A000204 -id:A000204 - cp's OEIS frontend

A203860 G.f.: Product_{n>=1} (1 - Lucas(n)x^n + (-1)^nx^(2*n)) where Lucas(n) = A000204(n).

1, -1, -4, -1, 1, 11, 7, 25, 18, -11, -1, 0, -325, -199, 122, -1364, -843, 550, 0, 11, 123, 0, 39650, 24476, -15126, 0, 271443, 164194, -103682, -1364, -1, -24476, 0, -9349, -123, -20633239, -12752043, 7881225, -843, 0, -226965629, -141422125, 88114450, 0, 1

Offset: 0

Paul D. Hanna, Jan 07 2012

sign

a(A093519(n)) = 0 where A093519 lists numbers that are not equal to the sum of two generalized pentagonal numbers.

			G.f.: A(x) = 1 - x - 4*x^2 - x^3 + x^4 + 11*x^5 + 7*x^6 + 25*x^7 +...
-log(A(x)) = x + 3*3*x^2/2 + 4*4*x^3/3 + 7*7*x^4/4 + 6*11*x^5/5 + 12*18*x^6/6 +...+ sigma(n)*A000204(n)*x^n/n +...
The g.f. equals the product:
A(x) = (1-x-x^2) * (1-3*x^2+x^4) * (1-4*x^3-x^6) * (1-7*x^4+x^8) * (1-11*x^5-x^10) * (1-18*x^6+x^12) *...* (1 - Lucas(n)*x^n + (-1)^n*x^(2*n)) *...
Positions of zeros form A093519:
[11,18,21,25,32,39,43,46,49,54,60,65,67,68,74,76,81,87,88,90,...]
which are numbers that are not the sum of two generalized pentagonal numbers.

Paul D. Hanna, Table of n, a(n) for n = 0..1000

Cf. A203861, A203850, A156234, A093519.

/* Subroutine used in PARI programs below: */
{Lucas(n)=fibonacci(n-1)+fibonacci(n+1)}

{a(n)=polcoeff(exp(sum(k=1, n, -sigma(k)*Lucas(k)*x^k/k)+x*O(x^n)), n)}

{a(n)=polcoeff(prod(m=1, n, 1 - Lucas(m)*x^m + (-1)^m*x^(2*m) +x*O(x^n)), n)}

G.f.: exp( Sum_{n>=1} -sigma(n) * A000204(n) * x^n/n ).

A233359 a(n) = |{0 < k < n: L(k) + q(n-k) is prime}|, where L(k) is the k-th Lucas number (A000204), and q(.) is the strict partition function (A000009).

Original entry on oeis.org

0, 1, 1, 2, 3, 1, 2, 4, 2, 2, 3, 3, 2, 4, 3, 5, 1, 4, 5, 3, 1, 3, 3, 7, 3, 3, 4, 5, 2, 2, 9, 2, 4, 4, 9, 2, 6, 6, 6, 3, 3, 1, 5, 7, 4, 4, 5, 7, 4, 9, 5, 6, 4, 1, 5, 6, 11, 9, 4, 2, 5, 5, 4, 6, 8, 9, 12, 3, 7, 5, 4, 10, 6, 7, 6, 3, 5, 8, 4, 4, 4, 4, 7, 7, 5, 1, 4, 9, 7, 4, 8, 7, 6, 5, 2, 3, 7, 11, 5, 5

Offset: 1

Zhi-Wei Sun, Dec 08 2013

nonn

Conjecture: a(n) > 0 for all n > 1.
We have verified this for n up to 60000.
Note that for n = 19976 there is no k = 0,...,n such that F(k) + q(n-k) is prime, where F(0), F(1), ... are the Fibonacci numbers.

			a(7) = 2 since L(1) + q(6) = 1 + 4 = 5 and L(6) + q(1) = 18 + 1 = 19 are both prime.
a(17) = 1 since L(13) + q(4) = 521 + 2 = 523 is prime.
a(21) = 1 since L(5) + q(16) = 11 + 32 = 43 is prime.
a(42) = 1 since L(22) + q(20) = 39603 + 64 = 39667 is prime.
a(54) = 1 since L(8) + q(46) = 47 + 2304 = 2351 is prime.
a(86) = 1 since L(67) + q(19) = 100501350283429 + 54 = 100501350283483 is prime.

Zhi-Wei Sun, Table of n, a(n) for n = 1..5000

Cf. A000009, A000032, A000040, A000204, A233307, A233346, A233360.

a[n_]:=Sum[If[PrimeQ[LucasL[k]+PartitionsQ[n-k]],1,0],{k,1,n-1}]
Table[a[n],{n,1,100}]

A272909 Numbers that are the product of two Lucas numbers L(i), for i >= 1, using the Lucas numbers as defined in A000204.

Original entry on oeis.org

1, 3, 4, 7, 9, 11, 12, 16, 18, 21, 28, 29, 33, 44, 47, 49, 54, 72, 76, 77, 87, 116, 121, 123, 126, 141, 188, 198, 199, 203, 228, 304, 319, 322, 324, 329, 369, 492, 517, 521, 522, 532, 597, 796, 836, 841, 843, 846, 861, 966, 1288, 1353, 1363, 1364, 1368, 1393

Offset: 1

Clark Kimberling, May 10 2016

Conjecture: if c and d are consecutive terms, then d - c is a product of two Lucas numbers or a product of two Fibonacci numbers.

Clark Kimberling, Table of n, a(n) for n = 1..1000

Cf. A049997 (Fibonacci(i)*Fibonacci(j)), A000204.

Take[Union@Flatten@Table[LucasL[i] LucasL[j], {i, 0, 15}, {j, i}], 60] (* adapted by Vincenzo Librandi, Sep 04 2016 *)

A101033 Triangle read by rows giving the coefficients of general sum formulas of n-th Lucas numbers (A000204). The k-th row (k>=1) contains T(i,k) for i=1 to 2k-1, where T(i,k) satisfies L(n) = Sum_{k=1..n} Sum_{i=1..2k-1} T(i,k) * C(n-k,i-1) * n^(n-k) / (n-1)!.

Original entry on oeis.org

1, 1, -2, -3, 2, 15, 51, 65, 27, 6, -148, -945, -2292, -2776, -1680, -405, 24, 2290, 19580, 71965, 145525, 175244, 125950, 50085, 8505, 120, -41676, -473072, -2340400, -6676835, -12132890, -14587261, -11619692, -5290005, -1752030, -229635, 720, 943908, 13132532, 81977672, 303352938, 740797855

Offset: 1

André F. Labossière, Nov 30 2004

			L(7)= (1/(7-1)!) * [ 7^(7-1) -{-1+2*(7-2)+3*C(7-2,2)}*7^(7-2) +{2+15*(7-3)+51*C(7-3,2)+65*C(7-3,3) +27*C(7-3,4)}*7^(7-3) -{-6+148*(7-4)+945*C(7-4,2)+2292*C(7-4,3)}*7^(7-4) +... ]
= (1/6!) * [ 7^6 -{-1+10+30}*7^5 +{2+60+306+260+27}*7^4 -{-6+444+2835+2292}*7^3 +{24+4580+19580}*7^2 -{-120+41676}*7 +{720} ] = (1/6!) * [ 7^6 -39*7^5 +655*7^4 -5565*7^3 +24184*7^2 -41556*7 +720 ]
= (1/720) * [ 117649 -655473 +1572655 -1908795 +1185016 -290892 +720 ] = 20880/720 = 29.

Ron Knott, The Lucas Numbers in Pascal's Triangle.

Cf. A101032, A000204, A100492, A099731, A000045, A094216, A094638, A000108.

A104765 Triangle T(n,k) read by rows: row n contains the first n Lucas numbers A000204.

Original entry on oeis.org

1, 1, 3, 1, 3, 4, 1, 3, 4, 7, 1, 3, 4, 7, 11, 1, 3, 4, 7, 11, 18, 1, 3, 4, 7, 11, 18, 29, 1, 3, 4, 7, 11, 18, 29, 47, 1, 3, 4, 7, 11, 18, 29, 47, 76, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, 1, 3, 4, 7, 11

Offset: 1

Gary W. Adamson, Mar 24 2005

Reading rows from the right to the left yields A104764.

Sequence B is called a reluctant sequence of sequence A, if B is triangle array read by rows: row number k coincides with first k elements of the sequence A. Sequence A104765 is the reluctant sequence of A000204. - Boris Putievskiy, Dec 14 2012

			First few rows of the triangle are:
  1;
  1, 3;
  1, 3, 4;
  1, 3, 4, 7;
  1, 3, 4, 7, 11;
  1, 3, 4, 7, 11, 18;
  ...

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
Boris Putievskiy, Transformations Integer Sequences And Pairing Functions, arXiv:1212.2732 [math.CO], 2012.

Cf. A000204, A104765, A104762, A104763.

Cf. A027961 (row sums).

Table[LucasL[k], {n, 1, 10}, {k, 1, n}] // Flatten (* G. C. Greubel, Dec 21 2017 *)
Module[{nn=20,luc},luc=LucasL[Range[nn]];Table[Take[luc,n],{n,nn}]]//Flatten (* Harvey P. Dale, Jul 10 2024 *)

for(n=1,10, for(k=1,n, print1(fibonacci(k+1) + fibonacci(k-1), ", "))) \\ G. C. Greubel, Dec 21 2017

T(n,k) = A000204(k), 1<=k<=n.
T(n,k) = A104764(n,n-k+1).
a(n) = A000204(m), where m = n-t(t+1)/2, t = floor((-1+sqrt(8*n-7))/2). - Boris Putievskiy, Dec 14 2012
G.f.: (x*y*(2*x*y+1))/((x-1)*(x^2*y^2+x*y-1)). - Vladimir Kruchinin, Jun 21 2025

Edited and extended by R. J. Mathar, Jul 23 2008

A153175 a(n) = L(7*n)/L(n) where L(n) = Lucas number A000204(n).

Original entry on oeis.org

29, 281, 6119, 101521, 1875749, 33281921, 599786069, 10745088481, 192933544679, 3461223997001, 62114818827629, 1114566304366081, 20000347407134669, 358889844987430121, 6440029487834912999, 115561554399692896321

Offset: 1

Artur Jasinski, Dec 20 2008

nonn

All numbers in this sequence are:
congruent to 9 mod 10 (iff n is odd),
congruent to 1 mod 10 (iff n is even).

G. C. Greubel, Table of n, a(n) for n = 1..795
Index entries for linear recurrences with constant coefficients, signature (13,104,-260,-260,104,13,-1).

Cf. A000032, A000204, A110391, A153173.

Cf. A153177, A153179, A153180. [From R. J. Mathar, Oct 22 2010]

[Lucas(7*n)/Lucas(n): n in [0..30]]; // G. C. Greubel, Dec 21 2017

Table[LucasL[7*n]/LucasL[n], {n, 1, 50}]

{lucas(n) = fibonacci(n+1) + fibonacci(n-1)};
for(n=0,30, print1( lucas(7*n)/lucas(n), ", ")) \\ G. C. Greubel, Dec 21 2017

From R. J. Mathar, Oct 22 2010: (Start)
a(n) = +13*a(n-1) +104*a(n-2) -260*a(n-3) -260*a(n-4) +104*a(n-5) +13*a(n-6) -a(n-7).
G.f.: -x*(-29+96*x+550*x^2-290*x^3-200*x^4+16*x^5+x^6) / ( (1+x)*(x^2-3*x+1)*(x^2-18*x+1)*(x^2+7*x+1) ).
a(n) = A005248(n) +A087215(n) -(-1)^n*A056854(n) - (-1)^n. (End)

A153177 a(n) = L(9*n)/L(n) where L(n) = Lucas number A000204(n).

Original entry on oeis.org

76, 1926, 109801, 4769326, 230701876, 10716675201, 505618944676, 23714405408926, 1114769987764201, 52357935173823126, 2459933168462154076, 115560463558534156801, 5428954301161174383676, 255043991670277234750326

Offset: 1

Artur Jasinski, Dec 20 2008

nonn

All numbers in this sequence are:
congruent to 1 mod 100 (iff n is congruent to 0 mod 3),
congruent to 26 mod 100 (iff n is congruent to 2 or 4 mod 6),
congruent to 76 mod 100 (iff n is congruent to 1 or 5 mod 6).

G. C. Greubel, Table of n, a(n) for n = 1..595
Index entries for linear recurrences with constant coefficients, signature (34,714,-4641,-12376,12376,4641,-714,-34,1).

Cf. A000032, A000204, A110391, A153173, A153175.

Cf. A153179, A153180. - R. J. Mathar, Oct 22 2010

[Lucas(9*n)/Lucas(n): n in [0..30]]; // G. C. Greubel, Dec 21 2017

Table[LucasL[9*n]/LucasL[n], {n, 1, 50}]
LinearRecurrence[{34,714,-4641,-12376,12376,4641,-714,-34,1},{76,1926,109801,4769326,230701876,10716675201,505618944676,23714405408926,1114769987764201},20] (* Harvey P. Dale, Aug 12 2012 *)

{lucas(n) = fibonacci(n+1) + fibonacci(n-1)};
for(n=0,30, print1( lucas(9*n)/lucas(n), ", ")) \\ G. C. Greubel, Dec 21 2017

From R. J. Mathar, Oct 22 2010: (Start)
a(n) = 34*a(n-1) +714*a(n-2) -4641*a(n-3) -12376*a(n-4) +12376*a(n-5) +4641*a(n-6) -714*a(n-7) -34*a(n-8) +a(n-9).
G.f.: -x*(76-658*x-9947*x^2+13644*x^3+26020*x^4-5306*x^5-1372*x^6+42*x^7 +x^8) / ((x-1)*(x^2+18*x+1)*(x^2-47*x+1)*(x^2+3*x+1)*(x^2-7*x+1)).
a(n) = 1-(-1)^n*A087215(n) -(-1)^n*A005248(n) +A056854(n) +A087265(n). (End)

A153179 a(n) = L(11*n)/L(n) where L(n) = A000204(n).

Original entry on oeis.org

199, 13201, 1970299, 224056801, 28374454999, 3450736132801, 426236170575799, 52337681992411201, 6441140796368008699, 792018481913198430001, 97420733208491869044199, 11981539981561372141075201

Offset: 1

Artur Jasinski, Dec 20 2008

nonn

All numbers in this sequence are:
congruent to 99 mod 100 (iff n is odd),
congruent to 1 mod 100 (iff n is even).

G. C. Greubel, Table of n, a(n) for n = 1..475
Index entries for linear recurrences with constant coefficients, signature (89,4895,-83215,-582505,1514513,1514513,-582505,-83215,4895,89,-1).

Cf. A000032, A000204, A110391, A153173, A153175, A153177.

[Lucas(11*n)/Lucas(n): n in [0..30]]; // G. C. Greubel, Dec 21 2017

Table[LucasL[11*n]/LucasL[n], {n, 1, 50}]

{lucas(n) = fibonacci(n+1) + fibonacci(n-1)};
for(n=0,30, print1( lucas(11*n)/lucas(n), ", ")) \\ G. C. Greubel, Dec 21 2017

From R. J. Mathar, Oct 22 2010: (Start)
a(n) = +89*a(n-1) +4895*a(n-2) -83215*a(n-3) -582505*a(n-4) +1514513*a(n-5) +1514513*a(n-6) -582505*a(n-7) -83215*a(n-8) +4895*a(n-9) +89*a(n-10) -a(n-11).
G.f.: -1 -1/(1+x) +(-2-47*x)/(x^2+47*x+1) +(2-3*x)/(x^2-3*x+1) +(-2-7*x)/(x^2+7*x+1) +(2-123*x)/(x^2-123*x+1) +(2-18*x)/(x^2-18*x+1).
a(n) = -(-1)^n -(-1)^n*A087265(n) +A005248(n) -(-1)^n*A056854(n) +A065705(n) +A087215(n). (End)

A166168 G.f.: exp( Sum_{n>=1} Lucas(n^2)*x^n/n ) where Lucas(n) = A000204(n).

Original entry on oeis.org

1, 1, 4, 29, 585, 34212, 5600397, 2490542953, 2968152042068, 9416588994339205, 79216509536543420965, 1762508872870620792746360, 103525263562786817866762466405, 16031370626878431551103688398524485

Offset: 0

Paul D. Hanna, Oct 08 2009

nonn

Conjectured to consist entirely of integers.
The Lucas numbers (A000204) forms the logarithmic derivative of the Fibonacci numbers (A000045).
Note that Lucas(n^2) = [(1+sqrt(5))/2]^(n^2) + [(1-sqrt(5))/2]^(n^2).

			G.f.: A(x) = 1 + x + 4*x^2 + 29*x^3 + 585*x^4 + 34212*x^5 +...
log(A(x)) = x + 7*x^2/2 + 76*x^3/3 + 2207*x^4/4 + 167761*x^5/5 + 33385282*x^6/6 +...+ Lucas(n^2)*x^n/n +...

G. C. Greubel, Table of n, a(n) for n = 0..60
Sawian Jaidee, Patrick Moss, Tom Ward, Time-changes preserving zeta functions, arXiv:1809.09199 [math.DS], 2018.

Cf. A166169, A156216, A155200, A000204, A000045.

with(combinat): seq(coeff(series(exp(add((fibonacci(k^2-1)+fibonacci(k^2+1))*x^k/k,k=1..n)),x,n+1), x, n), n = 0 .. 15); # Muniru A Asiru, Dec 18 2018

CoefficientList[Series[Exp[Sum[LucasL[n^2]*x^n/n, {n, 1, 200}]], {x, 0, 50}], x](* G. C. Greubel, May 06 2016 *)

{a(n)=polcoeff(exp(sum(m=1,n,(fibonacci(m^2-1)+fibonacci(m^2+1))*x^m/m)+x*O(x^n)),n)}

a(n) = (1/n)*Sum_{k=1..n} Lucas(k^2)*a(n-k), a(0)=1.

Logarithmic derivative yields A166169.

A191988 Ordered sums 2f+5g, where f and g are Lucas numbers beginning at 1 (A000204).

Original entry on oeis.org

7, 11, 13, 17, 19, 21, 22, 23, 26, 27, 28, 29, 34, 37, 41, 42, 43, 49, 51, 56, 57, 61, 63, 69, 71, 73, 77, 78, 91, 92, 93, 96, 98, 99, 104, 109, 112, 113, 114, 126, 129, 147, 148, 149, 151, 153, 157, 159, 167, 172, 181, 184, 187, 203, 207, 237, 239, 241, 242

Offset: 1

Clark Kimberling, Jun 20 2011

nonn

Cf. A000204, A191989, A191990, A191991.

c = 2; d = 5; f[n_] := LucasL[n];
g[n_] := c*f[n]; h[n_] := d*f[n];
t[i_, j_] := h[i] + g[j];
u = Table[t[i, j], {i, 1, 20}, {j, 1, 20}];
v = Union[Flatten[u]]    (* A191988 *)
t1[i_, j_] := If[g[i] - h[j] > 0, g[i] - h[j], 0]
u1 = Table[t1[i, j], {i, 1, 20}, {j, 1, 20}];
v1 = Union[Flatten[u1]]  (* A191989: c*f(i)-d*f(j) *)
g1[n_] := d*f[n]; h1[n_] := c*f[n];
t2[i_, j_] := If[g1[i] - h1[j] > 0, g1[i] - h1[j], 0]
u2 = Table[t2[i, j], {i, 1, 20}, {j, 1, 20}];
v2 = Union[Flatten[u2]]  (* A191990: d*f(i)-c*f(j) *)
v3 = Union[v1, v2]       (* A191991 *)

cp's OEIS Frontend

A203860 G.f.: Product_{n>=1} (1 - Lucas(n)*x^n + (-1)^n*x^(2*n)) where Lucas(n) = A000204(n).

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A233359 a(n) = |{0 < k < n: L(k) + q(n-k) is prime}|, where L(k) is the k-th Lucas number (A000204), and q(.) is the strict partition function (A000009).

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A272909 Numbers that are the product of two Lucas numbers L(i), for i >= 1, using the Lucas numbers as defined in A000204.

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A101033 Triangle read by rows giving the coefficients of general sum formulas of n-th Lucas numbers (A000204). The k-th row (k>=1) contains T(i,k) for i=1 to 2*k-1, where T(i,k) satisfies L(n) = Sum_{k=1..n} Sum_{i=1..2*k-1} T(i,k) * C(n-k,i-1) * n^(n-k) / (n-1)!.

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A104765 Triangle T(n,k) read by rows: row n contains the first n Lucas numbers A000204.

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A153175 a(n) = L(7*n)/L(n) where L(n) = Lucas number A000204(n).

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A153177 a(n) = L(9*n)/L(n) where L(n) = Lucas number A000204(n).

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A153179 a(n) = L(11*n)/L(n) where L(n) = A000204(n).

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A203860 G.f.: Product_{n>=1} (1 - Lucas(n)x^n + (-1)^nx^(2*n)) where Lucas(n) = A000204(n).

A101033 Triangle read by rows giving the coefficients of general sum formulas of n-th Lucas numbers (A000204). The k-th row (k>=1) contains T(i,k) for i=1 to 2k-1, where T(i,k) satisfies L(n) = Sum_{k=1..n} Sum_{i=1..2k-1} T(i,k) * C(n-k,i-1) * n^(n-k) / (n-1)!.

A191988 Ordered sums 2f+5g, where f and g are Lucas numbers beginning at 1 (A000204).