A114525 -id:A114525 - cp's OEIS frontend

A286840 One of the two successive approximations up to 13^n for 13-adic integer sqrt(-1). Here the 5 (mod 13) case (except for n=0).

0, 5, 70, 239, 239, 143044, 1999509, 6826318, 6826318, 822557039, 85658552023, 1188526486815, 11941488851037, 291518510320809, 2108769149874327, 13920898306972194, 13920898306972194, 2675587335039691558, 63228498770709057089, 513050126578538629605

Offset: 0

Seiichi Manyama, Aug 01 2017

Seiichi Manyama, Table of n, a(n) for n = 0..897
Wikipedia, Hensel's Lemma.

The two successive approximations up to p^n for p-adic integer sqrt(-1): A048898 and A048899 (p=5), this sequence and A286841 (p=13), A286877 and A286878 (p=17).

Cf. A034944, A114525, A286838.

{0}~Join~Table[#&@@Select[PowerModList[-1, 1/2, 13^k], Mod[#, 13] == 5 &], {k, 20}]  (* Giorgos Kalogeropoulos, Oct 21 2022 *)

a(n) = truncate(sqrt(-1+O(13^n))); \\ Michel Marcus, Aug 04 2017

def A(k, m, n):
    ary=[0]
    a, mod = k, m
    for i in range(n):
          b=a%mod
          ary.append(b)
          a=b**m
          mod*=m
    return ary
def a286840(n):
    return A(5, 13, n)
print(a286840(100)) # Indranil Ghosh, Aug 03 2017, after Ruby

def A(k, m, n)
  ary = [0]
  a, mod = k, m
  n.times{
    b = a % mod
    ary << b
    a = b ** m
    mod *= m
  }
  ary
end
def A286840(n)
  A(5, 13, n)
end
p A286840(100)

a(0) = 0 and a(1) = 5, a(n) = a(n-1) + 9 * (a(n-1)^2 + 1) mod 13^n for n > 1.

a(n) == L(13^n,5) (mod 13^n) == ((5 + sqrt(29))/2)^(13^n) + ((5 - sqrt(29))/2)^(13^n) (mod 13^n), where L(n,x) denotes the n-th Lucas polynomial of A114525. - Peter Bala, Nov 20 2022

A286841 One of the two successive approximations up to 13^n for 13-adic integer sqrt(-1). Here the 8 (mod 13) case (except for n=0).

Original entry on oeis.org

0, 8, 99, 1958, 28322, 228249, 2827300, 55922199, 808904403, 9781942334, 52199939826, 603633907222, 11356596271444, 11356596271444, 1828607235824962, 37264994707118563, 651495710876207647, 5974828584341646375, 49226908181248336040

Offset: 0

Seiichi Manyama, Aug 01 2017

Seiichi Manyama, Table of n, a(n) for n = 0..899
Wikipedia, Hensel's Lemma.

The two successive approximations up to p^n for p-adic integer sqrt(-1): A048898 and A048899 (p=5), A286840 and this sequence (p=13), A286877 and A286878 (p=17).

Cf. A114525, A286839.

{0}~Join~Table[#&@@Select[PowerModList[-1, 1/2, 13^k], Mod[#, 13] == 8 &], {k, 18}] (* Giorgos Kalogeropoulos, Oct 22 2022 *)

a(n) = if (n, 13^n - truncate(sqrt(-1+O(13^n))), 0); \\ Michel Marcus, Aug 04 2017

def A(k, m, n):
    ary=[0]
    a, mod = k, m
    for i in range(n):
          b=a%mod
          ary.append(b)
          a=b**m
          mod*=m
    return ary
def a286841(n):
    return A(8, 13, n)
print(a286841(100)) # Indranil Ghosh, Aug 03 2017, after Ruby

def A(k, m, n)
  ary = [0]
  a, mod = k, m
  n.times{
    b = a % mod
    ary << b
    a = b ** m
    mod *= m
  }
  ary
end
def A286841(n)
  A(8, 13, n)
end
p A286841(100)

If n > 0, a(n) = 13^n - A286840(n).
a(0) = 0 and a(1) = 8, a(n) = a(n-1) + 4 * (a(n-1)^2 + 1) mod 13^n for n > 1.
a(n) == L(13^n,8) (mod 13^n) == (4 + sqrt(17))^(13^n) + (4 - sqrt(17))^(13^n) (mod 13^n), where L(n,x) denotes the n-th Lucas polynomial of A114525. - Peter Bala, Nov 20 2022

A001655 Fibonomial coefficients: a(n) = F(n+1) * F(n+2) * F(n+3)/2, where F() = Fibonacci numbers A000045.

Original entry on oeis.org

1, 3, 15, 60, 260, 1092, 4641, 19635, 83215, 352440, 1493064, 6324552, 26791505, 113490195, 480752895, 2036500788, 8626757644, 36543528780, 154800876945, 655747029795, 2777789007071, 11766903040368, 49845401197200, 211148507782800, 894439432403425

Offset: 0

N. J. A. Sloane

In a triangle having sides of F(n+1), 2*F(n+2) and F(n+3), the product of the area and circumradius will be a(n). For example: a triangle having sides of 5, 16 and 13 will have an area of 4*sqrt(51), a circumradius of 65*sqrt(51)/51, and the product is 4*65 = 260. - Gary Detlefs, Dec 14 2010

Explanation of this comment: if a triangle with sides (a, b, c) has a circumradius R and an area A, then A*R = abc/4; here, with a = F(n+1), b=2*F(n+2) and c=F(n+3), this gives a(n)= A*R. - Bernard Schott, Jan 26 2023

			G.f. = 1 + 3*x + 15*x^2 + 60*x^3 + 260*x^4 + 1092*x^5 + 4641*x^6 + ...

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..200
A. Brousseau, A sequence of power formulas, Fib. Quart., 6 (1968), 81-83.
A. Brousseau, Fibonacci and Related Number Theoretic Tables, Fibonacci Association, San Jose, CA, 1972, p. 74.
Kenneth Edwards and Michael A. Allen, A new combinatorial interpretation of the Fibonacci numbers cubed, Fib. Q. 58:5 (2020) 128-134.
Milan Janjic and B. Petkovic, A Counting Function, arXiv preprint arXiv:1301.4550 [math.CO], 2013. - From _N. J. A. Sloane_, Feb 13 2013
Ronald Orozco López, Generating Functions of Generalized Simplicial Polytopic Numbers and (s,t)-Derivatives of Partial Theta Function, arXiv:2408.08943 [math.CO], 2024. See p. 13.
Ronald Orozco López, Simplicial d-Polytopic Numbers Defined on Generalized Fibonacci Polynomials, arXiv:2501.11490 [math.CO], 2025. See p. 10.
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
David Treeby, Further Physical Derivations of Fibonacci Summations, Fibonacci Quart. 54 (2016), no. 4, 327-334.
Index entries for linear recurrences with constant coefficients, signature (3,6,-3,-1).

Cf. A066258 (first differences), A215037 (partial sums), A363753 (alternating sums).

Cf. A000045, A055870, A065563, A079586, A114525, A256178.

[Fibonacci(n+3)*Fibonacci(n+2)*Fibonacci(n+1)/2: n in [0..30]]; // Vincenzo Librandi, May 09 2016

A001655:=1/(z**2-z-1)/(z**2+4*z-1); # Simon Plouffe in his 1992 dissertation.

Table[(Fibonacci[n+3]*Fibonacci[n+2]*Fibonacci[n+1])/2, {n,0,40}] (* Vladimir Joseph Stephan Orlovsky, Nov 23 2009 *)
LinearRecurrence[{3, 6, -3, -1}, {1, 3, 15, 60}, 25] (* Jean-François Alcover, Sep 23 2017 *)

b(n, k)=prod(j=1, k, fibonacci(n+j)/fibonacci(j)); vector(20, n, b(n-1, 3))  \\ Joerg Arndt, May 08 2016

G.f.: 1/(1-3*x-6*x^2+3*x^3+x^4) = 1/((1+x-x^2)*(1-4*x-x^2)) (see Comments to A055870).
a(n) = A010048(n+3, 3) = fibonomial(n+3, 3).
a(n) = (1/2) * A065563(n).
a(n) = 4*a(n-1) + a(n-2) + ((-1)^n)*F(n+1), n >= 2; a(0)=1, a(1)=3.
a(n) = (F(n+3)^3 - F(n+2)^3 - F(n+1)^3)/6. - Gary Detlefs, Dec 24 2010
a(n-1) = Sum_{k=0..n} F(k+1)*F(k)^2, n >= 1. - Wolfdieter Lang, Aug 01 2012
From Wolfdieter Lang, Aug 09 2012: (Start)
a(n-1)*(-1)^n =  Sum_{k=0..n} (-1)^k*F(k+1)^2*F(k), n >= 1. See the link under A215037, eq. (25).
a(n) = (F(3*(n+2)) + 2*(-1)^n*F(n+2))/10, n >= 0. See the same link, eq. (32). (End)
a(n) = -a(-4-n)*(-1)^n for all n in Z. - Michael Somos, Sep 19 2014
0 = a(n)*(-a(n+1) - a(n+2)) + a(n+1)*(-3*a(n+1) + a(n+2)) for all n in Z. - Michael Somos, Sep 19 2014
O.g.f.: exp( Sum_{n >= 1} L(n)*L(2*n)*x^n/n ), where L(n) = A000032(n) is a Lucas number. Cf. A114525, A256178. - Peter Bala, Mar 18 2015
Sum_{n>=0} (-1)^n/a(n) = 2 * A079586 - 6. - Amiram Eldar, Oct 04 2020
The formula by Gary Detlefs above is valid for all sequences of the Fibonacci type f(n) = f(n-1) + f(n-2): 3*f(n+2)*f(n+1)*f(n) = f(n+2)^3 - f(n+1)^3 - f(n)^3. - Klaus Purath, Mar 25 2021
a(n) = sqrt(Sum_{j=1..n+1} F(j)^3*F(j+1)^3). See Treeby link. - Michel Marcus, Apr 10 2022
a(n) = Sum_{k=1..n+1} A000129(k)*A056570(n+2-k). - Michael A. Allen, Jan 25 2023
G.f.: exp( Sum_{k>=1} F(4*k)/F(k) * x^k/k ), where F(n) = A000045(n). - Seiichi Manyama, May 07 2025

A090305 a(n) = 16*a(n-1) + a(n-2), starting with a(0) = 2 and a(1) = 16.

Original entry on oeis.org

2, 16, 258, 4144, 66562, 1069136, 17172738, 275832944, 4430499842, 71163830416, 1143051786498, 18359992414384, 294902930416642, 4736806879080656, 76083812995707138, 1222077814810394864, 19629328849962024962

Offset: 0

Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Jan 25 2004

Lim_{n-> infinity} a(n)/a(n+1) = 0.0622577... = 1/(8+sqrt(65)) = (sqrt(65)-8).

Lim_{n-> infinity} a(n+1)/a(n) = 16.0622577... = (8+sqrt(65)) = 1/(sqrt(65)-8).

			a(4) = 16*a(3) + a(2) = 16*4144 + 258 = (8+sqrt(65))^4 + (8-sqrt(65))^4 = 66561.99998497... + 0.00001502... = 66562.

G. C. Greubel, Table of n, a(n) for n = 0..500
Tanya Khovanova, Recursive Sequences
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)
Index entries for linear recurrences with constant coefficients, signature (16,1).

Cf. A002070, A015910.
Lucas polynomials: A114525.
Lucas polynomials Lucas(n,m): A000032 (m=1), A002203 (m=2), A006497 (m=3), A014448 (m=4), A087130 (m=5), A085447 (m=6), A086902 (m=7), A086594 (m=8), A087798 (m=9), A086927 (m=10), A001946 (m=11), A086928 (m=12), A088316 (m=13), A090300 (m=14), A090301 (m=15), this sequence (m=16), A090306 (m=17), A090307 (m=18), A090308 (m=19), A090309 (m=20), A090310 (m=21), A090313 (m=22), A090314 (m=23), A090316 (m=24), A330767 (m=25), A087281 (m=29), A087287 (m=76), A089772 (m=199).

m:=16;; a:=[2,m];; for n in [3..20] do a[n]:=m*a[n-1]+a[n-2]; od; a; # G. C. Greubel, Dec 31 2019

m:=16; I:=[2,m]; [n le 2 select I[n] else m*Self(n-1) +Self(n-2): n in [1..20]]; // G. C. Greubel, Dec 31 2019

seq(simplify(2*(-I)^n*ChebyshevT(n, 8*I)), n = 0..20); # G. C. Greubel, Dec 31 2019

LinearRecurrence[{16,1},{2,16},40] (* or *) With[{c=Sqrt[65]}, Simplify/@ Table[(c-8)((8+c)^n-(8-c)^n (129+16c)),{n,20}]] (* Harvey P. Dale, Oct 31 2011 *)
LucasL[Range[20]-1, 16] (* G. C. Greubel, Dec 31 2019 *)

vector(21, n, 2*(-I)^(n-1)*polchebyshev(n-1, 1, 8*I) ) \\ G. C. Greubel, Dec 31 2019

[2*(-I)^n*chebyshev_T(n, 8*I) for n in (0..20)] # G. C. Greubel, Dec 31 2019

a(n) = 16*a(n-1) + a(n-2), starting with a(0) = 2 and a(1) = 16.
a(n) = (8+sqrt(65))^n + (8-sqrt(65))^n.
a(n)^2 = a(2n) - 2 if n = 1, 3, 5, ...;
a(n)^2 = a(2n) + 2 if n = 2, 4, 6, ....
G.f.: (2-16*x)/(1-16*x-x^2). - Philippe Deléham, Nov 02 2008
a(n) = Lucas(n, 16) = 2*(-i)^n * ChebyshevT(n, 8*i). - G. C. Greubel, Dec 31 2019
E.g.f.: 2*exp(8*x)*cosh(sqrt(65)*x). - Stefano Spezia, Jan 01 2020

More terms from Ray Chandler, Feb 14 2004

A286877 One of the two successive approximations up to 17^n for 17-adic integer sqrt(-1). Here the 4 (mod 17) case (except for n=0).

Original entry on oeis.org

0, 4, 38, 2928, 27493, 1029745, 23747457, 313398285, 3596107669, 94280954402, 450044583893, 28673959190179, 28673959190179, 3524407382568745, 13428985415474682, 13428985415474682, 42949774758062711577, 91610966633729580058, 6709533061724423693474

Offset: 0

Seiichi Manyama, Aug 02 2017

x = ...GC5A24,

x^2 = ...GGGGGG = -1.

			a(1) = (   4)_17 = 4,
a(2) = (  24)_17 = 38,
a(3) = ( A24)_17 = 2928,
a(4) = (5A24)_17 = 27493.

Seiichi Manyama, Table of n, a(n) for n = 0..813
Peter Bala, Using Lucas polynomials to find the p-adic square roots of -1, -2 and -3, Dec 2022.
Wikipedia, Hensel's Lemma.

The two successive approximations up to p^n for p-adic integer sqrt(-1): A048898 and A048899 (p=5), A286840 and A286841 (p=13), this sequence and A286878 (p=17).

a(n) = truncate(sqrt(-1+O(17^n))); \\ Michel Marcus, Aug 04 2017

def A(k, m, n):
    ary=[0]
    a, mod = k, m
    for i in range(n):
          b=a%mod
          ary.append(b)
          a=b**m
          mod*=m
    return ary
def a286877(n):
    return A(4, 17, n)
print(a286877(100)) # Indranil Ghosh, Aug 03 2017

def A(k, m, n)
  ary = [0]
  a, mod = k, m
  n.times{
    b = a % mod
    ary << b
    a = b ** m
    mod *= m
  }
  ary
end
def A286877(n)
  A(4, 17, n)
end
p A286877(100)

a(0) = 0 and a(1) = 4, a(n) = a(n-1) + 2 * (a(n-1)^2 + 1) mod 17^n for n > 1.

a(n) == L(17^n,4) (mod 17^n) == (2 + sqrt(5))^(17^n) + (2 - sqrt(5))^(17^n) (mod 17^n), where L(n,x) denotes the n-th Lucas polynomial of A114525. - Peter Bala, Dec 02 2022

A290803 One of the two successive approximations up to 7^n for the 7-adic integer sqrt(-3). These are the numbers congruent to 2 mod 7 (except for the initial 0).

Original entry on oeis.org

0, 2, 37, 37, 2095, 14100, 47714, 165363, 988906, 29812911, 271934553, 1401835549, 9311142521, 36993716923, 133882727330, 2846775018726, 12341898038612, 145273620317016, 1308426190253051, 1308426190253051, 35505111746372480, 354674176936820484

Offset: 0

Seiichi Manyama, Aug 11 2017

x = ...256052,

x^2 = ...666664 = -3.

			a(1) =     2_7 = 2,
a(2) =    52_7 = 37,
a(3) =    52_7 = 37,
a(4) =  6052_7 = 2095,
a(5) = 56052_7 = 14100.

Seiichi Manyama, Table of n, a(n) for n = 0..1184
Peter Bala, Using Lucas polynomials to find the p-adic square roots of -1, -2 and -3, Dec 2022.
Wikipedia, Hensel's Lemma.

Cf. A114525, A290796, A290804.

a(n) = if (n, truncate(sqrt(-3+O(7^(n)))), 0)

a(0) = 0 and a(1) = 2, a(n) = a(n-1) + 5 * (a(n-1)^2 + 3) mod 7^n for n > 1.

a(n) = L(7^n,2) (mod 7^n) = ( (1 + sqrt(2))^(7^n) + (1 - sqrt(2))^(7^n) ) (mod 7^n), where L(n,x) denotes the n-th Lucas polynomial of A114525. - Peter Bala, Nov 28 2022

A290804 One of the two successive approximations up to 7^n for the 7-adic integer sqrt(-3). These are the numbers congruent to 5 mod 7 (except for the initial 0).

Original entry on oeis.org

0, 5, 12, 306, 306, 2707, 69935, 658180, 4775895, 10540696, 10540696, 575491194, 4530144680, 59895293484, 544340345519, 1900786491217, 20891032530989, 87356893670191, 319987407657398, 10090468995120092, 44287154551239521, 203871687146463523

Offset: 0

Seiichi Manyama, Aug 11 2017

x = ...410615,

x^2 = ...666664 = -3.

			a(1) =     5_7 = 5,
a(2) =    15_7 = 12,
a(3) =   615_7 = 306,
a(4) =   615_7 = 306,
a(5) = 10615_7 = 2707.

Seiichi Manyama, Table of n, a(n) for n = 0..1183
Peter Bala, Using Lucas polynomials to find the p -adic square roots of -1, -2 and -3, Dec 2022.
Wikipedia, Hensel's Lemma.

Cf. A114525, A290797, A290803.

a(n) = if (n, 7^n - truncate(sqrt(-3+O(7^(n)))), 0)

a(0) = 0 and a(1) = 5, a(n) = a(n-1) + 2 * (a(n-1)^2 + 3) mod 7^n for n > 1.
If n > 0, a(n) = 7^n - A290803(n).
a(n) = L(7^n,5) (mod 7^n) = ( ((5 + sqrt(29))/2)^(7^n) + ((5 - sqrt(29))/2)^(7^n) ) (mod 7^n), where L(n,x) denotes the n-th Lucas polynomial of A114525. - Peter Bala, Nov 28 2022

A110391 a(n) = L(3*n)/L(n), where L(n) = Lucas number.

Original entry on oeis.org

1, 4, 6, 19, 46, 124, 321, 844, 2206, 5779, 15126, 39604, 103681, 271444, 710646, 1860499, 4870846, 12752044, 33385281, 87403804, 228826126, 599074579, 1568397606, 4106118244, 10749957121, 28143753124, 73681302246, 192900153619, 505019158606, 1322157322204

Offset: 0

Amarnath Murthy, Jul 27 2005

Subsidiary sequences: a(n) = L((2k+1)*n)/L(n) for k = 2,3, etc. This is the sequence for k = 1.

			a(1) = L(3)/L(1) = 4/1 = 4.

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

Cf. A000204, A000032, A002878, A114525, A153173, A153175, A153177, A153179, A153180.

[Lucas(3*n)/Lucas(n): n in [0..30]]; // G. C. Greubel, Dec 17 2017

with(combinat): L:=n->fibonacci(n+2)-fibonacci(n-2): seq(L(3*n)/L(n),n=0..30); # Emeric Deutsch, Jul 31 2005

Table[LucasL[3 n]/LucasL[n], {n, 0, 27}] (* Michael De Vlieger, Mar 18 2015 *)
LinearRecurrence[{2,2,-1},{1,4,6},40] (* Harvey P. Dale, Aug 20 2020 *)

Vec((1+2*x-4*x^2)/((1+x)*(x^2-3*x+1)) + O(x^30)) \\ Colin Barker, Jun 03 2016

for(n=0,30, print1((fibonacci(3*n+1) + fibonacci(3*n-1))/( fibonacci(n+1) + fibonacci(n-1)), ", ")) \\ G. C. Greubel, Dec 17 2017

From R. J. Mathar, Oct 18 2010: (Start)
a(n) = A005248(n) - (-1)^n.
a(n) = +2*a(n-1) +2*a(n-2) -a(n-3).
G.f.: ( 1+2*x-4*x^2 ) / ( (1+x)*(x^2-3*x+1) ). (End)
Exp( Sum_{n >= 1} a(n)*t^n/n ) = 1 + 4*t + 11*t^2 + 29*t^3 + ... is the o.g.f. for A002878. This is the case x = 1 of the general result exp( Sum_{n >= 1} L(3*n,x)/L(n,x)*t^n/n ) = Sum_{n >= 0} L(2*n + 1,x)*t^n, where L(n,x) is the n-th Lucas polynomial of A114525. - Peter Bala, Mar 18 2015
a(n) = 2^(-n)*(-(-2)^n+(3-sqrt(5))^n+(3+sqrt(5))^n). - Colin Barker, Jun 03 2016

Corrected and extended by Emeric Deutsch and Erich Friedman, Jul 31 2005

A374439 Triangle read by rows: the coefficients of the Lucas-Fibonacci polynomials. T(n, k) = T(n - 1, k) + T(n - 2, k - 2) with initial values T(n, k) = k + 1 for k < 2.

Original entry on oeis.org

1, 1, 2, 1, 2, 1, 1, 2, 2, 2, 1, 2, 3, 4, 1, 1, 2, 4, 6, 3, 2, 1, 2, 5, 8, 6, 6, 1, 1, 2, 6, 10, 10, 12, 4, 2, 1, 2, 7, 12, 15, 20, 10, 8, 1, 1, 2, 8, 14, 21, 30, 20, 20, 5, 2, 1, 2, 9, 16, 28, 42, 35, 40, 15, 10, 1, 1, 2, 10, 18, 36, 56, 56, 70, 35, 30, 6, 2

Offset: 0

Peter Luschny, Jul 22 2024

There are several versions of Lucas and Fibonacci polynomials in this database. Our naming follows the convention of calling polynomials after the values of the polynomials at x = 1. This assumes a regular sequence of polynomials, that is, a sequence of polynomials where degree(p(n)) = n. This view makes the coefficients of the polynomials (the terms of a row) a refinement of the values at the unity.
A remarkable property of the polynomials under consideration is that they are dual in this respect. This means they give the Lucas numbers at x = 1 and the Fibonacci numbers at x = -1 (except for the sign). See the example section.
The Pell numbers and the dual Pell numbers are also values of the polynomials, at the points x = -1/2 and x = 1/2 (up to the normalization factor 2^n). This suggests a harmonized terminology: To call 2^n*P(n, -1/2) = 1, 0, 1, 2, 5, ... the Pell numbers (A000129) and 2^n*P(n, 1/2) = 1, 4, 9, 22, ... the dual Pell numbers (A048654).
Based on our naming convention one could call A162515 (without the prepended 0) the Fibonacci polynomials. In the definition above only the initial values would change to: T(n, k) = k + 1 for k < 1. To extend this line of thought we introduce A374438 as the third triangle of this family.
The triangle is closely related to the qStirling2 numbers at q = -1. For the definition of these numbers see A333143. This relates the triangle to A065941 and A103631.

			Triangle starts:
  [ 0] [1]
  [ 1] [1, 2]
  [ 2] [1, 2, 1]
  [ 3] [1, 2, 2,  2]
  [ 4] [1, 2, 3,  4,  1]
  [ 5] [1, 2, 4,  6,  3,  2]
  [ 6] [1, 2, 5,  8,  6,  6,  1]
  [ 7] [1, 2, 6, 10, 10, 12,  4,  2]
  [ 8] [1, 2, 7, 12, 15, 20, 10,  8,  1]
  [ 9] [1, 2, 8, 14, 21, 30, 20, 20,  5,  2]
  [10] [1, 2, 9, 16, 28, 42, 35, 40, 15, 10, 1]
.
Table of interpolated sequences:
  |  n | A039834 & A000045 | A000032 |   A000129   |   A048654  |
  |  n |     -P(n,-1)      | P(n,1)  |2^n*P(n,-1/2)|2^n*P(n,1/2)|
  |    |     Fibonacci     |  Lucas  |     Pell    |    Pell*   |
  |  0 |        -1         |     1   |       1     |       1    |
  |  1 |         1         |     3   |       0     |       4    |
  |  2 |         0         |     4   |       1     |       9    |
  |  3 |         1         |     7   |       2     |      22    |
  |  4 |         1         |    11   |       5     |      53    |
  |  5 |         2         |    18   |      12     |     128    |
  |  6 |         3         |    29   |      29     |     309    |
  |  7 |         5         |    47   |      70     |     746    |
  |  8 |         8         |    76   |     169     |    1801    |
  |  9 |        13         |   123   |     408     |    4348    |

Paolo Xausa, Rows n = 0..150 of the triangle, flattened
Peter Luschny, Illustrating the polynomials.

Triangles related to Lucas polynomials: A034807, A114525, A122075, A061896, A352362.
Triangles related to Fibonacci polynomials: A162515, A053119, A168561, A049310, A374441.
Sums include: A000204 (Lucas numbers, row), A000045 & A212804 (even sums, Fibonacci numbers), A006355 (odd sums), A039834 (alternating sign row).
Type m^n*P(n, 1/m): A000129 & A048654 (Pell, m=2), A108300 & A003688 (m=3), A001077 & A048875 (m=4).
Adding and subtracting the values in a row of the table (plus halving the values obtained in this way): A022087, A055389, A118658, A052542, A163271, A371596, A324969, A212804, A077985, A069306, A215928.
Columns include: A040000 (k=1), A000027 (k=2), A005843 (k=3), A000217 (k=4), A002378 (k=5).
Diagonals include: A000034 (k=n), A029578 (k=n-1), abs(A131259) (k=n-2).
Cf. A029578 (subdiagonal), A124038 (row reversed triangle, signed).
Cf. A039961, A065941 (qStirling2), A103631, A108299, A131259. A133607, A194005, A333143, A374438 (m=3).

function T(n,k) // T = A374439
  if k lt 0 or k gt n then return 0;
  elif k le 1 then return k+1;
  else return T(n-1,k) + T(n-2,k-2);
  end if;
end function;
[T(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jan 23 2025

A374439 := (n, k) -> ifelse(k::odd, 2, 1)*binomial(n - irem(k, 2) - iquo(k, 2), iquo(k, 2)):
# Alternative, using the function qStirling2 from A333143:
T := (n, k) -> 2^irem(k, 2)*qStirling2(n, k, -1):
seq(seq(T(n, k), k = 0..n), n = 0..10);

A374439[n_, k_] := (# + 1)*Binomial[n - (k + #)/2, (k - #)/2] & [Mod[k, 2]];
Table[A374439[n, k], {n, 0, 10}, {k, 0, n}]//Flatten (* Paolo Xausa, Jul 24 2024 *)

from functools import cache
@cache
def T(n: int, k: int) -> int:
    if k > n: return 0
    if k < 2: return k + 1
    return T(n - 1, k) + T(n - 2, k - 2)

from math import comb as binomial
def T(n: int, k: int) -> int:
    o = k & 1
    return binomial(n - o - (k - o) // 2, (k - o) // 2) << o

def P(n, x):
    if n < 0: return P(n, x)
    return sum(T(n, k)*x**k for k in range(n + 1))
def sgn(x: int) -> int: return (x > 0) - (x < 0)
# Table of interpolated sequences
print("|  n | A039834 & A000045 | A000032 |   A000129   |   A048654  |")
print("|  n |     -P(n,-1)      | P(n,1)  |2^n*P(n,-1/2)|2^n*P(n,1/2)|")
print("|    |     Fibonacci     |  Lucas  |     Pell    |    Pell*   |")
f = "| {0:2d} | {1:9d}         |  {2:4d}   |   {3:5d}     |    {4:4d}    |"
for n in range(10): print(f.format(n, -P(n, -1), P(n, 1), int(2**n*P(n, -1/2)), int(2**n*P(n, 1/2))))

from sage.combinat.q_analogues import q_stirling_number2
def A374439(n,k): return (-1)^((k+1)//2)*2^(k%2)*q_stirling_number2(n+1, k+1, -1)
print(flatten([[A374439(n, k) for k in range(n+1)] for n in range(13)])) # G. C. Greubel, Jan 23 2025

T(n, k) = 2^k' * binomial(n - k' - (k - k') / 2, (k - k') / 2) where k' = 1 if k is odd and otherwise 0.
T(n, k) = (1 + (k mod 2))*qStirling2(n, k, -1), see A333143.
2^n*P(n, -1/2) = A000129(n - 1), Pell numbers, P(-1) = 1.
2^n*P(n, 1/2) = A048654(n), dual Pell numbers.
T(2*n, n) = (1/2)*(-1)^n*( (1+(-1)^n)*A005809(n/2) - 2*(1-(-1)^n)*A045721((n-1)/2) ). - G. C. Greubel, Jan 23 2025

A162514 Triangle of coefficients of polynomials defined by the Binet form P(n,x) = U^n + L^n, where U = (x + d)/2, L = (x - d)/2, d = (4 + x^2)^(1/2). Decreasing powers of x.

Original entry on oeis.org

2, 1, 0, 1, 0, 2, 1, 0, 3, 0, 1, 0, 4, 0, 2, 1, 0, 5, 0, 5, 0, 1, 0, 6, 0, 9, 0, 2, 1, 0, 7, 0, 14, 0, 7, 0, 1, 0, 8, 0, 20, 0, 16, 0, 2, 1, 0, 9, 0, 27, 0, 30, 0, 9, 0, 1, 0, 10, 0, 35, 0, 50, 0, 25, 0, 2, 1, 0, 11, 0, 44, 0, 77, 0, 55, 0, 11, 0, 1, 0, 12, 0, 54, 0, 112, 0, 105, 0, 36, 0, 2, 1, 0, 13, 0

Offset: 0

Clark Kimberling, Jul 05 2009

For a signed version of this triangle corresponding to the row reversed version of the triangle A127672 see A244422. - Wolfdieter Lang, Aug 07 2014

The row reversed triangle is A114525. - Paolo Bonzini, Jun 23 2016

			Triangle begins
   2;  == 2
   1, 0;  == x + 0
   1, 0,  2;  == x^2 + 2
   1, 0,  3, 0;  == x^3 + 3*x + 0
   1, 0,  4, 0,  2;
   1, 0,  5, 0,  5, 0;
   1, 0,  6, 0,  9, 0,  2;
   1, 0,  7, 0, 14, 0,  7, 0;
   1, 0,  8, 0, 20, 0, 16, 0,  2;
   1, 0,  9, 0, 27, 0, 30, 0,  9, 0;
   1, 0, 10, 0, 35, 0, 50, 0, 25, 0, 2;
   ...
From _Wolfdieter Lang_, Aug 07 2014: (Start)
The row polynomials R(n, x) are:
  R(0, x) = 2, R(1, x) = 1 =   x*P(1,1/x),  R(2, x) = 1 + 2*x^2 = x^2*P(2,1/x), R(3, x) = 1 + 3*x^2 = x^3*P(3,1/x), ...
(End)

G. C. Greubel, Rows n=0..100 of triangle, flattened

Cf. A000032, A114525, A162515, A162516, A162517.

Table[Reverse[CoefficientList[LucasL[n, x], x]], {n, 0, 12}]//Flatten  (* G. C. Greubel, Nov 05 2018 *)

P(n)=
{
    local(U, L, d, r, x);
    if ( n<0, return(0) );
    x = 'x+O('x^(n+1));
    d=(4 + x^2)^(1/2);
    U=(x+d)/2;  L=(x-d)/2;
    r = U^n+L^n;
    r = truncate(r);
    return( r );
}
for (n=0, 10, print(Vec(P(n))) ); /* show triangle */
/* Joerg Arndt, Jul 24 2011 */

P(n,x) = x*P(n-1,x) + P(n-2,x) for n >= 2, P(0,x) = 2, P(1,x) = x.
From Wolfdieter Lang, Aug 07 2014: (Start)
T(n,m) = [x^(n-m)] P(n,x), m = 0, 1, ..., n and  n >= 0.
G.f. of polynomials P(n,x): (2 - x*z)/(1 - x*z - z^2).
G.f. of row polynomials R(n,x) = Sum_{m=0..n} T(n,m)*x^m:  (2 - z)/(1 - z - (x*z)^2) (rows for P(n,x) reversed).
(End)
For n > 0, T(n,2*m+1) = 0, T(n,2*m) = A034807(n,m). - Paolo Bonzini, Jun 23 2016

Name clarified by Wolfdieter Lang, Aug 07 2014

cp's OEIS Frontend

A286840 One of the two successive approximations up to 13^n for 13-adic integer sqrt(-1). Here the 5 (mod 13) case (except for n=0).

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A286841 One of the two successive approximations up to 13^n for 13-adic integer sqrt(-1). Here the 8 (mod 13) case (except for n=0).

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A001655 Fibonomial coefficients: a(n) = F(n+1) * F(n+2) * F(n+3)/2, where F() = Fibonacci numbers A000045.

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A090305 a(n) = 16*a(n-1) + a(n-2), starting with a(0) = 2 and a(1) = 16.

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A286877 One of the two successive approximations up to 17^n for 17-adic integer sqrt(-1). Here the 4 (mod 17) case (except for n=0).

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A290803 One of the two successive approximations up to 7^n for the 7-adic integer sqrt(-3). These are the numbers congruent to 2 mod 7 (except for the initial 0).

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