A001105 -id:A001105 - cp's OEIS frontend

A200139 Triangle T(n,k), read by rows, given by (1,1,0,0,0,0,0,0,0,...) DELTA (1,0,0,0,0,0,0,0,0,...) where DELTA is the operator defined in A084938.

1, 1, 1, 2, 3, 1, 4, 8, 5, 1, 8, 20, 18, 7, 1, 16, 48, 56, 32, 9, 1, 32, 112, 160, 120, 50, 11, 1, 64, 256, 432, 400, 220, 72, 13, 1, 128, 576, 1120, 1232, 840, 364, 98, 15, 1, 256, 1280, 2816, 3584, 2912, 1568, 560, 128, 17, 1, 512, 2816, 6912, 9984, 9408, 6048, 2688, 816, 162, 19, 1

Offset: 0

Philippe Deléham, Nov 13 2011

Riordan array ((1-x)/(1-2x),x/(1-2x)).
Product A097805*A007318 as infinite lower triangular arrays.
Product A193723*A130595 as infinite lower triangular arrays.
T(n,k) is the number of ways to place n unlabeled objects into any number of labeled bins (with at least one object in each bin) and then designate k of the bins. - Geoffrey Critzer, Nov 18 2012
Apparently, rows of this array are unsigned diagonals of A028297. - Tom Copeland, Oct 11 2014
Unsigned A118800, so my conjecture above is true. - Tom Copeland, Nov 14 2016

			Triangle begins:
   1
   1,   1
   2,   3,   1
   4,   8,   5,   1
   8,  20,  18,   7,   1
  16,  48,  56,  32,   9,   1
  32, 112, 160, 120,  50,  11,   1

Cf. A118800 (signed version), A081277, A039991, A001333 (antidiagonal sums), A025192 (row sums); diagonals: A000012, A005408, A001105, A002492, A072819l; columns: A011782, A001792, A001793, A001794, A006974, A006975, A006976.

Cf. A007318, A028297, A059260, A097805, A118801, A130595.

nn=15;f[list_]:=Select[list,#>0&];Map[f,CoefficientList[Series[(1-x)/(1-2x-y x) ,{x,0,nn}],{x,y}]]//Grid  (* Geoffrey Critzer, Nov 18 2012 *)

T(n,k) = 2*T(n-1,k)+T(n-1,k-1) with T(0,0)=T(1,0)=T(1,1)=1 and T(n,k)=0 for k<0 or for n

T(n,k) = A011782(n-k)*A135226(n,k) = 2^(n-k)*(binomial(n,k)+binomial(n-1,k-1))/2.

Sum_{k, 0<=k<=n} T(n,k)*x^k = A000007(n), A011782(n), A025192(n), A002001(n), A005054(n), A052934(n), A055272(n), A055274(n), A055275(n), A052268(n), A055276(n), A196731(n) for n=-1,0,1,2,3,4,5,6,7,8,9,10 respectively.

G.f.: (1-x)/(1-(2+y)*x).

T(n,k) = Sum_j>=0 T(n-1-j,k-1)*2^j.

T = A007318*A059260, so the row polynomials of this entry are given umbrally by p_n(x) = (1 + q.(x))^n, where q_n(x) are the row polynomials of A059260 and (q.(x))^k = q_k(x). Consequently, the e.g.f. is exp[tp.(x)] = exp[t(1+q.(x))] = e^t exp(tq.(x)) = [1 + (x+1)e^((x+2)t)]/(x+2), and p_n(x) = (x+1)(x+2)^(n-1) for n > 0. - Tom Copeland, Nov 15 2016

T^(-1) = A130595*(padded A130595), differently signed A118801. Cf. A097805. - Tom Copeland, Nov 17 2016

The n-th row polynomial in descending powers of x is the n-th Taylor polynomial of the rational function (1 + x)/(1 + 2*x) * (1 + 2*x)^n about 0. For example, for n = 4, (1 + x)/(1 + 2*x) * (1 + 2*x)^4 = (8*x^4 + 20*x*3 + 18*x^2 + 7*x + 1) + O(x^5). - Peter Bala, Feb 24 2018

A327329 Twice the sum of all divisors of all positive integers <= n.

Original entry on oeis.org

2, 8, 16, 30, 42, 66, 82, 112, 138, 174, 198, 254, 282, 330, 378, 440, 476, 554, 594, 678, 742, 814, 862, 982, 1044, 1128, 1208, 1320, 1380, 1524, 1588, 1714, 1810, 1918, 2014, 2196, 2272, 2392, 2504, 2684, 2768, 2960, 3048, 3216, 3372, 3516, 3612, 3860, 3974, 4160, 4304, 4500, 4608, 4848, 4992

Offset: 1

Omar E. Pol, Sep 25 2019

nonn

a(n) has a symmetric representation. Using two opposite quadrants, where in each quadrant there is the Dyck path related to partitions described in the n-th row of triangle A237593, a(n) is the total area (or the total number of cells) of the structure (see the example).

a(n) is also the total area of the horizontal faces in the stepped pyramid with n levels described in A245092 (that is the total area of the terraces plus the area of the base). - Omar E. Pol, Dec 15 2021

			Illustration of a(8) = 112 using a symmetric structure constructed with the Dyck path related to partitions described in the 8th row of triangle A237593.
                           _ _ _ _ _
                          |         |
                          |         |_
                          |           |_ _
                          |               |
                          |     56        |
                          |               |
                          |               |
           _ _ _ _ _ _ _ _|_ _ _ _ _ _ _ _|
          |               |
          |               |
          |               |
          |       56      |
          |_ _            |
              |_          |
                |         |
                |_ _ _ _ _|

Omar E. Pol, Perspective view of the stepped pyramid (first 16 levels)

Partial sums of A074400.

Cf. A001105, A006218, A013661, A024916, A067436, A222548, A236104, A237591, A237593, A243980, A245092, A262626.

Accumulate[2*DivisorSigma[1,Range[60]]] (* Harvey P. Dale, Sep 25 2021 *)

a(n) = 2*sum(k=1, n, sigma(k)); \\ Michel Marcus, Dec 20 2021

from sympy import divisor_sigma
from itertools import accumulate
def f(, n): return  + 2*divisor_sigma(n, 1)
def aupton(terms): return list(accumulate(range(terms+1), f))[1:]
print(aupton(55)) # Michael S. Branicky, Dec 16 2021

from math import isqrt
def A327329(n): return -(s:=isqrt(n))**2*(s+1)+sum((q:=n//k)*((k<<1)+q+1) for k in range(1,s+1)) # Chai Wah Wu, Oct 22 2023

a(n) = 2*A024916(n).
a(n) = A243980(n)/2.
a(n) = A006218(n) + A222548(n).
a(n) = A001105(n) - A067436(n).
lim_{n->infinity} a(n)/(n^2) = Pi^2/6 = zeta(2) (cf. A013661). - Omar E. Pol, Dec 16 2021

A016910 a(n) = (6*n)^2.

Original entry on oeis.org

0, 36, 144, 324, 576, 900, 1296, 1764, 2304, 2916, 3600, 4356, 5184, 6084, 7056, 8100, 9216, 10404, 11664, 12996, 14400, 15876, 17424, 19044, 20736, 22500, 24336, 26244, 28224, 30276, 32400, 34596, 36864, 39204, 41616, 44100, 46656, 49284, 51984, 54756, 57600, 60516, 63504, 66564, 69696, 72900

Offset: 0

N. J. A. Sloane

Areas A of two classes of triangles with integer sides (a,b,c) where a = 9k, b=10k and c = 17k, or a = 3k, b = 25k and c = 26k for k=0,1,2,... These areas are given by Heron's formula A = sqrt(s(s-a)(s-b)(s-c)) = (6k)^2, with the semiperimeter s = (a+b+c)/2. This sequence is a subsequence of A188158. - Michel Lagneau, Oct 11 2013

Sequence found by reading the line from 0, in the direction 0, 36, ..., in the square spiral whose vertices are the generalized 20-gonal numbers A218864. - Omar E. Pol, May 13 2018.

Vincenzo Librandi, Table of n, a(n) for n = 0..2000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A089491, A188158, A243941.

Cf. similar sequences of the type k*n^2: A000290 (k=1), A001105 (k=2), A033428 (k=3), A016742 (k=4), A033429 (k=5), A033581 (k=6), A033582 (k=7), A139098 (k=8), A016766 (k=9), A033583 (k=10), A033584 (k=11), A135453 (k=12), A152742 (k=13), A144555 (k=14), A064761 (k=15), A016802 (k=16), A244630 (k=17), A195321 (k=18), A244631 (k=19), A195322 (k=20), A064762 (k=21), A195323 (k=22), A244632 (k=23), A195824 (k=24), A016850 (k=25), A244633 (k=26), A244634 (k=27), A064763 (k=28), A244635 (k=29), A244636 (k=30).

[(6*n)^2: n in [0..40]]; // Vincenzo Librandi, May 03 2011

(6*Range[0,40])^2 (* or *) LinearRecurrence[{3,-3,1},{0,36,144},40] (* Harvey P. Dale, Dec 25 2017 *)
Table[36 n^2, {n, 0, 45}] (* Omar E. Pol, Jun 07 2018 *)

a(n)=36*n^2 \\ Charles R Greathouse IV, Jun 10 2016

From Ilya Gutkovskiy, Jun 09 2016: (Start)
O.g.f.: 36*x*(1 + x)/(1 - x)^3.
E.g.f.: 36*x*(1 + x)*exp(x).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
Sum_{n>=1} 1/a(n) = Pi^2/216 = A086726. (End)
Product_{n>=1} a(n)/A136017(n) = Pi/3. - Fred Daniel Kline, Jun 09 2016
a(n) = t(9*n) - 9*t(n), where t(i) = i*(i+k)/2 for any k. Special case (k=1): a(n) = A000217(9*n) - 9*A000217(n). - Bruno Berselli, Aug 31 2017
a(n) = 36*A000290(n) = 18*A001105(n) = 12*A033428 = 9*A016742(n) = 6*A033581(n) = 4*A016766(n) = 3*A135453(n) = 2*A195321(n). - Omar E. Pol, Jun 07 2018
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/432. - Amiram Eldar, Jun 27 2020
From Amiram Eldar, Jan 25 2021: (Start)
Product_{n>=1} (1 + 1/a(n)) = sinh(Pi/6)/(Pi/6).
Product_{n>=1} (1 - 1/a(n)) = sin(Pi/6)/(Pi/6) = 3/Pi (A089491). (End)

A078644 a(n) = tau(2*n^2)/2.

Original entry on oeis.org

1, 2, 3, 3, 3, 6, 3, 4, 5, 6, 3, 9, 3, 6, 9, 5, 3, 10, 3, 9, 9, 6, 3, 12, 5, 6, 7, 9, 3, 18, 3, 6, 9, 6, 9, 15, 3, 6, 9, 12, 3, 18, 3, 9, 15, 6, 3, 15, 5, 10, 9, 9, 3, 14, 9, 12, 9, 6, 3, 27, 3, 6, 15, 7, 9, 18, 3, 9, 9, 18, 3, 20, 3, 6, 15, 9, 9, 18, 3, 15, 9, 6, 3, 27, 9, 6, 9, 12, 3, 30, 9, 9, 9, 6, 9

Offset: 1

Vladeta Jovovic, Dec 13 2002

Inverse Moebius transform of A068068. Number of elements in the set {(x,y): x is odd, x|n, y|n, gcd(x,y)=1}.
The number of Pythagorean points (x,y), 0 < x < y, located on the hyperbola y = 2n(x-n)/(x-2n) and having "excess" x+y-z = 2n. - Seppo Mustonen, Jun 07 2005
a(n) is the number of Pythagorean triangles with radius of the inscribed circle equal to n. For number of primitive Pythagorean triangles having inradius n, see A068068(n). - Ant King, Mar 06 2006
Dirichlet convolution of A048691 and A154269. - R. J. Mathar, Jun 01 2011
Number of distinct L-shapes of thickness n where the L area equals the rectangular area that it "contains". Visually can be thought as those areas of A156688 (surrounded by equal border of thickness n: 2xy = (x+2n)(y+2n), x and y positive integers) where both x and y are even, so they can be split into L-shapes. So L-shapes have formula: 2xy = (x+n)(y+n). - Juhani Heino, Jul 23 2012

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
S. Mustonen, Visualization and characterization of Pythagorean triples
Seppo Mustonen, Visualization and characterization of Pythagorean triples [Local copy]
T. Omland, How many Pythagorean triples with given inradius?, J. Numb. Theory 170 (2017) 1-2.

Cf. A000005, A001105, A048691.

[NumberOfDivisors(2*n^2)/2 : n in [1..100]]; // Vincenzo Librandi, Aug 14 2018

with(numtheory): seq(add(mobius(2*d)^2*tau(n/d), d in divisors(n)), n=1..100); # Ridouane Oudra, Nov 17 2019

Table[DivisorSigma[0, 2 n^2] / 2, {n, 100}] (* Vincenzo Librandi, Aug 14 2018 *)

a(n) = numdiv(2*n^2)/2; \\ Michel Marcus, Oct 04 2013

[sigma(2*n^2,0)/2 for n in range(1,100)] # Joerg Arndt, May 12 2014

Multiplicative with a(2^e) = e+1, a(p^e) = 2*e+1, p > 2. a(n) = tau(n^2) if n is odd, a(n) = tau(n^2) - a(n/2) if n is even.
Dirichlet g.f.: zeta^3(s)/(zeta(2s)*(1+1/2^s)). - R. J. Mathar, Jun 01 2011
Sum_{k=1..n} a(k) ~ 2*n / (9*Pi^2) * (9*log(n)^2 + 6*log(n) * (-3 + 9*g + log(2) - 36*Pi^(-2)*z1) + 18 + 54*g^2 + 18*g * (log(2) - 3) - 6*log(2) - log(2)^2 - 54*sg1 + 2592*z1^2/Pi^4 - 72*Pi^-2*(9*g*z1 + (log(2) - 3)*z1 + 3*z2)), where g is the Euler-Mascheroni constant A001620, sg1 is the first Stieltjes constant A082633, z1 = Zeta'(2) = A073002, z2 = Zeta''(2) = A201994. - Vaclav Kotesovec, Feb 02 2019
a(n) = Sum_{d|n} mu(2d)^2*tau(n/d), Dirichlet convolution of A323239 and A000005. - Ridouane Oudra, Nov 17 2019
a(n) = A361689(n)/2. - R. J. Mathar, Mar 21 2023

A194715 15 times triangular numbers.

Original entry on oeis.org

0, 15, 45, 90, 150, 225, 315, 420, 540, 675, 825, 990, 1170, 1365, 1575, 1800, 2040, 2295, 2565, 2850, 3150, 3465, 3795, 4140, 4500, 4875, 5265, 5670, 6090, 6525, 6975, 7440, 7920, 8415, 8925, 9450, 9990, 10545, 11115, 11700, 12300, 12915, 13545, 14190, 14850, 15525

Offset: 0

Omar E. Pol, Oct 03 2011

Sequence found by reading the line from 0, in the direction 0, 15, ... and the same line from 0, in the direction 0, 45, ..., in the square spiral whose vertices are the generalized 17-gonal numbers.
Sum of the numbers from 7*n to 8*n. - Wesley Ivan Hurt, Dec 23 2015
Also the number of 4-cycles in the (n+6)-triangular honeycomb obtuse knight graph. - Eric W. Weisstein, Jul 28 2017

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
M. Janjic and B. Petkovic, A Counting Function, arXiv 1301.4550 [math.CO], 2013.
Eric Weisstein's World of Mathematics, Graph Cycle.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A028895, A035008, A045943, A069128, A163756.

Cf. A001105 (3-cycles in the triangular honeycomb obtuse knight graph), A290391 (5-cycles), A290392 (6-cycles). - Eric W. Weisstein, Jul 29 2017

[15*n*(n+1)/2: n in [0..50]]; // Vincenzo Librandi, Oct 04 2011

A194715:=n->15*n*(n+1)/2: seq(A194715(n), n=0..60); # Wesley Ivan Hurt, Dec 23 2015

15*Accumulate[Range[0, 60]] (* Harvey P. Dale, Feb 12 2012 *)
Table[15 n (n + 1)/2, {n, 0, 60}] (* Wesley Ivan Hurt, Dec 23 2015 *)
15 Binomial[Range[20], 2] (* Eric W. Weisstein, Jul 28 2017 *)
15 PolygonalNumber[Range[0, 20]] (* Eric W. Weisstein, Jul 28 2017 *)

a(n)=15*n*(n+1)/2 \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 15*n*(n+1)/2 = 15*A000217(n) = 5*A045943(n) = 3*A028895(n) = A069128(n+1) - 1.
From Wesley Ivan Hurt, Dec 23 2015: (Start)
G.f.: 15*x/(1-x)^3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2.
a(n) = Sum_{i=7*n..8*n} i. (End)
From Amiram Eldar, Feb 21 2023: (Start)
Sum_{n>=1} 1/a(n) = 2/15.
Sum_{n>=1} (-1)^(n+1)/a(n) = (4*log(2) - 2)/15.
Product_{n>=1} (1 - 1/a(n)) = -(15/(2*Pi))*cos(sqrt(23/15)*Pi/2).
Product_{n>=1} (1 + 1/a(n)) = (15/(2*Pi))*cos(sqrt(7/15)*Pi/2). (End)
E.g.f.: 15*exp(x)*x*(2 + x)/2. - Elmo R. Oliveira, Dec 25 2024

A244636 a(n) = 30*n^2.

Original entry on oeis.org

0, 30, 120, 270, 480, 750, 1080, 1470, 1920, 2430, 3000, 3630, 4320, 5070, 5880, 6750, 7680, 8670, 9720, 10830, 12000, 13230, 14520, 15870, 17280, 18750, 20280, 21870, 23520, 25230, 27000, 28830, 30720, 32670, 34680, 36750, 38880, 41070, 43320, 45630, 48000, 50430

Offset: 0

Vincenzo Librandi, Jul 03 2014

Sequence found by reading the line from 0, in the direction 0, 30, ..., in the square spiral whose vertices are the generalized 17-gonal numbers. - Omar E. Pol, Jul 03 2014

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

Cf. similar sequences listed in A244630.

Cf. A000290, A001105, A033428, A033429, A033581, A033583, A064761, A249674, A330451.

Magma
```
[30*n^2: n in [0..40]];
```

A244636:=n->30*n^2: seq(A244636(n), n=0..50); # Wesley Ivan Hurt, Jul 04 2014

Table[30 n^2, {n, 0, 40}]
CoefficientList[Series[30x (1+x)/(1-x)^3,{x,0,50}],x] (* or *) LinearRecurrence[ {3,-3,1},{0,30,120},50] (* Harvey P. Dale, Dec 02 2021 *)

a(n)=30*n^2 \\ Charles R Greathouse IV, Jun 17 2017

G.f.: 30*x*(1 + x)/(1 - x)^3. [corrected by Bruno Berselli, Jul 03 2014]
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2.
a(n) = 30*A000290(n) = 15*A001105(n) = 10*A033428(n) = 6*A033429(n) = 5*A033581(n) = 3*A033583(n) = 2*A064761(n). - Omar E. Pol, Jul 03 2014
From Elmo R. Oliveira, Dec 02 2024: (Start)
E.g.f.: 30*x*(1 + x)*exp(x).
a(n) = n*A249674(n) = A330451(3*n). (End)

A268581 a(n) = 2n^2 + 8n + 5.

Original entry on oeis.org

5, 15, 29, 47, 69, 95, 125, 159, 197, 239, 285, 335, 389, 447, 509, 575, 645, 719, 797, 879, 965, 1055, 1149, 1247, 1349, 1455, 1565, 1679, 1797, 1919, 2045, 2175, 2309, 2447, 2589, 2735, 2885, 3039, 3197, 3359, 3525, 3695, 3869, 4047, 4229, 4415, 4605

Offset: 0

Juri-Stepan Gerasimov, Apr 10 2016

Also, numbers m such that 2*m + 6 is a square.

All the terms end with a digit in {5, 7, 9}, or equivalently, are congruent to {5, 7, 9} mod 10. - Stefano Spezia, Aug 05 2021

Stefano Spezia, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. numbers n such that 2*n + k is a perfect square: A093328 (k=-6), A097080 (k=-5), no sequence (k=-4), A051890 (k=-3), A058331 (k=-2), A001844 (k=-1), A001105 (k=0), A046092 (k=1), A056222 (k=2), A142463 (k=3), A054000 (k=4), A090288 (k=5), this sequence (k=6), A059993 (k=7), A147973 (k=8), A139570 (k=9), no sequence (k=10), A222182 (k=11), A152811 (k=12), A181570 (k=13).

Magma
```
[2*n^2+8*n+5: n in [0..60]];
```
Magma
```
[n: n in [0..6000] | IsSquare(2*n+6)];
```

Table[2 n^2 + 8 n + 5, {n, 0, 50}] (* Vincenzo Librandi, Apr 13 2016 *)
LinearRecurrence[{3,-3,1},{5,15,29},50] (* Harvey P. Dale, Jan 18 2017 *)

lista(nn) = for(n=0, nn, print1(2*n^2+8*n+5, ", ")); \\ Altug Alkan, Apr 10 2016

[2*n^2 + 8*n + 5 for n in [0..46]] # Stefano Spezia, Aug 04 2021

From Vincenzo Librandi, Apr 13 2016: (Start)
G.f.: (5-x^2)/(1-x)^3.
a(n) = 2*(n+2)^2 - 3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). (End)
E.g.f.: exp(x)*(5 + 10*x + 2*x^2). - Stefano Spezia, Aug 03 2021

Changed offset from 1 to 0, adapted formulas and programs by Bruno Berselli, Apr 13 2016

A271625 a(n) = = 2*(n+1)^2 - 5.

Original entry on oeis.org

3, 13, 27, 45, 67, 93, 123, 157, 195, 237, 283, 333, 387, 445, 507, 573, 643, 717, 795, 877, 963, 1053, 1147, 1245, 1347, 1453, 1563, 1677, 1795, 1917, 2043, 2173, 2307, 2445, 2587, 2733, 2883, 3037, 3195, 3357, 3523, 3693, 3867, 4045, 4227, 4413, 4603, 4797, 4995, 5197, 5403, 5613, 5827

Offset: 1

Juri-Stepan Gerasimov, Apr 11 2016

Numbers n such that 2*n + 10 is a perfect square.

G. C. Greubel, Table of n, a(n) for n = 1..5000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000290, A201713.

Numbers h such that 2*h + k is a perfect square: A294774 (k=-9), A255843 (k=-8), A271649 (k=-7), A093328 (k=-6), A097080 (k=-5), A271624 (k=-4), A051890 (k=-3), A058331 (k=-2), A001844 (k=-1), A001105 (k=0), A046092 (k=1), A056222 (k=2), A142463 (k=3), A054000 (k=4), A090288 (k=5), A268581 (k=6), A059993 (k=7), (-1)*A147973 (k=8), A139570 (k=9), this sequence (k=10), A222182 (k=11), A152811 (k=12), A181510 (k=13), A161532 (k=14), no sequence (k=15).

Magma
```
[ 2*n^2 + 4*n - 3: n in [1..60]];
```

[ n: n in [1..6000] | IsSquare(2*n+10)];

Table[2 n^2 + 4 n - 3, {n, 53}] (* Michael De Vlieger, Apr 11 2016 *)
LinearRecurrence[{3,-3,1},{3,13,27},60] (* Harvey P. Dale, Jun 08 2023 *)
2*Range[2,60]^2 -5 (* G. C. Greubel, Jan 21 2025 *)

x='x+O('x^99); Vec(x*(3+4*x-3*x^2)/(1-x)^3) \\ Altug Alkan, Apr 11 2016

def A271625(n): return 2*pow(n+1,2) - 5
print([A271625(n) for n in range(1,61)]) # G. C. Greubel, Jan 21 2025

G.f.: x*(3 + 4*x - 3*x^2)/(1 - x)^3. - Ilya Gutkovskiy, Apr 11 2016
Sum_{n>=1} 1/a(n) = 13/30 - Pi*cot(sqrt(5/2)*Pi)/(2*sqrt(10)) = 0.5627678459924... . - Vaclav Kotesovec, Apr 11 2016
From Elmo R. Oliveira, Nov 17 2024: (Start)
E.g.f.: exp(x)*(2*x^2 + 6*x - 3) + 3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 3. (End)
a(n) = 2*A000290(n+1) - 5. - G. C. Greubel, Jan 21 2025

Name simplified by G. C. Greubel, Jan 21 2025

A302174 Smallest solution x of x^n + y^(n+1) = z^(n+2), x, y, z >= 1.

Original entry on oeis.org

1, 2, 27, 256, 472392, 262144, 13759414272, 4294967296, 4057816381784064

Offset: 0

Jacques Tramu, Apr 07 2018

From M. F. Hasler, Apr 13 2018: (Start)
Proofs for the upper limits given in the formula section:
For odd n = 2k-1, x = 2^(2*k^2) yields a solution with x^(2k-1) = y^(2k) = 1/2*z^(2k+1), y = 2^(k(2k-1)) and z = 2^(2k(k-1)+1), because 2*k^2*(2k-1) + 1 = (2k+1)*(2k(k-1)+1).
For even n = 2k, x = 2^(k*(2k+1))*3^(2k+2) yields a solution, with y = 2^(2*k^2)*3^(2k+1) and z = 2^(2*k^2-k+1)*3^(2k), because for the exponents of 3, (2k+1)^2 = (2k+2)2k + 1 and the factor 1+3 = 2^2 adds 2 to the (identical) exponent of 2 in x^(2k) and y^(2k+1), to factor as 2*k^2(2k+1) + 2 = (2k+2)(k(2k-1)+1). (End)

			1^0 + 3^1 = 2^2, therefore a(0) = 1.
2^1 + 5^2 = 3^3, so a(1) = 2. (No solution can have x = 1 because z^3 - 1 = (z - 1)(z^2 + z + 1) cannot be a square: if a = z - 1, then z^2 + z + 1 = a^2 + 3a + 3 is congruent to 3 modulo any factor of a, and a = 3b yields z^3 - 1 = 9*b*(3*b^2 + 3b + 1), the last factor being congruent to 1 modulo any factor of b, and cannot be a square.)
27^2 + 18^3 = 9^4, so a(2) = 27.
256^3 + 64^4 = 32^5, so a(3) = 256.
472392^4 + 52488^5 = 8748^6, so a(4) = 472392.

Cf. A001105 (2*k^2), A060757 (4^k^2 = 2^(2k^2)), A000244 (3^k).
Conjectured to be a subsequence of A003586 (2^i*3^j).
Cf. A300564, A300565, A300566, A300567, A300568 (z^4 = x^2 + y^3, ..., z^8 = x^6 + y^7).

For odd n, a(n) <= 2^((n+1)^2/2); for even n, a(n) <= 2^(n*(n+1)/2)*3^(n+2).

We may conjecture that, for n > 4, a(n) is given by these upper limits.

Extended to a(0) = 1 by M. F. Hasler, Apr 13 2018

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