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

A273866 Coefficients a(k,m) of polynomials a{k}(h) appearing in the product Product_{k >= 1} (1 - a{k}(h)x^k) = 1 - hx/(1-x).

1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 2, 2, 2, 1, 1, 3, 5, 5, 3, 1, 1, 3, 6, 7, 6, 3, 1, 1, 4, 9, 13, 13, 9, 4, 1, 1, 4, 10, 17, 20, 17, 10, 4, 1, 1, 5, 15, 30, 42, 42, 30, 15, 5, 1, 1, 5, 16, 36, 57, 66, 57, 36, 16, 5, 1

Offset: 1

Gevorg Hmayakyan, Jun 01 2016

The a(k,m) form a table where each row has k-1 elements starting from 2 and the a(1,1) = 1.

			a{1}(h) = h,
a{2}(h) = h,
a{3}(h) = h^2 + h,
a{4}(h) = h^3 + h^2 + h,
a{5}(h) = h^4 + 2*h^3 + 2*h^2 + h,
a{6}(h) = h^5 + 2*h^4 + 2*h^3 + 2*h^2 + h,
a{7}(h) = h^6 + 3*h^5 + 5*h^4 + 5*h^3 + 3*h^2 + h,
a{8}(h) = h^7 + 3*h^6 + 6*h^5 + 7*h^4 + 6*h^3 + 3*h^2 + h,
a{9}(h) = h^8 + 4*h^7 + 9*h^6 + 13*h^5 + 13*h^4 + 9*h^3 + 4*h^2 + h
...
and the corresponding a(k,m) table is:
  1,
  1,
  1,  1,
  1,  1,  1,
  1,  2,  2,  1,
  1,  2,  2,  2,  1,
  1,  3,  5,  5,  3,  1,
  1,  3,  6,  7,  6,  3,  1,
  1,  4,  9, 13, 13,  9,  4,  1,
  ...
a(7,3) = 5 because there are six strict trees contributing positive one {{5,1},1}, {{4,2},1}, {{4,1},2}, {{3,2},2}, {4,{2,1}}, {{3,1},3} and there is one strict tree contributing negative one {4,2,1}. - _Gus Wiseman_, Nov 14 2016

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. A196545, A220418, A220420, A273866, A273873, A279785, A290261, A290262, A290971, A295632.

with(ListTools), with(numtheory), with(combinat);
L := product(1-a[k]*x^k, k = 1 .. 600);
S := Flatten([seq(-h, i = 1 .. 100)]);
Sabs := Flatten([seq(i, i = 1 .. 100)]);
seq(assign(a[i] = solve(coeff(L, x^i) = `if`(is(i in Sabs), S[Search(i, Sabs)], 0), a[i])), i = 1 .. 20);
map(coeffs, [seq(simplify(a[i]), i = 1 .. 20)]);

strictrees[n_Integer?Positive]:=Prepend[Join@@Function[ptn,Tuples[strictrees/@ptn]]/@Select[IntegerPartitions[n],And[Length[#]>1,UnsameQ@@#]&],n];
Table[Sum[(-1)^(Count[tree,,{0,Infinity}]-1),{tree,Select[strictrees[n],Length[Flatten[{#}]]===m&]}],{n,1,9},{m,1,n-1/.(0->1)}] (* _Gus Wiseman, Nov 14 2016 *)
(* second program *)
A[m_, n_] :=
  A[m, n] =
   Which[m == 1, -h, m > n >= 1, 0, True,
    A[m - 1, n] - A[m - 1, m - 1]*A[m, n - m + 1]];
a[n_] := Expand[-A[n, n]];
a /@ Range[1, 25] (* Petros Hadjicostas, Oct 04 2019, courtesy of Jean-François Alcover *)

a(k,m) = a(k, k-m).
For prime p: Sum_{m = 1..p-1} a(p, m) = (2^p - 2)/p.
a{k}(h) satisfies Sum_{d|k} (1/d)*(a{k/d}(h))^d = ((h+1)^k - 1)/k. [Corrected by Petros Hadjicostas, Oct 04 2019]
For prime p: a{p}(h) = ((h+1)^p - h^p - 1)/p.
See A273873 for the definition of strict tree. Then a(n,m) = Sum_t (-1)^{v(t)-1} where the sum is over all strict trees of weight n with m leaves, and v(t) is the number of nodes in t (including the leaves, which are positive integers). See example 2 and the first Mathematica program. - Gus Wiseman, Nov 14 2016

A013961 a(n) = sigma_13(n), the sum of the 13th powers of the divisors of n.

Original entry on oeis.org

1, 8193, 1594324, 67117057, 1220703126, 13062296532, 96889010408, 549822930945, 2541867422653, 10001220711318, 34522712143932, 107006334784468, 302875106592254, 793811662272744, 1946196290656824, 4504149450301441, 9904578032905938, 20825519793796029, 42052983462257060

Offset: 1

N. J. A. Sloane

If the canonical factorization of n into prime powers is the product of p^e(p) then sigma_k(n) = Product_{p prime} ((p^((e(p)+1)*k)) - 1)/(p^k - 1).
Sum_{d|n} 1/d^k is equal to sigma_k(n)/n^k. So sequences A017665-A017712 also give the numerators and denominators of sigma_k(n)/n^k for k = 1..24. The power sums sigma_k(n) are in sequences A000203 (k=1), A001157-A001160 (k=2,3,4,5), A013954-A013972 for k = 6,7,...,24. - Ahmed Fares (ahmedfares(AT)my-deja.com), Apr 05 2001
By Fermat's little theorem n^13 == n (mod 13). Hence sigma_13(n) == sigma_(1) (mod 13). In fact, sigma_13(n) == sigma_(1) (mod 2730), where 2730 = 2*3*5*7*13 = the numerator of Bernoulli(12). - Peter Bala, Jan 12 2025

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for sequences related to sigma(n).

Cf. A000203, A001157-A001160, A013672, A013954-A013972, A017665-A017712.

[DivisorSigma(13, n): n in [1..20]]; // Vincenzo Librandi, Sep 10 2016

A013961 := proc(n)
    numtheory[sigma][13](n) ;
end proc: # R. J. Mathar, Sep 21 2017

DivisorSigma[13, Range[30]] (* Vincenzo Librandi, Sep 10 2016 *)

my(N=99, q='q+O('q^N)); Vec(sum(n=1, N, n^13*q^n/(1-q^n))) \\ Altug Alkan, Sep 10 2016

a(n) = sigma(n, 13); \\ Michel Marcus, Sep 10 2016

[sigma(n,13)for n in range(1,16)] # Zerinvary Lajos, Jun 04 2009

G.f.: Sum_{k>=1} k^13*x^k/(1-x^k). - Benoit Cloitre, Apr 21 2003
Dirichlet g.f.: zeta(s-13)*zeta(s). - Ilya Gutkovskiy, Sep 10 2016
Empirical: Sum_{n>=1} a(n)/exp(2*Pi*n) = 1/24. - Simon Plouffe, Mar 01 2021
From Amiram Eldar, Oct 29 2023: (Start)
Multiplicative with a(p^e) = (p^(13*e+13)-1)/(p^13-1).
Sum_{k=1..n} a(k) = zeta(14) * n^14 / 14 + O(n^15). (End)

A087215 Lucas(6n): a(n) = 18a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 18.

Original entry on oeis.org

2, 18, 322, 5778, 103682, 1860498, 33385282, 599074578, 10749957122, 192900153618, 3461452808002, 62113250390418, 1114577054219522, 20000273725560978, 358890350005878082, 6440026026380244498, 115561578124838522882, 2073668380220713167378

Offset: 0

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

a(n+1)/a(n) converges to 9 + sqrt(80) = 17.9442719... a(0)/a(1) = 2/18; a(1)/a(2) = 18/322; a(2)/a(3) = 322/5778; a(3)/a(4) = 5778/103682; etc.
Lim_{n -> oo} a(n)/a(n+1) = 0.05572809000084... = 1/(9 + sqrt(80)) = 9 - sqrt(80).
From Peter Bala, Oct 13 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^6) = 1.0555459720... = 1 + 1/(18 + 1/(322 + 1/(5778 + ...))).
Also F(-Phi^6) = 0.9444348576... has the continued fraction representation 1 - 1/(18 - 1/(322 - 1/(5788 - ...))) and the simple continued fraction expansion 1/(1 + 1/((18 - 2) + 1/(1 + 1/((322 - 2) + 1/(1 + 1/((5788 - 2) + 1/(1 + ...))))))).
F(Phi^6)*F(-Phi^6) = 0.9968944099... has the simple continued fraction expansion 1/(1 + 1/((18^2 - 4) + 1/(1 + 1/((322^2 - 4) + 1/(1 + 1/((5788^2 - 4) + 1/(1 + ...))))))).
1/2 + (1/2)*F(Phi^6)/F(-Phi^6) = 1.0588241282... has the simple continued fraction expansion 1 + 1/((18 - 2) + 1/(1 + 1/((5778 - 2) + 1/(1 + 1/(1860498 - 2) + 1/(1 + ...))))). (End)

			a(4) = 103682 = 18*a(3) - a(2) = 18*5778 - 322 = (9 + sqrt(80))^4 + (9 - sqrt(80))^4 = 103681.99999035512... + 0.00000964487... = 103682.

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.

Colin Barker, Table of n, a(n) for n = 0..750
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
P. Bhadouria, D. Jhala, and B. Singh, Binomial Transforms of the k-Lucas Sequences and its Properties, The Journal of Mathematics and Computer Science (JMCS), Volume 8, Issue 1, Pages 81-92; sequence R_4.
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 (18,-1).

Cf. A074919.
Row 2 * 2 of array A188645.
Cf. Lucas(k*n): A000032 (k = 1), A005248 (k = 2), A014448 (k = 3), A056854 (k = 4), A001946 (k = 5), A087281 (k = 7), A087265 (k = 8), A087287 (k = 9), A065705 (k = 10), A089772 (k = 11), A089775 (k = 12).

[ Lucas(6*n) : n in [0..100]]; // Vincenzo Librandi, Apr 14 2011

a[0] = 2; a[1] = 18; a[n_] := 18a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 15}] (* Robert G. Wilson v, Jan 30 2004 *)
Table[LucasL[6n], {n, 0, 18}]  (* or *) CoefficientList[Series[2*(1 - 9*x)/(1 - 18*x + x^2), {x, 0, 17}], x] (* Indranil Ghosh, Mar 15 2017 *)

Vec(2*(1-9*x)/(1-18*x+x^2) + O(x^20)) \\ Colin Barker, Jan 24 2016

a(n) = if(n<2, 17^n + 1, 18*a(n - 1) - a(n - 2));
for(n=0, 17, print1(a(n),", ")) \\ Indranil Ghosh, Mar 15 2017

a(n) = A000032(6*n).
a(n) = 18*a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 18.
a(n) = (9 + sqrt(80))^n + (9 - sqrt(80))^n.
G.f.: 2*(1-9*x)/(1-18*x+x^2). - Philippe Deléham, Nov 17 2008
a(n) = 2*A023039(n). - R. J. Mathar, Oct 22 2010
From Peter Bala, Oct 13 2019: (Start)
a(n) = F(6*n+6)/F(6) - F(6*n-6)/F(6) = A049660(n+1) - A049660(n-1).
a(n) = trace(M^n), where M is the 2 X 2 matrix [0, 1; 1, 1]^6 = [5, 8; 8, 13].
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).
16*Sum_{n >= 1} 1/(a(n) - 20/a(n)) = 1: (20 = Lucas(6) + 2 and 16 = Lucas(6) - 2)
20*Sum_{n >= 1} (-1)^(n+1)/(a(n) + 16/a(n)) = 1.
Series acceleration formulas for sum of reciprocals:
Sum_{n >= 1} 1/a(n) = 1/16 - 20*Sum_{n >= 1} 1/(a(n)*(a(n)^2 - 20)).
Sum_{n >= 1} (-1)^(n+1)/a(n) = 1/20 + 16*Sum_{n >= 1} (-1)^(n+1)/(a(n)*(a(n)^2 + 16)).
Sum_{n >= 1} 1/a(n) = ( (theta_3(9-4*sqrt(5)))^2 - 1 )/4 and
Sum_{n >= 1} (-1)^(n+1)/a(n) = ( 1 - (theta_3(4*sqrt(5)-9))^2 )/4,
where theta_3(x) = 1 + 2*Sum_{n >= 1} x^(n^2) (see A000122). Cf. A153415 and A003499.
x*exp(Sum_{n >= 1} a(n)*x^/n) = x + 18*x^2 + 323*x^3 + ... is the o.g.f. for A049660. (End)
E.g.f.: 2*exp(9*x)*cosh(4*sqrt(5)*x). - Stefano Spezia, Oct 18 2019
a(n) = L(2n-1)^2 * F(2n+1) + L(2n+1)^2 * F(2n-1), where F(n) = A000045(n) and L(n) = A000032(n). - Diego Rattaggi, Nov 12 2020
From  Peter Bala, Apr 16 2025: (Start)
a(n) = Lucas(2*n)^3 - 3*Lucas(2*n) = 2*T(3, (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(9 - sqrt(80))^2 - 1) and
Sum_{n >= 1} (-1)^(n+1)/a(n) = (1/4) * (1 - theta_3(sqrt(80) - 9)^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)

A001638 A Fielder sequence: a(n) = a(n-1) + a(n-3) + a(n-4), n >= 4.

Original entry on oeis.org

4, 1, 1, 4, 9, 11, 16, 29, 49, 76, 121, 199, 324, 521, 841, 1364, 2209, 3571, 5776, 9349, 15129, 24476, 39601, 64079, 103684, 167761, 271441, 439204, 710649, 1149851, 1860496, 3010349, 4870849, 7881196, 12752041, 20633239, 33385284, 54018521, 87403801

Offset: 0

N. J. A. Sloane

For n > 1, a(n) is the number of ways of choosing a subset of vertices of an n-cycle so that every vertex of the n-cycle is adjacent to one of the chosen vertices. (Note that this is not the same as the number of dominating sets of the n-cycle, which is given by A001644.) - Joel B. Lewis, Sep 12 2010
For n >= 3, a(n) is also the number of total dominating sets in the n-cycle graph. - Eric W. Weisstein, Apr 10 2018
For n > 0, a(n) is the number of ways to tile a strip of length n with squares, trominos, and quadrominos where the inital tile (of length 1, 3, or 4) can take on 1, 3, or 4 colors respectively. - Greg Dresden and Yuan Shen, Aug 10 2024

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..1000
Daniel C. Fielder, Special integer sequences controlled by three parameters, Fibonacci Quarterly 6, 1968, 64-70.
Fern Gossow, Lyndon-like cyclic sieving and Gauss congruence, arXiv:2410.05678 [math.CO], 2024. See p. 26.
Middle European Math Olympiad 2010, Team Problem 3. Available online at the Art of Problem Solving. - _Joel B. Lewis_, Sep 12 2010
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
Eric Weisstein's World of Mathematics, Cycle Graph
Eric Weisstein's World of Mathematics, Total Dominating Set
Wikipedia, Companion matrix.
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 to sequences related to Olympiads.
Index entries for linear recurrences with constant coefficients, signature (1,0,1,1).

Cf. A000032, A001609, A001634 - A001636, A001638 - A001645, A001648, A001649.

I:=[4,1,1,4]; [n le 4 select I[n] else Self(n-1) + Self(n-3) + Self(n-4): n in [1..30]]; // G. C. Greubel, Jan 09 2018

A001638:=-(z+1)*(4*z**2-z+1)/(z**2+z-1)/(z**2+1); # conjectured by Simon Plouffe in his 1992 dissertation; gives sequence except for the initial 4

LinearRecurrence[{1, 0, 1, 1}, {4, 1, 1, 4}, 50] (* T. D. Noe, Aug 09 2012 *)
Table[LucasL[n] + 2 Cos[n Pi/2], {n, 0, 20}] (* Eric W. Weisstein, Apr 10 2018 *)
CoefficientList[Series[(-4 + 3 x + x^3)/(-1 + x + x^3 + x^4), {x, 0, 20}], x] (* Eric W. Weisstein, Apr 10 2018 *)

a(n)=if(n<0,0,fibonacci(n+1)+fibonacci(n-1)+simplify(I^n+(-I)^n))

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

G.f.: (1-x)*(4+x+x^2)/((1+x^2)*(1-x-x^2)).
a(n) = L(n) + i^n + (-i)^n, a(2n) = L(n)^2, a(2n+1) = L(2n+1) where L() is Lucas sequence A000032.
a(n) = a(n-1) + a(n-3) + a(n-4). - Eric W. Weisstein, Apr 10 2018
a(n) = Trace(M^n), where M = [0, 0, 0, 1; 1, 0, 0, 1; 0, 1, 0, 0; 0, 0, 1, 1] is the companion matrix to the monic polynomial x^4 - x^3 - 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, Jan 08 2023

Edited by Michael Somos, Feb 17 2002 and Nov 02 2002

A015911 Numbers k such that 2^k mod k is odd.

Original entry on oeis.org

25, 45, 55, 91, 95, 99, 125, 135, 143, 153, 155, 161, 175, 187, 225, 235, 245, 247, 261, 273, 275, 279, 285, 289, 297, 319, 329, 333, 335, 355, 363, 369, 387, 391, 403, 407, 413, 423, 425, 429, 435, 437, 441, 459, 473, 477, 481, 483, 493, 507, 517, 525, 529

Offset: 1

Robert G. Wilson v

nonn

All terms are composite: due to Fermat's little theorem, 2^p == 2 (mod p) when p is prime. - M. F. Hasler, May 10 2021

Zak Seidov, Table of n, a(n) for n = 1..10000
Wikipedia, Fermat's little theorem.

Cf. A015910, A226221.

q:= n-> is(2&^n mod n, odd):
select(q, [$1..1000])[];  # Alois P. Heinz, May 10 2021

Select[Range@532, OddQ@PowerMod[2, #, # ] &]

is(n)=lift(Mod(2,n)^n)%2 \\ Charles R Greathouse IV, May 31 2013

A064535 a(n) = (2^prime(n)-2)/prime(n); a(0) = 0 by convention.

Original entry on oeis.org

0, 1, 2, 6, 18, 186, 630, 7710, 27594, 364722, 18512790, 69273666, 3714566310, 53634713550, 204560302842, 2994414645858, 169947155749830, 9770521225481754, 37800705069076950, 2202596307308603178, 33256101992039755026, 129379903640264252430

Offset: 0

Shane Findley, Oct 09 2001

nonn

As a corollary to Fermat's little theorem, (2^p - 2)/p is always an integer for p prime. - Alonso del Arte, May 04 2013

			a(3) = 6, because prime(3) = 5, and (2^5 - 2)/5 = 30/5 = 6.
a(4) = 18, because prime(4) = 7, and (2^7  - 2)/7 = 126/7 = 18.

Harry J. Smith, Table of n, a(n) for n = 0..100

Cf. A007663, A056743, A225101 (superset).

[0] cat [(2^NthPrime(n)-2)/NthPrime(n): n in [1..25]]; // Vincenzo Librandi, Sep 14 2018

A064535 := proc(n) ( 2^ithprime(n) - 2 )/ithprime(n); end;

Table[(2^Prime[n] - 2)/Prime[n], {n, 50}] (* Alonso del Arte, Apr 28 2013 *)

{ for (n=0, 100, if (n, a=(2^prime(n) - 2)/prime(n), a=0); write("b064535.txt", n, " ", a) ) } \\ Harry J. Smith, Sep 17 2009

a(n) = A001037(prime(n)) for n >= 1. - Hilko Koning, Sep 10 2018

a(n) = 2*A007663(n) for n > 1. - Jeppe Stig Nielsen, May 16 2021

A069051 Primes p such that p-1 divides 2^p-2.

Original entry on oeis.org

2, 3, 7, 19, 43, 127, 163, 379, 487, 883, 1459, 2647, 3079, 3943, 5419, 9199, 11827, 14407, 16759, 18523, 24967, 26407, 37339, 39367, 42463, 71443, 77659, 95923, 99079, 113779, 117307, 143263, 174763, 175447, 184843, 265483, 304039, 308827

Offset: 1

Benoit Cloitre, Apr 03 2002

nonn

These are the prime values of n such that 2^n == 2 (mod n*(n-1)). - V. Raman, Sep 17 2012
These are the prime values p such that n^(2^(p-1)) is congruent to n or -n (mod p) for all n in Z/pZ, the commutative ring associated with each term. This results follows from Fermat's little theorem. - Philip A. Hoskins, Feb 08 2013
A prime p is in this sequence iff p-1 belongs to A014741. For p>2, this is equivalent to (p-1)/2 belonging to A014945. - Max Alekseyev, Aug 31 2016
From Thomas Ordowski, Nov 20 2018: (Start)
Conjecture: if n-1 divides 2^n-2, then (2^n-2)/(n-1) is squarefree.
Numbers n such that b^n == b (mod (n-1)*n) for every integer b are 2, 3, 7, and 43; i.e., only prime numbers of the form A014117(k) + 1. (End)
These are primes p such that p^2 divides b^(2^p-2) - 1 for every b coprime to p. - Thomas Ordowski, Jul 01 2024

Charles R Greathouse IV, Table of n, a(n) for n = 1..3314 (First 553 terms from V. Raman)
Mersenne Forum, Prime Conjecture

Cf. A216822, A217468, A211203, A211349, A330382.

a(n)-1 form subsequence of A014741; (a(n)-1)/2 for n>1 forms a subsequence of A014945.

Filtered([1..350000],p->IsPrime(p) and (2^p-2) mod (p-1)=0); # Muniru A Asiru, Dec 03 2018

[p : p in PrimesUpTo(310000) | IsZero((2^p-2) mod (p-1))]; // Vincenzo Librandi, Dec 03 2018

Select[Prime[Range[10000]], Mod[2^# - 2, # - 1] == 0 &] (* T. D. Noe, Sep 19 2012 *)
Join[{2,3},Select[Prime[Range[30000]],PowerMod[2,#,#-1]==2&]] (* Harvey P. Dale, Apr 17 2022 *)

isA069051(p)=Mod(2,p-1)^p==2 && isprime(p); \\ Charles R Greathouse IV, Sep 19 2012

from sympy import prime
for n in range(1,350000):
    if (2**prime(n)-2) % (prime(n)-1)==0:
        print(prime(n)) # Stefano Spezia, Dec 07 2018

a(1) added by Charles R Greathouse IV, Sep 19 2012

A170912 Write cos(x) = Product_{n >= 1} (1 + g_nx^(2n)); a(n) = numerator(g_n).

Original entry on oeis.org

-1, 1, 7, 131, 1843, 97261, 4683059, 1331727679, 568285777, 9521655609199, 175554688130609, 11334988388673161, 3457026400678609391, 6594042537777612027841, 249248595232521829462213, 268938575250382935485761673113, 3929672369519648081411955883, 4719016202742955262333630268611

Offset: 1

N. J. A. Sloane, Jan 30 2010

			-1/2, 1/24, 7/360, 131/13440, 1843/453600, 97261/47900160, ...

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. A170913.

t1:=cos(x);
L:=100;
t0:=series(t1,x,L):
g:=[]; M:=40; t2:=t0:
for n from 1 to M do
t3:=coeff(t2,x,n); t2:=series(t2/(1+t3*x^n),x,L); g:=[op(g),t3];
od:
g;
h:=[seq(g[2*n],n=1..nops(g)/2)];
h1:=map(numer,h);
h2:=map(denom,h);

A[m_, n_] :=
  A[m, n] =
   Which[m == 1, (-1)^n/(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 04 2019, courtesy of Jean-François Alcover *)

A170913 Write cos(x) = Product_{n>=1} (1 + g_nx^(2n)); a(n) = denominator(g_n).

Original entry on oeis.org

2, 24, 360, 13440, 453600, 47900160, 5448643200, 2988969984000, 3126159036000, 101370917007360000, 4390627842881280000, 552984315270266880000, 393839317506450816000000, 1465809349094778175488000000, 129517997955171415349760000000, 263130836933693530167218012160000000

Offset: 1

N. J. A. Sloane, Jan 30 2010

			-1/2, 1/24, 7/360, 131/13440, 1843/453600, 97261/47900160, ...

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. A170912.

t1:=cos(x);
L:=100;
t0:=series(t1, x, L):
g:=[]; M:=40; t2:=t0:
for n from 1 to M do
t3:=coeff(t2, x, n); t2:=series(t2/(1+t3*x^n), x, L); g:=[op(g), t3];
od:
g;
h:=[seq(g[2*n], n=1..nops(g)/2)];
h1:=map(numer, h);
h2:=map(denom, h);

A[m_, n_] :=
  A[m, n] =
   Which[m == 1, (-1)^n/(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 04 2019, courtesy of  Jean-François Alcover *)

A353583 Numerators of coefficients c(n) in product expansion of 1 + tan x = Product_{k>=1} 1 + c(k)*x^k.

Original entry on oeis.org

1, 0, 1, -1, 7, -7, 199, -71, 484, -368, 187909, -610103, 2068657, -63614, 1530164189, -1715846683, 7628902283, -125125345078, 9521826231889, -17921564328719, 291162274608871, -47147385565688, 552647133893696333, -36898601487519532, 4761064630028162378

Offset: 1

M. F. Hasler, May 07 2022

See A353584 for the denominators, and A353586 for the analog for (tan x)/x.

			1 + tan x = (1 + x)(1 + 1/3*x^3)(1 - 1/3*x^4/3)(1 + 7/15*x^5)(1 - 7/15*x^5)(...),
and this sequence lists the numerators of (1, 0, 1/3, -1/3, 7/15, -7/15, ...).

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. A353584 (denominators), A353586 / A353587 (similar for (tan x)/x).

Cf. A170918 / A170919 for a variant.

t=1+tan(x+O(x)^29); vector(#t-1,n,c=polcoef(t,n);t/=1+c*x^n;numerator(c))

cp's OEIS Frontend

A273866 Coefficients a(k,m) of polynomials a{k}(h) appearing in the product Product_{k >= 1} (1 - a{k}(h)*x^k) = 1 - h*x/(1-x).

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A013961 a(n) = sigma_13(n), the sum of the 13th powers of the divisors of n.

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A087215 Lucas(6*n): a(n) = 18*a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 18.

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A001638 A Fielder sequence: a(n) = a(n-1) + a(n-3) + a(n-4), n >= 4.

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A015911 Numbers k such that 2^k mod k is odd.

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A064535 a(n) = (2^prime(n)-2)/prime(n); a(0) = 0 by convention.

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A273866 Coefficients a(k,m) of polynomials a{k}(h) appearing in the product Product_{k >= 1} (1 - a{k}(h)x^k) = 1 - hx/(1-x).

A087215 Lucas(6n): a(n) = 18a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 18.

A170912 Write cos(x) = Product_{n >= 1} (1 + g_nx^(2n)); a(n) = numerator(g_n).

A170913 Write cos(x) = Product_{n>=1} (1 + g_nx^(2n)); a(n) = denominator(g_n).