(1860498 + ...))).
Also F(-Phi^10) = 0.9918699143... has the continued fraction representation 1 - 1/(123 - 1/(15127 - 1/(1860498 - ...))) and the simple continued fraction expansion 1/(1 + 1/((123 - 2) + 1/(1 + 1/((15127 - 2) + 1/(1 + 1/((1860498 - 2) + 1/(1 + ...))))))).
F(Phi^10)*F(-Phi^10) = 0.9999338930... has the simple continued fraction expansion 1/(1 + 1/((123^2 - 4) + 1/(1 + 1/((15127^2 - 4) + 1/(1 + 1/((1860498^2 - 4) + 1/(1 + ...))))))).
1/2 + (1/2)*F(Phi^10)/F(-Phi^10) = 1.0081967213... has the simple continued fraction expansion 1 + 1/((123 - 2) + 1/(1 + 1/((1860498 - 2) + 1/(1 + 1/(28143753123 - 2) + 1/(1 + ...))))). (End)

			a(4) = 228826127 = 123*a(3) - a(2) = 123*1860498 - 15127=((123+sqrt(15125))/2)^4 + ( (123-sqrt(15125))/2)^4 =228826126.99999999562986 + 0.00000000437013 = 228826127.
a(4) = L(10 * 4) = L(40) = 228826127. - _Indranil Ghosh_, Feb 08 2017

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..477
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 (123,-1).

Cf. A000032: a(n) = A000032(10*n).

Cf. Lucas(k*n): A005248 (k = 2), A014448 (k = 3), A056854 (k = 4), A001946 (k = 5), A087215 (k = 6), A087281 (k = 7), A087265 (k = 8), A087287 (k = 9), A089772 (k = 11), A089775 (k = 12).

[Lucas(10*n): n in [0..90]]; // Vincenzo Librandi, Apr 14 2011

LucasL[10*Range[0, 20]] (* Paolo Xausa, Mar 04 2024 *)

a(n) = 123*a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 123.
a(n) = ((123 + sqrt(15125))/2)^n + ((123 - sqrt(15125))/2)^n.
a(n)^2 = a(2*n) + 2.
G.f.: (2 - 123*x)/(1 - 123*x + x^2). -  Philippe Deléham, Nov 18 2008
From Peter Bala, Oct 14 2019: (Start)
a(n) = F(10*n+10)/F(10) - F(10*n-10)/F(10) = A049670(n+1) - A049670(n-1).
a(n) = trace(M^n), where M is the 2 X 2 matrix [0, 1; 1, 1]^10 = [34, 55; 55, 89].
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).
121*Sum_{n >= 1} 1/(a(n) - 125/a(n)) = 1: (125 = Lucas(10) + 2 and 121 = Lucas(10) - 2)
125*Sum_{n >= 1} (-1)^(n+1)/(a(n) + 121/a(n)) = 1.
x*exp(Sum_{n >= 1} a(n)*x^/n) = x + 123*x^2 + 15128*x^3 + ... is the o.g.f. for A049670. (End)
E.g.f.: exp((1/2)*(123 - 55*sqrt(5))*x)*(1 + exp(55*sqrt(5)*x)). - Stefano Spezia, Oct 18 2019
From Peter Bala, Apr 16 2025: (Start)
a(n) = Lucas(2*n)^5 - 5*Lucas(2*n)^3 + 5*Lucas(2*n) = 2*T(5, (1/2)*Lucas(2*n)), where T(k, x) denotes the k-th Chebyshev polynomial of the first kind.
Sum_{n >= 1} 1/a(n) = (1/4) * (theta_3( (123 - sqrt(15125))/2 )^2 - 1) and
Sum_{n >= 1} (-1)^(n+1)/a(n) = (1/4) * (1 - theta_3( (sqrt(15125) - 123)/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)

A191677 Numbers n such that 1^(n-1)+2^(n-1)+...+n^(n-1) == 0 (mod n).

Original entry on oeis.org

1, 4, 8, 12, 16, 20, 24, 28, 32, 35, 36, 40, 44, 48, 52, 55, 56, 60, 64, 68, 72, 76, 77, 80, 84, 88, 92, 95, 96, 100, 104, 108, 112, 115, 116, 119, 120, 124, 128, 132, 136, 140, 143, 144, 148, 152, 155, 156, 160, 161, 164, 168, 172, 176, 180, 184, 187, 188, 192, 196, 200, 203, 204

Offset: 1

José María Grau Ribas, Jun 10 2011

nonn

Fermat's little theorem shows that this sequence contains no primes. Related to Giuga's conjecture that the sum is -1 iff n is prime. - Charles R Greathouse IV, Jun 10 2011

Is this is the disjoint union of all multiples of 4 and {1} and A121707 (n^3 divides Sum_{kM. F. Hasler, Jul 22 2019

Ivan Neretin, Table of n, a(n) for n = 1..10000

Cf. A121707 (n^3 divides Sum_{k

is191677[n_]:=Mod[Sum[PowerMod[k, n - 1, n], {k, 1, n - 1}], n] == 0;
Select[Range[300], is191677]

select( is_A191677(n)=!sum(k=1,n-1,Mod(k,n)^(n-1)), [1..200]) \\ M. F. Hasler, Jul 22 2019

A213382 Numbers n such that n^n mod (n + 2) = n.

Original entry on oeis.org

1, 4, 7, 13, 16, 19, 31, 37, 49, 55, 61, 67, 85, 91, 109, 121, 127, 139, 157, 175, 181, 193, 196, 199, 211, 217, 235, 247, 265, 289, 301, 307, 313, 319, 325, 337, 379, 391, 397, 409, 415, 445, 451, 469, 487, 499, 517, 535, 541, 571, 577, 589, 595, 631, 667, 679

Offset: 1

Alex Ratushnyak, Jun 10 2012

nonn

Equivalently, numbers n such that (n^n+2)/(n+2) is an integer. Derek Orr, May 23 2014
It was conjectured that A176003 is a subsequence.
Terms that do not appear in A176003: 16, 61, 193, 196, 313, 397, 691, 729, 769 ...
The conjecture is correct: verify the cases 1 and 3, then it suffices to show that (3p-2)^(3p-2) = 3p-2 mod 3 and mod p. Mod 3 the congruence is 1^(3p-2) = 1, and mod p the congruence is (-2)^(3p-2) = -2 which is true by Fermat's little theorem. - Charles R Greathouse IV, Sep 12 2012
a(62) = 729 is the first number not congruent to 1 mod 3. - Derek Orr, May 23 2014

			A213381(n) = 7^7 mod 9 = 7, so 7 is in the sequence.

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Cf. A213381 : a(n) = n^n mod (n+2).

Cf. A176003.

Select[Range[700],PowerMod[#,#,#+2]==#&] (* Harvey P. Dale, Oct 03 2015 *)

is(n)=Mod(n,n+2)^n==n \\ Charles R Greathouse IV, Sep 12 2012

for n in range(999):
    x = n**n % (n+2)
    if x==n:
        print(n, end=", ")

A002604 a(n) = n^6 + 1.

Original entry on oeis.org

1, 2, 65, 730, 4097, 15626, 46657, 117650, 262145, 531442, 1000001, 1771562, 2985985, 4826810, 7529537, 11390626, 16777217, 24137570, 34012225, 47045882, 64000001, 85766122, 113379905, 148035890

Offset: 0

N. J. A. Sloane

Because of Fermat's little theorem, a(n) is never divisible by 7. - Altug Alkan, Apr 08 2016

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

Equals A001014 + 1. Cf. A024004, A002522.

[n^6 + 1: n in [0..50]]; // Vincenzo Librandi, May 02 2011

Table[n^6+1,{n,0,40}] (* Vladimir Joseph Stephan Orlovsky, Apr 15 2011 *)
LinearRecurrence[{7,-21,35,-35,21,-7,1},{1,2,65,730,4097,15626,46657},30] (* Harvey P. Dale, Jul 28 2021 *)

PARI
```
a(n)=n^6+1
```

G.f. (-1 + 5*x - 72*x^2 - 282*x^3 - 317*x^4 - 51*x^5 - 2*x^6) / (x - 1)^7. - R. J. Mathar, Aug 06 2012
Sum_{n>=0} 1/a(n) = 1/2 + Pi * (coth(Pi) + (sinh(Pi) + sqrt(3)*sin(sqrt(3)*Pi)) / (cosh(Pi) - cos(sqrt(3)*Pi))) / 6 = 1.5171007340332164261529... . - Vaclav Kotesovec, Feb 14 2015
Sum_{n>=0} (-1)^n/a(n) = 1/2 + Pi/(6*sinh(Pi)) + Pi * (sqrt(3)*cosh(Pi/2) * sin((sqrt(3)*Pi)/2) + cos((sqrt(3)*Pi)/2) * sinh(Pi/2)) / (3*(cosh(Pi) - cos(sqrt(3)*Pi))) = 0.514210347292695053493... . - Vaclav Kotesovec, Feb 14 2015

A170919 a(n) = denominator of the coefficient c(n) of x^n in (tan x)/Product_{k=1..n-1} 1 + c(k)*x^k, n = 1, 2, 3, ...

Original entry on oeis.org

1, 1, 3, 3, 5, 45, 105, 315, 2835, 14175, 5775, 467775, 6081075, 2837835, 212837625, 70945875, 3618239625, 97692469875, 206239658625, 9280784638125, 1031198293125, 142924083427125, 322279795963125, 101111706320625, 136968913284328125, 161872352063296875

Offset: 1

N. J. A. Sloane, Jan 30 2010

			1, -1, 7/3, -14/3, 54/5, -1112/45, 6574/105, -48488/315, 1143731/2835, ...

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.
Giedrius Alkauskas, Algebraic functions with Fermat property, eigenvalues of transfer operator and Riemann zeros, and other open problems, arXiv:1609.09842 [math.NT], 2016.
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.
Wolfdieter Lang, Recurrences for the general problem.

Cf. A170918 (numerators), A170910-A170917.
Cf. A353583 / A353584 for power product expansion of 1 + tan x.
Cf. A353586 / A353587 for power product expansion of (tan x)/x.

L := 28: g := NULL:
t := series(tan(x), x, L):
for n from 1 to L-2 do
   c := coeff(t, x, n);
   t := series(t/(1 + c*x^n), x, L);
   g := g, c;
od: map(denom, [g]); # Based on Maple in A170918. - Peter Luschny, Oct 05 2019

Following a suggestion from Ilya Gutkovskiy, name corrected so that it matches the data by Peter Luschny, May 12 2022

A201560 a(n) = (Sum(m^(n-1), m=1..n-1) + 1) modulo n.

Original entry on oeis.org

0, 0, 0, 1, 0, 4, 0, 1, 7, 6, 0, 1, 0, 8, 11, 1, 0, 10, 0, 1, 15, 12, 0, 1, 21, 14, 19, 1, 0, 16, 0, 1, 23, 18, 1, 1, 0, 20, 27, 1, 0, 22, 0, 1, 22, 24, 0, 1, 43, 26, 35, 1, 0, 28, 1, 1, 39, 30, 0, 1, 0, 32, 43, 1, 53, 34, 0, 1, 47, 36, 0, 1, 0, 38, 51, 1, 1

Offset: 1

Jonathan Sondow, Jan 11 2012

nonn

Equals 0 if n is 1 or a prime, by Fermat's little theorem. It is conjectured that the converse is also true; see A055030, A055032, A204187 and note that a(n) = 0 <==> A055032(n) = 1 <==> A204187(n) = n-1.

			Sum(m^3, m=1..3) + 1 = 1^3 + 2^3 + 3^3 + 1 = 37 == 1 (mod 4), so a(4) = 1.

R. K. Guy, Unsolved Problems in Number Theory, A17.

Ivan Neretin, Table of n, a(n) for n = 1..10000

Cf. A055023, A055030, A055031, A055032, A204187.

Table[Mod[Plus @@ PowerMod[Range[n - 1], n - 1, n] + 1, n], {n, 77}] (* Ivan Neretin, Sep 23 2016 *)

a(prime) = 0 and a(4n) = 1.

a(n) == A204187(n) + 1 (mod n).

A220420 Express the Sum_{n>=0} p(n)x^n, where p(n) is the partition function, as a product Product_{k>=1} (1 + a(k)x^k).

Original entry on oeis.org

1, 2, 1, 4, 1, 0, 1, 14, 1, -4, 1, -8, 1, -16, 1, 196, 1, -54, 1, -92, 1, -184, 1, 144, 1, -628, 1, -1040, 1, -2160, 1, 41102, 1, -7708, 1, -12932, 1, -27592, 1, 54020, 1, -98496, 1, -173720, 1, -364720, 1, 853624, 1, -1341970, 1, -2383916, 1, -4918536, 1

Offset: 1

Michel Marcus, Dec 14 2012

sign

This is the PPE (power product expansion) of A000041.
When n is odd, a(n) = 1.
When n is even, a(n) = 2, 4, 0, 14, -4, -8, -16, 196, -54, -92, -184, 144, -628, -1040, -2160, 41102, ...
Alkauskas (2016, Problem 3, p. 3) conjectured that a(8*k+2), a(8*k+4), and a(8*k+6) are all negative, and a(8*k) is positive for k >= 1. [This statement is not wholly true for k = 0.] - Petros Hadjicostas, Oct 07 2019

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.
Giedrius Alkauskas, Algebraic functions with Fermat property, eigenvalues of transfer operator and Riemann zeros, and other open problems, arXiv:1609.09842 [math.NT], 2016.
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, Canadian Journal of Mathematics 47(6) (1995), 1219-1239.
W. Lang, Recurrences for the general problem.

Cf. A000041, A147541, A170908, A170909, A170910, A170911, A170912, A170913, A170914, A170915, A170916, A170917, A220418, A290261.

terms = 55; sol[0] = {};
sol[m_] := sol[m] = Join[sol[m - 1], If[OddQ[m], {a[m] -> 1}, First @ Solve[Thread[Table[PartitionsP[n], {n, 0, m}] == CoefficientList[ (Product[1 + a[n]*x^n, {n, 1, m}] /. sol[m - 1]) + O[x]^(m + 1), x]]]]];
Array[a, terms] /. sol[terms] (* Jean-François Alcover, Dec 06 2018, corrected Oct 03 2019 *)
(* Second program: *)
A[m_, n_] := A[m, n] = Which[m == 1, PartitionsP[n], 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] (* Jean-François Alcover, Oct 03 2019, using the formula given by Petros Hadjicostas *)

a(m) = {default(seriesprecision, m+1); ak = vector(m); pol = 1 / eta(x + x * O(x^m)); ak[1] = polcoeff(pol, 1); for (k=2, m, pol = taylor(pol / (1+ak[k-1]*x^(k-1)), x); ak[k] = polcoeff(pol, k, x);); for (k=1, m, print1(ak[k], ", "););}

From Petros Hadjicostas, Oct 04 2019: (Start)
Define (A(m,n): n,m >= 1) by A(m=1,n) = p(n) = A000041(n) 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).]
a(n) = Sum_{s|n} s/n + Sum_{s|n, s > 1} (-a(n/s))^s/s. [Eq. (1) in Alkauskas (2008, 2009).]
(End)

A227470 Least k such that n divides sigma(n*k).

Original entry on oeis.org

1, 3, 2, 3, 8, 1, 4, 7, 10, 4, 43, 2, 9, 2, 8, 21, 67, 5, 37, 6, 20, 43, 137, 5, 149, 9, 34, 1, 173, 4, 16, 21, 27, 64, 76, 22, 73, 37, 6, 3, 163, 10, 257, 43, 6, 137, 281, 11, 52, 76, 67, 45, 211, 17, 109, 4, 49, 173, 353, 2, 169, 8, 32, 93, 72, 27, 401, 67

Offset: 1

Alex Ratushnyak, Jul 12 2013

nonn

Theorem: a(n) always exists.

Proof: If n is a power of a prime, say n = p^a, then, by Euler's generalization of Fermat's little theorem and the multiplicative property of sigma, one can take k = x^(p^a-p^(a-1)-1) where x is a different prime from p. Similarly. if n = p^a*q^b, then take k = x^(p^a-p^(a-1)-1)*y^(q^b-q^(b-1)-1) where {x,y} are primes different from {p,q}. And so on. These k's have the desired property, and so there is always at least one candidate for the minimal k. - N. J. A. Sloane, May 01 2016

			Least k such that 9 divides sigma(9*k) is k = 10: sigma(90) = 234 = 9*26. So a(9) = 10.
Least k such that 89 divides sigma(89*k) is k = 1024: sigma(89*1024) = 184230 = 89*2070. So a(89) = 1024.

R. J. Mathar, Table of n, a(n) for n = 1..1000

Indices of 1's: A007691.
Cf. A000203, A227302, A227303, A097018.
See A272349 for the sequence [n*a(n)]. - N. J. A. Sloane, May 01 2016

A227470 := proc(n)
    local k;
    for k from 1 do
        if modp(numtheory[sigma](k*n),n) =0 then
            return k;
        end if;
    end do:
end proc: # R. J. Mathar, May 06 2016

lknds[n_]:=Module[{k=1},While[!Divisible[DivisorSigma[1,k*n],n],k++];k]; Array[lknds,70] (* Harvey P. Dale, Jul 10 2014 *)

a227470(n) = {k=1; while(sigma(n*k)%n != 0, k++); k} \\ Michael B. Porter, Jul 15 2013

a(n) = A272349(n)/n. - R. J. Mathar, May 06 2016

A328186 Write 1/(1 + sin x) = Product_{n>=1} (1 + f_n x^n); a(n) = denominator(f_n).

Original entry on oeis.org

1, 1, 6, 6, 120, 360, 5040, 2520, 72576, 1814400, 39916800, 59875200, 1245404160, 21794572800, 1307674368000, 81729648000, 71137485619200, 3201186852864000, 121645100408832000, 12164510040883200, 10218188434341888000, 281000181944401920000, 25852016738884976640000

Offset: 1

Petros Hadjicostas, Oct 06 2019

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/(1 + sin(x)). For that coefficient, we use a modification of a formula by Peter Luschny in the documentation of sequences A099612 and A099617.
Write 1 + sin x = Product_{n>=1} (1 + g_n * x^n). We have A170914(n) = numerator(g_n) and A170915(n) = denominator(g_n).
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.]
Wolfdieter Lang (see the link below) examined inverse power product expansions both for ordinary g.f.'s and for exponential g.f.'s. He connects inverse power product expansions to unital series associated to (infinite dimensional) Witt vectors and to the so-called "Somos transformation".
There are more formulas for f_n and g_n in the references listed below. In all cases, we assume the g.f.'s are unital, i.e., the g.f.'s start with a constant 1.

			f_n = -1, 1, 1/6, 5/6, 19/120, -47/360, 659/5040, 1837/2520, 7675/72576, -154729/1814400, 3578279/39916800, 3984853/59875200, 95259767/1245404160, ...

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 A328191.

Cf. A099612, A099617, A170914, A170915, A279107.

# Calculates the fractions f_n (choose L much larger than M):
PPE := 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/(1 + sin(x));
t0 := series(t1, x, L);
f := []; t2 := t0;
for n to M do
t3 := coeff(t2, x, n);
t2 := series(t2/(1 + t3*x^n), x, L);
f := [op(f), t3];
end do;
end if;
[seq(f[n], n = 1 .. nops(f))];
end proc;
# Calculates the denominators of f_n:
h := map(denom, PPE(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, 2*(-1)^n*I^(n + 2)*PolyLog[-(n + 1), -I]/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 using a program by Jean-François Alcover and a formula from A099612 and A099617 *)

a(2*n + 1) = A170915(2*n + 1) for n >= 0.
Define (A(m,n): n,m >= 1) by A(m=1, n) = 2 * (-1)^n * i^(n + 2) * PolyLog(-(n + 1), -i)/n! for n >= 1 (with i := sqrt(-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 f_n = A(n,n) and thus a(n) = denominator(A(n,n)).
If we write 1 + sin x = Product_{n>=1} (1 + g_n * x^n) and we know (g_n: n >= 1), then f_n = -g_n + Sum_{s|n, s > 1} (1/s) * ((-f_{n/s})^s + (-g_{n/s})^s). This proves of course that f_n = -g_n for n odd.

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A065705 a(n) = Lucas(10*n).

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A074058 Reflected tetranacci numbers A073817.

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A191677 Numbers n such that 1^(n-1)+2^(n-1)+...+n^(n-1) == 0 (mod n).

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A213382 Numbers n such that n^n mod (n + 2) = n.

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A002604 a(n) = n^6 + 1.

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A170919 a(n) = denominator of the coefficient c(n) of x^n in (tan x)/Product_{k=1..n-1} 1 + c(k)*x^k, n = 1, 2, 3, ...

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A201560 a(n) = (Sum(m^(n-1), m=1..n-1) + 1) modulo n.

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A220420 Express the Sum_{n>=0} p(n)x^n, where p(n) is the partition function, as a product Product_{k>=1} (1 + a(k)x^k).