A081016 -id:A081016 - cp's OEIS frontend

A172968 a(n) = 7*a(n-1) - a(n-2) for n>1, a(0)=1, a(1)=2.

1, 2, 13, 89, 610, 4181, 28657, 196418, 1346269, 9227465, 63245986, 433494437, 2971215073, 20365011074, 139583862445, 956722026041, 6557470319842, 44945570212853, 308061521170129, 2111485077978050, 14472334024676221

Offset: 0

Artur Jasinski, Feb 06 2010

G. C. Greubel, Table of n, a(n) for n = 0..1000
Marta Na Chen, Wenchang Chu, Four classes of symmetric sums over cyclically binomial products, Symmetry 17 (2025) no 2
Index entries for linear recurrences with constant coefficients, signature (7,-1).

Cf. A000032, A000045, A081016.

Essentially the same as A033891.

List([0..30], n-> Fibonacci(4*n-1)); # G. C. Greubel, Jul 15 2019

[n le 2 select n else 7*Self(n-1)-Self(n-2): n in [1..30]]; // Bruno Berselli, Mar 29 2016

with(combinat):F:= n-> fibonacci(n):L:=n-> 2*F(n+1)-F(n):
seq(1/2*(L(4*n)-F(4*n)), n=0..20); # Gary Detlefs, Nov 28 2010

Table[Sqrt[1-2m+5m^2]/.m ->Fibonacci[2n+1]Fibonacci[2n+2], {n, -1, 30}]
CoefficientList[Series[(1-5x)/(1-7x+x^2), {x, 0, 30}], x] (* Michael De Vlieger, Mar 29 2016 *)
Fibonacci[4*Range[0, 30] -1] (* G. C. Greubel, Jul 15 2019 *)

x='x+O('x^30); Vec((1-5*x)/(1-7*x+x^2)) \\ Altug Alkan, Mar 29 2016

[fibonacci(4*n-1) for n in (0..30)] # G. C. Greubel, Jul 15 2019

a(n) = (1/10)*((5+sqrt(5))*((7-3*sqrt(5))/2)^n + ((5-sqrt(5))*((7+3*sqrt(5))/2)^n)).
a(n) = sqrt(1 - 2*F(2n+1)*F(2n+2) + 5*(F(2n+1)*F(2n+2))^2), where F = A000045.
a(n) = sqrt(1 - 2*A081016(n) + 5*A081016(n)^2).
a(n) = A033891(n-1), n>0. - R. J. Mathar, Feb 08 2010
a(n) = (Lucas(4*n) - Fibonacci(4*n))/2, where Lucas = A000032. - Gary Detlefs, Nov 28 2010
G.f.: (1 - 5*x)/(1 - 7*x + x^2). - Bruno Berselli, Mar 29 2016
a(n) = Fibonacci(4*n-1). - G. C. Greubel, Jul 15 2019
a(n) = (a(n-1)^2 + 9)/a(n-2). - Klaus Purath, Aug 30 2020

A276472 Modified Pascal's triangle read by rows: T(n,k) = T(n-1,k) + T(n-1,k-1), 12. T(n,n) = T(n,n-1) + T(n-1,n-1), n>1. T(1,1) = 1, T(2,1) = 1. n>=1.

Original entry on oeis.org

1, 1, 2, 4, 3, 5, 11, 7, 8, 13, 29, 18, 15, 21, 34, 76, 47, 33, 36, 55, 89, 199, 123, 80, 69, 91, 144, 233, 521, 322, 203, 149, 160, 235, 377, 610, 1364, 843, 525, 352, 309, 395, 612, 987, 1597, 3571, 2207, 1368, 877, 661, 704, 1007, 1599, 2584, 4181

Offset: 1

Yuriy Sibirmovsky, Sep 12 2016

The recurrence relations for the border terms are the only way in which this differs from Pascal's triangle.
Column T(2n,n+1) appears to be divisible by 4 for n>=2; T(2n-1,n) divisible by 3 for n>=2; T(2n,n-2) divisible by 2 for n>=3.
The symmetry of T(n,k) can be observed in a hexagonal arrangement (see the links).
Consider T(n,k) mod 3 = q. Terms with q = 0 show reflection symmetry with respect to the central column T(2n-1,n), while q = 1 and q = 2 are mirror images of each other (see the link).

			Triangle T(n,k) begins:
n\k 1    2    3    4   5    6    7    8    9
1   1
2   1    2
3   4    3    5
4   11   7    8    13
5   29   18   15   21   34
6   76   47   33   36   55   89
7   199  123  80   69   91   144 233
8   521  322  203  149  160  235 377  610
9   1364 843  525  352  309  395 612  987  1597
...
In another format:
__________________1__________________
_______________1_____2_______________
____________4_____3_____5____________
________11_____7_____8_____13________
____29_____18_____15____21_____34____
_76_____47____33_____36____55_____89_

Yuriy Sibirmovsky, T(n,k), read by rows as a linear sequence a(j) for j = 1..5050
Yuriy Sibirmovsky, Symmetrical hexagonal arrangement for initial terms of T(n,k)
Yuriy Sibirmovsky, T(n,k) compared with Pascal's triangle
Yuriy Sibirmovsky, Illustration for T(n,k) mod 3

Cf. A000045, A000204, A001519, A001906, A002878, A005248, A054441, A088305, A122367, A258109, A026671, A026726, A026732, A004187, A049685, A033891, A206351, A092521, A081018, A049684, A058038, A081016, A003482, A049629, A133273, A049664, A049683, A119032, A023039, A007805, A033889.

Nm=12;
T=Table[0,{n,1,Nm},{k,1,n}];
T[[1,1]]=1;
T[[2,1]]=1;
T[[2,2]]=2;
Do[T[[n,1]]=T[[n-1,1]]+T[[n,2]];
T[[n,n]]=T[[n-1,n-1]]+T[[n,n-1]];
If[k!=1&&k!=n,T[[n,k]]=T[[n-1,k]]+T[[n-1,k-1]]],{n,3,Nm},{k,1,n}];
{Row[#,"\t"]}&/@T//Grid

T(n,k) = if (k==1, if (n==1, 1, if (n==2, 1, T(n-1,1) + T(n,2))), if (kMichel Marcus, Sep 14 2016

Conjectures:
Relations with other sequences:
T(n+1,1) = A002878(n-1), n>=1.
T(n,n) = A001519(n) = A122367(n-1), n>=1.
T(n+1,2) = A005248(n-1), n>=1.
T(n+1,n) = A001906(n) = A088305(n), n>=1.
T(2n-1,n) = 3*A054441(n-1), n>=2. [the central column].
Sum_{k=1..n} T(n,k) = 3*A105693(n-1), n>=2. [row sums].
Sum_{k=1..n} T(n,k)-T(n,1)-T(n,n) = 3*A258109(n), n>=2.
T(2n,n+1) - T(2n,n) = A026671(n), n>=1.
T(2n,n-1) - T(2n,n) = 2*A026726(n-1), n>=2.
T(n,ceiling(n/2)) - T(n-1,floor(n/2)) = 2*A026732(n-3), n>=3.
T(2n+1,2n) = 3*A004187(n), n>=1.
T(2n+1,2) = 3*A049685(n-1), n>=1.
T(2n+1,2n) + T(2n+1,2) = 3*A033891(n-1), n>=1.
T(2n+1,3) = 5*A206351(n), n>=1.
T(2n+1,2n)/3 - T(2n+1,3)/5 = 4*A092521(n-1), n>=2.
T(2n,1) = 1 + 5*A081018(n-1), n>=1.
T(2n,2) = 2 + 5*A049684(n-1), n>=1.
T(2n+1,2) = 3 + 5*A058038(n-1), n>=1.
T(2n,3) = 3 + 5*A081016(n-2), n>=2.
T(2n+1,1) = 4 + 5*A003482(n-1), n>=1.
T(3n,1) = 4*A049629(n-1), n>=1.
T(3n,1) = 4 + 8*A119032(n), n>=1.
T(3n+1,3) = 8*A133273(n), n>=1.
T(3n+2,3n+2) = 2 + 32*A049664(n), n>=1.
T(3n,3n-2) = 4 + 32*A049664(n-1), n>=1.
T(3n+2,2) = 2 + 16*A049683(n), n>=1.
T(3n+2,2) = 2*A023039(n), n>=1.
T(2n-1,2n-1) = A033889(n-1), n>=1.
T(3n-1,3n-1) = 2*A007805(n-1), n>=1.
T(5n-1,1) = 11*A097842(n-1), n>=1.
T(4n+5,3) - T(4n+1,3) = 15*A000045(8n+1), n>=1.
T(5n+4,3) - T(5n-1,3) = 11*A000204(10n-2), n>=1.
Relations between left and right sides:
T(n,1) = T(n,n) - T(n-2,n-2), n>=3.
T(n,2) = T(n,n-1) - T(n-2,n-3), n>=4.
T(n,1) + T(n,n) = 3*T(n,n-1), n>=2.

A240836 Numbers n such that n^3 = xyz where 2 <= x <= y <= z , n^3+1 = (x-1)(y+1)(z+1).

Original entry on oeis.org

2, 12, 80, 546, 3740, 25632, 175682, 1204140, 8253296, 56568930, 387729212, 2657535552, 18215019650, 124847601996, 855718194320, 5865179758242, 40200540113372, 275538601035360, 1888569667134146, 12944449068903660, 88722573815191472, 608113567637436642

Offset: 1

Naohiro Nomoto, Apr 12 2014

Also, z/y approx = y/x approx = golden ratio.

			546^3 = 338 * 546 * 882, 546^3 + 1 = 337 * 547 * 883.
25632^3 = 15842 * 25632 * 41472, 25632^3 + 1 = 15841 * 25633 * 41473.

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Cf. A000045, A079472, A081016, A110035, A126116.

F:=Fibonacci;; List([1..30], n-> 2*F(2*n)*F(2*n-1) ); # G. C. Greubel, Jul 15 2019

F:=Fibonacci; [2*F(2*n)*F(2*n-1): n in [1..30]]; // G. C. Greubel, Jul 15 2019

with(combinat); A240836:=n->2*fibonacci(2*n)*fibonacci(2*n-1); seq(A240836(n), n=1..30); # Wesley Ivan Hurt, Apr 13 2014

Table[2Fibonacci[2n]Fibonacci[2n-1], {n, 30}] (* Wesley Ivan Hurt, Apr 13 2014 *)

vector(30, n, f=fibonacci; 2*f(2*n)*f(2*n-1)) \\ G. C. Greubel, Jul 15 2019

f=fibonacci; [2*f(2*n)*f(2*n-1) for n in (1..30)] # G. C. Greubel, Jul 15 2019

a(n) = 2*F(2n)*F(2n-1) where F(n) are the Fibonacci numbers (A000045).
G.f.: 2*x*(1-2*x)/((1-x)*(1-7*x+x^2)). - Colin Barker, Apr 13 2014
a(n) = 2 * A081016(n-1). - Wesley Ivan Hurt, Apr 13 2014

More terms from Colin Barker, Apr 13 2014

A356716 a(n) is the integer w such that (c(n)^2, -d(n)^2, -w) is a primitive solution to the Diophantine equation 2x^3 + 2y^3 + z^3 = 11^3, where c(n) = F(n+2) + (-1)^n * F(n-3), d(n) = F(n+1) + (-1)^n * F(n-4) and F(n) is the n-th Fibonacci number (A000045).

Original entry on oeis.org

5, 19, 31, 101, 179, 655, 1189, 4451, 8111, 30469, 55555, 208799, 380741, 1431091, 2609599, 9808805, 17886419, 67230511, 122595301, 460804739, 840280655, 3158402629, 5759369251, 21648013631, 39475304069, 148377692755, 270567759199, 1016995835621, 1854499010291

Offset: 1

XU Pingya, Aug 24 2022

Conjecture:
(i) For all k > 2, 2*x^3 + 2*y^3 + z^3 = A089270(k)^3 have primitive solutions form (c(n)^2, -d(n)^2, -w(n)) with d(n) = 3*d(n-2) - d(n-4), c(n) = d(n+2) - d(n) and w(n) = 8*w(n-2) - 8*w(n-4) + w(n-6).
(ii) This sequence is a subsequence of A089270.
From XU Pingya, Jun 07 2024: (Start)
Several positive examples of conjecture:
When A089270(4,5,6,7) = {19,29,31,41}, d(n) can be taken as:
(1/2) * (F(n+3) + (-1)^n * F(n-6));
((1-(-1)^n)/2) * (F(n+3) + F(n-4)) + ((1+(-1)^n)/2) * (F(n+3) - F(n-4));
((1-(-1)^n)/2) * (2*F(n-1) + 3*F(n-3)) + ((1+(-1)^n)/2) * (3*F(n-2) + 2*F(n-4));
and
((1-(-1)^n)/2) * (2*F(n+1) + F(n-5)) + ((1+(-1)^n)/2) * (F(n+2) + 2*F(n-4)).
When A089270(17) = 121, d(n) can be taken as d(1,2,3,4) = {-3,0,7,11}. (End)
From XU Pingya, Jul 17 2024: (Start)
Furthermore, we observe that if (x, y) (y < x/2) is the solution of the Diophantine equation x^2 + x * y - y^2 = A089270(k). Let
d(2*n-1) = x * F(2*n-2) - y * F(2*n-3), c(2*n-1) = d(2*n+1) - d(2*n-1);
d(2*n) = x * F(2*n-2) + y * F(2*n-1), c(2*n) = d(2*n+2) - d(2*n).
Then such c(n) and d(n) satisfy the conjecture. (End)

			For n=3, 2 * ((F(5) - F(0))^2)^3 + 2 * (-(F(4) - F(-1))^2)^3 + (-31)^3 = 2 * 25^3 - 2 * 4^3 - 31^3 = 1331, a(3) = 31.

Index entries for linear recurrences with constant coefficients, signature (1,7,-7,-1,1).

Cf. A000045, A081016, A089270, A206351, A228208, A237132.

Cf. also A337928, A354336, A356717.

Table[(-1331+2*((Fibonacci[n+2]+(-1)^n*Fibonacci[n-3]))^6-2*(Fibonacci[n+1]+(-1)^n*Fibonacci[n-4])^6)^(1/3), {n,28}]

a(n) = (-1331 + 2 * A237132(n)^6 - 2 * A228208(n-1)^6)^(1/3).
a(n) = ((1-(-1)^n)/2) * (-1 + 6 * Sum_{k=0..n-1} Fibonacci(4*k-1) + 14 * Sum_{k=0..n-2} Fibonacci(4*k+1)) + ((1+(-1)^n)/2) * (-1 + 6 * Sum_{k=0..n-1} Fibonacci(4*k-1) + 14 * Sum_{k=0..n-1} Fibonacci(4*k+1)).
a(n) = ((1-(-1)^n)/2) * (-1 + 6*A206351(n) + 14*A081016(n-2)) + ((1+(-1)^n)/2) * (-1 + 6*A206351(n) + 14*A081016(n-1)).
From Stefano Spezia, Aug 25 2022: (Start)
G.f.: x*(5 + 14*x - 23*x^2 - 28*x^3 - x^4)/((1 - x)*(1 - 3*x + x^2)*(1 + 3*x + x^2)).
a(n) = a(n-1) + 7*a(n-2) - 7*a(n-3) - a(n-4) + a(n-5) for n > 5. (End)
From XU Pingya, Jul 17 2024: (Start)
a(2*n-1) = (F(2*n) + F(2*n-2) + F(2*n-5))^2 + (F(2*n) + F(2*n-2) + F(2*n-5)) * (F(2*n-2) + F(2*n-4) + F(2*n-7)) - (F(2*n-2) + F(2*n-4) + F(2*n-7))^2;
a(2*n) = (F(2*n+2) + F(2*n-3))^2 + (F(2*n+2) + F(2*n-3)) * (F(2*n) + F(2*n-5)) - (F(2*n) + F(2*n-5))^2. (End)

A356717 a(n) is the integer w such that (c(n)^2, -d(n)^2, w) is a primitive solution to the Diophantine equation 2x^3 + 2y^3 + z^3 = 11^3, where c(n) = F(n+2) + (-1)^n * F(n-3), d(n) = F(n+3) + (-1)^n * F(n-2) and F(n) is the n-th Fibonacci number (A000045).

Original entry on oeis.org

1, 29, 59, 241, 445, 1691, 3089, 11629, 21211, 79745, 145421, 546619, 996769, 3746621, 6831995, 25679761, 46827229, 176011739, 320958641, 1206402445, 2199883291, 8268805409, 15078224429, 56675235451, 103347687745, 388457842781, 708355589819, 2662529664049

Offset: 1

XU Pingya, Aug 24 2022

			For n=3, 2 * ((F(5) - F(0))^2)^3 + 2 * (-(F(6) - F(1))^2)^3 + 59^3 = 2 * 25^3 - 2 * 49^3 + 59^3 = 1331, a(3) = 59.

Index entries for linear recurrences with constant coefficients, signature (1,7,-7,-1,1).

Cf. A000045, A081016, A081018, A089270, A228208, A237132.

Cf. also A337929, A354337, A356716.

Table[(1331-2*((Fibonacci[n+2]+(-1)^n*Fibonacci[n-3]))^6+2*(Fibonacci[n+3]+(-1)^n*Fibonacci[n-2])^6)^(1/3), {n,28}]

a(n) = (1331 - 2 * A237132(n)^6 + 2 * A228208(n+1)^6)^(1/3).
a(n) = ((1-(-1)^n)/2) * (-5 + 14 * Sum_{k=1..n-1} Fibonacci(4*k-1) + 6 * Sum_{k=0..n-1} Fibonacci(4*k+1)) + ((1+(-1)^n)/2) * (-5 + 14 * Sum_{k=1..n} Fibonacci(4*k-1) + 6 * Sum_{k=0..n-1} Fibonacci(4*k+1)).
a(n) = ((1-(-1)^n)/2) * (-5 + 14 * A081018(n-1) + 6 * A081016(n-1)) + ((1+(-1)^n)/2) * (-5 + 14 * A081018(n) + 6 * A081016(n-1)).
From Stefano Spezia, Aug 25 2022: (Start)
G.f.: x*(1 + 28*x + 23*x^2 - 14*x^3 - 5*x^4)/((1 - x)*(1 - 3*x + x^2)*(1 + 3*x + x^2)).
a(n) = a(n-1) + 7*a(n-2) - 7*a(n-3) - a(n-4) + a(n-5) for n > 5. (End)

cp's OEIS Frontend

A172968 a(n) = 7*a(n-1) - a(n-2) for n>1, a(0)=1, a(1)=2.

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A276472 Modified Pascal's triangle read by rows: T(n,k) = T(n-1,k) + T(n-1,k-1), 12. T(n,n) = T(n,n-1) + T(n-1,n-1), n>1. T(1,1) = 1, T(2,1) = 1. n>=1.

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A240836 Numbers n such that n^3 = x*y*z where 2 <= x <= y <= z , n^3+1 = (x-1)*(y+1)*(z+1).

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A356716 a(n) is the integer w such that (c(n)^2, -d(n)^2, -w) is a primitive solution to the Diophantine equation 2*x^3 + 2*y^3 + z^3 = 11^3, where c(n) = F(n+2) + (-1)^n * F(n-3), d(n) = F(n+1) + (-1)^n * F(n-4) and F(n) is the n-th Fibonacci number (A000045).

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A356717 a(n) is the integer w such that (c(n)^2, -d(n)^2, w) is a primitive solution to the Diophantine equation 2*x^3 + 2*y^3 + z^3 = 11^3, where c(n) = F(n+2) + (-1)^n * F(n-3), d(n) = F(n+3) + (-1)^n * F(n-2) and F(n) is the n-th Fibonacci number (A000045).

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A240836 Numbers n such that n^3 = xyz where 2 <= x <= y <= z , n^3+1 = (x-1)(y+1)(z+1).

A356716 a(n) is the integer w such that (c(n)^2, -d(n)^2, -w) is a primitive solution to the Diophantine equation 2x^3 + 2y^3 + z^3 = 11^3, where c(n) = F(n+2) + (-1)^n * F(n-3), d(n) = F(n+1) + (-1)^n * F(n-4) and F(n) is the n-th Fibonacci number (A000045).

A356717 a(n) is the integer w such that (c(n)^2, -d(n)^2, w) is a primitive solution to the Diophantine equation 2x^3 + 2y^3 + z^3 = 11^3, where c(n) = F(n+2) + (-1)^n * F(n-3), d(n) = F(n+3) + (-1)^n * F(n-2) and F(n) is the n-th Fibonacci number (A000045).