A033429 -id:A033429 - cp's OEIS frontend

A131242 Partial sums of A059995: a(n) = sum_{k=0..n} floor(k/10).

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 100, 105, 110, 115, 120, 125, 130, 135, 140, 145, 150, 156, 162, 168, 174, 180, 186, 192, 198

Offset: 0

Hieronymus Fischer, Jun 21 2007

Complementary with A130488 regarding triangular numbers, in that A130488(n)+10*a(n)=n(n+1)/2=A000217(n).

			As square array :
    0,   0,   0,   0,   0,   0,   0,   0,   0,    0
    1,   2,   3,   4,   5,   6,   7,   8,   9,   10
   12,  14,  16,  18,  20,  22,  24,  26,  28,   30
   33,  36,  39,  42,  45,  48,  51,  54,  57,   60
   64,  68,  72,  76,  80,  84,  88,  92,  96,  100
  105, 110, 115, 120, 125, 130, 135, 140, 145,  150
  156, 162, 168, 174, 180, 186, 192, 198, 204,  210
... - _Philippe Deléham_, Mar 27 2013

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,-1,0,0,0,0,0,0,0,1,-2,1).

Cf. A008728, A059995, A010879, A002266, A130488, A000217, A002620, A130518, A130519, A130520, A174709, A174738, A118729, A218470.

Table[(1/2)*Floor[n/10]*(2*n - 8 - 10*Floor[n/10]), {n,0,50}] (* G. C. Greubel, Dec 13 2016 *)
Accumulate[Table[FromDigits[Most[IntegerDigits[n]]],{n,0,110}]] (* or *) LinearRecurrence[{2,-1,0,0,0,0,0,0,0,1,-2,1},{0,0,0,0,0,0,0,0,0,0,1,2},120] (* Harvey P. Dale, Apr 06 2017 *)

for(n=0,50, print1((1/2)*floor(n/10)*(2n-8-10*floor(n/10)), ", ")) \\ G. C. Greubel, Dec 13 2016

a(n)=my(k=n\10); k*(n-5*k-4) \\ Charles R Greathouse IV, Dec 13 2016

a(n) = (1/2)*floor(n/10)*(2n-8-10*floor(n/10)).
a(n) = A059995(n)*(2n-8-10*A059995(n))/2.
a(n) = (1/2)*A059995(n)*(n-8+A010879(n)).
a(n) = (n-A010879(n))*(n+A010879(n)-8)/20.
G.f.: x^10/((1-x^10)(1-x)^2).
From Philippe Deléham, Mar 27 2013: (Start)
a(10n)   = A051624(n).
a(10n+1) = A135706(n).
a(10n+2) = A147874(n+1).
a(10n+3) = 2*A005476(n).
a(10n+4) = A033429(n).
a(10n+5) = A202803(n).
a(10n+6) = A168668(n).
a(10n+7) = 2*A147875(n).
a(10n+8) = A135705(n).
a(10n+9) = A124080(n). (End)
a(n) = A008728(n-10) for n>= 10. - Georg Fischer, Nov 03 2018

A185019 a(n) = n(14n-3).

Original entry on oeis.org

0, 11, 50, 117, 212, 335, 486, 665, 872, 1107, 1370, 1661, 1980, 2327, 2702, 3105, 3536, 3995, 4482, 4997, 5540, 6111, 6710, 7337, 7992, 8675, 9386, 10125, 10892, 11687, 12510, 13361, 14240, 15147, 16082, 17045, 18036, 19055, 20102, 21177, 22280, 23411, 24570

Offset: 0

Bruno Berselli, Oct 14 2011 - based on remarks and sequences by Omar E. Pol

Bruno Berselli, Table of n, a(n) for n = 0..1000
Bruno Berselli, Illustration of initial terms: An origin of A185019.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A013656, A033429, A049452, A049453, A147296, A195025.
Cf. A000384, A195023, A195024, A135703.
Cf. A195020 (vertices of the numerical spiral in link).

Magma
```
[n*(14*n-3): n in [0..42]];
```

I:=[0,11,50]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+Self(n-3): n in [1..50]]; // Vincenzo Librandi, Aug 18 2013

CoefficientList[Series[x (11 + 17 x)/(1 - x)^3, {x, 0, 45}], x] (* Vincenzo Librandi, Aug 18 2013 *)

PARI
```
for(n=0, 42, print1(n*(14*n-3)", "));
```

G.f.: x*(11+17*x)/(1-x)^3.
a(n) = A195025(-n).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n >= 3. - Wesley Ivan Hurt, Dec 18 2020
From Elmo R. Oliveira, Dec 30 2024: (Start)
E.g.f.: exp(x)*x*(11 + 14*x).
a(n) = n + A195023(n). (End)

A139273 a(n) = n(8n - 3).

Original entry on oeis.org

0, 5, 26, 63, 116, 185, 270, 371, 488, 621, 770, 935, 1116, 1313, 1526, 1755, 2000, 2261, 2538, 2831, 3140, 3465, 3806, 4163, 4536, 4925, 5330, 5751, 6188, 6641, 7110, 7595, 8096, 8613, 9146, 9695, 10260, 10841, 11438, 12051, 12680

Offset: 0

Omar E. Pol, Apr 26 2008

Sequence found by reading the line from 0, in the direction 0, 5, ..., in the square spiral whose vertices are the triangular numbers A000217. Opposite numbers to the members of A139277 in the same spiral.
Also, sequence of numbers of the form d*A000217(n-1) + 5*n with generating functions x*(5+(d-5)*x)/(1-x)^3; the inverse binomial transform is 0,5,d,0,0,.. (0 continued). See Crossrefs. - Bruno Berselli, Feb 11 2011
Even decagonal numbers divided by 2. - Omar E. Pol, Aug 19 2011

G. C. Greubel, Table of n, a(n) for n = 0..5000
Omar E. Pol, Determinacion geometrica de los numeros primos y perfectos.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A014634, A014635, A033585, A033586, A033587, A035008, A051870, A069129, A085250, A072279, A139272, A139274, A139275, A139276, A139278, A139279, A139280, A139281, A139282.

Cf. numbers of the form n*(d*n+10-d)/2: A008587, A056000, A028347, A140090, A014106, A028895, A045944, A186029, A007742, A022267, A033429, A022268, A049452, A186030, A135703, A152734.

[ n*(8*n-3) : n in [0..40] ];  // Bruno Berselli, Feb 11 2011

Table[n (8 n - 3), {n, 0, 40}] (* or *) LinearRecurrence[{3, -3, 1}, {0, 5, 26}, 40] (* Harvey P. Dale, Feb 02 2012 *)

a(n)=n*(8*n-3) \\ Charles R Greathouse IV, Sep 24 2015

a(n) = 8*n^2 - 3*n.
Sequences of the form a(n) = 8*n^2 + c*n have generating functions x{c+8+(8-c)x} / (1-x)^3 and recurrence a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). The inverse binomial transform is 0, c+8, 16, 0, 0, ... (0 continued). This applies to A139271-A139278, positive or negative c. - R. J. Mathar, May 12 2008
a(n) = 16*n + a(n-1) - 11 for n>0, a(0)=0. - Vincenzo Librandi, Aug 03 2010
From Bruno Berselli, Feb 11 2011: (Start)
G.f.: x*(5 + 11*x)/(1 - x)^3.
a(n) = 4*A000217(n) + A051866(n). (End)
a(n) = A028994(n)/2. - Omar E. Pol, Aug 19 2011
a(0)=0, a(1)=5, a(2)=26; for n>2, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Harvey P. Dale, Feb 02 2012
E.g.f.: (8*x^2 + 5*x)*exp(x). - G. C. Greubel, Jul 18 2017
Sum_{n>=1} 1/a(n) = 4*log(2)/3 - (sqrt(2)-1)*Pi/6 - sqrt(2)*arccoth(sqrt(2))/3. - Amiram Eldar, Jul 03 2020

A244630 a(n) = 17*n^2.

Original entry on oeis.org

0, 17, 68, 153, 272, 425, 612, 833, 1088, 1377, 1700, 2057, 2448, 2873, 3332, 3825, 4352, 4913, 5508, 6137, 6800, 7497, 8228, 8993, 9792, 10625, 11492, 12393, 13328, 14297, 15300, 16337, 17408, 18513, 19652, 20825, 22032, 23273, 24548, 25857, 27200, 28577, 29988

Offset: 0

Vincenzo Librandi, Jul 03 2014

First bisection of A195047. - Bruno Berselli, Jul 03 2014

Norms of purely imaginary numbers in Z[sqrt(-17)] (for example, 3*sqrt(-17) has norm 153). - Alonso del Arte, Jun 23 2018

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

Cf. A008599, A195047.

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), this sequence (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).

List([0..45],n->17*n^2); # Muniru A Asiru, Jun 29 2018

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

seq(17*n^2,n=0..45); # Muniru A Asiru, Jun 29 2018

Mathematica
```
Table[17 n^2, {n, 0, 40}]
```

a(n)=17*n^2 \\ Charles R Greathouse IV, Oct 07 2015

for (i <- 0 to 50) yield 17 * (i * i) // Alonso del Arte, Jun 29 2018

G.f.: 17*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) = 17*A000290(n). - Omar E. Pol, Jul 03 2014
a(n) = a(-n). - Muniru A Asiru, Jun 29 2018
From Elmo R. Oliveira, Dec 02 2024: (Start)
E.g.f.: 17*x*(1 + x)*exp(x).
a(n) = n*A008599(n) = A195047(2*n). (End)

A016850 a(n) = (5*n)^2.

Original entry on oeis.org

0, 25, 100, 225, 400, 625, 900, 1225, 1600, 2025, 2500, 3025, 3600, 4225, 4900, 5625, 6400, 7225, 8100, 9025, 10000, 11025, 12100, 13225, 14400, 15625, 16900, 18225, 19600, 21025, 22500, 24025, 25600, 27225, 28900, 30625, 32400, 34225, 36100, 38025, 40000, 42025

Offset: 0

N. J. A. Sloane

If we define C(n) = (5*n)^2 (n > 0), the sequence is the first "square-sequence" such that for every n there exists p such that C(n) = C(p) + C(p+n). We observe in fact that p = 3*n because 25 = 3^2 + 4^2. The sequence without 0 is linked with the first nontrivial solution (trivial: n^2 = 0^2 + n^2) of the equation X^2 = 2*Y^2 + 2*n^2 where X = 2*k and Y = 2*p + n which is equivalent to k^2 = p^2 + (p+n)^2 for n given. The second such "square-sequence" is (29*n)^2 (n > 0) because 29^2 = 20^2 + 21^2 and with this relation we obtain (29*n)^2 = (20*n)^2 + (20*n+n)^2. - Richard Choulet, Dec 23 2007

Vincenzo Librandi, Table of n, a(n) for n = 0..800
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000290, A033429, A053742 (first differences), A008587, A008607.

Similar sequences listed in A244630.

[(5*n)^2: n in [0..50]]; // Vincenzo Librandi, Apr 26 2011

(5Range[0, 31])^2 (* Alonso del Arte, Oct 08 2017 *)

a(n)=(5*n)^2 \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 25*n^2 = 25*A000290(n) = 5*A033429(n). - Omar E. Pol, Jul 03 2014
From Amiram Eldar, Jan 25 2021: (Start)
Sum_{n>=1} 1/a(n) = Pi^2/150.
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/300.
Product_{n>=1} (1 + 1/a(n)) = sinh(Pi/5)/(Pi/5).
Product_{n>=1} (1 - 1/a(n)) = sin(Pi/5)/(Pi/5) = 5*sqrt((5-sqrt(5))/2)/(2*Pi). (End)
a(n) = Sum_{i=0..n-1} A053742(i). - John Elias, Jun 30 2021
G.f.: 25*x*(1 + x)/(1 - x)^3. - Stefano Spezia, Jul 08 2023
From Elmo R. Oliveira, Nov 30 2024: (Start)
E.g.f.: 25*x*(1 + x)*exp(x).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2.
a(n) = n*A008607(n) = A000290(A008587(n)) = A008587(n)^2. (End)

A061167 a(n) = n^5 - n.

Original entry on oeis.org

0, 0, 30, 240, 1020, 3120, 7770, 16800, 32760, 59040, 99990, 161040, 248820, 371280, 537810, 759360, 1048560, 1419840, 1889550, 2476080, 3199980, 4084080, 5153610, 6436320, 7962600, 9765600, 11881350, 14348880, 17210340, 20511120, 24299970, 28629120

Offset: 0

Henry Bottomley, Apr 18 2001

(b^2+c^2)/(bc+1) is an integer if {b,c} are of the form {0,n}, {n,n^3}, {n^3,n^5-n}, {n^5-n,n^7-2n^3}, {n^7-2n^3,n^9-3n^5+n}, etc. for some n, in which case the division results in n^2. Cf. A052530.

Convolution of A033429 by A033581. - R. J. Mathar, Aug 19 2008

			a(2) = 32 - 2 = 30.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
D. Zagier, Problems posed at the St Andrews Colloquium, 1996
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Subsequence of A249674.

Cf. A000584, A024002, A033429, A033455, A033581, A052530.

[n^5-n: n in [0..40]]; // Vincenzo Librandi, May 02 2011

Table[n^5 - n, {n, 0, 40}] (* Vladimir Joseph Stephan Orlovsky, Feb 20 2012 *)

a(n) = 30*A033455(n-1). [Corrected by Bernard Schott, Mar 16 2021]
a(n) = -n*A024002(n).
a(n) = A000584(n) - n.
O.g.f.: 30x^2(1+x)^2/(1-x)^6. - R. J. Mathar, Aug 19 2008
a(n) = n * (n-1) * (n+1) * (n^2+1). - Bernard Schott, Mar 16 2021
E.g.f.: exp(x)*x^2*(15 + 25*x + 10*x^2 + x^3). - Stefano Spezia, Dec 27 2021

A195322 a(n) = 20*n^2.

Original entry on oeis.org

0, 20, 80, 180, 320, 500, 720, 980, 1280, 1620, 2000, 2420, 2880, 3380, 3920, 4500, 5120, 5780, 6480, 7220, 8000, 8820, 9680, 10580, 11520, 12500, 13520, 14580, 15680, 16820, 18000, 19220, 20480, 21780, 23120, 24500, 25920, 27380, 28880, 30420, 32000, 33620, 35280

Offset: 0

Omar E. Pol, Sep 16 2011

Sequence found by reading the line from 0, in the direction 0, 20, ..., in the square spiral whose vertices are the generalized dodecagonal numbers A195162. Semiaxis opposite to A195317 in the same spiral.
a(n) is the sum of all the integers less than 10*n which are not multiple of 2 or 5. a(2) = (1 + 3 + 7 + 9) + (11 + 13 + 17 + 19) = 20 + 60 = 80 = 20 * 2^2. (Link Crux Mathematicorum). - Bernard Schott, May 15 2017
Number of terms less than 10^k (k=0, 1, 2, ...): 1, 1, 3, 8, 23, 71, 224, 708, 2237, 7072, 22361, 70711, ... - Muniru A Asiru, Feb 01 2018

			From _Muniru A Asiru_, Feb 01 2018: (Start)
n=0, a(0) = 20*0^2 = 0.
n=1, a(1) = 20*1^2 = 20.
n=1, a(2) = 20*2^2 = 80.
n=1, a(3) = 20*3^2 = 180.
n=1, a(4) = 20*4^2 = 320.
...
(End)

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Léo Sauvé, Problem 53, Crux Mathematicorum, Vol. 1, Nov. 1975, page 88.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Bisection of A195148.
Cf. A016802, A033581, A033583, A135453, A139098, A144555, A195321, A195323.
Cf. A000290, A001105, A005899, A010014, A016742, A033429, A195162, A195317.
Cf. A008602.

List([0..10^3],n->20*n^2); # Muniru A Asiru, Feb 01 2018

[20*n^2: n in [0..40]]; // Vincenzo Librandi, Sep 20 2011

a := n -> 20*n^2; seq(a(n), n=0..10^3); # Muniru A Asiru, Feb 01 2018

20 Range[0, 40]^2 (* or *) LinearRecurrence[{3, -3, 1}, {0, 20, 80}, 50] (* Harvey P. Dale, Jan 18 2013 *)

a(n) = 20*n^2 \\ Charles R Greathouse IV, Oct 07 2015

a(n) = 20*A000290(n) = 10*A001105(n) = 5*A016742(n) = 4*A033429(n) = 2*A033583(n).
a(0)=0, a(1)=20, a(2)=80; for n > 2, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Harvey P. Dale, Jan 18 2013
a(n) = A010014(n) - A005899(n) for n > 0. - R. J. Cano, Sep 29 2015
From Elmo R. Oliveira, Nov 30 2024: (Start)
G.f.: 20*x*(1 + x)/(1-x)^3.
E.g.f.: 20*x*(1 + x)*exp(x).
a(n) = n*A008602(n) = A195148(2*n). (End)

A316466 a(n) = 2n(7*n - 3).

Original entry on oeis.org

0, 8, 44, 108, 200, 320, 468, 644, 848, 1080, 1340, 1628, 1944, 2288, 2660, 3060, 3488, 3944, 4428, 4940, 5480, 6048, 6644, 7268, 7920, 8600, 9308, 10044, 10808, 11600, 12420, 13268, 14144, 15048, 15980, 16940, 17928, 18944, 19988, 21060, 22160, 23288, 24444, 25628, 26840

Offset: 0

Bruno Berselli, Jul 04 2018

This is the case k = 9 of Sum_{i = 2..k} P(i,n) = (k - 1)*n*((k - 2)*n - (k - 6))/4, where P(k,n) = n*((k - 2)*n - (k - 4))/2 (see Crossrefs for similar sequences and "Square array in A139600" in Links section).

14*x + 9 is a square for x = a(n) or x = a(-n).

Colin Barker, Table of n, a(n) for n = 0..1000
Bruno Berselli, Square array in A139600.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A139600, A218471.

Similar sequences (see the first comment): A000096 (k = 3), A045943 (k = 4), A049451 (k = 5), A033429 (k = 6), A167469 (k = 7), A152744 (k = 8), this sequence (k = 9), A152994 (k = 10).

GAP
```
List([0..50], n -> 2*n*(7*n-3));
```
Julia
```
[2*n*(7*n-3) for n in 0:50] |> println
```
Magma
```
[2*n*(7*n-3): n in [0..50]];
```

Table[2 n (7 n - 3), {n, 0, 50}]
LinearRecurrence[{3,-3,1},{0,8,44},50] (* Harvey P. Dale, Jan 24 2021 *)

Maxima
```
makelist(2*n*(7*n-3), n, 0, 50);
```
PARI
```
vector(50, n, n--; 2*n*(7*n-3))
```

concat(0, Vec(4*x*(2 + 5*x)/(1 - x)^3 + O(x^40))) \\ Colin Barker, Jul 05 2018

Python
```
[2*n*(7*n-3) for n in range(50)]
```
Sage
```
[2*n*(7*n-3) for n in (0..50)]
```

O.g.f.: 4*x*(2 + 5*x)/(1 - x)^3.
E.g.f.: 2*x*(4 + 7*x)*exp(x).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
a(n) = 4*A218471(n).

A064761 a(n) = 15*n^2.

Original entry on oeis.org

0, 15, 60, 135, 240, 375, 540, 735, 960, 1215, 1500, 1815, 2160, 2535, 2940, 3375, 3840, 4335, 4860, 5415, 6000, 6615, 7260, 7935, 8640, 9375, 10140, 10935, 11760, 12615, 13500, 14415, 15360, 16335, 17340, 18375, 19440, 20535, 21660, 22815

Offset: 0

Roberto E. Martinez II, Oct 18 2001

Number of edges in a complete 6-partite graph of order 6n, K_n,n,n,n,n,n.

Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A000290, A008585, A008587, A022272, A033428, A033429, A033581, A033583.

Cf. A008597, A195046.

Table[15*n^2, {n, 0, 45}] (* Amiram Eldar, Feb 03 2021 *)

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

a(n) = 15*A000290(n) = 5*A033428(n) = 3*A033429(n). - Omar E. Pol, Dec 13 2008
a(n) = A008587(n)*A008585(n). - Reinhard Zumkeller, Apr 12 2010
a(n) = a(n-1) + 30*n - 15 for n > 0, a(0)=0. - Vincenzo Librandi, Dec 15 2010
a(n) = A022272(n) + A022272(-n). - Bruno Berselli, Mar 31 2015
a(n) = t(6*n) - 6*t(n), where t(i) = i*(i+k)/2 for any k. Special case (k=1): a(n) = A000217(6*n) - 6*A000217(n). - Bruno Berselli, Aug 31 2017
From Amiram Eldar, Feb 03 2021: (Start)
Sum_{n>=1} 1/a(n) = Pi^2/90.
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/180.
Product_{n>=1} (1 + 1/a(n)) = sqrt(15)*sinh(Pi/sqrt(15))/Pi.
Product_{n>=1} (1 - 1/a(n)) = sqrt(15)*sin(Pi/sqrt(15))/Pi. (End)
From Elmo R. Oliveira, Nov 29 2024: (Start)
G.f.: 15*x*(1 + x)/(1 - x)^3.
E.g.f.: 15*x*(1 + x)*exp(x).
a(n) = n*A008597(n) = A195046(2*n).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)

A069011 Triangle with T(n,k) = n^2 + k^2.

Original entry on oeis.org

0, 1, 2, 4, 5, 8, 9, 10, 13, 18, 16, 17, 20, 25, 32, 25, 26, 29, 34, 41, 50, 36, 37, 40, 45, 52, 61, 72, 49, 50, 53, 58, 65, 74, 85, 98, 64, 65, 68, 73, 80, 89, 100, 113, 128, 81, 82, 85, 90, 97, 106, 117, 130, 145, 162, 100, 101, 104, 109, 116, 125, 136, 149, 164, 181, 200

Offset: 0

Henry Bottomley, Apr 02 2002

For any i,j >=0 a(i)*a(j) is a member of this sequence, since (a^2 + b^2)*(c^2 + d^2) = (a*c + b*d)^2 + (a*d - b*c)^2. - Boris Putievskiy, May 05 2013
A227481(n) = number of squares in row n. - Reinhard Zumkeller, Oct 11 2013
Norm of the complex numbers n +- i*k and k +- i*n, where i denotes the imaginary unit. - Stefano Spezia, Aug 07 2025

			Triangle T(n,k) begins:
    0;
    1,   2;
    4,   5,   8;
    9,  10,  13,  18;
   16,  17,  20,  25,  32;
   25,  26,  29,  34,  41,  50;
   36,  37,  40,  45,  52,  61,  72;
   49,  50,  53,  58,  65,  74,  85,  98;
   64,  65,  68,  73,  80,  89, 100, 113, 128;
   81,  82,  85,  90,  97, 106, 117, 130, 145, 162;
  100, 101, 104, 109, 116, 125, 136, 149, 164, 181, 200;
  ...

Reinhard Zumkeller, Rows n = 0..120 of triangle, flattened

Cf. A001481 for terms in this sequence, A000161 for number of times each term appears, A048147 for square array.
Column k=0 gives A000290.
Main diagonal gives A001105.
Row sums give A132124.
T(2n,n) gives A033429.

a069011 n k = a069011_tabl !! n !! k
a069011_row n = a069011_tabl !! n
a069011_tabl = map snd $ iterate f (1, [0]) where
   f (i, xs@(x:_)) = (i + 2, (x + i) : zipWith (+) xs [i + 1, i + 3 ..])
-- Reinhard Zumkeller, Oct 11 2013

Table[n^2 + k^2, {n, 0, 12}, {k, 0, n}] (* Paolo Xausa, Aug 07 2025 *)

T(n+1,k+1) = T(n,k) + 2*(n+k+1), k=0..n; T(n+1,0) = T(n,0) + 2*n + 1. - Reinhard Zumkeller, Oct 11 2013

G.f.: x*(1 + 2*y + 5*x^3*y^2 - x^2*y*(2 + 5*y) + x*(1 - 4*y + 2*y^2))/((1 - x)^3*(1 - x*y)^3). - Stefano Spezia, Aug 04 2025

cp's OEIS Frontend

A131242 Partial sums of A059995: a(n) = sum_{k=0..n} floor(k/10).

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PARI

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A185019 a(n) = n*(14*n-3).

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Magma

Magma

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PARI

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A139273 a(n) = n*(8*n - 3).

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Magma

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A244630 a(n) = 17*n^2.

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GAP

Magma

Maple

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PARI

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A016850 a(n) = (5*n)^2.

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Magma

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A061167 a(n) = n^5 - n.

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Magma

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A195322 a(n) = 20*n^2.

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A185019 a(n) = n(14n-3).

A139273 a(n) = n(8n - 3).

A316466 a(n) = 2n(7*n - 3).