A060626 -id:A060626 - cp's OEIS frontend

A033996 8 times triangular numbers: a(n) = 4n(n+1).

0, 8, 24, 48, 80, 120, 168, 224, 288, 360, 440, 528, 624, 728, 840, 960, 1088, 1224, 1368, 1520, 1680, 1848, 2024, 2208, 2400, 2600, 2808, 3024, 3248, 3480, 3720, 3968, 4224, 4488, 4760, 5040, 5328, 5624, 5928, 6240, 6560, 6888, 7224, 7568, 7920, 8280

Offset: 0

N. J. A. Sloane, Dec 11 1999

Write 0, 1, 2, ... in a clockwise spiral; sequence gives numbers on one of 4 diagonals.
Also, least m > n such that T(m)*T(n) is a square and more precisely that of A055112(n). {T(n) = A000217(n)}. - Lekraj Beedassy, May 14 2004
Also sequence found by reading the line from 0, in the direction 0, 8, ... and the same line from 0, in the direction 0, 24, ..., in the square spiral whose vertices are the generalized decagonal numbers A074377. Axis perpendicular to A195146 in the same spiral. - Omar E. Pol, Sep 18 2011
Number of diagonals with length sqrt(5) in an (n+1) X (n+1) square grid. Every 1 X 2 rectangle has two such diagonals. - Wesley Ivan Hurt, Mar 25 2015
Imagine a board made of squares (like a chessboard), one of whose squares is completely surrounded by square-shaped layers made of adjacent squares. a(n) is the total number of squares in the first to n-th layer. a(1) = 8 because there are 8 neighbors to the unit square; adding them gives a 3 X 3 square. a(2) = 24 = 8 + 16 because we need 16 more squares in the next layer to get a 5 X 5 square: a(n) = (2*n+1)^2 - 1 counting the (2n+1) X (2n+1) square minus the central square. - R. J. Cano, Sep 26 2015
The three platonic solids (the simplex, hypercube, and cross-polytope) with unit side length in n dimensions all have rational volume if and only if n appears in this sequence, after 0. - Brian T Kuhns, Feb 26 2016
The number of active (ON, black) cells in the n-th stage of growth of the two-dimensional cellular automaton defined by "Rule 645", based on the 5-celled von Neumann neighborhood. - Robert Price, May 19 2016
The square root of a(n), n>0, has continued fraction [2n; {1,4n}] with whole number part 2n and periodic part {1,4n}. - Ron Knott, May 11 2017
Numbers k such that k+1 is a square and k is a multiple of 4. - Bruno Berselli, Sep 28 2017
a(n) is the number of vertices of the octagonal network O(n,n); O(m,n) is defined by Fig. 1 of the Siddiqui et al. reference. - Emeric Deutsch, May 13 2018
a(n) is the number of vertices in conjoined n X n octagons which are arranged into a square array, a.k.a. truncated square tiling. - Donghwi Park, Dec 20 2020
a(n-2) is the number of ways to place 3 adjacent marks in a diagonal, horizontal, or vertical row on an n X n tic-tac-toe grid. - Matej Veselovac, May 28 2021

			Spiral with 0, 8, 24, 48, ... along lower right diagonal:
.
  36--37--38--39--40--41--42
   |                       |
  35  16--17--18--19--20  43
   |   |               |   |
  34  15   4---5---6  21  44
   |   |   |       |   |   |
  33  14   3   0   7  22  45
   |   |   |   | \ |   |   |
  32  13   2---1   8  23  46
   |   |           | \ |   |
  31  12--11--10---9  24  47
   |                   | \ |
  30--29--28--27--26--25  48
                            \
[Reformatted by _Jon E. Schoenfield_, Dec 25 2016]

Stuart M. Ellerstein, J. Recreational Math. 29 (3) 188, 1998.
R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 2nd ed., 1994, p. 99.
Stephen Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 170.

Ivan Panchenko, Table of n, a(n) for n = 0..1000
M. K. Siddiqui, M. Naeem, N. A. Rahman, and M. Imran, Computing topological indices of certain networks, J. of Optoelectronics and Advanced Materials, 18, No. 9-10, 2016, 884-892.
N. J. A. Sloane, On the Number of ON Cells in Cellular Automata, arXiv:1503.01168 [math.CO], 2015.
Leo Tavares, Illustration: Centroid Diamonds.
Eric Weisstein's World of Mathematics, Elementary Cellular Automaton.
Eric Weisstein's World of Mathematics, Hamiltonian Path.
Eric Weisstein's World of Mathematics, Knight Graph.
Stephen Wolfram, A New Kind of Science
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
Index entries for sequences related to cellular automata
Index to 2D 5-Neighbor Cellular Automata.
Index to Elementary Cellular Automata.

Cf. A000217, A016754, A002378, A024966, A027468, A028895, A028896, A045943, A046092, A049598, A088538, A124080, A008590 (first differences), A130809 (partial sums).
Sequences from spirals: A001107, A002939, A002943, A007742, A033951, A033952, A033953, A033954, A033988, A033989, A033990, A033991, A033996. - Omar E. Pol, Dec 12 2008
Sequences on the four axes of the square spiral: Starting at 0: A001107, A033991, A007742, A033954; starting at 1: A054552, A054556, A054567, A033951.
Sequences on the four diagonals of the square spiral: Starting at 0: A002939 = 2*A000384, A016742 = 4*A000290, A002943 = 2*A014105, A033996 = 8*A000217; starting at 1: A054554, A053755, A054569, A016754.
Sequences obtained by reading alternate terms on the X and Y axes and the two main diagonals of the square spiral: Starting at 0: A035608, A156859, A002378 = 2*A000217, A137932 = 4*A002620; starting at 1: A317186, A267682, A002061, A080335.

[ 4*n*(n+1) : n in [0..50] ]; // Wesley Ivan Hurt, Jun 09 2014

seq(8*binomial(n+1, 2), n=0..46); # Zerinvary Lajos, Nov 24 2006
[seq((2*n+1)^2-1, n=0..46)];

Table[(2n - 1)^2 - 1, {n, 50}] (* Alonso del Arte, Mar 31 2013 *)

nsqm1(n) = { forstep(x=1,n,2, y = x*x-1; print1(y, ", ") ) }

a(n) = 4*n^2 + 4*n = (2*n+1)^2 - 1.
G.f.: 8*x/(1-x)^3.
a(n) = A016754(n) - 1 = 2*A046092(n) = 4*A002378(n). - Lekraj Beedassy, May 25 2004
a(n) = A049598(n) - A046092(n); a(n) = A124080(n) - A002378(n). - Zerinvary Lajos, Mar 06 2007
a(n) = 8*A000217(n). - Omar E. Pol, Dec 12 2008
a(n) = A005843(n) * A163300(n). - Juri-Stepan Gerasimov, Jul 26 2009
a(n) = a(n-1) + 8*n (with a(0)=0). - Vincenzo Librandi, Nov 17 2010
For n > 0, a(n) = A058031(n+1) - A062938(n-1). - Charlie Marion, Apr 11 2013
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Wesley Ivan Hurt, Mar 25 2015
a(n) = A000578(n+1) - A152618(n). - Bui Quang Tuan, Apr 01 2015
a(n) - a(n-1) = A008590(n), n > 0. - Altug Alkan, Sep 26 2015
From Ilya Gutkovskiy, May 19 2016: (Start)
E.g.f.: 4*x*(2 + x)*exp(x).
Sum_{n>=1} 1/a(n) = 1/4. (End)
Product_{n>=1} a(n)/A016754(n) = Pi/4. - Daniel Suteu, Dec 25 2016
a(n) = A056220(n) + A056220(n+1). - Bruce J. Nicholson, May 29 2017
sqrt(a(n)+1) - sqrt(a(n)) = (sqrt(n+1) - sqrt(n))^2. - Seiichi Manyama, Dec 23 2018
a(n)*a(n+k) + 4*k^2 = m^2 where m = (a(n) + a(n+k))/2 - 2*k^2; for k=1, m = 4*n^2 + 8*n + 2 = A060626(n). - Ezhilarasu Velayutham, May 22 2019
Sum_{n>=1} (-1)^n/a(n) = 1/4 - log(2)/2. - Vaclav Kotesovec, Dec 21 2020
From Amiram Eldar, Feb 21 2023: (Start)
Product_{n>=1} (1 - 1/a(n)) = -(4/Pi)*cos(Pi/sqrt(2)).
Product_{n>=1} (1 + 1/a(n)) = 4/Pi (A088538). (End)

A091823 a(n) = 2n^2 + 3n - 1.

Original entry on oeis.org

4, 13, 26, 43, 64, 89, 118, 151, 188, 229, 274, 323, 376, 433, 494, 559, 628, 701, 778, 859, 944, 1033, 1126, 1223, 1324, 1429, 1538, 1651, 1768, 1889, 2014, 2143, 2276, 2413, 2554, 2699, 2848, 3001, 3158, 3319, 3484, 3653, 3826, 4003, 4184, 4369, 4558, 4751

Offset: 1

Howard A. Landman, Mar 08 2004

a(n) is the position of the row in Pascal's triangle (A007318) in which three consecutive entries appear in the ratio n: n+1: n+2. (Even valid for n = 0 if you allow for a position of -1 to have value 0.) The solution is unique for each n.
The row numbers are given by A060626.
This sequence plus 1 (i.e., a(n) = 2*n^2 + 3*n) is the sequence A014106. - Howard A. Landman, Mar 28 2004
If Y and Z are a 2-blocks of a 2n-set X then, for n>=2, a(n-1) is the number of (2n-2)-subsets of X intersecting Y. - Milan Janjic, Nov 18 2007
One might prepend an initial -1: "-1, 4, 13, 26, 43, ..." - Vladimir Joseph Stephan Orlovsky, Oct 25 2008 (This would require too many other changes. - N. J. A. Sloane, Mar 27 2014)

			Entries in the ratio 1:2:3 appear in row 14 of Pascal's triangle (A007318) starting at position 4 (1001, 2002, 3003). Entries in the ratio 2:3:4 appear in row 34 of Pascal's triangle starting at position 13 (927983760, 1391975640, 1855967520); and so on (row 62, pos. 26; row 98, pos. 43; ...).

Bruno Berselli, Table of n, a(n) for n = 1..1000
Guo-Niu Han, Enumeration of Standard Puzzles, 2011, p. 21. [Cached copy]
Guo-Niu Han, Enumeration of Standard Puzzles, arXiv:2006.14070 [math.CO], 2020.
Milan Janjic, Two Enumerative Functions. [Broken link]
Leo Tavares, Illustration: Square/Triangular Union
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A007318, A014106, A060626.

Cf. A000290, A000217.

[2*n^2+3*n-1: n in [1..50]]; // Bruno Berselli, Mar 28 2014

A091823:=n->2*n^2 + 3*n - 1; seq(A091823(n), n=1..100); # Wesley Ivan Hurt, Mar 27 2014

Table[2 n^2 + 3 n - 1, {n, 50}] (* Bruno Berselli, Mar 28 2014 *)

a(n)=2*n^2+3*n-1 \\ Charles R Greathouse IV, Sep 24 2015

#!/usr/bin/perl $a = 1; while (1) { $k = $a*(2*$a + 3) - 1; print "$k,"; $a ++; }

a(n) = n + 4*binomial(2+n, n), with offset 0. - Zerinvary Lajos, May 12 2006
G.f.: x*(4 + x - x^2)/(1 - x)^3. - Vincenzo Librandi, Mar 28 2014
a(n) = A000290(n+1) + 2*A000217(n) - 2. - Leo Tavares, Aug 31 2023
E.g.f.: 1 + exp(x)*(2*x^2 + 5*x - 1). - Stefano Spezia, Jun 16 2024

A132411 a(0) = 0, a(1) = 1 and a(n) = n^2 - 1 with n >= 2.

Original entry on oeis.org

0, 1, 3, 8, 15, 24, 35, 48, 63, 80, 99, 120, 143, 168, 195, 224, 255, 288, 323, 360, 399, 440, 483, 528, 575, 624, 675, 728, 783, 840, 899, 960, 1023, 1088, 1155, 1224, 1295, 1368, 1443, 1520, 1599, 1680, 1763, 1848, 1935, 2024, 2115, 2208, 2303, 2400, 2499, 2600

Offset: 0

Mohamed Bouhamida, Nov 12 2007

X values of solutions to the equation X^3 - (X + 1)^2 + X + 2 = Y^2.
To prove that X = 1 or X = n^2 - 1: Y^2 = X^3 - (X + 1)^2 + X + 2 = X^3 - X^2 - X + 1 = (X + 1)(X^2 - 2X + 1) = (X + 1)*(X - 1)^2 it means: X = 1 or (X + 1) must be a perfect square, so X = 1 or X = n^2 - 1 with n >= 1. Which gives: (X, Y) = (0, 1) or (X, Y) = (1, 0) or (X, Y) = (n^2 - 1, n*(n^2 - 2)) with n >= 2.
An equivalent technique of integer factorization would work for example for the equation X^3 + 3*X^2 - 9*X + 5 = (X+5)(X-1)^2 = Y^2, looking for perfect squares of the form X + 5 = n^2. Another example is X^3 + X^2 - 5*X + 3 = (X+3)*(X-1)^2 = Y^2 with solutions generated from perfect squares of the form X + 3 = n^2. - R. J. Mathar, Nov 20 2007
Sum of possible divisors of a prime number up to its square root, with duplicate entries removed. - Odimar Fabeny, Aug 25 2010
a(0) = 0, a(1) = 1 and a(n) is the smallest k different from n such that n divides k and n+1 divides k+1. - Michel Lagneau, Apr 27 2013
The identity (4*n^2-2)^2 - (n^2-1)*(4*n)^2 = 4 can be written as A060626(n+1)^2 - a(n+2)*A008586(n+2)^2 = 4. - Vincenzo Librandi, Jun 16 2014
Conjecture: the sequence terms are the exponents in the expansion of the q-series Sum_{n >= 0} q^(2*n) * Product_{k >= 2*n+2} 1 - (-q)^k = 1 + q^3 + q^8 + q^15 + q^24 + .... - Peter Bala, May 10 2025

			0^3 - 1^2 + 2 = 1^2, 1^3 - 2^2 + 3 = 0^2, 3^3 - 4^2 + 5 = 4^2.
For P(n) = 29 we have sqrt(29) = 5.3851... so possible divisors are 3 and 5; for P(n) = 53 we have sqrt(53) = 7.2801... so possible divisors are 3, 5 and 7. - _Odimar Fabeny_, Aug 25 2010

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

Cf. A002522, A005563, A008586, A028560, A060626, A170949.

[0,1] cat [n^2 - 1: n in [2..60]]; // Vincenzo Librandi, May 01 2014

a:= n-> `if`(n<2, n, n^2-1):
seq(a(n), n=0..55);  # Alois P. Heinz, Jan 24 2021

Join[{0, 1}, LinearRecurrence[{3, -3, 1}, {3, 8, 15}, 80]] (* and *) Table[If[n < 2, n, n^2 - 1], {n, 0, 80}] (* Vladimir Joseph Stephan Orlovsky, Feb 14 2012 *)
Join[{0,1},Range[2,50]^2-1] (* Harvey P. Dale, Feb 27 2013 *)
CoefficientList[Series[x + x^2 (-3 + x)/(-1 + x)^3, {x, 0, 60}], x] (* Vincenzo Librandi, May 01 2014 *)

concat(0, Vec(x+x^2*(-3+x)/(-1+x)^3 + O(x^100))) \\ Altug Alkan, Dec 18 2015

a(n)=if(n>1,n^2-1,n) \\ Charles R Greathouse IV, Dec 18 2015

a(n) = A005563(n-1), n > 1.
G.f.: x + x^2*(-3+x)/(-1+x)^3. - R. J. Mathar, Nov 20 2007
Starting (1, 3, 8, 15, 24, ...) = binomial transform of [1, 2, 3, -1, 1, -1, ...]. - Gary W. Adamson, May 12 2008
a(n) = A170949(A002522(n-1)) for n > 0. - Reinhard Zumkeller, Mar 08 2010
Sum_{n>0} 1/a(n) = 7/4. - Enrique Pérez Herrero, Dec 18 2015
Sum_{n>=1} (-1)^(n+1)/a(n) = 3/4. - Amiram Eldar, Sep 27 2022
From Elmo R. Oliveira, May 29 2025: (Start)
E.g.f.: exp(x)*(x^2 + x - 1) + x + 1.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n >= 5. (End)

Definition simplified by N. J. A. Sloane, Sep 05 2010

A188644 Array of (k^n + k^(-n))/2 where k = (sqrt(x^2-1) + x)^2 for integers x >= 1.

Original entry on oeis.org

1, 1, 1, 1, 7, 1, 1, 97, 17, 1, 1, 1351, 577, 31, 1, 1, 18817, 19601, 1921, 49, 1, 1, 262087, 665857, 119071, 4801, 71, 1, 1, 3650401, 22619537, 7380481, 470449, 10081, 97, 1, 1, 50843527, 768398401, 457470751, 46099201, 1431431, 18817, 127, 1

Offset: 0

Charles L. Hohn, Apr 06 2011

Conjecture: Given the function f(x,y) = (sqrt(x^2+y) + x)^2 and constant k=f(x,y), then for all integers x >= 1 and y=[+-]1, k may be irrational, but (k^n + k^(-n))/2 always produces integer sequences; y=-1 results shown here; y=1 results are A188645.

Also square array A(n,k), n >= 1, k >= 0, read by antidiagonals, where A(n,k) is Chebyshev polynomial of the first kind T_{2*k}(x), evaluated at x=n. - Seiichi Manyama, Dec 30 2018

			Row 2 gives {( (2+sqrt(3))^(2*n) + (2-sqrt(3))^(2*n) )/2}.
Square array begins:
     | 0    1       2          3             4
-----+---------------------------------------------
   1 | 1,   1,      1,         1,            1, ...
   2 | 1,   7,     97,      1351,        18817, ...
   3 | 1,  17,    577,     19601,       665857, ...
   4 | 1,  31,   1921,    119071,      7380481, ...
   5 | 1,  49,   4801,    470449,     46099201, ...
   6 | 1,  71,  10081,   1431431,    203253121, ...
   7 | 1,  97,  18817,   3650401,    708158977, ...
   8 | 1, 127,  32257,   8193151,   2081028097, ...
   9 | 1, 161,  51841,  16692641,   5374978561, ...
  10 | 1, 199,  79201,  31521799,  12545596801, ...
  11 | 1, 241, 116161,  55989361,  26986755841, ...
  12 | 1, 287, 164737,  94558751,  54276558337, ...
  13 | 1, 337, 227137, 153090001, 103182433537, ...
  14 | 1, 391, 305761, 239104711, 186979578241, ...
  15 | 1, 449, 403201, 362074049, 325142092801, ...
  ...

Row 2 is A011943, row 3 is A056771, row 8 is A175633, (row 2)*2 is A067902, (row 9)*2 is A089775.
Column 0-5 give A000012, A056220, A144130, A243132, A243134, A243136.
(column 1)*2 is A060626.
Cf. A188645 (f(x, y) as above with y=1).
Diagonals give A173129, A322899.
Cf. A188646, A322836.

max = 9; y = -1; t = Table[k = ((x^2 + y)^(1/2) + x)^2; ((k^n) + (k^(-n)))/2 // FullSimplify, {n, 0, max - 1}, {x, 1, max}]; Table[ t[[n - k + 1, k]], {n, 1, max}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jul 17 2013 *)

A(n,k) = (A188646(n,k-1) + A188646(n,k))/2.

A(n,k) = Sum_{j=0..k} binomial(2*k,2*j)*(n^2-1)^(k-j)*n^(2*j). - Seiichi Manyama, Jan 01 2019

Edited by Seiichi Manyama, Dec 30 2018

More terms from Seiichi Manyama, Jan 01 2019

A348692 Triangle whose n-th row lists the integers m such that A000178(n) / m! is a square, where A000178(n) = n$ = 1!2!...*n! is the superfactorial of n; if there is no such m, then n-th row = 0.

Original entry on oeis.org

1, 2, 0, 2, 0, 0, 0, 3, 4, 0, 0, 0, 6, 0, 8, 9, 0, 8, 9, 0, 7, 0, 10, 0, 0, 0, 12, 0, 0, 0, 14, 0, 0, 0, 15, 16, 0, 18, 0, 18, 0, 0, 0, 20, 0, 0, 0, 22, 0, 0, 0, 24, 25, 0, 0, 0, 26, 0, 0, 0, 28, 0, 0, 0, 30, 0, 32, 0, 32, 0, 0, 0, 34, 0, 0, 0, 35, 36, 0, 0, 0, 38, 0, 0, 0, 40

Offset: 1

Bernard Schott, Oct 30 2021

This sequence is the generalization of a problem proposed during the 17th Tournament of Towns (Spring 1996) and also during the first stage of the Moscow Mathematical Olympiad (1995-1996); the problem asked the question for n = 100 (see Andreescu-Gelca reference, Norman Do link, and Examples section).
Exhaustive results coming from Mabry-McCormick's link and adapted for OEIS:
-> n$ (A000178) is never a square if n > 1.
-> There is no solution if n is odd > 1, hence row(2q+1) = 0 when q > 0.
-> When n is even and there is a solution, then m belongs to {n/2 - 2, n/2 - 1, n/2, n/2 + 1, n/2 + 2}.
-> If 4 divides n (A008536), m = n/2 is always a solution because
    (n$) / (n/2)! = ( 2^(n/4) * Product_{j=1..n/2} ((2j-1)!) )^2.
-> For other cases, see Formula section.
-> When n is even, there are 0, 1 or 2 solutions, so, the maximal length of a row is 2.
-> It is not possible to get more than three consecutive 0 terms, and three consecutive 0 terms correspond to three consecutive rows such that (n, n+1, n+2) = (4u+1, 4u+2, 4u+3) for some u >= 1.

			For n = 4, 4$ / 3! = 48, 4$ / 4! = 12 but 4$ / 2! = 12^2, hence, m = 2.
For n = 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 n = 14, m_1 = 8 because 14$ / 8! = 1309248519599593818685440000000^2 and m_2 = 9 because 14$ / 9! = 436416173199864606228480000000^2.
For n = 16, m_1 = 8 because 16$ / 8! = 6848282921689337839624757371207680000000000^2 and m_2 = 9 because 16$ / 9! = 2282760973896445946541585790402560000000000^2.
For n = 18, m = 7 because 18$ / 7! = 29230177671473293820176594405114531928195727360000000000000^2 and there is no other solution.
For n = 100, m = 50, unique solution to the Olympiad problems.
Triangle begins:
    1;
    2;
    0;
    2;
    0;
    0;
    0;
    8,  9;
    0;
    ...

Titu Andreescu and Rǎzvan Gelca, Putnam and Beyond, New York, Springer, 2007, problem 725, pp. 253 and 686.
Peter J. Taylor and A. M. Storozhev, Tournament of Towns 1993-1997, Book 4, Tournament 17, Spring 1996, O Level, Senior questions, Australian Mathematics Trust, 1998, problem 3, p. 96.

Diophante, A1963 - Le vilain petit canard (in French).
Norman Do, Factorial fun, Puzzle Corner 13, Gaz. Aust. Math. Soc. 36, 2009, 176-179, page 178.
Rick Mabry and Laura McCormick, Square products of punctured sequences of factorials, Gaz. Aust. Math. Soc., 2009, pages 346-352.
Tournament of Towns, Tournament 17, 1995-1996, Spring 1996, O Level, Senior questions, question 3 (in Russian).
Index to sequences related to Olympiads and other Mathematical competitions.

Cf. A000178, A001541, A008536, A035008, A060626, A139098.

sf(n)=prod(k=2, n, k!); \\ A000178
row(n) = my(s=sf(n)); Vec(select(issquare, vector(n, k, s/k!), 1));
lista(nn) = {my(list = List()); for (n=1, nn, my(r=row(n)); if (#r, for (k=1, #r, listput(list, r[k])), listput(list, 0));); Vec(list);} \\ Michel Marcus, Oct 30 2021

When there are two such integers m, then m_1 < m_2.
If n = 8*q^2 (A139098), then m_1 = n/2 - 1 = 4q^2-1 (see example for n=8).
If n = 8q*(q+1) (A035008), then m_2 = n/2 + 1 = (2q+1)^2 (see example for n=16).
if n = 4q^2 - 2 (A060626), then m_1 = n/2 + 1 = 2q^2 (see example for n=14).
If n = 2q^2, q>1 in A001541, then m = n/2 - 2 = q^2-2 (see example for n=18).
If n = 2q^2-4, q>1 in A001541, then m_2 = n/2 + 2 = q^2 (see example for n=14).

A247850 The 5th Hermite Polynomial evaluated at n: H_5(n) = 32n^5 - 160n^3 + 120*n.

Original entry on oeis.org

0, -8, -16, 3816, 23008, 80600, 214992, 483784, 967616, 1774008, 3041200, 4941992, 7687584, 11531416, 16773008, 23761800, 32900992, 44651384, 59535216, 78140008, 101122400, 129211992, 163215184, 204019016, 252595008, 310003000, 377394992, 456018984

Offset: 0

Vincenzo Librandi, Sep 25 2014

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Eric Weisstein's World of Mathematics, Hermite Polynomial.
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. Hermite_k(n): A005843 (k=1), A060626 (k=2), A163322 (k=3), A163323 (k=4), this sequence (k=5), A247851 (k=6), A247852 (k=7), A247853 (k=8), A247854 (k=9), A247855 (k=10).

Table[32 n^5 - 160 n^3 + 120 n, {n, 0, 30}]

a(n)=32*n^5-160*n^3+120*n \\ Charles R Greathouse IV, Oct 07 2015

from sympy import hermite
def A247850(n): return hermite(5,n) # Chai Wah Wu, Jan 06 2022

G.f.: -x*(8-32*x-3792*x^2-32*x^3+8*x^4)/(x-1)^6.

a(n) = 6*a(n-1)-15*a(n-2)+20*a(n-3)-15*a(n-4)+6*a(n-5)-a(n-6).

A349079 Numbers k such that there exists m, 1 <= m <= k with the property that A000178(k) / m! is a square, where A000178(k) = k$ = 1!2!...*k! is the superfactorial of k.

Original entry on oeis.org

1, 2, 4, 8, 12, 14, 16, 18, 20, 24, 28, 32, 34, 36, 40, 44, 48, 52, 56, 60, 62, 64, 68, 72, 76, 80, 84, 88, 92, 96, 98, 100, 104, 108, 112, 116, 120, 124, 128, 132, 136, 140, 142, 144, 148, 152, 156, 160, 164, 168, 172, 176, 180, 184, 188, 192, 194, 196, 200, 204, 208, 212, 216, 220, 224, 228

Offset: 1

Bernard Schott, Nov 07 2021

nonn

If k is a term, then A348692(k) lists integers m such that k$ / m! is a square; and for each k, there exist only one (A349080) or two (A349081) such integers m.
See A348692 for further information, links and references about Olympiads.
Except for 1, all terms are even, and, when k is such an even term, corresponding m belong(s) to {k/2 - 2, k/2 - 1, k/2, k/2 + 1, k/2 + 2}.
This sequence is the union of {1} and of three infinite and disjoint subsequences:
-> A008586, so every positive multiple of 4 is a term and in this case, for k=4*q, (k$)/(k/2)! = ( 2^(k/4) * Product_{j=1..k/2} ((2j-1)!) )^2 (see example 4).
-> A060626, so every k = 4*q^2 - 2 (q >= 1) is a term (see examples 2 and 14).
-> 2*A055792 = {k = 2q^2 with q>1 in A001541} = {18, 578, ...} (see example 18).

			2 is a term as 2$ / 2! = 1^2.
4 is a term as 4$ / 2! = 12^2.
14 is a term as 14$ / 8! = 1309248519599593818685440000000^2 and also 14$ / 9! = 436416173199864606228480000000^2.
18 is a term as 18$ / 7! = 29230177671473293820176594405114531928195727360000000000000^2.

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 and other Mathematical competitions.

Cf. A001541, A000178, A055792, A348692, A349080, A349081.

Subsequences: A008586, A060626.

supfact[n_] := supfact[n] = BarnesG[n + 2]; fact[n_] := fact[n] = n!; q[k_] := AnyTrue[Range[k], IntegerQ @ Sqrt[supfact[k]/fact[#]] &]; Select[Range[230], q] (* Amiram Eldar, Nov 08 2021 *)

f(n) = prod(k=2, n, k!); \\ A000178
isok(k) = my(sf=f(k)); for (m=1, k, if (issquare(sf/m!), return(1))); \\ Michel Marcus, Nov 08 2021

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)

A349080 Numbers k for which there exists only one integer m with 1 <= m <= 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

1, 2, 4, 12, 18, 20, 24, 28, 34, 36, 40, 44, 52, 56, 60, 62, 64, 68, 76, 80, 84, 88, 92, 98, 100, 104, 108, 112, 116, 120, 124, 132, 136, 140, 142, 144, 148, 152, 156, 164, 168, 172, 176, 180, 184, 188, 192, 194, 196, 204, 208, 212, 216, 220, 224, 228, 232, 236, 244, 248, 252, 254, 256

Offset: 1

Bernard Schott, Nov 20 2021

nonn

This sequence is the union of {1} and of three infinite and disjoint subsequences.
-> Numbers k divisible by 4 but not of the form 8q^2 or 8q(q+1) = {4, 12, 20, 24, 28, ...} (see A182834). For these numbers, the corresponding unique m = k/2 (see example for k = 4).
-> Even numbers k not divisible by 4 and of the form k = 2*A055792 = 2*q^2, q>1 in A001541 = {18, 578, ...}. For these numbers, the corresponding unique m = k/2 - 2 = q^2-2 (see example for k = 18)
-> Even numbers k not divisible by 4, that are in A060626 but not of the form k=2q^2-4 with q>1 in A001541 = {2, 34, 62, 98, 142, 194, ...} (A349496). For these numbers, the corresponding unique m = k/2 + 1 (see example for k = 2).
See A348692 for further information.

			For k = 2, 2$ / 2! = 1^2, hence 2 is a term.
For k = 4, 4$ /1! = 288, 4$ / 3! = 48, 4$ / 4! = 12 but for m = 2, 4$ / 2! = 12^2, hence 4 is a term.
For k = 18 and m = 7, we have 18$ / 7! = 29230177671473293820176594405114531928195727360000000000000^2 and there is no other solution m, hence 18 is a term.

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

Cf. A000178, A001541, A060626, A348692, A182834, A349496.

Subsequence of A349079.

q[n_] := Count[BarnesG[n + 2]/Range[n]!, ?(IntegerQ@Sqrt[#] &)] == 1; Select[Range[100], q] (* _Amiram Eldar, Nov 20 2021 *)

sf(n) = prod(k=2, n, k!); \\ A000178
isok(m) = my(s=sf(m)); #select(issquare, vector(m, k, s/k!), 1) == 1; \\ Michel Marcus, Nov 20 2021

cp's OEIS Frontend

A349496 Numbers of the form 4*t^2-2 (A060626) when t >= 1 is an integer that is not a term in A001542.

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A033996 8 times triangular numbers: a(n) = 4*n*(n+1).

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A091823 a(n) = 2*n^2 + 3*n - 1.

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A132411 a(0) = 0, a(1) = 1 and a(n) = n^2 - 1 with n >= 2.

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A188644 Array of (k^n + k^(-n))/2 where k = (sqrt(x^2-1) + x)^2 for integers x >= 1.

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A348692 Triangle whose n-th row lists the integers m such that A000178(n) / m! is a square, where A000178(n) = n$ = 1!*2!*...*n! is the superfactorial of n; if there is no such m, then n-th row = 0.

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A033996 8 times triangular numbers: a(n) = 4n(n+1).

A091823 a(n) = 2n^2 + 3n - 1.

A348692 Triangle whose n-th row lists the integers m such that A000178(n) / m! is a square, where A000178(n) = n$ = 1!2!...*n! is the superfactorial of n; if there is no such m, then n-th row = 0.

A247850 The 5th Hermite Polynomial evaluated at n: H_5(n) = 32n^5 - 160n^3 + 120*n.

A349079 Numbers k such that there exists m, 1 <= m <= k with the property that A000178(k) / m! is a square, where A000178(k) = k$ = 1!2!...*k! is the superfactorial of k.

A349080 Numbers k for which there exists only one integer m with 1 <= m <= k such that A000178(k) / m! is a square, where A000178(k) = k$ = 1!2!...*k! is the superfactorial of k.