Bruno Berselli - cp's OEIS frontend

A327440 a(n) = floor(3*n/10).

0, 0, 0, 0, 1, 1, 1, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 8, 8, 8, 9, 9, 9, 9, 10, 10, 10, 11, 11, 11, 12, 12, 12, 12, 13, 13, 13, 14, 14, 14, 15, 15, 15, 15, 16, 16, 16, 17, 17, 17, 18, 18, 18, 18, 19, 19, 19, 20, 20, 20, 21, 21, 21, 21, 22, 22, 22, 23

Offset: 0

Bruno Berselli, Sep 11 2019

The sequence can be obtained from A008585 by deleting the last digit of each term.

Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,0,0,0,0,1,-1).

Cf. A008585.

Similar sequences with the formula floor(k*n/10): A059995 (k=1); A002266 (k=2); A057354 (k=4); A004526 (k=5); A057355 (k=6); A188511 (k=7); A090223 (k=8).

[div(3*n, 10) for n in 0:80] |> println

Table[Floor[3 n/10], {n, 0, 80}]
LinearRecurrence[{1, 0, 0, 0, 0, 0, 0, 0, 0, 1, -1}, {0, 0, 0, 0, 1, 1, 1, 2, 2, 2, 3}, 80]

PARI
```
vector(80, n, n--; floor(3*n/10))
```

O.g.f.: x^4*(1 + x^3 + x^6)/((1 + x)*(1 - x)^2*(1 - x + x^2 - x^3 + x^4)*(1 + x + x^2 + x^3 + x^4)) = (x^4 + x^7 + x^10)/(1 - x - x^10 + x^11).

a(n) = a(n-1) + a(n-10) - a(n-11) for n > 10.

A322489 Numbers k such that k^k ends with 4.

Original entry on oeis.org

2, 18, 22, 38, 42, 58, 62, 78, 82, 98, 102, 118, 122, 138, 142, 158, 162, 178, 182, 198, 202, 218, 222, 238, 242, 258, 262, 278, 282, 298, 302, 318, 322, 338, 342, 358, 362, 378, 382, 398, 402, 418, 422, 438, 442, 458, 462, 478, 482, 498, 502, 518, 522, 538, 542, 558

Offset: 1

Bruno Berselli, Dec 12 2018

Also numbers k == 2 (mod 4) such that 2^k and k^2 end with the same digit.

Numbers congruent to {2, 18} mod 20. - Amiram Eldar, Feb 27 2023

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

Cf. A004526, A056849.
Subsequence of A139544, A235700.
Numbers k such that k^k ends with d: A008592 (d=0), A017281 (d=1), A067870 (d=3), this sequence (d=4), A017329 (d=5), A271346 (d=6), A322490 (d=7), A017377 (d=9).

GAP
```
List([1..70], n -> 10*n+3*(-1)^n-5);
```

[10*n+3*(-1)^n-5 for n in 1:70] |> println

Magma
```
[10*n+3*(-1)^n-5: n in [1..70]];
```

select(n->n^n mod 10=4,[$1..558]); # Paolo P. Lava, Dec 18 2018

Mathematica
```
Table[10 n + 3 (-1)^n - 5, {n, 1, 60}]
```
Maxima
```
makelist(10*n+3*(-1)^n-5, n, 1, 70);
```

apply(A322489(n)=10*n+3*(-1)^n-5, [1..70]) \\ M. F. Hasler, Dec 14 2018

Vec(2*x*(1 + 8*x + x^2) / ((1 - x)^2*(1 + x)) + O(x^70)) \\ Colin Barker, Dec 13 2018

[10*n+3*(-1)**n-5 for n in range(1, 70)]

Sage
```
[10*n+3*(-1)^n-5 for n in (1..70)]
```

O.g.f.: 2*x*(1 + 8*x + x^2)/((1 + x)*(1 - x)^2).
E.g.f.: 2 + 3*exp(-x) + 5*(2*x - 1)*exp(x).
a(n) = -a(-n+1) = a(n-1) + a(n-2) - a(n-3).
a(n) = 10*n + 3*(-1)^n - 5. Therefore:
a(n) = 10*n - 8 for odd n;
a(n) = 10*n - 2 for even n.
a(n+2*k) = a(n) + 20*k.
Sum_{n>=1} (-1)^(n+1)/a(n) = tan(2*Pi/5)*Pi/20 = sqrt(5+2*sqrt(5))*Pi/20. - Amiram Eldar, Feb 27 2023

A322490 Numbers k such that k^k ends with 7.

Original entry on oeis.org

3, 17, 23, 37, 43, 57, 63, 77, 83, 97, 103, 117, 123, 137, 143, 157, 163, 177, 183, 197, 203, 217, 223, 237, 243, 257, 263, 277, 283, 297, 303, 317, 323, 337, 343, 357, 363, 377, 383, 397, 403, 417, 423, 437, 443, 457, 463, 477, 483, 497, 503, 517, 523, 537, 543, 557, 563

Offset: 1

Bruno Berselli, Dec 12 2018

Equivalently, numbers k such that k and (7^h)^k end with the same digit, where h == 1 (mod 4).
Also, numbers k such that k and (3^h)^k end with the same digit, where h == 3 (mod 4).
Numbers congruent to {3, 17} mod 20. - Amiram Eldar, Feb 27 2023

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

Cf. A004526, A056849.
Subsequence of A063226, A295009.
Similar sequences are listed in A322489.

GAP
```
List([1..70], n -> 10*n+2*(-1)^n-5);
```

[10*n+2*(-1)^n-5 for n in 1:70] |> println

Magma
```
[10*n+2*(-1)^n-5: n in [1..70]];
```

select(n->n^n mod 10=7,[$1..563]); # Paolo P. Lava, Dec 18 2018

Table[10 n + 2 (-1)^n - 5, {n, 1, 60}]
LinearRecurrence[{1,1,-1},{3,17,23},80] (* Harvey P. Dale, Sep 15 2019 *)

Maxima
```
makelist(10*n+2*(-1)^n-5, n, 1, 70);
```

apply(A322490(n)=10*n+2*(-1)^n-5, [1..70])

Vec(x*(3 + 14*x + 3*x^2) / ((1 + x)*(1 - x)^2) + O(x^55)) \\ Colin Barker, Dec 13 2018

[10*n+2*(-1)**n-5 for n in range(1, 70)]

Sage
```
[10*n+2*(-1)^n-5 for n in (1..70)]
```

O.g.f.: x*(3 + 14*x + 3*x^2)/((1 + x)*(1 - x)^2).
E.g.f.: 3 + 2*exp(-x) + 5*(2*x - 1)*exp(x).
a(n) = -a(-n+1) = a(n-1) + a(n-2) - a(n-3).
a(n) = 10*n + 2*(-1)^n - 5. Therefore:
a(n) = 10*n - 7 for odd n;
a(n) = 10*n - 3 for even n.
a(n+2*k) = a(n) + 20*k.
Sum_{n>=1} (-1)^(n+1)/a(n) = tan(7*Pi/20)*Pi/20. - Amiram Eldar, Feb 27 2023

A321383 Numbers k such that the concatenation k21 is a square.

Original entry on oeis.org

1, 15, 37, 79, 123, 193, 259, 357, 445, 571, 681, 835, 967, 1149, 1303, 1513, 1689, 1927, 2125, 2391, 2611, 2905, 3147, 3469, 3733, 4083, 4369, 4747, 5055, 5461, 5791, 6225, 6577, 7039, 7413, 7903, 8299, 8817, 9235, 9781, 10221, 10795, 11257, 11859, 12343, 12973, 13479

Offset: 1

Bruno Berselli, Nov 08 2018

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. A008805.

Numbers k such that the concatenation km is a square: A132356 (m = 1), A273365 (m = 4), A273366 (m = 5), A273367 (m = 6), A273368 (m = 9); missing sequence for m = 16; this sequence for m = 21; missing sequence for m = 24; A002378 (m = 25).

List([1..50], n -> (50*(n-1)*n+3*(2*n-1)*(-1)^n+11)/8);

[div((50*(n-1)*n+3*(2*n-1)*(-1)^n+11), 8) for n in 1:50] |> println

[(50*(n-1)*n+3*(2*n-1)*(-1)^n+11)/8: n in [1..50]];

Table[(50 (n - 1) n + 3 (2 n - 1) (-1)^n + 11)/8, {n, 1, 50}]

makelist((50*(n-1)*n+3*(2*n-1)*(-1)^n+11)/8, n, 1, 50);

vector(50, n, nn; (50*(n-1)*n+3*(2*n-1)*(-1)^n+11)/8)

Vec(x*(1 + 14*x + 20*x^2 + 14*x^3 + x^4) / ((1 - x)^3*(1 + x)^2) + O(x^50)) \\ Colin Barker, Nov 12 2018

[(50*(n-1)*n+3*(2*n-1)*(-1)**n+11)/8 for n in range(1, 50)]

[(50*(n-1)*n+3*(2*n-1)*(-1)^n+11)/8 for n in (1..50)]

G.f.: x*(1 + 14*x + 20*x^2 + 14*x^3 + x^4)/((1 + x)^2*(1 - x)^3).
a(n) = a(-n+1) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5).
a(n) = 2*a(n-2) - a(n-4) + 50.
a(n) = (50*(n - 1)*n + 3*(2*n - 1)*(-1)^n + 11)/8. Therefore:
a(n) = (25*n^2 - 22*n + 4)/4 for even n;
a(n) = (25*n^2 - 28*n + 7)/4 for odd n.

A319007 Sum of the next n nonnegative integers repeated (A004526).

Original entry on oeis.org

0, 1, 5, 14, 29, 51, 82, 124, 178, 245, 327, 426, 543, 679, 836, 1016, 1220, 1449, 1705, 1990, 2305, 2651, 3030, 3444, 3894, 4381, 4907, 5474, 6083, 6735, 7432, 8176, 8968, 9809, 10701, 11646, 12645, 13699, 14810, 15980, 17210, 18501, 19855, 21274, 22759, 24311, 25932

Offset: 1

Bruno Berselli, Sep 07 2018

After 29, all terms are composite.

			Next n nonnegative integers repeated:    Sums:
0,  ......................................   0
0, 1,  ...................................   1
1, 2, 2,  ................................   5
3, 3, 4, 4,  .............................  14
5, 5, 6, 6, 7,  ..........................  29
7, 8, 8, 9, 9, 10,  ......................  51, etc.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (4,-7,8,-7,4,-1).

Cf. A004526, A101455, A319006.

Sum of the next n nonnegative integers: A027480.

[Integers()! (n*(n^2-2)+(-(n mod 2))^(n*(n-1)/2))/4: n in [1..50]];

a := n -> (n^3 - 2*n + (-(n mod 2))^binomial(n,2))/4;
seq(a(n), n=1..47); # Peter Luschny, Sep 09 2018

Table[(2 n (n^2 - 2) + (1 - (-1)^n) (-1)^((n-1)/2))/8, {n, 1, 50}]

concat(0, Vec(x^2*(1 + x + x^2)/((1 + x^2)*(1 - x)^4) + O(x^50))) \\ Colin Barker, Sep 10 2018

G.f.: x^2*(1 + x + x^2)/((1 + x^2)*(1 - x)^4).
a(n) = -a(-n) = 4*a(n-1) - 7*a(n-2) + 8*a(n-3) - 7*a(n-4) + 4*a(n-5) - a(n-6).
a(n) = (2*n*(n^2 - 2) + (1 - (-1)^n)*(-1)^((n-1)/2))/8.
a(n) = A319006(n) - n.
a(n) = (n^3 - 2*n + Chi(n))/4 where Chi(n) = A101455(n). - Peter Luschny, Sep 09 2018

A319006 Sum of the next n positive integers repeated (A008619).

Original entry on oeis.org

1, 3, 8, 18, 34, 57, 89, 132, 187, 255, 338, 438, 556, 693, 851, 1032, 1237, 1467, 1724, 2010, 2326, 2673, 3053, 3468, 3919, 4407, 4934, 5502, 6112, 6765, 7463, 8208, 9001, 9843, 10736, 11682, 12682, 13737, 14849, 16020, 17251, 18543, 19898, 21318, 22804, 24357, 25979

Offset: 1

Bruno Berselli, Sep 07 2018

			Next n positive integers repeated:       Sums:
1,  ......................................   1
1, 2,  ...................................   3
2, 3, 3,  ................................   8
4, 4, 5,  5,  ............................  18
6, 6, 7,  7,  8,  ........................  34
8, 9, 9, 10, 10, 11,  ....................  57, etc.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (4,-7,8,-7,4,-1).

Cf. A008619, A101455, A319007.

Sum of the next n positive integers: A006003 (after 0).

[Integers()! (n*(n^2+2)+(-(n mod 2))^(n*(n-1)/2))/4: n in [1..50]];

a := n -> (n^3 + 2*n + (-(n mod 2))^binomial(n, 2))/4:
seq(a(n), n=1..47); # Peter Luschny, Sep 09 2018

Table[(2 n (n^2 + 2) + (1 - (-1)^n) (-1)^((n-1)/2))/8, {n, 1, 50}]
Module[{nn=50,lst},lst=Flatten[Table[{n,n},{n,(nn(nn+1))/2}]];Total/@ TakeList[lst,Range[nn]]] (* Requires Mathematica version 11 or later *) (* or *) LinearRecurrence[{4,-7,8,-7,4,-1},{1,3,8,18,34,57},50] (* Harvey P. Dale, Jul 10 2021 *)

Vec(x*(1 - x + 3*x^2 - x^3 + x^4)/((1 + x^2)*(1 - x)^4) + O(x^50)) \\ Colin Barker, Sep 10 2018

G.f.: x*(1 - x + 3*x^2 - x^3 + x^4)/((1 + x^2)*(1 - x)^4).
a(n) = -a(-n) = 4*a(n-1) - 7*a(n-2) + 8*a(n-3) - 7*a(n-4) + 4*a(n-5) - a(n-6).
a(n) = (2*n*(n^2 + 2) + (1 - (-1)^n)*(-1)^((n-1)/2))/8.
a(n) = A319007(n) + n.
a(n) = (n^3 + 2*n + Chi(n))/4 where Chi(n) = A101455(n). - Peter Luschny, Sep 09 2018

A318765 a(n) = (n + 2)*(n^2 + n - 1).

Original entry on oeis.org

-2, 3, 20, 55, 114, 203, 328, 495, 710, 979, 1308, 1703, 2170, 2715, 3344, 4063, 4878, 5795, 6820, 7959, 9218, 10603, 12120, 13775, 15574, 17523, 19628, 21895, 24330, 26939, 29728, 32703, 35870, 39235, 42804, 46583, 50578, 54795, 59240, 63919, 68838, 74003, 79420, 85095

Offset: 0

Bruno Berselli, Sep 04 2018

First differences are in A004538.

a(n) is divisible by 11 for n = 3, 7, 9, 14, 18, 20, 25, 29, 31, 36, 40, ... with formula (1/3)*(11*m + (1 + (m mod 3))*(-1)^((m-1) mod 3) + 8), m >= 0.

Colin Barker, Table of n, a(n) for n = 0..1000
Bruno Berselli, Table of similar sequences (row k=3, m>1).
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A004538.
Subsequence of A047216.
Similar sequences (see Table in Links section): A011379, A027444, A033445, A034262, A045991, A069778.

GAP
```
List([0..50], n -> (n+2)*(n^2+n-1));
```

[(n+2)*(n^2+n-1) for n in 0:50] |> println

Magma
```
[(n+2)*(n^2+n-1): n in [0..50]];
```

seq((n+2)*(n^2+n-1),n=0..43); # Paolo P. Lava, Sep 04 2018

Table[(n + 2) (n^2 + n - 1), {n, 0, 50}]

Maxima
```
makelist((n+2)*(n^2+n-1), n, 0, 50);
```
PARI
```
vector(50, n, n--; (n+2)*(n^2+n-1))
```
Python
```
[(n+2)*(n**2+n-1) for n in range(50)]
```
Sage
```
[(n+2)*(n^2+n-1) for n in (0..50)]
```

O.g.f.: (-2 + 11*x - 4*x^2 + x^3)/(1 - x)^4.
E.g.f.: (-2 + 5*x + 6*x^2 + x^3)*exp(x).
a(n) = -A033445(-n-1).
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n >= 5. - Wesley Ivan Hurt, Dec 18 2020

A316672 Numbers k for which 120*k + 169 is a square.

Original entry on oeis.org

-1, 0, 1, 3, 10, 14, 17, 22, 36, 43, 48, 56, 77, 87, 94, 105, 133, 146, 155, 169, 204, 220, 231, 248, 290, 309, 322, 342, 391, 413, 428, 451, 507, 532, 549, 575, 638, 666, 685, 714, 784, 815, 836, 868, 945, 979, 1002, 1037, 1121, 1158, 1183, 1221, 1312, 1352, 1379, 1420

Offset: 1

Bruno Berselli, Jul 10 2018

All terms of A303305 belong to this sequence.

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

Cf. A051869, A303305.
Subsequence of A047283.
Cf. Numbers k for which 8*(2*h+1)*k + (2*h-1)^2 is a square: A000217 (h=0), A001318 (h=1), A085787 (h=2), A118277 (h=3), A195160 (h=4), A195313 (h=5), A277082 (h=6), this sequence (h=7), A303813 (h=8), A303298 (h=9); A303815 (h=13).

[k: k in [0..1500] | IsSquare(120*k+169)];

select(k->issqr(120*k+169),[$-1..1500]); # Muniru A Asiru, Jul 10 2018

LinearRecurrence[{1, 0, 0, 2, -2, 0, 0, -1, 1}, {-1, 0, 1, 3, 10, 14, 17, 22, 36}, 60]

isok(n) = issquare(120*n+169); \\ Michel Marcus, Jul 11 2018

Vec(x*(-1 + x + x^2 + 2*x^3 + 9*x^4 + 2*x^5 + x^6 + x^7 - x^8)/((1 + x)^2*(1 - x)^3*(1 + x^2)^2) + O(x^40)) \\ Colin Barker, Jul 18 2018

print([k for k in (0..1500) if is_square(120*k+169)])

O.g.f.: x*(-1 + x + x^2 + 2*x^3 + 9*x^4 + 2*x^5 + x^6 + x^7 - x^8)/((1 + x)^2*(1 - x)^3*(1 + x^2)^2).
a(n) = a(1-n) = a(n-1) + 2*a(n-4) - 2*a(n-5) - a(n-8) + a(n-9).
a(n) = (30*n^2 - 2*(15 + 3*(-1)^n + 10*i^(n*(n+1)))*n + 2*(5 + (-1)^n)*i^(n*(n+1)) + 3*(-1)^n - 79)/64, with i = sqrt(-1). Therefore:
a(4*k+1) = (3*k + 2)*(5*k - 1)/2;
a(4*k+2) = k*(15*k + 13)/2, first bisection of A303305;
a(4*k+3) = (k + 1)*(15*k + 2)/2, second bisection of A303305 (see A051869);
a(4*k+4) = (3*k + 1)*(5*k + 6)/2.

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).

A316224 a(n) = n(2n + 1)(4n + 1).

Original entry on oeis.org

0, 15, 90, 273, 612, 1155, 1950, 3045, 4488, 6327, 8610, 11385, 14700, 18603, 23142, 28365, 34320, 41055, 48618, 57057, 66420, 76755, 88110, 100533, 114072, 128775, 144690, 161865, 180348, 200187, 221430, 244125, 268320, 294063, 321402, 350385, 381060, 413475, 447678, 483717

Offset: 0

Bruno Berselli, Jun 27 2018

Sums of the consecutive integers from A000384(n) to A000384(n+1)-1. This is the case s=6 of the formula n*(n*(s-2) + 1)*(n*(s-2) + 2)/2 related to s-gonal numbers.

The inverse binomial transform is 0, 15, 60, 48, 0, ... (0 continued).

			Row sums of the triangle:
|  0 |  ................................................................. 0
|  1 |  2  3  4  5  .................................................... 15
|  6 |  7  8  9 10 11 12 13 14  ........................................ 90
| 15 | 16 17 18 19 20 21 22 23 24 25 26 27  ........................... 273
| 28 | 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44  ............... 612
| 45 | 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65  .. 1155
...
where:
. first column is A000384,
. second column is A130883 (without 1),
. third column is A033816,
. diagonal is A014106,
. 0, 2, 8, 18, 32, 50, ... are in A001105.

Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

First bisection of A059270 and subsequence of A034828, A047866, A109900, A290168.

Sums of the consecutive integers from P(s,n) to P(s,n+1)-1, where P(s,k) is the k-th s-gonal number: A027480 (s=3), A055112 (s=4), A228888 (s=5).

GAP
```
List([0..40], n -> n*(2*n+1)*(4*n+1));
```

[n*(2*n+1)*(4*n+1) for n in 0:40] |> println

Magma
```
[n*(2*n+1)*(4*n+1): n in [0..40]];
```

seq(n*(2*n+1)*(4*n+1),n=0..40); # Muniru A Asiru, Jun 27 2018

Table[n (2 n + 1) (4 n + 1), {n, 0, 40}]

Maxima
```
makelist(n*(2*n+1)*(4*n+1), n, 0, 40);
```
PARI
```
vector(40, n, n--; n*(2*n+1)*(4*n+1))
```
Python
```
[n*(2*n+1)*(4*n+1) for n in range(40)]
```
Sage
```
[n*(2*n+1)*(4*n+1) for n in (0..40)]
```

O.g.f.: 3*x*(5 + 10*x + x^2)/(1 - x)^4.
E.g.f.: x*(15 + 30*x + 8*x^2)*exp(x).
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).
a(n) =  3*A258582(n).
a(n) = -3*A100157(-n).
Sum_{n>0} 1/a(n) = 2*(3 - log(4)) - Pi.
Sum_{n>=1} (-1)^(n+1)/a(n) = log(2) + 2*sqrt(2)*log(1+sqrt(2)) + (sqrt(2)-1/2)*Pi - 6. - Amiram Eldar, Sep 17 2022

cp's OEIS Frontend

User: Bruno Berselli

A327440 a(n) = floor(3*n/10).

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A322489 Numbers k such that k^k ends with 4.

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A322490 Numbers k such that k^k ends with 7.

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A321383 Numbers k such that the concatenation k21 is a square.

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A319007 Sum of the next n nonnegative integers repeated (A004526).

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

A316224 a(n) = n(2n + 1)(4n + 1).