A047221 -id:A047221 - cp's OEIS frontend

A273373 Squares ending in digit 6.

16, 36, 196, 256, 576, 676, 1156, 1296, 1936, 2116, 2916, 3136, 4096, 4356, 5476, 5776, 7056, 7396, 8836, 9216, 10816, 11236, 12996, 13456, 15376, 15876, 17956, 18496, 20736, 21316, 23716, 24336, 26896, 27556, 30276, 30976, 33856, 34596, 37636, 38416, 41616

Offset: 1

Vincenzo Librandi, May 21 2016

These are the only squares whose second last digit is odd. This implies that the only squares whose last two digits are the same are those ending with 0 or 4; those ending with 1, 5, and 9 are paired with even second last digits. - Waldemar Puszkarz, May 24 2016

Seiichi Manyama, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. A000290, A047221, A090773.
Cf. A017341 (numbers ending in 6), A017343 (cubes ending in 6).
Cf. squares with last digit k: A017270 (k=0), A273372 (k=1), A273375 (k=4), A017330 (k=5), this sequence (k=6), A273374 (k=9).

/* By definition: */ [n^2: n in [0..200] | Modexp(n,2,10) eq 6];

[(10*n - 3*(-1)^n - 5)^2/4: n in [1..50]];

seq(seq((10*i+j)^2,j=[4,6]),i=0..20); # Robert Israel, May 24 2016

Table[(10 n - 3 (-1)^n - 5)^2/4, {n, 1, 50}]
CoefficientList[Series[4 (4 + 5 x + 32 x^2 + 5 x^3 + 4 x^4) / ((1 + x)^2 (1 - x)^3), {x, 0, 50}], x]
Select[Range[250]^2,Mod[#,10]==6&] (* Harvey P. Dale, May 31 2020 *)

G.f.: 4*x*(4 + 5*x + 32*x^2 + 5*x^3 + 4*x^4)/((1 + x)^2*(1 - x)^3).
a(n) = 4*A047221(n)^2 = (10*n - 3*(-1)^n - 5)^2/4.
a(n) = A090773(n)^2. - Michel Marcus, May 25 2016
Sum_{n>=1} 1/a(n) = 2*Pi^2/(25*(5+sqrt(5))). - Amiram Eldar, Feb 16 2023

Corrected and extended by Bruno Berselli, May 23 2016

A292612 a(n) = F(n)^2 + 4(-1)^n = F(n+3)F(n-3), where F = A000045.

Original entry on oeis.org

4, -3, 5, 0, 13, 21, 68, 165, 445, 1152, 3029, 7917, 20740, 54285, 142133, 372096, 974173, 2550405, 6677060, 17480757, 45765229, 119814912, 313679525, 821223645, 2149991428, 5628750621, 14736260453, 38580030720, 101003831725, 264431464437, 692290561604, 1812440220357

Offset: 0

Bruno Berselli, Sep 20 2017

This is the case k=3 of the identity F(n)^2 - F(k)^2*(-1)^(n+k) = F(n+k)*F(n-k), known also as Catalan's identity.

Colin Barker, Table of n, a(n) for n = 0..1000
Eric Weisstein's World of Mathematics, Catalan's Identity.
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Cf. A000045, A001622, A005248: Lucas(2*n), A001654: F(n)*F(n+1).

Cf. A007598 (k=0), A059929 (k=1, without initial 1), A192883 (k=2, without initial -1), this sequence (k=3).

List([0..10^2],n ->Fibonacci(n)^2+4*(-1)^n); # Muniru A Asiru, Sep 26 2017

[Fibonacci(n)^2+4*(-1)^n: n in [0..40]];

with(combinat,fibonacci):  A292612:=seq(fibonacci(n)^2+4*(-1)^n, n=0..10^2); # Muniru A Asiru, Sep 26 2017

Table[Fibonacci[n]^2 + 4 (-1)^n, {n, 0, 40}]

for(n=0, 40, print1(fibonacci(n)^2+4*(-1)^n", "));

Vec((4-11*x+3*x^2)/((1+x)*(1-3*x+x^2))+O(x^30)) \\ Colin Barker, Sep 20 2017

[fibonacci(n)^2+4*(-1)^n for n in range(40)]

G.f.: (4 - 11*x + 3*x^2)/((1 + x)*(1 - 3*x + x^2)).
a(n) = a(-n) = 2*a(n-1) + 2*a(n-2) - a(n-3).
a(n) = 4*A001654(n+1) - 11*A001654(n) + 3*A001654(n-1) with A001654(-1)=0.
5*a(n) = Lucas(2*n) + 18*(-1)^n. Note that Lucas(2*n) + r*(-1)^n is divisible by 5 for r = -2, 3, -7, 8, -12, 13, -17, 18, -22, 23, -27, ... = (-1/4)*(3 + 5*(2*m+1)*(-1)^m) = (-1)^m*A047221(m). On the other hand, a(n) is divisible by 5 when n is a member of A047221.
a(n) = (1/5)*(18*(-1)^n + ((3-sqrt(5))/2)^n + ((3+sqrt(5))/2)^n). - Colin Barker, Sep 20 2017
Sum_{n>=4} 1/a(n) = 143/960. - Amiram Eldar, Oct 05 2020
Sum_{n>=4} (-1)^n/a(n) = 3/(4*phi) - 407/960, where phi is the golden ratio (A001622). - Amiram Eldar, Oct 06 2020

A153180 a(n) = L(13n)/L(n) where L(n) = Lucas number A000204(n).

Original entry on oeis.org

521, 90481, 35355581, 10525900321, 3489827263001, 1111126318086721, 359316586176453881, 115509240442846111681, 37216910406644366498621, 11980863523543017476802001, 3858153294795970321295258921

Offset: 1

Artur Jasinski, Dec 20 2008

nonn

All numbers in this sequence are:
congruent to 1 mod 10
congruent to 1 mod 100 (iff n is congruent to 0 mod 5),Q congruent to 21 mod 100 (iff n is congruent to 1 or 4 mod 5),
congruent to 81 mod 100 (iff n is congruent to 2 or 3 mod 5).Q

Index entries for linear recurrences with constant coefficients, signature (233, 33552, -1493064, -27372840, 186135312, 488605194, -488605194, -186135312, 27372840, 1493064, -33552, -233, 1).

A000032, A000204, A047221, A110391, A153173, A153175, A153177, A153179.

Table[LucasL[13 n]/LucasL[n], {n, 1, 150}]

a(n)= +233*a(n-1) +33552*a(n-2) -1493064*a(n-3) -27372840*a(n-4) +186135312*a(n-5) +488605194*a(n-6) -488605194*a(n-7) -186135312*a(n-8) +27372840*a(n-9) +1493064*a(n-10) -33552*a(n-11) -233*a(n-12) +a(n-13). G.f.: -1+ (-2-123*x)/(x^2+123*x+1) +(2-322*x)/(x^2-322*x+1) +(-2-3*x)/(x^2+3*x+1) +(2-7*x)/(x^2-7*x+1) +(2-47*x)/(x^2-47*x+1) -1/(x-1)+ (-2-18*x)/(x^2+18*x+1). [From R. J. Mathar, Oct 22 2010]

A212423 Frobenius pseudoprimes == 2,3 (mod 5) with respect to Fibonacci polynomial x^2 - x - 1.

Original entry on oeis.org

5777, 10877, 75077, 100127, 113573, 161027, 162133, 231703, 430127, 635627, 851927, 1033997, 1106327, 1256293, 1388903, 1697183, 2263127, 2435423, 2662277, 3175883, 3399527, 3452147, 3774377, 3900797, 4109363, 4226777, 4403027, 4828277, 4870847

Offset: 1

Max Alekseyev, May 16 2012

nonn

Grantham incorrectly claims that "the first Frobenius pseudoprime with respect to the Fibonacci polynomial x^2 - x - 1 is 5777". However n = 5777 is the first Frobenius pseudoprime with respect to x^2 - x - 1 that has Jacobi symbol (5/n) = -1, i.e., n == 2,3 (mod 5). Unrestricted version with the first term 4181 is given in A212424.
Intersection of A212424 and A047221.
Composite k == 2,3 (mod 5) such that Fibonacci(k) == -1 (mod k) and that k divides Fibonacci(k+1). - Jianing Song, Sep 12 2018

R. Crandall, C. B. Pomerance. Prime Numbers: A Computational Perspective. Springer, 2nd ed., 2005.

Amiram Eldar, Table of n, a(n) for n = 1..14070 (terms below 10^13 from Dana Jacobsen's site)
Jon Grantham, Frobenius Pseudoprimes, Mathematics of Computation, 7 (2000), 873-891.
Dana Jacobsen, Pseudoprime Statistics, Tables, and Data.
Eric Weisstein's World of Mathematics, Frobenius Pseudoprime.

Cf. A047221, A094063, A094395, A094411, A212424.

{ isFP23(n) = if(ispseudoprime(n),return(0)); t=Mod(x*Mod(1,n),(x^2-x-1)*Mod(1,n))^n; (kronecker(5,n)==-1 && t==1-x) }

A016907 (5n+4)^11.

Original entry on oeis.org

4194304, 31381059609, 4049565169664, 116490258898219, 1521681143169024, 12200509765705829, 70188843638032384, 317475837322472439, 1196683881290399744, 3909821048582988049, 11384956040305711104

Offset: 0

N. J. A. Sloane

If 5n+4 is a perfect square, that is for n=0,1,9,12,28,33,57,64,..., a(n)=A010810(j), where the indices j are listed in A047221. - R. J. Mathar, Apr 22 2008.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (12,-66,220,-495,792,-924,792,-495,220,-66,12,-1).

[(5*n+4)^11: n in [0..30]]; // Vincenzo Librandi, May 03 2011

G.f.: (x^11 +362797044*x^10 +280958106005*x^9 +14192390602560*x^8 +157922022917730*x^7 +565509776420088*x^6 +753319100251602*x^5 +384167580654720*x^4 +69965704049565*x^3 +3673269278420*x^2 +31330727961*x +4194304) / (x -1)^12. [Colin Barker, Feb 22 2013]

A318958 A(n, k) is a square array read in the decreasing antidiagonals, for n >= 0 and k >= 0.

Original entry on oeis.org

0, 0, 0, 0, -1, -1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 2, 2, 3, 2, 2, 0, 1, 3, 3, 4, 3, 3, 0, 3, 4, 6, 6, 7, 6, 6, 0, 2, 5, 6, 8, 8, 9, 8, 8, 0, 4, 6, 9, 10, 12, 12, 13, 12, 12, 0, 3, 7, 9, 12, 13, 15, 15, 16, 15, 15, 0, 5, 8, 12, 14, 17, 18, 20, 20, 21, 20, 20

Offset: 0

Paul Curtz, Sep 06 2018

			The array starts:
[n\k][0,   1,  2,  3,  4,  5,  6,  7,  8,  9, 10, 11, ...]
[0]   0,   0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0, ... = A000004
[1]   0,  -1,  1,  0,  2,  1,  3,  2,  4,  3,  5,  4, ... = A028242(n-2)
[2]  -1,   0,  1,  2,  3,  4,  5,  6,  7,  8,  9, 10, ... = A023443(n)
[3]   0,   0,  3,  3,  6,  6,  9,  9, 12, 12, 15, 15, ... = 3*A004526(n)
[4]   0,   2,  4,  6,  8, 10, 12, 14, 16, 18, 20, 22, ... = A005843(n)
[5]   2,   3,  7,  8, 12, 13, 17, 18, 22, 23, 27, 28, ... = A047221(n+1)
[6]   3,   6,  9, 12, 15, 18, 21, 24, 27, 30, 33, 36, ... = A008585(n+1)
[7]   6,   8, 13, 15, 20, 22, 27, 29, 34, 36, 41, 43, ... = A047336(n+2)
[8]   8,  12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, ... = A008586(n+2)
Successive columns: A198442(n-2), A198442(n-1), A004652(n), A198442(n+1), A198442(n+2), A079524(n), ... .
First subdiagonal: 0, 0, 3, 6, ... = A242477(n).
First upperdiagonal: 0, 1, 2, 6, 10, ... = A238377(n-1).
Array written as a triangle:
0;
0,  0;
0, -1, -1;
0,  1,  0, 0;
0,  0,  1, 0, 0;
0,  2,  2, 3, 2, 2;
etc.

Cf. A000004, A004526, A004652, A005843, A008585, A008586, A023443, A028242, A047221, A047336, A079524, A198442, A238377, A242477.

A := proc(n, k) option remember; local h;
h := n -> `if`(n<3, [0, 0, -1][n+1], iquo(n^2-4*n+3, 4));
if k = 0 then h(n) elif k = 1 then h(n+1) else A(n, k-2) + n fi end: # Peter Luschny, Sep 08 2018

h[n_] := If[n < 3, {0, 0, -1}[[n + 1]], Quotient[n^2 - 4 n + 3, 4]];
A[n_, k_] := A[n, k] = If[k == 0, h[n], If[k == 1, h[n+1], A[n, k-2] + n]];
Table[A[n - k, k], {n, 0, 11}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, Jul 22 2019, after Peter Luschny *)

Let h(n) = 0, 0, -1, A198442(1), A198442(2), A198442(3), ... Then A(n, 0) = h(n), A(n, 1) = h(n+1) and A(n, k) = A(n, k-2) + n otherwise.

cp's OEIS Frontend

A273373 Squares ending in digit 6.

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A292612 a(n) = F(n)^2 + 4*(-1)^n = F(n+3)*F(n-3), where F = A000045.

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A153180 a(n) = L(13n)/L(n) where L(n) = Lucas number A000204(n).

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A212423 Frobenius pseudoprimes == 2,3 (mod 5) with respect to Fibonacci polynomial x^2 - x - 1.

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A016907 (5n+4)^11.

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A318958 A(n, k) is a square array read in the decreasing antidiagonals, for n >= 0 and k >= 0.

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Maple

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A292612 a(n) = F(n)^2 + 4(-1)^n = F(n+3)F(n-3), where F = A000045.