A165900 -id:A165900 - cp's OEIS frontend

A214630 a(n) is the reduced numerator of 1/4 - 1/A109043(n)^2 = (1 - 1/A026741(n)^2)/4.

-1, 0, 0, 2, 3, 6, 2, 12, 15, 20, 6, 30, 35, 42, 12, 56, 63, 72, 20, 90, 99, 110, 30, 132, 143, 156, 42, 182, 195, 210, 56, 240, 255, 272, 72, 306, 323, 342, 90, 380, 399, 420, 110, 462, 483, 506, 132, 552, 575, 600, 156

Offset: 0

Paul Curtz, Jul 23 2012

The unreduced fractions are -1/0, 0/4, 0/1, 8/36, 3/16, 24/100, 2/9, 48/196, 15/64, 80/324, 6/25, ... = c(n)/A061038(n), say.
Note that c(n)=A061037(n) + (period of length 2: repeat 0, 3).
c(n) is a permutation of A198442(n). The corresponding ranks are (the 0's have been swapped for convenience) 0,2,1,6,4,10,... = A145979(n-2).
Define the following sequences, satisfying the recurrence a(n) = 2*a(n-4) - a(n-8),
e(n) = -1, 0, 0, 2, 1, 4, 1, 6, 3, 8, 2, 10, 5, ... (after -1, a permutation of A004526(n) or mix A026741(n-1), 2*n),
f(n) = 1, 2, 1, 4, 3, 6, 2, 8, 5, 10, 3, 12, 7, ..., (another permutation of A004526(n+2) or mix A026741(n+1), 2*n+2).
f(n) - e(n) = periodic of period length 4: repeat 2, 2, 1, 2.
e(n) + f(n) = 0, 2, 1, 6, 4, 10, ... = A145979(n-2).
Then c(n) = e(n)*f(n).
Note that A061038(n) - 4*c(n) = periodic of period length 4: repeat 4, 4, 1, 4.
After division (by period 2: repeat 1, 4, A010685(n)), the reduced fractions of c(n) are -1/0, 0/1 ?, 0/4 ?, 2/9, 3/16, 6/25, 2/9, 12/49, 15/64, 20/81, 6/25, ... = a(n)/b(n).
Note that a(1+4*n) + a(2+4*n) + a(3+4*n) = 2,20,56,... = A002378(1+3*n) = A194767(3*n).
A061037(n-2) - a(n-2) = 0, -3, 0, -3, 0, 3, 0, 15, 0, 33, 0, 57, ... = Fip(n-2).
Fip(n-2)/3 = 0,-1,0,-1,0,1,0,5,0,11,0,19,0,29, .... Without 0's: A165900(n) (a Fibonacci polynomial); also -A110331(n+1) (Pell numbers).
g(n) = -1, 0, 0, 1, 1, 2, 1, 3, 3, 4, ... = mix A026741(n-1), n.
h(n) = 1, 1, 1, 2, 3, 3, 2, 4, 5, 5, ... = mix A026741(n+1), n+1.
h(n) - g(n) = (period 2: repeat 2, 1, 1, 1 = A177704(n-1)).
k(n) = 1, 1, 0, 2, 3, 3, 1, 4, 5, 5, ... = mix A174239(n), n+1.
l(n) = -1, 0, 1, 1, 1, 2, 2, 3, 3, 4, ... .
k(n) - l(n) = period 4: repeat 2, 1, -1, 1.
2) By the second formula in the definition, we take first 1 - 1/A026741(n)^2.
Hence, using a convention for the first fraction, -1/0, 0/1, 0/1, 8/9, 3/4, 24/25, 8/9, 48/49, 15/16, 80/81, 24/25, ... = (A005563(n-1) - A033996(n))/A168077(n) = q(n)/A168077(n).
For a(n), we divide by 4.
Note that A214297 is the reduced numerator of  1/4 - 1/A061038(n).
Note also that A168077(n) = A026741(n)^2.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,3,0,0,0,-3,0,0,0,1).

Cf. A000466, A002378, A131723.

m:=50; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((2*x^9+3*x^8+6*x^7+2*x^6+6*x^5+6*x^4+2*x^3-1)/((1-x)^3*(x+1)^3*(x^2+1)^3))); // G. C. Greubel, Sep 20 2018

CoefficientList[Series[(2*x^9+3*x^8+6*x^7+2*x^6+6*x^5+6*x^4+2*x^3-1)/((1-x)^3*(x+1)^3*(x^2+1)^3), {x, 0, 50}], x] (* G. C. Greubel, Sep 20 2018 *)
LinearRecurrence[{0,0,0,3,0,0,0,-3,0,0,0,1},{-1,0,0,2,3,6,2,12,15,20,6,30},60] (* Harvey P. Dale, Jul 01 2019 *)

Vec(-(2*x^9+3*x^8+6*x^7+2*x^6+6*x^5+6*x^4+2*x^3-1)/((x-1)^3*(x+ 1)^3*(x^2+1)^3) + O(x^100)) \\ Colin Barker, Jan 22 2015

a(4*n) = 4*n^2-1 = (2*n-1)*(2*n+1), a(2*n+1) = a(4*n+2) = n(n+1).
a(n)= A198442(n)/(period of length 4: repeat 1,1,4,1=A010121(n+2)).
a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12). Is this the shortest possible recurrence? See A214297.
a(n+2) - a(n-2) = 0, 2, 4, 6, 2, 10, 12, 14, 4, ... = 2*A214392(n). a(-2)=a(-1)=0=a(1)=a(2).
a(n+4) - a(n-4) = 0, 4, 2, 12, 16, 20, 6, 28, 32, 36,... = 2*A188167(n). a(-4)=3=a(4), a(-3)=2=a(3).
a(n) = g(n) * h(n).
a(n) = k(n) * l(n).
G.f.: -(2*x^9+3*x^8+6*x^7+2*x^6+6*x^5+6*x^4+2*x^3-1) / ((x-1)^3*(x+1)^3*(x^2+1)^3). - Colin Barker, Jan 22 2015
From Luce ETIENNE, Apr 08 2017: (Start)
a(n) = (13*n^2-28-3*(n^2+4)*(-1)^n+3*(n^2-4)*((-1)^((2*n-1+(-1)^n)/4)+(-1)^((2*n+1-(-1)^n)/4)))/64.
a(n) = (13*n^2-28-3*(n^2+4)*cos(n*Pi)+6*(n^2-4)*cos(n*Pi/2))/64. (End)

Edited by N. J. A. Sloane, Aug 04 2012

A259925 a(n) = (n^2 - n - 1)^n.

Original entry on oeis.org

1, -1, 1, 125, 14641, 2476099, 594823321, 194754273881, 83733937890625, 45848500718449031, 31181719929966183601, 25804264053054077850709, 25542038069936263923006961, 29806575070123343006591796875, 40504199006061377874300161158921

Offset: 0

Ilya Gutkovskiy, Jul 09 2015

(n^2-n-1) is the Fibonacci polynomial; so (n^2 - n - 1)^n = 0 has a single root phi (A001622).

			For n = 0, a(0) = (0^2 - 0 - 1)^0 = (-1)^0 = 1.

Cf. A165900.

[(n^2 - n - 1)^n: n in [0..20]]; // Vincenzo Librandi, Jul 10 2015

A259925:=n->(n^2-n-1)^n: seq(A259925(n), n=0..20); # Wesley Ivan Hurt, Jul 10 2015

Mathematica
```
Table[(n^2 - n - 1)^n, {n, 0, 10}]
```

vector(20, n,  n--; (n^2 - n - 1)^n) \\ Michel Marcus, Aug 06 2015

[(n**2 - n - 1)**n for n in range(21)] # Anders Hellström, Jul 10 2015

a(n) = A165900(n)^n.

More terms from Vincenzo Librandi, Jul 10 2015

A319070 a(n) is the area of the surface made of the rectangles with vertices (d, n/d), (D, n/d), (D, n/D), (d, n/D) for all (d, D), pair of consecutive divisors of n.

Original entry on oeis.org

0, 1, 4, 4, 16, 7, 36, 12, 24, 19, 100, 17, 144, 39, 44, 32, 256, 33, 324, 41, 72, 103, 484, 40, 160, 147, 108, 65, 784, 57, 900, 80, 152, 259, 228, 66, 1296, 327, 204, 93, 1600, 99, 1764, 137, 160, 487, 2116, 92, 504, 165, 332, 185, 2704, 135, 388

Offset: 1

Luc Rousseau, Sep 09 2018

			The divisors of n=12 are {1, 2, 3, 4, 6, 12}. The widths of the rectangles from the definition are obtained by difference: {1, 1, 1, 2, 6}. By symmetry, their heights are the same, but in reverse order: {6, 2, 1, 1, 1}. The sought total area is the sum of products width*height of each rectangle, in other words it is the dot product 1*6 + 1*2 + 1*1 + 2*1 + 6*1. Result: 17. So, a(12)=17.

Rémy Sigrist, Table of n, a(n) for n = 1..10000
Luc Rousseau, Diagram illustrating a(10)=19.

Cf. A191743, A290110 (introducing factorization patterns of sequences of divisors).

Cf. A165900 (the Fibonacci polynomial).

a[n_] := Module[{x = Differences[Divisors[n]]}, Plus @@ (x*Reverse[x])];
Table[a[n], {n, 1, 55}]

arect(n, d, D) = (D-d)*(n/d - n/D);
a(n) = my(vd = divisors(n)); sum(k=1, #vd-1, arect(n, vd[k], vd[k+1])); \\ Michel Marcus, Oct 28 2018

a(1) = 0.
a(p) = (p-1)^2 for p a prime number.
a(p^k) = (p-1)^2*k*p^(k-1) for p^k a prime power.
a(p*q) = 2*(p-1)^2*q + (q-p)^2 for p and q primes (p < q).
a(n) = (n/2 - 1)^2 + 3 if n=2*p with p a prime greater than 2.
a(n) = (n/p + F(p-1))^2 + p^2 - F(p-1)^2 if n = p*q, p < q primes; where F denotes the Fibonacci polynomial, F(x) = x^2 - x - 1 (see A165900).
For more complex factorization patterns of n, the formula depends on the factorization pattern of the sequence of divisors of n (see A191743 or A290110), e.g.:
a(p^2*q) = 4*p*q*(p-1)^2 + (q-p^2)^2 if 1 < p < p^2 < q < p*q < p^2*q,
but
a(p^2*q) = 2*p*q*(p-1)^2 + 2*p*(q-p)^2 + (p^2-q)^2 if 1 < p < q < p^2 < p*q < p^2*q.
a(n) = Sum_{i=1..tau(n)-1} (d_[tau(n)-i+1] - d_[tau(n)-i])*(d_[i+1] - d_[i]), where {d_i}, i=1..tau(n) is the increasing sequence of divisors of n. - Ridouane Oudra, Oct 17 2021

A384329 Table read by rows: row n is the unique primitive Pythagorean triple (a,b,c) such that (a-b+c)/2 = A000217(n) and its long leg and hypotenuse are consecutive natural numbers, n >= 0.

Original entry on oeis.org

-1, 0, 1, 1, 0, 1, 5, 12, 13, 11, 60, 61, 19, 180, 181, 29, 420, 421, 41, 840, 841, 55, 1512, 1513, 71, 2520, 2521, 89, 3960, 3961, 109, 5940, 5941, 131, 8580, 8581, 155, 12012, 12013, 181, 16380, 16381, 209, 21840, 21841, 239, 28560, 28561, 271, 36720, 36721, 305, 46512, 46513, 341, 58140, 58141

Offset: 0

Miguel-Ángel Pérez García-Ortega, May 26 2025

Row n = 0 and n = 1 are included by convention and correspond to the Pythagorean triples (-1)^2 + 0^2 = 1^2 and 1^2 + 0^2 = 1^2.

			  n=0:     -1,     0,     1;
  n=1:      1,     0,     1;
  n=2:      5,    12,    13;
  n=3:     11,    60,    61;
  ...

José Miguel Blanco Casado and Miguel-Ángel Pérez García-Ortega, El Libro de las Ternas Pitagóricas

Cf. A000217, A165900 (short leg), A062392 (semiperimeter), A384498 (sum of the legs).

a=Table[(n(n+1))/2,{n,0,18}];Apply[Join,Map[{2#-1,2#^2-2#,2#^2-2#+1}&,a]]

row(n) = (2*T(n) - 1, 2*T(n)*(T(n) - 1), 2*T(n)*(T(n) - 1) + 1) where T(n) = A000217(n).

A307991 Fibonacci numbers of the form k^2 - k - 1 with integer k.

Original entry on oeis.org

1, 5, 55, 89

Offset: 1

Amiram Eldar, May 09 2019

The corresponding values of k are 2, 3, 8, 10.
Intersection of A000045 and A028387.
Also Fibonacci numbers whose reciprocals equal to Sum_{i>=1} F(i)/k^(i+1), where F(i) is the i-th Fibonacci number.
de Weger proved that there are no other terms.

			89 is in the sequence since 89 = 10^2 - 10 - 1 or equivalently 1/89 = 1/10^2 + 1/10^3 + 2/10^4 + 3/10^5 + 5/10^6 + ... This is why the first digits of the decimal expansion of 1/89 = 0.011235... are the first terms of the Fibonacci sequence.

Fenton Stancliff, A curious property of a_11, Scripta Math., Vol. 19 (1953), p. 126.

B. M. M. de Weger, A Curious Property of the Eleventh Fibonacci Number, The Rocky Mountain Journal of Mathematics, Vol. 25, No. 3 (1995), pp. 977-994.

Cf. A000045, A021093, A028387, A039595, A165900, A227875.

Select[Fibonacci[Range[2, 20]], IntegerQ[Sqrt[4# + 5]] &]

A355659 Martingale win/loss triangle, read by rows: T(n,k) = total number of dollars won (or lost) using the martingale method on all possible n trials that have exactly k losses and n-k wins, for 0 <= k <= n.

Original entry on oeis.org

0, 1, -1, 2, 1, -3, 3, 5, -1, -7, 4, 11, 7, -7, -15, 5, 19, 24, 4, -21, -31, 6, 29, 53, 38, -12, -51, -63, 7, 41, 97, 111, 41, -57, -113, -127, 8, 55, 159, 243, 187, 5, -163, -239, -255, 9, 71, 242, 458, 500, 248, -130, -394, -493, -511, 10, 89, 349, 784, 1084, 874, 202, -488, -878, -1003, -1023

Offset: 0

Greg Dresden and Max Winnick, Jul 12 2022

The martingale betting method is as follows: bet $1. If win, bet $1 on next trial. If lose, double your bet on next trial. Repeat for a total of n times.

We can use row n of the triangle to find the total expected value for n trials, if we assume that the probability of each win is p. The expected value is Sum_{k=0..n} T(n,k)*p^k*(1-p)^(n-k). In a "fair" game where p = 1/2, this equals 0, as expected.

			Triangle T(n,k) begins:
n\k| 0   1   2   3    4   5    6    7    8    9
---+-------------------------------------------
  0| 0
  1| 1  -1
  2| 2   1  -3
  3| 3   5  -1  -7
  4| 4  11   7  -7  -15
  5| 5  19  24   4  -21 -31
  6| 6  29  53  38  -12 -51  -63
  7| 7  41  97 111   41 -57 -113 -127
  8| 8  55 159 243  187   5 -163 -239 -255
  9| 9  71 242 458  500 248 -130 -394 -493 -511
Examples from triangle:
T(4,3) = -7: In this example, we consider all possibilities with 4 trials that result in 3 losses and one win. There are binomial(4,3) = 4 different combinations to consider (lllw, llwl, lwll, and wlll), which have net earnings of +1, 0, -2, -6 respectively when using the martingale method, giving a total of -7.
T(6,2) = 53: In this example, we have 6 trials and we consider the results with 2 losses and 4 wins. There are binomial(6,2) = 15 such combinations to consider (wwwwll, wwwlwl, wwwllw, wwlwwl, wwlwlw, wwllww, wlwwwl, wlwwlw, wlwlww, wllwww, lwwwwl, lwwwlw, lwwlww, lwlwww, llwwww), and summing over all 15 earnings gives us a total of 53.
T(2,0) = 2: In this example, we have 2 trials, with 0 losses and 2 wins. In this one single case, the martingale method gives us earnings of +1 and +1 with a total of 2.

Cf. A000225, A070313, A165900.

T(n,k) = T(n-1,k) + T(n-1,k-1) + binomial(n-1,k) for 0 < k < n.
Sum_{k=0..n} T(n,k) = 0 (the sum of each row equals 0).
The following six formulas describe the three leftmost columns and the three rightmost diagonals of the triangle drawn below.
T(n,0) = n (this is the scenario with n trials, 0 losses; since the martingale method has us bet 1 after each win, we end up with total earnings equal to n).
T(n,1) = n^2 - n - 1 (this scenario is when there are n trials with just 1 loss; calculations show that this equals n^2 - n - 1 = A165900(n)).
T(n,2) = (n^3 - 3n^2 - 2)/2.
T(n,n) = 1 - 2^n = A000225(n).
T(n,n-1)= 1 + 2*n - 2^n = A070313(n).
T(n,n-2) = (3*n^2 - n)/2 + 1 - 2^n.
G.f.: x*(1-y)*(1-x*y) / ((1 - x*(1+y))^2 * (1-2*x*y)). - Kevin Ryde, Aug 30 2022

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A214630 a(n) is the reduced numerator of 1/4 - 1/A109043(n)^2 = (1 - 1/A026741(n)^2)/4.

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A259925 a(n) = (n^2 - n - 1)^n.

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A319070 a(n) is the area of the surface made of the rectangles with vertices (d, n/d), (D, n/d), (D, n/D), (d, n/D) for all (d, D), pair of consecutive divisors of n.

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A384329 Table read by rows: row n is the unique primitive Pythagorean triple (a,b,c) such that (a-b+c)/2 = A000217(n) and its long leg and hypotenuse are consecutive natural numbers, n >= 0.

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A307991 Fibonacci numbers of the form k^2 - k - 1 with integer k.

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A355659 Martingale win/loss triangle, read by rows: T(n,k) = total number of dollars won (or lost) using the martingale method on all possible n trials that have exactly k losses and n-k wins, for 0 <= k <= n.

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