A035008 -id:A035008 - cp's OEIS frontend

A108099 a(n) = 8n^2 + 8n + 4.

4, 20, 52, 100, 164, 244, 340, 452, 580, 724, 884, 1060, 1252, 1460, 1684, 1924, 2180, 2452, 2740, 3044, 3364, 3700, 4052, 4420, 4804, 5204, 5620, 6052, 6500, 6964, 7444, 7940, 8452, 8980, 9524, 10084, 10660, 11252, 11860, 12484, 13124, 13780, 14452, 15140, 15844

Offset: 0

Dorthe Roel (dorthe_roel(AT)hotmail.com or dorthe.roel1(AT)skolekom.dk), Jun 07 2005

Also the number for Waterman [polyhedra] have a unit rhombic dodecahedron face so sqrt 4, sqrt 20, sqrt 52, etc...and a one-to-one match...that is, no omissions and no extras. - Steve Waterman and Roger Kaufman (swaterman(AT)watermanpolyhedron.com), Apr 02 2009. [This sentence makes no sense - some words must have been dropped. - N. J. A. Sloane, Jun 12 2014]
Also, sequence found by reading the segment (4,20) together with the line from 20, in the direction 20, 52, ..., in the square spiral whose vertices are the triangular numbers A000217. - Omar E. Pol, Sep 04 2011
Sum of consecutive even squares: (2*n)^2 + (2*n + 2)^2 = 8*n^2 + 8*n + 4. - Michel Marcus, Jan 27 2014

Ivan Panchenko, Table of n, a(n) for n = 0..1000
Adrian Rossiter, Antiprism.
Steve Waterman, Polyhedra Project.
Steve Waterman, Waterman's Polyhedral Mensuration chart.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A001844, A016742, A035008, A069129, A069894.

[ 8*n^2 + 8*n + 4 : n in [0..50] ]; // Wesley Ivan Hurt, Jun 09 2014

A108099:=n->8*n^2 + 8*n + 4; seq(A108099(n), n=0..50); # Wesley Ivan Hurt, Jun 09 2014

CoefficientList[Series[-(4*(z^2 + 2*z + 1))/(z - 1)^3, {z, 0, 100}], z] (* and *) Table[8*n*(n + 1) + 4, {n, 0, 100}] (* Vladimir Joseph Stephan Orlovsky, Jul 17 2011 *)

a(n)=8*n^2+8*n+4 \\ Charles R Greathouse IV, Jul 17 2011

a(n) = 8*n^2 + 8*n + 4.
G.f.: 4*(1+2*x+x^2)/(1-x)^3.
a(n) = 16*n + a(n-1), a(0)=4. - Vincenzo Librandi, Nov 13 2010
a(n) = A069129(n+1) + 3. - Omar E. Pol, Sep 04 2011
a(n) = A035008(n) + 4. - Omar E. Pol, Jun 12 2014
From Elmo R. Oliveira, Oct 27 2024: (Start)
E.g.f.: 4*(1 + 4*x + 2*x^2)*exp(x).
a(n) = 4*A001844(n) = 2*A069894(n).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)

A186851 Array read by antidiagonals: T(n,k) = number of n-step knight's tours on a (k+2)X(k+2) board summed over all starting positions.

Original entry on oeis.org

9, 16, 16, 25, 48, 16, 36, 96, 104, 16, 49, 160, 328, 208, 16, 64, 240, 664, 976, 400, 16, 81, 336, 1112, 2576, 2800, 800, 16, 100, 448, 1672, 5056, 9328, 8352, 1280, 16, 121, 576, 2344, 8320, 21480, 34448, 21664, 2208, 0, 144, 720, 3128, 12368, 39616, 91328

Offset: 1

R. H. Hardin and D. S. McNeil in the Sequence Fans Mailing List, Feb 27 2011

Here an n-step knight's tour is a directed path with n vertices (or a self-avoiding walk with n-1 steps). - Andrew Howroyd, Jan 02 2023

			Table starts:
   9   16     25      36      49      64      81     100    121    144 ...
  16   48     96     160     240     336     448     576    720    880 ...
  16  104    328     664    1112    1672    2344    3128   4024   5032 ...
  16  208    976    2576    5056    8320   12368   17200  22816  29216 ...
  16  400   2800    9328   21480   39616   63440   92656 127264 167264 ...
  16  800   8352   34448   91328  186544  322528  498320 712080 ...
  16 1280  21664  118480  372384  847520 1584576 2596480 ...
  16 2208  57392  405040 1508784 3846192 7777808 ...
   0 3184 135184 1290112 5807488 ...
   0 4640 317296 4089632 ...
  ...
Some n=3 solutions for 5X5:
  0 0 0 0 0    0 0 0 0 1    0 0 0 0 0    0 0 0 0 0
  0 0 0 0 0    0 0 2 0 0    0 0 0 0 0    0 0 0 0 0
  0 0 1 0 0    0 0 0 0 0    0 0 3 0 0    0 0 0 0 0
  0 0 0 3 0    0 3 0 0 0    2 0 0 0 0    3 0 0 0 1
  0 2 0 0 0    0 0 0 0 0    0 0 1 0 0    0 0 2 0 0

R. H. Hardin, Table of n, a(n) for n = 1..99

Rows 1..8 are A000290(n+2), A035008, A186852, A186853, A186854, A186855, A186856, A186857.
Column 6 is A186441.
Cf. A289204.

\\ G(n) gives polynomial valid for k >= 2*n-4.
Knights={[1,2; 1,-2; -1,2; -1,-2; 2,1; 2,-1; -2,1; -2,-1]}
G(n,f=i->'n-(i-2)) = {
  local(x=vector(n), y=vector(n));
  my(recurse(k)=
     forstep(i=2-k%2, k-1, 2, if(x[i]==x[k] && y[i]==y[k], return(0)));
     if (k==n, f(vecmax(x)-vecmin(x))*f(vecmax(y)-vecmin(y)), sum(i=1, 8, x[k+1] = x[k]+Knights[i,1]; y[k+1]=y[k]+Knights[i,2]; self()(k+1)) );
  );
  if(n==1, recurse(1), x[1]=1; y[1]=2; 8*recurse(2))
}
row(m,n)={my(p=if(n>=2*m-4, G(m,i->'x-(i-2)))); vector(n, k, if(k>=2*m-4, subst(p,'x, k), G(m, i->max(0, k+2-i))))} \\ Andrew Howroyd, Jan 02 2023

Empirical, for all rows: a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n>3,3,4,6,8,10 respectively for row in 1..6.
From Andrew Howroyd, Jan 02 2023: (Start)
The above empirical formula is correct. Equivalently T(m,n) for given m and n >= 2*m-4 is given by a quadratic polynomial in n. This is because a w X h rectangle can be placed on a k X k grid at integer coordinates in (k-w+1)*(k-h+1) ways when w and h are at most k and every knights path with m vertices spans such a rectangle with width and height at most 2*m - 1.
Sum_{i=2..(k+2)^2} T(i,k)/2 = A289204(k+2).
T(n,k) = 0 for n > (k-2)^2.
(End)

A262221 a(n) = 25n(n + 1)/2 + 1.

Original entry on oeis.org

1, 26, 76, 151, 251, 376, 526, 701, 901, 1126, 1376, 1651, 1951, 2276, 2626, 3001, 3401, 3826, 4276, 4751, 5251, 5776, 6326, 6901, 7501, 8126, 8776, 9451, 10151, 10876, 11626, 12401, 13201, 14026, 14876, 15751, 16651, 17576, 18526, 19501, 20501, 21526, 22576, 23651

Offset: 0

Bruno Berselli, Sep 15 2015

Also centered 25-gonal (or icosipentagonal) numbers.
This is the case k=25 of the formula (k*n*(n+1) - (-1)^k + 1)/2. See table in Links section for similar sequences.
For k=2*n, the formula shown above gives A011379.
Primes in sequence: 151, 251, 701, 1951, 3001, 4751, 10151, 12401, ...

E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 51 (23rd row of the table).

Bruno Berselli, Table of n, a(n) for n = 0..1000
Bruno Berselli, Table of numbers of the form (k*n*(n+1)-(-1)^k+1)/2.
Eric Weisstein's World of Mathematics, Centered Polygonal Numbers.
Index entries for sequences related to centered polygonal numbers.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A008607 (first differences), A011379, A034856, A054254, A080956, A123296, A186349, A255184.
Cf. centered polygonal numbers listed in A069190.
Similar sequences of the form (k*n*(n+1) - (-1)^k + 1)/2 with -1 <= k <= 26: A000004, A000124, A002378, A005448, A005891, A028896, A033996, A035008, A046092, A049598, A060544, A064200, A069099, A069125, A069126, A069128, A069130, A069132, A069174, A069178, A080956, A124080, A163756, A163758, A163761, A164136, A173307.

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

Table[25 n (n + 1)/2 + 1, {n, 0, 50}]
25*Accumulate[Range[0,50]]+1 (* or *) LinearRecurrence[{3,-3,1},{1,26,76},50] (* Harvey P. Dale, Jan 29 2023 *)

PARI
```
vector(50, n, n--; 25*n*(n+1)/2+1)
```
Sage
```
[25*n*(n+1)/2+1 for n in (0..50)]
```

G.f.: (1 + 23*x + x^2)/(1 - x)^3.
a(n) = a(-n-1) = 3*a(n-1) - 3*a(n-2) + a(n-3).
a(n) = A123296(n) + 1.
a(n) = A000217(5*n+2) - 2.
a(n) = A034856(5*n+1).
a(n) = A186349(10*n+1).
a(n) = A054254(5*n+2) with n>0, a(0)=1.
a(n) = A000217(n+1) + 23*A000217(n) + A000217(n-1) with A000217(-1)=0.
Sum_{i>=0} 1/a(i) = 1.078209111... = 2*Pi*tan(Pi*sqrt(17)/10)/(5*sqrt(17)).
From Amiram Eldar, Jun 21 2020: (Start)
Sum_{n>=0} a(n)/n! = 77*e/2.
Sum_{n>=0} (-1)^(n+1) * a(n)/n! = 23/(2*e). (End)
E.g.f.: exp(x)*(2 + 50*x + 25*x^2)/2. - Elmo R. Oliveira, Dec 24 2024

A139277 a(n) = n(8n+5).

Original entry on oeis.org

0, 13, 42, 87, 148, 225, 318, 427, 552, 693, 850, 1023, 1212, 1417, 1638, 1875, 2128, 2397, 2682, 2983, 3300, 3633, 3982, 4347, 4728, 5125, 5538, 5967, 6412, 6873, 7350, 7843, 8352, 8877, 9418, 9975, 10548, 11137, 11742, 12363, 13000

Offset: 0

Omar E. Pol, Apr 26 2008

Sequence found by reading the line from 0, in the direction 0, 13, ..., in the square spiral whose vertices are the triangular numbers A000217. Opposite numbers to the members of A139273 in the same spiral.

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.

Cf. A139271, A139272, A139273, A139274, A139275, A139276, A139278, A139279, A139280, A139281, A139282.

Table[n (8 n + 5), {n, 0, 50}] (* Bruno Berselli, Aug 22 2018 *)
LinearRecurrence[{3,-3,1},{0,13,42},50] (* Harvey P. Dale, Dec 04 2018 *)

a(n)=n*(8*n+5) \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 8*n^2 + 5*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) - 3 for n > 0, a(0)=0. - Vincenzo Librandi, Aug 03 2010
Sum_{n>=1} 1/a(n) = (sqrt(2)-1)*Pi/10 - 4*log(2)/5 + sqrt(2)*log(sqrt(2)+1)/5 + 8/25. - Amiram Eldar, Mar 18 2022
E.g.f.: exp(x)*x*(13 + 8*x). - Elmo R. Oliveira, Dec 15 2024

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

A176027 Binomial transform of A005563.

Original entry on oeis.org

0, 3, 14, 48, 144, 400, 1056, 2688, 6656, 16128, 38400, 90112, 208896, 479232, 1089536, 2457600, 5505024, 12255232, 27131904, 59768832, 131072000, 286261248, 622854144, 1350565888, 2919235584, 6291456000, 13522436096

Offset: 0

Paul Curtz, Dec 06 2010

The numbers appear on the diagonal of a table T(n,k), where the left column contains the elements of A005563, and further columns are recursively T(n,k) = T(n,k-1)+T(n-1,k-1):
....0....-1.....0.....0.....0.....0.....0.....0.....0.....0.
....3.....3.....2.....2.....2.....2.....2.....2.....2.....2.
....8....11....14....16....18....20....22....24....26....28.
...15....23....34....48....64....82...102...124...148...174.
...24....39....62....96...144...208...290...392...516...664.
...35....59....98...160...256...400...608...898..1290..1806.
...48....83...142...240...400...656..1056..1664..2562..3852.
...63...111...194...336...576...976..1632..2688..4352..6914.
...80...143...254...448...784..1360..2336..3968..6656.11008.
...99...179...322...576..1024..1808..3168..5504..9472.16128.
..120...219...398...720..1296..2320..4128..7296.12800.22272.
The second column is A142463, the third A060626, the fourth essentially A035008 and the fifth essentially A016802. Transposing the array gives A005563 and its higher order differences in the individual rows.

Vincenzo Librandi, Table of n, a(n) for n = 0..3000
Index entries for linear recurrences with constant coefficients, signature (6,-12,8).

Cf. A001792, A001793, A005563, A058396, A084266, A127276.

[2^(n-2)*n*(5+n) : n in [0..30]]; // Vincenzo Librandi, Oct 08 2011

LinearRecurrence[{6,-12,8},{0,3,14},30] (* Harvey P. Dale, Oct 19 2015 *)

a(n)=n*(n+5)<<(n-2) \\ Charles R Greathouse IV, Sep 21 2017

G.f.: x*(-3+4*x)/(2*x-1)^3. - R. J. Mathar, Dec 11 2010
a(n) = 2^(n-2)*n*(5+n). - R. J. Mathar, Dec 11 2010
a(n) = A127276(n) - A127276(n+1).
a(n+1)-a(n) = A084266(n+1).
a(n+2) = 16*A058396(n) for n > 0.
a(n) = 2*a(n-1) + A001792(n).
a(n) = A001793(n) - 2^(n-1) for n > 0. - Brad Clardy, Mar 02 2012
a(n) = Sum_{k=0..n-1} Sum_{i=0..n-1} (k+3) * C(n-1,i). - Wesley Ivan Hurt, Sep 20 2017
From Amiram Eldar, Aug 13 2022: (Start)
Sum_{n>=1} 1/a(n) = 1322/75 - 124*log(2)/5.
Sum_{n>=1} (-1)^(n+1)/a(n) = 132*log(3/2)/5 - 782/75. (End)

A185869 (Odd,even)-polka dot array in the natural number array A000027; read by antidiagonals.

Original entry on oeis.org

2, 7, 9, 16, 18, 20, 29, 31, 33, 35, 46, 48, 50, 52, 54, 67, 69, 71, 73, 75, 77, 92, 94, 96, 98, 100, 102, 104, 121, 123, 125, 127, 129, 131, 133, 135, 154, 156, 158, 160, 162, 164, 166, 168, 170, 191, 193, 195, 197, 199, 201, 203, 205, 207, 209, 232, 234, 236, 238, 240, 242, 244, 246, 248, 250, 252, 277, 279, 281, 283, 285, 287, 289, 291, 293, 295, 297, 299, 326, 328, 330, 332, 334, 336, 338, 340, 342, 344, 346, 348, 350, 379, 381, 383, 385, 387, 389, 391, 393, 395, 397, 399, 401, 403, 405

Offset: 1

Clark Kimberling, Feb 05 2011

This is the second of four polka dot arrays; see A185868.
row 1: A130883;
row 2: A100037;
row 3: A100038;
row 4: A100039;
col 1: A014107;
col 2: A033537;
col 3: A100040;
col 4: A100041;
diag (2,18,...): A077591;
diag (7,31,...): A157914;
diag (16,48,...): A035008;
diag (29,69,...): A108928;
antidiagonal sums: A033431;
antidiagonal sums: 2*(1^3, 2^3, 3^3, 4^3,...) = 2*A000578.
A060432(n) + n is odd if and only if n is in this sequence. - Peter Kagey, Feb 03 2016

			Northwest corner:
  2....7....16...29...46
  9....18...31...48...69
  20...33...50...71...96
  35...52...73...98...127

Peter Kagey, Table of n, a(n) for n = 1..10000

Cf. A000027 (as an array), A060432, A185868, A185870, A185871.

a185869 n = a185869_list !! (n - 1)
a185869_list = scanl (+) 2 $ a' 1
  where  a' n = 2 * n + 3 : replicate n 2 ++ a' (n + 1)
-- Peter Kagey, Sep 02 2016

f[n_,k_]:=2n-1+(2n+2k-3)(n+k-1);
TableForm[Table[f[n,k],{n,1,10},{k,1,15}]]
Table[f[n-k+1,k],{n,14},{k,n,1,-1}]//Flatten

from math import isqrt, comb
def A185869(n):
    a = (m:=isqrt(k:=n<<1))+(k>m*(m+1))
    x = n-comb(a,2)
    y = a-x+1
    return y*((y+(c:=x<<1)<<1)-5)+x*(c-3)+2 # Chai Wah Wu, Jun 18 2025

T(n,k) = 2n-1+(n+k-1)*(2n+2k-3), k>=1, n>=1.

A349081 Numbers k for which there exist two integers m with 1 <= m_1 < m_2 <= k such that A000178(k) / m! is a square, where A000178(k) = k$ = 1!2!...*k! is the superfactorial of k.

Original entry on oeis.org

8, 14, 16, 32, 48, 72, 96, 128, 160, 200, 240, 288, 336, 392, 448, 512, 574, 576, 648, 720, 800, 880, 968, 1056, 1152, 1248, 1352, 1456, 1568, 1680, 1800, 1920, 2048, 2176, 2312, 2448, 2592, 2736, 2888, 3040, 3200, 3360, 3528, 3696, 3872, 4048, 4232, 4416, 4608, 4800, 5000

Offset: 1

Bernard Schott, Dec 01 2021

nonn

This sequence is the union of three infinite and disjoint subsequences:
-> Numbers k = 8t^2 > 0 (A139098); for these numbers, m_1 = k/2 - 1 = 4t^2-1 < m_2 = k/2 = 4t^2  (see example for k = 8).
-> Numbers k = 8t*(t+1) (A035008); for these numbers, m_1 = k/2 = 4t(t+1) < m_2 =  k/2 + 1 = (2t+1)^2 (see example for k = 16).
-> Even numbers of the form 2t^2-4, t>1 in A001541 (A349766); for these numbers, m_1 = k/2 + 1 = t^2 - 1 <  m_2 = k/2 + 2 = t^2 (see example for k = 14).
See A348692 for further information.

			For k = 8, 8$ / 2! is not a square, but m_1 = 3 because 8$ / 3! = 29030400^2 and m_2 = 4 because 8$ / 4! = 14515200^2.
For k = 14, m_1 = 8 because 14$ / 8! = 1309248519599593818685440000000^2 and m_2 = 9 because 14$ / 9! = 436416173199864606228480000000^2.
For k = 16, m_1 = 8 because 16$ / 8! = 6848282921689337839624757371207680000000000^2 and m_2 = 9 because 16$ / 9! = 2282760973896445946541585790402560000000000^2.

Rick Mabry and Laura McCormick, Square products of punctured sequences of factorials, Gaz. Aust. Math. Soc., 2009, pages 346-352.

Cf. A001541, A035008, A139098, A348692, A349080.

Subsequence of A349079.

Do[j=0;l=1;g=BarnesG[k+2];While[j<2&&l<=k,If[IntegerQ@Sqrt[g/l!],j++];l++];If[j==2,Print@k],{k,5000}] (* Giorgos Kalogeropoulos, Dec 02 2021 *)

sf(n) = prod(k=2, n, k!); \\ A000178
isok(m) = if (!(m%2), my(s=sf(m)); #select(issquare, vector(4, k, s/(m/2+k-2)!), 1) == 2); \\ Michel Marcus, Dec 04 2021

A141478 a(n) = binomial(n+2,3)*4^3.

Original entry on oeis.org

64, 256, 640, 1280, 2240, 3584, 5376, 7680, 10560, 14080, 18304, 23296, 29120, 35840, 43520, 52224, 62016, 72960, 85120, 98560, 113344, 129536, 147200, 166400, 187200, 209664, 233856, 259840, 287680, 317440, 349184, 382976, 418880, 456960, 497280, 539904, 584896

Offset: 1

Zerinvary Lajos, Aug 09 2008

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A000292, A035008, A038231 (3rd subdiagonal), A210440.

[Binomial(n+2,3)*4^3: n in [1..34]];  // Bruno Berselli, Apr 07 2011

I:=[64, 256, 640, 1280]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..50]]; // Vincenzo Librandi, Jun 29 2012

Maple
```
seq(binomial(n+2,3)*4^3, n=1..36);
```

CoefficientList[Series[64/(1-x)^4,{x,0,40}],x] (* Vincenzo Librandi, Jun 29 2012 *)

G.f.: 64*x/(1-x)^4.
a(n) = 32*n*(n+1)*(n+2)/3 = 64*A000292(n).
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Vincenzo Librandi, Jun 29 2012
From Amiram Eldar, Aug 29 2022: (Start)
Sum_{n>=1} 1/a(n) = 3/128.
Sum_{n>=1} (-1)^(n+1)/a(n) = 3*log(2)/16 - 15/128. (End)
From Elmo R. Oliveira, Aug 19 2025: (Start)
E.g.f.: 32*x*(6 + 6*x + x^2)*exp(x)/3.
a(n) = 16*A210440(n). (End)

Offset adapted to the g.f. by Bruno Berselli, Apr 07 2011

A349766 Numbers of the form 2*t^2-4 when t > 1 is a term in A001541.

Original entry on oeis.org

14, 574, 19598, 665854, 22619534, 768398398, 26102926094, 886731088894, 30122754096398, 1023286908188734, 34761632124320654, 1180872205318713598, 40114893348711941774, 1362725501650887306814, 46292552162781456489998, 1572584048032918633353214, 53421565080956452077519374

Offset: 1

Bernard Schott, Dec 04 2021

nonn

Equivalently: integers k such that k$ / (k/2+1)! and k$ / (k/2+2)! are both squares when A000178 (k) = k$ = 1!*2!*...*k! is the superfactorial of k (see A348692 for further information).

The 3 subsequences of A349081 are A035008, A139098 and this one.

			A001541(1) = 3, then for t = 3, 2*t^2-4 = 14; also for k = 14, 14$ / 8! = 1309248519599593818685440000000^2 and 14$ / 9! = 436416173199864606228480000000^2. Hence, 14 is a term.

Rick Mabry and Laura McCormick, Square products of punctured sequences of factorials, Gaz. Aust. Math. Soc., 2009, pages 346-352.
Index to sequences related to Olympiads.
Index entries for linear recurrences with constant coefficients, signature (35, -35, 1).

Cf. A000178, A035008, A139098, A348692, A349079, A349080, A349081.

Subsequence of A060626.

with(orthopoly):
sequence = (2*T(n,3)^2-4, n=1..20);

(2*#^2 - 4) & /@ LinearRecurrence[{6, -1}, {3, 17}, 17] (* Amiram Eldar, Dec 04 2021 *)
LinearRecurrence[{35, -35, 1},{14, 574, 19598},17] (* Ray Chandler, Mar 01 2024 *)

a(n) = my(t=subst(polchebyshev(n), 'x, 3)); 2*t^2-4; \\ Michel Marcus, Dec 04 2021

a(n) = 2*(cosh(2*n*arcsinh(1)))^2 - 4.

a(n) = 16*A001110(n) - 2. - Hugo Pfoertner, Dec 04 2021

cp's OEIS Frontend

A108099 a(n) = 8*n^2 + 8*n + 4.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Formula

A186851 Array read by antidiagonals: T(n,k) = number of n-step knight's tours on a (k+2)X(k+2) board summed over all starting positions.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

Formula

A262221 a(n) = 25*n*(n + 1)/2 + 1.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Sage

Formula

A139277 a(n) = n*(8*n+5).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A194715 15 times triangular numbers.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Formula

A176027 Binomial transform of A005563.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A185869 (Odd,even)-polka dot array in the natural number array A000027; read by antidiagonals.

Views

Author

Keywords

Comments

A108099 a(n) = 8n^2 + 8n + 4.

A262221 a(n) = 25n(n + 1)/2 + 1.

A139277 a(n) = n(8n+5).

A349081 Numbers k for which there exist two integers m with 1 <= m_1 < m_2 <= k such that A000178(k) / m! is a square, where A000178(k) = k$ = 1!2!...*k! is the superfactorial of k.