A132411 -id:A132411 - cp's OEIS frontend

A157636 Triangle read by rows: T(n, k) = 1 if k=0 or k=n, otherwise = nk(n-k)/2.

1, 1, 1, 1, 1, 1, 1, 3, 3, 1, 1, 6, 8, 6, 1, 1, 10, 15, 15, 10, 1, 1, 15, 24, 27, 24, 15, 1, 1, 21, 35, 42, 42, 35, 21, 1, 1, 28, 48, 60, 64, 60, 48, 28, 1, 1, 36, 63, 81, 90, 90, 81, 63, 36, 1, 1, 45, 80, 105, 120, 125, 120, 105, 80, 45, 1

Offset: 0

Roger L. Bagula, Mar 03 2009

			Triangle begins as:
  1;
  1,  1;
  1,  1,  1;
  1,  3,  3,   1;
  1,  6,  8,   6,   1;
  1, 10, 15,  15,  10,   1;
  1, 15, 24,  27,  24,  15,   1;
  1, 21, 35,  42,  42,  35,  21,   1;
  1, 28, 48,  60,  64,  60,  48,  28,  1;
  1, 36, 63,  81,  90,  90,  81,  63, 36,  1;
  1, 45, 80, 105, 120, 125, 120, 105, 80, 45, 1;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A000578, A002415, A107985, A132411.

A157636:= func< n,k | k eq 0 or k eq n select 1 else n*k*(n-k)/2 >;
[A157636(n,k): k in [0..n], n in [0..15]]; // G. C. Greubel, Dec 13 2021

T[n_, k_] = If[n*k*(n-k)==0, 1, n*k*(n-k)/2];
Table[T[n, k], {n,0,15}, {k,0,n}]//Flatten

def A157636(n,k): return 1 if (k==0 or k==n) else n*k*(n-k)/2
flatten([[A157636(n,k) for k in (0..n)] for n in (0..15)]) # G. C. Greubel, Dec 13 2021

T(n, k) = 1 if k=0 or k=n, otherwise = n*k*(n-k)/2.
Sum_{k=0..n} T(n, k) = 2 + n^2*(n^2 - 1)/12 = 2 + A002415(n) if n>0.
From G. C. Greubel, Dec 13 2021: (Start)
T(n, k) = T(n, n-k).
T(n, 1) = [n<2] + binomial(n, 2).
T(n, 2) = A132411(n-1), for n >= 2.
T(2*n, n) = [n=0] + A000578(n). (End)

A180357 a(n) = n^7 + 7*n.

Original entry on oeis.org

0, 8, 142, 2208, 16412, 78160, 279978, 823592, 2097208, 4783032, 10000070, 19487248, 35831892, 62748608, 105413602, 170859480, 268435568, 410338792, 612220158, 893871872, 1280000140, 1801088688, 2494358042

Offset: 0

Odimar Fabeny, Aug 30 2010

Index entries for linear recurrences with constant coefficients, signature (8,-28,56,-70,56,-28,8,-1).

Cf. A132411, A079908, A180354, A180355, A180356.

Table[n^7+7n,{n,0,30}] (* Harvey P. Dale, Sep 10 2010 *)

From Chai Wah Wu, Oct 15 2016: (Start)
a(n) = 8*a(n-1) - 28*a(n-2) + 56*a(n-3) - 70*a(n-4) + 56*a(n-5) - 28*a(n-6) + 8*a(n-7) - a(n-8) for n > 7.
G.f.: 2*x*(4*x^6 + 39*x^5 + 648*x^4 + 1138*x^3 + 648*x^2 + 39*x + 4)/(x - 1)^8. (End)

First term corrected and additional terms from Harvey P. Dale, Sep 10 2010

A265611 a(n) = a(n-1) + floor((n-1)/2) - (-1)^n + 2 for n>=2, a(0)=1, a(1)=3.

Original entry on oeis.org

1, 3, 4, 8, 10, 15, 18, 24, 28, 35, 40, 48, 54, 63, 70, 80, 88, 99, 108, 120, 130, 143, 154, 168, 180, 195, 208, 224, 238, 255, 270, 288, 304, 323, 340, 360, 378, 399, 418, 440, 460, 483, 504, 528, 550, 575, 598, 624, 648, 675, 700, 728, 754, 783, 810, 840

Offset: 0

Peter Luschny, Dec 17 2015

Enrique Pérez Herrero, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).
Peter Luschny, Looking for an interpretation, seqfan mailing list.

Cf. A002620, A005563, A028552, A052928, A132411, A198442, A217748.
Cf. A084964 and A097065, after the first 3: a(n+1) - a(n) for n>0.
Cf. A055998, after 3: a(n+1) + a(n) for n>0.
Cf. A063929: a(2*n+1) gives the second column of the triangle; for n>0, a(2*n) gives the third column.

[1] cat [(2*n*(n+6)-5*(-1)^n+5)/8: n in [1..60]]; // Bruno Berselli, Dec 18 2015

A265611 := proc(n) iquo(n+1,2); %*(%+irem(n+1,2)+2)+0^n end:
seq(A265611(n), n=0..55);

Join[{1}, Table[(2 n (n + 6) - 5 (-1)^n + 5)/8, {n, 1, 60}]] (* Bruno Berselli, Dec 18 2015 *)

Vec((x^4-2*x^3+2*x^2-x-1)/(x^4-2*x^3+2*x-1) + O(x^1000)) \\ Altug Alkan, Dec 18 2015

# The initial values x, y = 0, 1 give the quarter-squares A002620.
def A265611():
    x, y = 1, 2
    while True:
       yield x
       x, y = x + y, x//y + 1
a = A265611(); print([next(a) for i in range(60)])

O.g.f.: (x^4-2*x^3+2*x^2-x-1)/(x^4-2*x^3+2*x-1).
E.g.f.: 1-(5/8)*exp(-x)+(1/8)*(5+14*x+2*x^2)*exp(x).
a(2*n) = n*(n+3) + 0^n = A028552(n) + 0^n. [Here 0^0 = 1, otherwise 0^s = 0. - N. J. A. Sloane, Aug 26 2022]
a(2*n+1) = (n+1)*(n+3) = A005563(n+1).
a(n+1) - a(n) = floor(n/2) + 2 + (-1)^n - 0^n.
a(n) = a(-n-6) = (2*n*(n+6) - 5*(-1)^n + 5)/8 for n>0, a(0)=1. [Bruno Berselli, Dec 18 2015]
For n>0, a(n) = n + 1 + Sum_{i=1..n+1} floor(i/2) + (-1)^i =  n + floor((n+1)^2/4) + (1 - (-1)^n)/2. - Enrique Pérez Herrero, Dec 18 2015
Sum_{n>=0} 1/a(n) = 85/36. - Enrique Pérez Herrero, Dec 18 2015
a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4) for n>5. - R. H. Hardin, Dec 21 2015, proved by Susanne Wienand for the algorithm sent to the seqfan mailing list and used in the Sage script below.
a(n) = A002620(n+1) + A052928(n+1) for n>=1. (Note A198442(n) = A002620(n+2) - A052928(n+2) for n>=1.) - Peter Luschny, Dec 22 2015
a(n) = (floor((n+3)/2)-1)*(ceiling((n+3)/2)+1) for n>0. - Wesley Ivan Hurt, Mar 30 2017

A180358 n^8+8n.

Original entry on oeis.org

0, 9, 272, 6585, 65568, 390665, 1679664, 5764857, 16777280, 43046793, 100000080, 214358969, 429981792, 815730825, 1475789168, 2562890745, 4294967424, 6975757577, 11019960720, 16983563193, 25600000160, 37822859529, 54875873712

Offset: 0

Odimar Fabeny, Aug 30 2010

Index entries for linear recurrences with constant coefficients, signature (9, -36, 84, -126, 126, -84, 36, -9, 1).

Cf.: A132411, A079908, A180354, A180355, A180356, A180357.

Table[n^8+8n,{n,0,30}] (* or *) LinearRecurrence[ {9,-36,84,-126,126,-84,36,-9,1},{0,9,272,6585,65568,390665,1679664,5764857,16777280},30] (* Harvey P. Dale, Jan 03 2012 *)

a(n)= +9*a(n-1) -36*a(n-2) +84*a(n-3) -126*a(n-4) +126*a(n-5) -84*a(n-6) +36*a(n-7) -9*a(n-8) +a(n-9). G.f.: x*(-9-191*x-4461*x^2-15339*x^3-15899*x^4-4125*x^5-303*x^6+7*x^7) / (x-1)^9 . [From R. J. Mathar, Sep 19 2010]

a(0) corrected by R. J. Mathar, Sep 19 2010

A356879 Numbers k such that the sum k^x + k^y can be a square with {x, y} >= 0.

Original entry on oeis.org

0, 2, 3, 8, 15, 18, 24, 32, 35, 48, 50, 63, 72, 80, 98, 99, 120, 128, 143, 162, 168, 195, 200, 224, 242, 255, 288, 323, 338, 360, 392, 399, 440, 450, 483, 512, 528, 575, 578, 624, 648, 675, 722, 728, 783, 800, 840, 882, 899, 960, 968, 1023, 1058, 1088, 1152, 1155, 1224

Offset: 0

Karl-Heinz Hofmann, Sep 12 2022

nonn

Characteristics of the terms:
  - Any x combined with any y is a solution.
      This special case is valid only for k = 0 (exception: x = y = 0).
  - Any x is possible and if x is odd: y = x. If x is even: y = x + 3.
      This special case is valid only for k = 2 (see A356880).
  - Only even x combined with y = x + 1 gives a solution.
      Those terms are the terms of A132411.
  - Only odd x combined with y = x gives a solution.
      Those terms are the terms of A001105.
  - Any x is possible and if x is odd: y = x. If x is even: y = x + 1.
      Those terms are the terms of A132592.

			Squares that can be produced with k = 8: 8^0 + 8^1 = 9; 8^1 + 8^1 = 16; 8^2 + 8^3 = 576; 8^3 + 8^3 = 1024; 8^4 + 8^5 = 36864; 8^5 + 8^5 = 65536; 8^6 + 8^7 = 2359296, ....

Cf. A132411 is a subsequence (except A132411(1)), A001105 is a subsequence.
Cf. A132592 is a subsequence.
Cf. A356880 (k = 2), A270473 (k = 3).

Select[Range[0, 1225], IntegerQ[Sqrt[# + 1]] || IntegerQ[Sqrt[#/2]] &] (* Amiram Eldar, Sep 18 2022 *)

from gmpy2 import is_square
print([n for n in range(0,1225) if is_square(n+1) or (n % 2 == 0 and is_square(n//2))])

A133475 Integers n such that n^3 + n^2 - 9*n + 16 is a square.

Original entry on oeis.org

-4, -3, -1, 0, 1, 3, 5, 11, 15, 28, 47, 55, 81, 549, 1799, 8361

Offset: 1

Mohamed Bouhamida, Nov 29 2007

The set of x values of integral points on the elliptic curve y^2 = x^3 + x^2 - 9*x + 16.

			0^3 + (-5)^2 + (-9) = 4^2, 1^3 + (-4)^2 + (-8) = 3^2, 3^3 + (-2)^2 + (-6) = 5^2

Cf. A117950, A132411, A132414, A002522, A028872.

P := PolynomialRing(Integers()); {x: x in Sort([ p[1] : p in IntegralPoints(EllipticCurve(n^3 + n^2 - 9*n + 16)) ])};

ok[x_] := Reduce[{y^2 == x^3 + x^2 - 9*x + 16, y >= 0}, y, Integers] =!= False; Select[Table[x, {x, -4, 10000}], ok ] (* Jean-François Alcover, Sep 07 2011 *)

is(n)=issquare(n^3+n^2-9*n+16) \\ Charles R Greathouse IV, Sep 06 2016

EllipticCurve([0,1,0,-9,16]).integral_points()

Edited by Max Alekseyev, Nov 13 2009

A180359 n^9+9n.

Original entry on oeis.org

0, 10, 530, 19710, 262180, 1953170, 10077750, 40353670, 134217800, 387420570, 1000000090, 2357947790, 5159780460, 10604499490, 20661046910, 38443359510, 68719476880, 118587876650, 198359290530, 322687697950, 512000000180

Offset: 0

Odimar Fabeny, Aug 30 2010

Index entries for linear recurrences with constant coefficients, signature (10, -45, 120, -210, 252, -210, 120, -45, 10, -1).

Cf. A132411, A079908, A180354, A180355, A180356, A180357, A180358.

a(n)= +10*a(n-1) -45*a(n-2) +120*a(n-3) -210*a(n-4) +252*a(n-5) -210*a(n-6) +120*a(n-7) -45*a(n-8) +10*a(n-9) -a(n-10). G.f.: 10*x*(1+43*x+1486*x^2+8773*x^3+15682*x^4+8773*x^5+1486*x^6+43*x^7+x^8)/(x -1)^10. [From R. J. Mathar, Sep 19 2010]

a(0) corrected by R. J. Mathar, Sep 19 2010

A343009 a(n) = (n^(2n)-1)/(n^2-1) for n > 1, a(1) = 1.

Original entry on oeis.org

1, 5, 91, 4369, 406901, 62193781, 14129647351, 4467856773185, 1876182941212489, 1010101010101010101, 678356244890331342611, 555922008415320588345745, 546031727340884622966664381, 633213824057681722185793753109, 856031514432518244055765015738351

Offset: 1

Thomas Ordowski, Apr 02 2021

nonn

Conjecture: for n > 2, a(n) is a Fermat pseudoprime to base n.
If p is an odd prime, then a(p) is a Cipolla pseudoprime to base p.
Is a(m) a Fermat pseudoprime to base m for every composite m?
Amiram Eldar confirmed this up to m = 3800.
From Jianing Song, Aug 28 2022: (Start)
a(n) = Product_{d|(2n),d>2} Phi(d,n), where Phi(n,x) is the d-th cyclotomic polynomial. Note that Phi(n,x) > 1 for x >= 2 unless (n,x) = (1,2): suppose that n >= 3 and x >= 2, then Phi(n,x) = Product_{1<=j<=n,gcd(j,n)=1} (x - exp(2*j*Pi*i/n)) = Product_{1<=j<=n/2,gcd(j,n)=1} (x^2 - 2*cos(2*j*Pi/n)*x + 1) = Product_{1<=j<=n/2,gcd(j,n)=1} ((x - cos(2*j*Pi/n))^2 + (sin(2*j*Pi/n))^2) > 1 since x - cos(2*j*Pi/n) > 1. This shows that a(n) is composite for n > 2.
For n > 2, a(n) is a Fermat pseudoprime to base n, since n^(2*n) == 1 (mod a(n)) and 2*n divides a(n)-1 = n^2*(n^(2*n-2)-1)/(n^2-1): if n is even, then 2*n | n^2; if n is odd, then n | n^2 and 2 | n^2+1 = (n^4-1)/(n^2-1) | (n^(2*n-2)-1)/(n^2-1). (End)

			a(10) = (10^20-1)/99 = 1010101010101010101.

Cf. A005563, A023037, A117812, A124899, A132411, A210461.

a[n_] := (n^(2*n)-1)/(n^2-1); a[1] = 1; Array[a, 15] (* Amiram Eldar, Apr 02 2021 *)

a(n) = Sum_{k=0..n-1} n^(2*k). - Davide Rotondo, Aug 28 2022
From Alois P. Heinz, Aug 28 2022: (Start)
a(n) = A117812(n)/A005563(n-1) = A117812(n)/A132411(n-1) for n>=2.
Limit_{n -> 1} (n^(2*n)-1)/(n^2-1) = 1. (End).

More terms from Amiram Eldar, Apr 02 2021

a(1)=1 prepended and name adapted by Alois P. Heinz, Aug 28 2022

cp's OEIS Frontend

A157636 Triangle read by rows: T(n, k) = 1 if k=0 or k=n, otherwise = n*k*(n-k)/2.

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A180357 a(n) = n^7 + 7*n.

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A265611 a(n) = a(n-1) + floor((n-1)/2) - (-1)^n + 2 for n>=2, a(0)=1, a(1)=3.

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A180358 n^8+8n.

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A356879 Numbers k such that the sum k^x + k^y can be a square with {x, y} >= 0.

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Python

A133475 Integers n such that n^3 + n^2 - 9*n + 16 is a square.

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Author

Keywords

Comments

Examples

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Programs

Magma

Mathematica

PARI

Sage

Extensions

A180359 n^9+9n.

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Extensions

A343009 a(n) = (n^(2n)-1)/(n^2-1) for n > 1, a(1) = 1.

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A157636 Triangle read by rows: T(n, k) = 1 if k=0 or k=n, otherwise = nk(n-k)/2.