A006497 -id:A006497 - cp's OEIS frontend

A080039 a(n) = floor((1+sqrt(2))^n).

1, 2, 5, 14, 33, 82, 197, 478, 1153, 2786, 6725, 16238, 39201, 94642, 228485, 551614, 1331713, 3215042, 7761797, 18738638, 45239073, 109216786, 263672645, 636562078, 1536796801, 3710155682, 8957108165, 21624372014, 52205852193

Offset: 0

Mario Catalani (mario.catalani(AT)unito.it), Jan 21 2003

a(n) = P(n) - (1+(-1)^n)/2, where P(n) is the Pell sequence (A000129) with initial conditions 2, 2.

For n>0 a(n) is the maximum element in the continued fraction for P(n)*sqrt(2) where P=A000129 - Benoit Cloitre, Jun 19 2005

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,2,-2,-1).

Cf. A001622, A006497, A014176, A098316.

CoefficientList[Series[(1-t^2+2t^3)/(1-2t-2t^2+2t^3+t^4), {t, 0, 30}], t]
Floor[(1+Sqrt[2])^Range[0,40]] (* or *) LinearRecurrence[{2,2,-2,-1},{1,2,5,14},40] (* Harvey P. Dale, Aug 10 2021 *)

t='t+O('t^50); Vec((1-t^2+2*t^3)/(1-2*t-2*t^2+2*t^3+t^4)) \\ G. C. Greubel, Jul 05 2017

G.f.: (1-t^2+2*t^3)/(1-2*t-2*t^2+2*t^3+t^4).
From Hieronymus Fischer, Jan 02 2009: (Start)
The fractional part of (1+sqrt(2))^n equals (1+sqrt(2))^(-n), if n odd. For even n, the fractional part of (1+sqrt(2))^n is equal to 1-(1+sqrt(2))^(-n).
fract((1+sqrt(2))^n) = (1/2)*(1+(-1)^n)-(-1)^n*(1+sqrt(2))^(-n) = (1/2)*(1+(-1)^n)-(1-sqrt(2))^n.
See A001622 for a general formula concerning the fractional parts of powers of numbers x>1, which satisfy x-x^(-1)=floor(x).
a(n) = (sqrt(2)+1)^n - (1/2) + (-1)^n*((sqrt(2)-1)^n - (1/2)) for n>0. (End)

A180148 a(n) = 3*a(n-1) + a(n-2) with a(0)=2 and a(1)=5.

Original entry on oeis.org

2, 5, 17, 56, 185, 611, 2018, 6665, 22013, 72704, 240125, 793079, 2619362, 8651165, 28572857, 94369736, 311682065, 1029415931, 3399929858, 11229205505, 37087546373, 122491844624, 404563080245, 1336181085359, 4413106336322, 14575500094325, 48139606619297

Offset: 0

Johannes W. Meijer, Aug 13 2010

Inverse binomial transform of A052961 (without the leading 1).

For n >= 1, also the number of matchings in the n-alkane graph. - Eric W. Weisstein, Jul 14 2021

Alois P. Heinz, Table of n, a(n) for n = 0..1000
Eric Weisstein's World of Mathematics, Alkane Graph
Eric Weisstein's World of Mathematics, Matching
Eric Weisstein's World of Mathematics, Maximum Independent Edge Set
Index entries for linear recurrences with constant coefficients, signature (3,1).

Cf. A003688, A052906.
Appears in A180142.
Cf. A000602 (more information on n-alkanes).

a:= n-> (<<0|1>, <1|3>>^n. <<2, 5>>)[1,1]:
seq(a(n), n=0..27);  # Alois P. Heinz, Jul 14 2021

LinearRecurrence[{3, 1}, {5, 7}, 20] (* Eric W. Weisstein, Jul 14 2021 *)
CoefficientList[Series[(2 - x)/(1 - 3 x - x^2), {x, 0, 20}], x] (* Eric W. Weisstein, Jul 14 2021 *)

a(n)=([0,1;1,3]^n*[2;5])[1,1] \\ Charles R Greathouse IV, Oct 13 2016

G.f.: (2-x)/(1-3*x-x^2).
a(n) = 3*a(n-1) + a(n-2) with a(0)=2 and a(1)=5.
a(n) = ((4+7*A)*A^(-n-1) + (4+7*B)*B^(-n-1))/13 with A = (-3+sqrt(13))/2 and B = (-3-sqrt(13))/2.
Lim_{k->infinity} a(n+k)/a(k) = (-1)^n*2/(A006497(n) - A006190(n)*sqrt(13)).
a(n) = 2 * Sum_{k=0..n-2} A168561(n-2,k)*3^k + 5 * Sum_{k=0..n-1} A168561(n-1,k)*3^k, n>0. - R. J. Mathar, Feb 14 2024
a(n) = 2*A006190(n+1) - A006190(n). - R. J. Mathar, Feb 14 2024

A335670 Odd composite integers m such that A014448(m) == 4 (mod m).

Original entry on oeis.org

9, 85, 161, 341, 705, 897, 901, 1105, 1281, 1853, 2465, 2737, 3745, 4181, 4209, 4577, 5473, 5611, 5777, 6119, 6721, 9701, 9729, 10877, 11041, 12209, 12349, 13201, 13481, 14981, 15251, 16185, 16545, 16771, 19669, 20591, 20769, 20801, 21845, 23323, 24465, 25345

Offset: 1

Ovidiu Bagdasar, Jun 17 2020

nonn

If p is a prime, then A014448(p)==4 (mod p).
This sequence contains the odd composite integers for which the congruence holds.
The generalized Pell-Lucas sequence of integer parameters (a,b) defined by V(n+2)=a*V(n+1)-b*V(n) and V(0)=2, V(1)=a, satisfy the identity V(p)==a (mod p) whenever p is prime and b=-1,1.
For a=4, b=-1, V(n) recovers A014448(n) (even Lucas numbers).

			9 is the first odd composite integer for which A014448(9)=439204==4 (mod 9).

D. Andrica, O. Bagdasar, Recurrent Sequences: Key Results, Applications and Problems. Springer (to appear, 2020).

Chai Wah Wu, Table of n, a(n) for n = 1..1000
D. Andrica and O. Bagdasar, On some new arithmetic properties of the generalized Lucas sequences, preprint for Mediterr. J. Math. 18, 47 (2021).

Cf. A006497, A005845 (a=1), A330276 (a=2), A335669 (a=3), A335671 (a=5).

M:= <<4|1>,<1|0>>:
f:= proc(n) uses LinearAlgebra:-Modular;
local A;
A:= Mod(n,M,integer[8]);
A:= MatrixPower(n,A,n);
2*A[1,1] - 4*A[1,2] mod n;
end proc:
select(t -> f(t) = 4 and not isprime(t), [seq(i,i=3..10^5,2)]); # Robert Israel, Jun 19 2020

Select[Range[3, 25000, 2], CompositeQ[#] && Divisible[LucasL[3#] - 4, #] &] (* Amiram Eldar, Jun 18 2020 *)

More terms from Jinyuan Wang, Jun 17 2020

A335671 Odd composite integers m such that A087130(m) == 5 (mod m).

Original entry on oeis.org

9, 27, 65, 121, 145, 377, 385, 533, 1035, 1189, 1305, 1885, 2233, 2465, 4081, 5089, 5993, 6409, 6721, 7107, 10877, 11281, 11285, 13281, 13369, 13741, 13833, 14705, 15457, 16721, 17545, 18901, 19601, 19951, 20329, 20705, 22881, 24769, 25345, 26599, 26937, 28741, 29161

Offset: 1

Ovidiu Bagdasar, Jun 17 2020

nonn

If p is a prime, then A087130(p)==5 (mod p).
This sequence contains the odd composite integers for which the congruence holds.
The generalized Pell-Lucas sequence of integer parameters (a,b) defined by V(n+2)=a*V(n+1)-b*V(n) and V(0)=2, V(1)=a, satisfy the identity V(p)==a (mod p) whenever p is prime and b=-1,1.
For a=5, b=-1, V(n) recovers A087130(n).

			9 is the first odd composite integer for which A087130(9)=2744420==5 (mod 9).

D. Andrica, O. Bagdasar, Recurrent Sequences: Key Results, Applications and Problems. Springer (to appear, 2020).

Robert Israel, Table of n, a(n) for n = 1..1000
D. Andrica and O. Bagdasar, On some new arithmetic properties of the generalized Lucas sequences, preprint for Mediterr. J. Math. 18, 47 (2021).

Cf. A006497, A005845 (a=1), A330276 (a=2), A335669 (a=3), A335670 (a=4).

M:= <<5|1>,<1|0>>:
f:= proc(n) uses LinearAlgebra:-Modular;
local A;
A:= Mod(n,M,integer[8]);
A:= MatrixPower(n,A,n);
2*A[1,1] - 5*A[1,2] mod n;
end proc:
select(t -> f(t) = 5 and not isprime(t), [seq(i,i=3..10^5,2)]); # Robert Israel, Jun 19 2020

Select[Range[3, 30000, 2], CompositeQ[#] && Divisible[LucasL[#, 5] - 5, #] &] (* Amiram Eldar, Jun 18 2020 *)

More terms from Jinyuan Wang, Jun 17 2020

A100230 Main diagonal of triangle A100229.

Original entry on oeis.org

1, 2, 10, 35, 118, 392, 1297, 4286, 14158, 46763, 154450, 510116, 1684801, 5564522, 18378370, 60699635, 200477278, 662131472, 2186871697, 7222746566, 23855111398, 78788080763, 260219353690, 859446141836, 2838557779201

Offset: 0

Paul D. Hanna, Nov 29 2004

nonn

Let F(x) = Product_{n >= 1} (1 + x^(4*n + 1))/(1 - x^(4*n + 3)). Let alpha = (1/2)*(3 - sqrt(13)). This sequence occurs as partial numerators in the simple continued fraction expansion of the real number F(alpha) = 1.34372 29374 22358 27049 ... = 1 + 1/(2 + 1/(1 + 1/(10 + 1/(35 + 1/(1 + 1/(118 + 1/(392 + 1/(1 + ...)))))))). - Peter Bala, Oct 17 2019

Index entries for linear recurrences with constant coefficients, signature (4,-2,-1).

Cf. A100228, A100229.

Equals A006497(n) - 1.

LinearRecurrence[{4,-2,-1},{1,2,10},30] (* Harvey P. Dale, May 06 2012 *)

a(n)=if(n==0,1,n*polcoeff(log((1-x)/(1-3*x-x^2)+x*O(x^n)),n))

a(n) = 3*a(n-1) + a(n-2) + 3 for n>1, with a(0)=1, a(1)=2.
G.f.: Sum_{n>=1} a(n)*x^n/n = log((1-x)/(1-3*x-x^2)).
a(0)=1, a(1)=2, a(2)=10, a(n)=4*a(n-1)-2*a(n-2)-a(n-3). [Harvey P. Dale, May 06 2012]

A309220 Square array A read by antidiagonals: the columns are given by A(n,1)=1, A(n,2)=n+1, A(n,3) = n^2+2n+3, A(n,4) = n^3+3n^2+6n+4, A(n,5) = n^4+4n^3+10n^2+12*n+7, ..., whose coefficients are given by A104509 (see also A118981).

Original entry on oeis.org

1, 1, 2, 1, 3, 6, 1, 4, 11, 14, 1, 5, 18, 36, 34, 1, 6, 27, 76, 119, 82, 1, 7, 38, 140, 322, 393, 198, 1, 8, 51, 234, 727, 1364, 1298, 478, 1, 9, 66, 364, 1442, 3775, 5778, 4287, 1154, 1, 10, 83, 536, 2599, 8886, 19602, 24476, 14159, 2786, 1, 11, 102, 756, 4354, 18557

Offset: 1

N. J. A. Sloane, Aug 12 2019, based on R. J. Mathar's 2011 analysis of A118980

As pointed out by Peter Munn, A117938 gives the same triangle, except that it has an additional diagonal at the right. - N. J. A. Sloane, Aug 13 2019

			The first few antidiagonals are:
1,
1,2,
1,3,6,
1,4,11,14,
1,5,18,36,34,
1,6,27,76,119,82,
1,7,38,140,322,393,198,
...
The first nine rows of A are
1, 2, 6, 14, 34, 82, 198, 478, 1154, 2786, 6726, 16238, ...
1, 3, 11, 36, 119, 393, 1298, 4287, 14159, 46764, 154451, 510117, ...
1, 4, 18, 76, 322, 1364, 5778, 24476, 103682, 439204, 1860498, 7881196, ...
1, 5, 27, 140, 727, 3775, 19602, 101785, 528527, 2744420, 14250627, 73997555, ...
1, 6, 38, 234, 1442, 8886, 54758, 337434, 2079362, 12813606, 78960998, 486579594, ...
1, 7, 51, 364, 2599, 18557, 132498, 946043, 6754799, 48229636, 344362251, 2458765393, ...
1, 8, 66, 536, 4354, 35368, 287298, 2333752, 18957314, 153992264, 1250895426, 10161155672, ...
1, 9, 83, 756, 6887, 62739, 571538, 5206581, 47430767, 432083484, 3936182123, 35857722591, ...
1, 10, 102, 1030, 10402, 105050, 1060902, 10714070, 108201602, 1092730090, 11035502502, 111447755110, ...

Cf. A104509, A117938, A118980, A118981, A099425 (top row), A006497 (essentially the 2nd row), A014448 (essentially the 3rd row), A087130 (essentially the 4th row).

M := 12;
A:=Array(1..2*M,1..2*M,0):
for i from 1 to M do A[i,1]:=1; od:
S := series((1 + x^2)/(1-x-x^2 + x*y), x, 120): # this is g.f. for A104509
for n from 1 to M do
R2 := expand(coeff(S, x, n));
R3 := [seq(abs(coeff(R2,y,n-i)),i=0..n)];
f := k-> add( R3[i]*k^(n-i+1), i=1..nops(R3) ): # this is the formula for the (n+1)-st column
s1 := [seq(f(i),i=1..M)];
for i from 1 to M do A[i,n+1]:=s1[i]; od:
od:
for i from 1 to M do lprint([seq(A[i,j],j=1..M)]); od:
# alternative by R. J. Mathar, Aug 13 2019 :
A104509 := proc(n,k)
    (1+x^2)/(1-x-x^2+x*y) ;
    coeftayl(%,x=0,n) ;
    coeftayl(%,y=0,k) ;
end proc:
A309220 := proc(n::integer,k::integer)
    local x;
    add( abs(A104509(k-1,i))*x^i,i=0..k-1) ;
    subs(x=n,%) ;
end proc:
seq( seq(A309220(d-k,k),k=1..d-1),d=2..13) ;

A352362 Array read by ascending antidiagonals. T(n, k) = L(k, n) where L are the Lucas polynomials.

Original entry on oeis.org

2, 2, 0, 2, 1, 2, 2, 2, 3, 0, 2, 3, 6, 4, 2, 2, 4, 11, 14, 7, 0, 2, 5, 18, 36, 34, 11, 2, 2, 6, 27, 76, 119, 82, 18, 0, 2, 7, 38, 140, 322, 393, 198, 29, 2, 2, 8, 51, 234, 727, 1364, 1298, 478, 47, 0, 2, 9, 66, 364, 1442, 3775, 5778, 4287, 1154, 76, 2

Offset: 0

Peter Luschny, Mar 18 2022

			Array starts:
n\k 0, 1,  2,   3,    4,     5,      6,       7,        8, ...
--------------------------------------------------------------
[0] 2, 0,  2,   0,    2,     0,      2,       0,        2, ... A010673
[1] 2, 1,  3,   4,    7,    11,     18,      29,       47, ... A000032
[2] 2, 2,  6,  14,   34,    82,    198,     478,     1154, ... A002203
[3] 2, 3, 11,  36,  119,   393,   1298,    4287,    14159, ... A006497
[4] 2, 4, 18,  76,  322,  1364,   5778,   24476,   103682, ... A014448
[5] 2, 5, 27, 140,  727,  3775,  19602,  101785,   528527, ... A087130
[6] 2, 6, 38, 234, 1442,  8886,  54758,  337434,  2079362, ... A085447
[7] 2, 7, 51, 364, 2599, 18557, 132498,  946043,  6754799, ... A086902
[8] 2, 8, 66, 536, 4354, 35368, 287298, 2333752, 18957314, ... A086594
[9] 2, 9, 83, 756, 6887, 62739, 571538, 5206581, 47430767, ... A087798
A007395|A059100|
    A001477 A079908

Eric Weisstein's World of Mathematics, Lucas Polynomial

Rows: A010673, A000032, A002203, A006497, A014448, A087130, A085447, A086902, A086594, A087798.
Columns: A007395, A001477, A059100, A079908.
Cf. A320570 (main diagonal), A114525, A309220 (variant), A117938 (variant), A352361 (Fibonacci polynomials), A350470 (Jacobsthal polynomials).

T := (n, k) -> (n/2 + sqrt((n/2)^2 + 1))^k + (n/2 - sqrt((n/2)^2 + 1))^k:
seq(seq(simplify(T(n - k, k)), k = 0..n), n = 0..10);

Table[LucasL[k, n], {n, 0, 9}, {k, 0, 9}] // TableForm
(* or *)
T[ 0, k_] := 2 Mod[k+1, 2]; T[n_, 0] := 2;
T[n_, k_] := n^k Hypergeometric2F1[1/2 - k/2, -k/2, 1 - k, -4/n^2];
Table[T[n, k], {n, 0, 9}, {k, 0, 8}] // TableForm

T(n, k) = ([0, 1; 1, k]^n*[2; k])[1, 1] ;
export(T)
for(k = 0, 9, print(parvector(10, n, T(n - 1, k))))

T(n, k) = Sum_{j=0..floor(k/2)} binomial(k-j, j)*(k/(k-j))*n^(k-2*j) for k >= 1.
T(n, k) = (n/2 + sqrt((n/2)^2 + 1))^k + (n/2 - sqrt((n/2)^2 + 1))^k.
T(n, k) = [x^k] ((2 - n*x)/(1 - n*x - x^2)).
T(n, k) = n^k*hypergeom([1/2 - k/2, -k/2], [1 - k], -4/n^2) for n,k >= 1.

A209493 Indices of primes in sequence A006190.

Original entry on oeis.org

2, 5, 11, 17, 61, 103, 167, 193, 293, 643, 647, 911, 11243, 29437, 55021, 80141

Offset: 1

Vaclav Kotesovec, Mar 09 2012

Index must be prime. The indices greater than 293 yield probable primes.

Cf. A006190, A006497, A000129, A096650.

seq=RecurrenceTable[{a[n]==3*a[n-1]+a[n-2], a[1]==1, a[2]==3}, a, {n, 1000}]; Select[Range[1000], PrimeQ[seq[[#]]]&]

a(14)-a(15) from Vaclav Kotesovec, Sep 08 2013

a(16) from Michael S. Branicky, Nov 03 2024

A051928 Number of independent sets of vertices in graph K_3 X C_n (n > 2).

Original entry on oeis.org

4, 1, 13, 34, 121, 391, 1300, 4285, 14161, 46762, 154453, 510115, 1684804, 5564521, 18378373, 60699634, 200477281, 662131471, 2186871700, 7222746565, 23855111401, 78788080762, 260219353693, 859446141835, 2838557779204, 9375119479441, 30963916217533

Offset: 0

Stephen G Penrice, Dec 19 1999

Colin Barker, Table of n, a(n) for n = 0..1000
C. Bautista-Ramos and C. Guillen-Galvan, Fibonacci numbers of generalized Zykov sums, J. Integer Seq., 15 (2012), Article 12.7.8.
Index entries for linear recurrences with constant coefficients, signature (2,4,1).

Row 3 of A287376.

LinearRecurrence[{2,4,1},{4,1,13},30] (* Harvey P. Dale, Nov 20 2021 *)

Vec((4-7*x-5*x^2)/((1+x)*(1-3*x-x^2)) + O(x^30)) \\ Colin Barker, May 11 2017

a(n) = 2*a(n-1) + 4*a(n-2) + a(n-3).
G.f.: (4-7*x-5*x^2)/((1+x)*(1-3*x-x^2)). - Colin Barker, May 22 2012
a(n) = 2*(-1)^n + ((3-sqrt(13))/2)^n + ((3+sqrt(13))/2)^n. - Colin Barker, May 11 2017
a(n) = A006497+2*(-1)^n. - R. J. Mathar, Oct 20 2017

A253247 Pisano period of A006190(n^2) divided by Pisano period of A006190(n).

Original entry on oeis.org

1, 2, 3, 4, 5, 1, 7, 8, 9, 5, 11, 4, 13, 7, 5, 16, 17, 9, 19, 10, 21, 11, 23, 8, 25, 13, 27, 7, 29, 5, 31, 32, 33, 17, 35, 36, 37, 19, 39, 40, 41, 7, 43, 11, 45, 23, 47, 16, 49, 25, 51, 26, 53, 27, 55, 14, 57, 29, 59, 10, 61, 31, 63, 64, 65, 11, 67, 17, 69, 35, 71, 72

Offset: 1

Eric Chen, Apr 09 2015

nonn

For all n, a(n)|n.
Conjecture: a(n) = 1 only for n = 1 and 6. (This conjecture is true if and only if the generalized Wall's conjecture to A006190 is true.)
If there exists any prime p such that A175182(p^2) = A175182(p), then the conjecture fails.
For any prime p, these three statements are equivalent:
(1) A175182(p^2) = A175182(p).
(2) A006190(p-(p|13)) = 3 (mod p^2).
(3) A006497(p) = 1 (mod p^2).
Since A175182(241^2) = A175182(241) = 484, so the prime 241 is a Wall-Sun-Sun prime to A006190 (Lucas (P, Q) = (3, -1)) and no others < 10^8, so the conjecture is true for all primes < 10^8 except 241.
All of Wall's theorems are true for A175182. For example, let P(n) = A175182(n), p and q are primes, then P(pq) = lcm(P(p), P(q)), and for every prime p, P(p)|(p-1) if (p|13) = 1, P(p)|(2p+2) if (p|13) = -1 (P(13) = 52, which if divisible by 13), while (p|13) is the Legendre symbol, and the fixed points of A175182 are 1, 6, and 12*13^k, k>0.

Eric Chen, Table of n, a(n) for n = 1..1000

Cf. A237517, A175182.

F := proc(k, n) option remember; if n <= 1 then n; else k*procname(k, n-1)+procname(k, n-2) ; end if; end proc:
Pper := proc(k, m) local cha, zer, n, fmodm ; cha := [] ; zer := [] ; for n from 0 do fmodm := F(k, n) mod m ; cha := [op(cha), fmodm] ; if fmodm = 0 then zer := [op(zer), n] ; end if; if nops(zer) = 5 then break; end if; end do ; if [op(1..zer[2], cha) ] = [ op(zer[2]+1..zer[3], cha) ] and [op(1..zer[2], cha)] = [ op(zer[3]+1..zer[4], cha) ] and [op(1..zer[2], cha)] = [ op(zer[4]+1..zer[5], cha) ] then return zer[2] ; elif [op(1..zer[3], cha) ] = [ op(zer[3]+1..zer[5], cha) ] then return zer[3] ; else return zer[5] ; end if; end proc:
k := 3 ; seq( Pper(k, m^2) div Pper(k, m), m=1..300) ;

A006190[n_] := Fibonacci[n, 3];
A175182[n_] := Module[{k=1}, While[Mod[A006190[k], n] != 0 || Mod[A006190[k+1]-1, n] != 0, k++]; k];
Table[A175182[n^2] / A175182[n], {n, 72}] (* corrected by Jason Yuen, Jun 28 2025 *)

fibmod(n, m)=((Mod([3, 1; 1, 0], m))^n)[1, 2]
entry_p(p)=my(k=1, c=Mod(1, p), o); while(c, [o, c]=[c, 3*c+o]; k++); k
entry(n)=if(n==1, return(1)); my(f=factor(n), v); v=vector(#f~, i, if(f[i, 1]>1e8 && f[i, 1] != 241, entry_p(f[i, 1]^f[i, 2]), entry_p(f[i, 1])*f[i, 1]^(f[i, 2] - 1))); if(f[1, 1]==2&&f[1, 2]>1, v[1]=3<

a(n) = A175182(n^2) / A175182(n).

cp's OEIS Frontend

A080039 a(n) = floor((1+sqrt(2))^n).

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A180148 a(n) = 3*a(n-1) + a(n-2) with a(0)=2 and a(1)=5.

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A335670 Odd composite integers m such that A014448(m) == 4 (mod m).

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A335671 Odd composite integers m such that A087130(m) == 5 (mod m).

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A100230 Main diagonal of triangle A100229.

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A309220 Square array A read by antidiagonals: the columns are given by A(n,1)=1, A(n,2)=n+1, A(n,3) = n^2+2n+3, A(n,4) = n^3+3*n^2+6*n+4, A(n,5) = n^4+4*n^3+10*n^2+12*n+7, ..., whose coefficients are given by A104509 (see also A118981).

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A352362 Array read by ascending antidiagonals. T(n, k) = L(k, n) where L are the Lucas polynomials.

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A309220 Square array A read by antidiagonals: the columns are given by A(n,1)=1, A(n,2)=n+1, A(n,3) = n^2+2n+3, A(n,4) = n^3+3n^2+6n+4, A(n,5) = n^4+4n^3+10n^2+12*n+7, ..., whose coefficients are given by A104509 (see also A118981).