A014206 -id:A014206 - cp's OEIS frontend

A127408 Negative value of coefficient of x^(n-3) in the characteristic polynomial of a certain n X n integer circulant matrix.

18, 144, 625, 1980, 5145, 11648, 23814, 45000, 79860, 134640, 217503, 338884, 511875, 752640, 1080860, 1520208, 2098854, 2850000, 3812445, 5031180, 6558013, 8452224, 10781250, 13621400, 17058600, 21189168, 26120619, 31972500, 38877255

Offset: 3

Paul Max Payton, Jan 14 2007

The n X n circulant matrix used here has first row 1 through n and each successive row is a circular rotation of the previous row to the right by one element.

The coefficient of x^(n-3) exists only for n>2, so the sequence starts with a(3). In order to obtain a nonnegative sequence the coefficient (which is negative for all n>2) is multiplied by -1.

			The circulant matrix for n = 5 is
[1 2 3 4 5]
[5 1 2 3 4]
[4 5 1 2 3]
[3 4 5 1 2]
[2 3 4 5 1]
The characteristic polynomial of this matrix is x^5 - 5*x^4 -100*x^3 - 625*x^2 - 1750*x - 1875. The coefficient of x^(n-3) is -625, hence a(5) = 625.

Daniel Zwillinger, Ed., "CRC Standard Mathematical Tables and Formulae", 31st Edition, ISBN 1-58488-291, Section 2.6.2.25 (page 141) and Section 2.6.11.3 (page 152).

Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

Cf. A000142 (factorial numbers), A014206 (n^2+n+2), A127407, A127409, A127410, A127411, A127412.

[ -Coefficient(CharacteristicPolynomial(Matrix(IntegerRing(), n, n, [< i, j, 1 + (j-i) mod n > : i, j in [1..n] ] )), n-3) : n in [3..31] ] ; // Klaus Brockhaus, Jan 26 2007

[ (n-2) * (n-1) * n^3 * (2*(n-2) + 14) / (2 * Factorial(4)) : n in [3..31] ] ; // Klaus Brockhaus, Jan 26 2007

n * (n+1) * (n+2)^3 * (2*n + 14) / (2 * factorial(4)); % Paul Max Payton, Jan 14 2007

a(n) = {-polcoef(charpoly(matrix(n,n,i,j,(j-i)%n+1),x),n-3)} \\ Klaus Brockhaus, Jan 26 2007

a(n) = {(n^6+2*n^5-13*n^4+10*n^3)/4!} \\ Klaus Brockhaus, Jan 26 2007

a(n+2) = n*(n+1)*(n+2)^3*(2n+14)/(2 * 4!) for n>=1.
a(n) = (n^6+2*n^5-13*n^4+10*n^3)/4! for n>=3.
G.f.: x^3*(3-x)*(6+8*x+x^2)/(1-x)^7. - Colin Barker, May 13 2012

Edited and extended by Klaus Brockhaus, Jan 26 2007

A127409 Negative value of coefficient of x^(n-4) in the characteristic polynomial of a certain n X n integer circulant matrix.

Original entry on oeis.org

160, 1750, 10044, 40817, 132608, 367416, 903000, 2020458, 4191264, 8168446, 15107092, 26719875, 45473792, 74834816, 119567664, 186098388, 282948000, 421245846, 615331948, 883458037, 1248597504, 1739375000, 2391126920

Offset: 4

Paul Max Payton, Jan 14 2007

The n X n circulant matrix used here has first row 1 through n and each successive row is a circular rotation of the previous row to the right by one element.

The coefficient of x^(n-4) exists only for n>3, so the sequence starts with a(4). In order to obtain a nonnegative sequence the coefficient (which is negative for all n>3) is multiplied by -1.

			The circulant matrix for n = 5 is
[1 2 3 4 5]
[5 1 2 3 4]
[4 5 1 2 3]
[3 4 5 1 2]
[2 3 4 5 1]
The characteristic polynomial of this matrix is x^5 - 5*x^4 -100*x^3 - 625*x^2 - 1750*x - 1875. The coefficient of x^(n-4) is -1750, hence a(5) = 1750.

Daniel Zwillinger, ed., "CRC Standard Mathematical Tables and Formulae", 31st Edition, ISBN 1-58488-291, Section 2.6.2.25 (page 141) and Section 2.6.11.3 (page 152).

Index entries for linear recurrences with constant coefficients, signature (9,-36,84,-126,126,-84,36,-9,1).

Cf. A000142 (factorial numbers), A014206 (n^2+n+2), A127407, A127408, A127410, A127411, A127412.

[ -Coefficient(CharacteristicPolynomial(Matrix(IntegerRing(), n, n, [< i, j, 1 + (j-i) mod n > : i, j in [1..n] ] )), n-4) : n in [4..26] ];  // Klaus Brockhaus, Jan 27 2007

[ (n-3)*(n-2)*(n-1)*n^4*(3*n+13) / (2 * Factorial(5)) : n in [4..26] ]; // Klaus Brockhaus, Jan 27 2007

n * (n+1) * (n+2) * (n+3)^4 * (3*n + 22) / (2 * factorial(5)); % Paul Max Payton, Jan 14 2007

a(n) = {-polcoeff(charpoly(matrix(n,n,i,j,(j-i)%n+1),x),n-4)} \\ Klaus Brockhaus, Jan 27 2007

a(n) = {(3*n^8 - 5*n^7 - 45*n^6 + 125*n^5 - 78*n^4)/(2*5!)} \\ Klaus Brockhaus, Jan 27 2007

a(n+3) = n*(n+1)*(n+2)*(n+3)^4*(3*n+22)/(2*5!) for n>=1.
a(n) = (3*n^8-5*n^7-45*n^6+125*n^5-78*n^4)/(2*5!) for n>=4.
G.f.: x^4*(160+310*x+54*x^2-19*x^3-x^4)/(1-x)^9. - Colin Barker, May 13 2012

Edited, corrected and extended by Klaus Brockhaus, Jan 27 2007

A127412 Triangular table containing values of coefficients of the characteristic polynomial of a certain n x n circulant matrix, read by rows.

Original entry on oeis.org

1, 1, -1, 1, -2, -3, 1, -3, -15, -18, 1, -4, -44, -144, -160, 1, -5, -100, -625, -1750, -1875, 1, -6, -195, -1980, -10044, -25920, -27216, 1, -7, -343, -5145, -40817, -184877, -453789, -470596, 1, -8, -560, -11648, -132608, -917504, -3866624, -9175040, -9437184, 1, -9, -864, -23814, -367416, -3582306

Offset: 0

Paul Max Payton, Feb 09 2007

This is a lower triangular table.

			The third row represents the coefficients of the characteristic polynomial of [1 2 3; 3 1 2; 2 3 1], which is x^3 - 3*x^2 - 15*x - 18. Thus the row reads 1,-3,-15,-18.

Daniel Zwillinger, ed., "CRC Standard Mathematical Tables and Formulae", 31st Edition, ISBN 1-58488-291, Section 2.6.2.25 (page 141) and Section 2.6.11.3 (page 152).

Cf. A000142, A014206, A127407, A127408, A127409, A127410, A127411.

First column is unity. Second column (A127407) is a(n+1) = n*(n+1)^2*(n+8)/(2*3!) for n>=1. Third column (A127408) is a(n+2) = n*(n+1)*(n+2)^3*(2n+14)/(2 * 4!) for n>=1. In general, k-th column is given by a(n+(k-1)) = n*(n+1)*(n+2)*...*(n+(k-1))^k*((k-1)n+S(k))/(2 * (k+1)!) for n>=1, where S(k) is the k-th term of A014206.

A241199 Numbers n such that 4 consecutive terms of binomial(n,k) satisfy a quadratic relation for 0 <= k <= n/2.

Original entry on oeis.org

14, 19, 31, 38, 54, 63, 83, 94, 118, 131, 159, 174, 206, 223, 259, 278, 318, 339, 383, 406, 454, 479, 531, 558, 614, 643, 703, 734, 798, 831, 899, 934, 1006, 1043, 1119, 1158, 1238, 1279, 1363, 1406, 1494, 1539, 1631, 1678, 1774, 1823, 1923, 1974, 2078, 2131

Offset: 1

T. D. Noe, Apr 17 2014

From Robert Israel, Apr 28 2015: (Start)
Numbers n >= 14 such that 3*n + 7 is a square.
This is because
C(n,i+3) - 3*C(n,i+2) + 3*C(n,i+1) - C(n,i) = n!/((n-i)!*(i+3)!) * g(n,i)
where g(n,i) = (n-3-2*i) * ((n-3-2*i)^2 - 3*n - 7). (End)

			Binomial(14,k) = (1, 14, 91, 364, 1001, 2002, 3003, 3432) for k = 0..7. The 4 terms beginning with 91 equal 182 - 273*x + 182*x^2 for x = 1..4.

Robert Israel, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Sequence A241200 gives the position of the first of the 4 terms. Sequence A008865 gives the terms greater than 2 for which 3 consecutive terms satisfy a linear relation.
A014206 is a related sequence. - Avi Friedlich, Apr 28 2015
Cf. A062730 (3 terms in arithmetic progression in Pascal's triangle).

map(k -> (3*k^2+8*k+3,3*k^2+10*k+6),[$1..100]); # Robert Israel, Apr 28 2015

Select[Range[2500], MemberQ[Differences[Binomial[#, Range[0, #/2]], 3], 0] &]
LinearRecurrence[{1,2,-2,-1,1},{14,19,31,38,54},50] (* Harvey P. Dale, Oct 29 2017 *)

Vec(-x*(6*x^4-3*x^3-16*x^2+5*x+14)/((x-1)^3*(x+1)^2) + O(x^100)) \\ Colin Barker, Apr 29 2015

a(n)=(6*n^2+42*n+55-(-1)^n*(2*n+7))/8 \\ Charles R Greathouse IV, Apr 15 2016

a(n) = (55-7*(-1)^n-2*(-21+(-1)^n)*n+6*n^2)/8. G.f.: -x*(6*x^4-3*x^3-16*x^2+5*x+14) / ((x-1)^3*(x+1)^2). - Colin Barker, Apr 18 2014 and Apr 29 2015
The terms appear to satisfy a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5), with initial terms 14, 19, 31, 38, 54. - T. D. Noe, Apr 18 2014
Numbers are of the form A200182(3n+1) and A200182(3n-1). - Avi Friedlich, Apr 25 2015
a(2*k-1) = 3*k^2 + 8*k + 3, a(2*k) = 3*k^2 + 10*k + 6. - Robert Israel, Apr 28 2015

A271649 a(n) = 2*(n^2 - n + 2).

Original entry on oeis.org

4, 8, 16, 28, 44, 64, 88, 116, 148, 184, 224, 268, 316, 368, 424, 484, 548, 616, 688, 764, 844, 928, 1016, 1108, 1204, 1304, 1408, 1516, 1628, 1744, 1864, 1988, 2116, 2248, 2384, 2524, 2668, 2816, 2968, 3124, 3284, 3448, 3616, 3788, 3964, 4144, 4328, 4516, 4708, 4904, 5104, 5308, 5516

Offset: 1

Juri-Stepan Gerasimov, Apr 11 2016

Numbers n such that 2*n - 7 is a perfect square.

Galois numbers for three-dimensional vector space, defined as the total number of subspaces in a three-dimensional vector space over GF(n-1), when n-1 is a power of a prime. - Artur Jasinski, Aug 31 2016, corrected by Robert Israel, Sep 23 2016

			a(1) = 2*(1^2 - 1 + 2) = 4.

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

Cf. A000124, A014206, A137882.

Numbers h such that 2*h + k is a perfect square: no sequence (k=-9), A255843 (k=-8), this sequence (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), A271625 (k=10), A222182 (k=11), A152811 (k=12), A181510 (k=13), A161532 (k=14), no sequence (k=15).

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

[ n: n in [1..6000] | IsSquare(2*n-7)];

A271649:=n->2*(n^2-n+2): seq(A271649(n), n=1..60); # Wesley Ivan Hurt, Aug 31 2016

Table[2 (n^2 - n + 2), {n, 53}] (* or *)
Select[Range@ 5516, IntegerQ@ Sqrt[2 # - 7] &] (* or *)
Table[SeriesCoefficient[(-4 (1 - x + x^2))/(-1 + x)^3, {x, 0, n}], {n, 0, 52}] (* Michael De Vlieger, Apr 11 2016 *)
LinearRecurrence[{3,-3,1},{4,8,16},60] (* Harvey P. Dale, Jun 14 2022 *)

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

a(n) = 4*A000124(n).
a(n) = 2*A014206(n).
a(n) = A137882(n), n > 1. - R. J. Mathar, Apr 12 2016
Sum_{n>=1} 1/a(n) = tanh(sqrt(7)*Pi/2)*Pi/(2*sqrt(7)). - Amiram Eldar, Jul 30 2024
From Elmo R. Oliveira, Nov 18 2024: (Start)
G.f.: 4*x*(1 - x + x^2)/(1 - x)^3.
E.g.f.: 2*(exp(x)*(x^2 + 2) - 2).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 3. (End)

A300401 Array T(n,k) = n(binomial(k, 2) + 1) + k(binomial(n, 2) + 1) read by antidiagonals.

Original entry on oeis.org

0, 1, 1, 2, 2, 2, 3, 4, 4, 3, 4, 7, 8, 7, 4, 5, 11, 14, 14, 11, 5, 6, 16, 22, 24, 22, 16, 6, 7, 22, 32, 37, 37, 32, 22, 7, 8, 29, 44, 53, 56, 53, 44, 29, 8, 9, 37, 58, 72, 79, 79, 72, 58, 37, 9, 10, 46, 74, 94, 106, 110, 106, 94, 74, 46, 10, 11, 56, 92, 119

Offset: 0

Franck Maminirina Ramaharo, Mar 05 2018

Antidiagonal sums are given by 2*A055795.
Rows/columns n are binomial transform of {n, A152947(n+1), n, 0, 0, 0, ...}.
Some primes in the array are
n = 1: {2, 7, 11, 29, 37, 67, 79, 137, 191, 211, 277, 379, ...} = A055469, primes of the form k*(k + 1)/2 + 1;
n = 3: {3, 7, 37, 53, 479, 653, 1249, 1619, 2503, 3727, 4349, 5737, 7109, 8179, 9803, 11839, 12107, ...};
n = 4: {11, 37, 79, 137, 211, 821, 991, 1597, 1831, 2081, 2347, ...} = A188382, primes of the form 8*(2*k - 1)^2 + 2*(2*k - 1) + 1.

			The array T(n,k) begins
0     1    2    3    4     5     6     7     8     9    10    11  ...
1     2    4    7   11    16    22    29    37    46    56    67  ...
2     4    8   14   22    32    44    58    74    92   112   134  ...
3     7   14   24   37    53    72    94   119   147   178   212  ...
4    11   22   37   56    79   106   137   172   211   254   301  ...
5    16   32   53   79   110   146   187   233   284   340   401  ...
6    22   44   72  106   146   192   244   302   366   436   512  ...
7    29   58   94  137   187   244   308   379   457   542   634  ...
8    37   74  119  172   233   302   379   464   557   658   767  ...
9    46   92  147  211   284   366   457   557   666   784   911  ...
10   56  112  178  254   340   436   542   658   784   920  1066  ...
11   67  134  212  301   401   512   634   767   911  1066  1232  ...
12   79  158  249  352   467   594   733   884  1047  1222  1409  ...
13   92  184  289  407   538   682   839  1009  1192  1388  1597  ...
14  106  212  332  466   614   776   952  1142  1346  1564  1796  ...
15  121  242  378  529   695   876  1072  1283  1509  1750  2006  ...
16  137  274  427  596   781   982  1199  1432  1681  1946  2227  ...
17  154  308  479  667   872  1094  1333  1589  1862  2152  2459  ...
18  172  344  534  742   968  1212  1474  1754  2052  2368  2702  ...
19  191  382  592  821  1069  1336  1622  1927  2251  2594  2956  ...
20  211  422  653  904  1175  1466  1777  2108  2459  2830  3221  ...
...
The inverse binomial transforms of the columns are
0     1    2    3    4     5     6     7     8     9    10    11  ...  A001477
1     1    2    4    7    11    22    29    37    45    56    67  ...  A152947
0     1    2    3    4     5     6     7     8     9    10    11  ...  A001477
0     0    0    0    0     0     0     0     0     0     0     0  ...
0     0    0    0    0     0     0     0     0     0     0     0  ...
0     0    0    0    0     0     0     0     0     0     0     0  ...
...

Miklós Bóna, Introduction to Enumerative Combinatorics, McGraw-Hill, 2007.
L. Comtet, Advanced Combinatorics: The Art of Finite and Infinite Expansions, Reidel Publishing Company, 1974.
R. P. Stanley, Enumerative Combinatorics, second edition, Cambridge University Press, 2011.

Michael De Vlieger, Table of n, a(n) for n = 0..11475 (rows 0 <= n <= 150, flattened).
Cheyne Homberger, Patterns in Permutations and Involutions: A Structural and Enumerative Approach, arXiv preprint 1410.2657 [math.CO], 2014.
Franck Ramaharo, A generating polynomial for the two-bridge knot with Conway's notation C(n,r), arXiv:1902.08989 [math.CO], 2019.

Cf. A000124, A001477, A006000, A008815, A014206, A051601, A055469, A077028, A081436, A084849, A131074, A134394, A139600, A141387, A179000, A188377, A188382, A273465.

T := (n, k) -> n*(binomial(k, 2) + 1) + k*(binomial(n, 2) + 1);
for n from 0 to 20 do seq(T(n, k), k = 0 .. 20) od;

T[n_, k_] := n (Binomial[k, 2] + 1) + k (Binomial[n, 2] + 1);
Table[T[n - k, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Dec 07 2018 *)

T(n, k) := n*(binomial(k, 2) + 1) + k*(binomial(n, 2) + 1)$
for n:0 thru 20 do
  print(makelist(T(n, k), k, 0, 20));

T(n, k) = n*(binomial(k,2) + 1) + k*(binomial(n,2) + 1);
tabl(nn) = for (n=0, nn, for (k=0, nn, print1(T(n, k), ", ")); print); \\ Michel Marcus, Mar 12 2018

T(n,k) = T(k,n) = n*A152947(k+1) + k*A152947(n+1).
T(n,0) = A001477(n).
T(n,1) = A000124(n).
T(n,2) = A014206(n).
T(n,3) = A273465(3*n+2).
T(n,4) = A084849(n+1).
T(n,n) = A179000(n-1,n), n >= 1.
T(2*n,2*n) = 8*A081436(n-1), n >= 1.
T(2*n+1,2*n+1) = 2*A006000(2*n+1).
T(n,n+1) = A188377(n+3).
T(n,n+2) = A188377(n+2), n >= 1.
Sum_{k=0..n} T(k,n-k) = 2*(binomial(n, 4) + binomial(n, 2)).
G.f.: -((2*x*y - y - x)*(2*x*y - y - x + 1))/(((x - 1)*(y - 1))^3).
E.g.f.: (1/2)*(x + y)*(x*y + 2)*exp(x + y).

A027712 Numbers k such that k^2+k+2 is a palindrome.

Original entry on oeis.org

0, 1, 2, 4, 6, 14, 15, 21, 50, 92, 201, 203, 292, 479, 897, 1424, 1530, 1654, 2001, 2106, 2183, 16780, 20001, 20993, 28377, 89777, 200001, 219083, 501474, 1620660, 1651754, 2000001, 14842995, 17101809, 20000001, 21147906, 21855108, 22012038, 29287052

Offset: 1

Patrick De Geest

Giovanni Resta, Table of n, a(n) for n = 1..65
P. De Geest, Palindromic Quasi_Over_Squares of the form n^2+(n+X)

Cf. A027713, A014206, A028413, A027714.

palQ[n_] := Block[{d = IntegerDigits[n]}, d == Reverse[d]]; f[n_] := n^2 + n + 2; Select[Range[0, 10^5], palQ@ f@ # &] (* Giovanni Resta, Aug 29 2018 *)

More terms from Giovanni Resta, Aug 28 2018

A059214 Square array T(k,n) = C(n-1,k) + Sum_{i=0..k} C(n,i) read by antidiagonals (k >= 1, n >= 1).

Original entry on oeis.org

2, 2, 4, 2, 4, 6, 2, 4, 8, 8, 2, 4, 8, 14, 10, 2, 4, 8, 16, 22, 12, 2, 4, 8, 16, 30, 32, 14, 2, 4, 8, 16, 32, 52, 44, 16, 2, 4, 8, 16, 32, 62, 84, 58, 18, 2, 4, 8, 16, 32, 64, 114, 128, 74, 20, 2, 4, 8, 16, 32, 64, 126, 198, 186, 92, 22, 2, 4, 8, 16, 32, 64

Offset: 1

N. J. A. Sloane, Feb 15 2001

For k > 1, gives maximal number of regions into which k-space can be divided by n hyperspheres.

The maximum number of subsets of a set of n points in k-space that can be formed by intersecting it with a hyperplane. - Günter Rote, Dec 18 2018

			Array begins
   2 4 6  8 10 12 ...
   2 4 8 14 22 32 ...
   2 4 8 16 30 52 ...

L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 73, Problem 4.

Paolo Xausa, Table of n, a(n) for n = 1..11325 (antidiagonals 1..150 of the array, flattened).
E. F. Harding, The number of partitions of a set of n points in k dimensions induced by hyperplanes, Proc. Edinburgh Math. Soc., 15 (1967), 285-289.

Cf. A014206 (dim 2), A046127 (dim 3), A059173 (dim 4), A059174 (dim 5).
Equals twice A216274.
Apart from left border, same as A059250. A178522 is probably the best version.

A059214[k_,n_]:=Binomial[n-1,k]+Sum[Binomial[n,i],{i,0,k}];
Table[A059214[k-n+1,n],{k,10},{n,k}] (* Paolo Xausa, Dec 29 2023 *)

T(k,n) = C(n-1, k) + Sum_{i=0..k} C(n, i).

A220508 T(n,k) = n^2 + k if k <= n, otherwise T(n,k) = k*(k + 2) - n; square array T(n,k) read by ascending antidiagonals (n >= 0, k >= 0).

Original entry on oeis.org

0, 1, 3, 4, 2, 8, 9, 5, 7, 15, 16, 10, 6, 14, 24, 25, 17, 11, 13, 23, 35, 36, 26, 18, 12, 22, 34, 48, 49, 37, 27, 19, 21, 33, 47, 63, 64, 50, 38, 28, 20, 32, 46, 62, 80, 81, 65, 51, 39, 29, 31, 45, 61, 79, 99, 100, 82, 66, 52, 40, 30, 44, 60, 78, 98, 120

Offset: 0

Omar E. Pol, Feb 09 2013

This sequence consists of 0 together with a permutation of the natural numbers. The nonnegative integers (A001477) are arranged in the successive layers from T(0,0) = 0. The n-th layer start with T(n,1) = n^2. The n-th layer is formed by the first n+1 elements of row n and the first n elements in increasing order of the column n.
The first antidiagonal is formed by odd numbers: 1, 3. The second antidiagonal is formed by even numbers: 4, 2, 8. The third antidiagonal is formed by odd numbers: 9, 5, 7, 15. And so on.
It appears that in the n-th layer there is at least a prime number <= g and also there is at least a prime number > g, where g is the number on the main diagonal, the n-th oblong number A002378(n), if n >= 1.

			The second layer is [4, 5, 6, 7, 8] which looks like this:
  .  .  8
  .  .  7,
  4, 5, 6,
Square array T(0,0)..T(10,10) begins:
    0,   3,   8,  15,  24,  35,  48,  63,  80,  99, 120,...
    1,   2,   7,  14,  23,  34,  47,  62,  79,  98, 119,...
    4,   5,   6,  13,  22,  33,  46,  61,  78,  97, 118,...
    9,  10,  11,  12,  21,  32,  45,  60,  77,  96, 117,...
   16,  17,  18,  19,  20,  31,  44,  59,  76,  95, 118,...
   25,  26,  27,  28,  29,  30,  43,  58,  75,  94, 117,...
   36,  37,  38,  39,  40,  41,  42,  57,  74,  93, 114,...
   49,  50,  51,  52,  53,  54,  55,  56,  73,  92, 113,...
   64,  65,  66,  67,  68,  69,  70,  71,  72,  91, 112,...
   81,  82,  83,  84,  85,  86,  87,  88,  89,  90, 111,...
  100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110,...
  ...

Index entries for sequences that are permutations of the natural numbers

Column 1 is A000290. Main diagonal is A002378. Column 2 is essentially A002522. Row 1 is A005563. Row 2 gives the absolute terms of A008865.

Cf. A000005, A000040, A002061, A002620, A014206, A014209, A027688, A028387, A028552, A060736, A220516.

From Petros Hadjicostas, Mar 10 2021: (Start)
T(n,k) = (A342354(n,k) - 1)/2.
O.g.f.: (x^4*y^3 + 3*x^3*y^4 + x^4*y^2 - 10*x^3*y^3 - x^2*y^4 + 3*x^3*y^2 + x^2*y^3 - 4*x^3*y + 8*x^2*y^2 + 3*x^2*y + x*y^2 + x^2 - 10*x*y - y^2 + x + 3*y)/((1 - x)^3*(1 - y)^3*(1 - x*y)^2). (End)

Name edited by Petros Hadjicostas, Mar 10 2021

A336535 a(n) = (m(n)^2 + 3)(m(n)^2 + 7)/32, where m(n) = 2n - 1.

Original entry on oeis.org

1, 6, 28, 91, 231, 496, 946, 1653, 2701, 4186, 6216, 8911, 12403, 16836, 22366, 29161, 37401, 47278, 58996, 72771, 88831, 107416, 128778, 153181, 180901, 212226, 247456, 286903, 330891, 379756, 433846, 493521, 559153, 631126, 709836, 795691, 889111, 990528, 1100386, 1219141, 1347261, 1485226, 1633528, 1792671

Offset: 1

Jeff Brown, Jul 24 2020

For m(n) = 3,5,11, and 181, the perfect numbers (A000396), 6, 28, 496, and 33550336 are produced, respectively. 3,5,11, and 181 are the numbers m(n) such that (m(n)^2+7) is a power of 2. cf A038198.

The unique primitive Pythagorean triple whose inradius T(n) and its long leg and hypotenuse are consecutive natural numbers is (2*T(n)+1, 2*T(n)*(T(n)+1), 2*T(n)*(T(n)+1)+1) and its semiperimeter is (T(n)+1)*(2*T(n)+1) where T(n) = A000217(n). - Miguel-Ángel Pérez García-Ortega, May 16 2025

			m(2) = 2*2-1 = 3 and (3^2+3)*(3^2+7)/32 = 6, so 6 is in the sequence.

David M. Burton, Elementary Number Theory, McGraw-Hill (2011), 25.

Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000396, A038198.
Cf. A002061, A014206.
Cf. A000217, A058919, A383833, A383834.

Table[((n^2+3)*(n^2+7))/32, {n,1,100,2}]
a=Table[(n(n+1))/2,{n,0,40}];Apply[Join,Map[{(#+1)(2#+1)}&,a]] (* Miguel-Ángel Pérez García-Ortega, May 16 2025 *)

a(n)=(1-n+n^2)*(2-n+n^2)/2 \\ Charles R Greathouse IV, Oct 21 2022

From Stefano Spezia, Jul 25 2020: (Start)
O.g.f.: x*(1 + x + 8*x^2 + x^3 + x^4)/(1 - x)^5.
a(n) = (1 - n + n^2)*(2 - n + n^2)/2.
a(n) = A002061(n)*A014206(n-1)/2.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n > 5. (End)
a(n) = (A000217(n-1)+1)*(2*A000217(n-1)+1). - Miguel-Ángel Pérez García-Ortega, May 16 2025

cp's OEIS Frontend

A127408 Negative value of coefficient of x^(n-3) in the characteristic polynomial of a certain n X n integer circulant matrix.

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A127412 Triangular table containing values of coefficients of the characteristic polynomial of a certain n x n circulant matrix, read by rows.

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A241199 Numbers n such that 4 consecutive terms of binomial(n,k) satisfy a quadratic relation for 0 <= k <= n/2.

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A271649 a(n) = 2*(n^2 - n + 2).

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A300401 Array T(n,k) = n*(binomial(k, 2) + 1) + k*(binomial(n, 2) + 1) read by antidiagonals.

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A300401 Array T(n,k) = n(binomial(k, 2) + 1) + k(binomial(n, 2) + 1) read by antidiagonals.

A336535 a(n) = (m(n)^2 + 3)(m(n)^2 + 7)/32, where m(n) = 2n - 1.