A010000 -id:A010000 - cp's OEIS frontend

A197985 a(n) = round((n+1/n)^2).

4, 6, 11, 18, 27, 38, 51, 66, 83, 102, 123, 146, 171, 198, 227, 258, 291, 326, 363, 402, 443, 486, 531, 578, 627, 678, 731, 786, 843, 902, 963, 1026, 1091, 1158, 1227, 1298, 1371, 1446, 1523, 1602, 1683, 1766, 1851, 1938, 2027

Offset: 1

Vincenzo Librandi, Oct 20 2011

Shifted variant of A102305. - R. J. Mathar, Oct 20 2011

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A059100, A102305, A183199, A160842, A010000.

Magma
```
[Round((n+1/n)^2): n in [1..60]];
```

Table[Floor[(n+1/n)^2+1/2],{n,50}] (* Harvey P. Dale, Aug 12 2012 *)
Join[{4}, 2+Range[2,50]^2] (* G. C. Greubel, Feb 04 2024 *)

[4]+[n^2+2 for n in range(2,51)] # G. C. Greubel, Feb 04 2024

a(n) = n^2 + 2, n > 1.
a(n) = a(n-1) + 2*n - 1, n > 2.
From G. C. Greubel, Feb 04 2024: (Start)
G.f.: x*(4 - 6*x + 5*x^2 - x^3)/(1 - x)^3.
E.g.f.: -2 + x + (2 + x + x^2)*exp(x). (End)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 4. - Chai Wah Wu, May 09 2024

A089000 Square table, read by antidiagonals, of coefficients T(k,n) (row k; column n) defined by: T(k,n) = k*T(k,n-1)+ n; T(k,0) = 0.

Original entry on oeis.org

0, 1, 0, 2, 1, 0, 3, 3, 1, 0, 4, 6, 4, 1, 0, 5, 10, 11, 5, 1, 0, 6, 15, 26, 18, 6, 1, 0, 7, 21, 57, 58, 27, 7, 1, 0, 8, 28, 120, 179, 112, 38, 8, 1, 0, 9, 36, 247, 543, 453, 194, 51, 9, 1, 0, 10, 45, 502, 1636, 1818, 975, 310, 66, 10, 1, 0

Offset: 0

Philippe Deléham, Nov 02 2003

Rows begin:
{0, 1, 2, 3, 4, 5, 6, 7, 8, ...}:see A001477
{0, 1, 3, 6, 10, 15, 21, 28, ...} : see A000217
{0, 1, 4, 11, 26, 57, 120, 247, 502, ...} : see A000295
{0, 1, 5, 18, 58, 179, 543, 1636, ...} : see A000340
{0, 1, 6, 27, 112, 453, 1818, 7279, ...} : see A014825
{0, 1, 7, 38, 194, 975, 4881, 24412, ...} : see A014827
{0, 1, 8, 51, 310, 1865, 11196, 67183, ...}: see diagonals of triangle A088990
Diagonal begin:
{0, 1, 4, 18, 112, 975, 11196, ... } :see A062805
{0, 1, 5, 27, 194, 1865, ...} : see A023811
Column {3, 6, 11, 18, 27, 38, 51, ...} : see A010000

Unprotect[Power]; 0^0=1; T[n_,k_]:=Sum[j*k^(n-j),{j,0,n}]; Table[T[n-k,k],{n,0,10},{k,0,n}]//Flatten (* Stefano Spezia, Apr 19 2025 *)

T(k, n)= (k^(n+1)- (k-1)*n - k)/(k-1)^2. T(k, n) = Sum(j, 0<=j<=n; j*k^(n-j)).

A104563 A floretion-generated sequence relating to centered square numbers.

Original entry on oeis.org

0, 1, 3, 5, 8, 13, 19, 25, 32, 41, 51, 61, 72, 85, 99, 113, 128, 145, 163, 181, 200, 221, 243, 265, 288, 313, 339, 365, 392, 421, 451, 481, 512, 545, 579, 613, 648, 685, 723, 761, 800, 841, 883, 925, 968, 1013, 1059, 1105, 1152, 1201, 1251

Offset: 0

Creighton Dement, Mar 15 2005

Floretion Algebra Multiplication Program, FAMP Code: a(n) = 1vesrokseq[A*B] with A = - .5'i - .5i' + .5'ii' + .5e, B = + .5'ii' - .5'jj' + .5'kk' + .5e. RokType: Y[sqa.Findk()] = Y[sqa.Findk()] + Math.signum(Y[sqa.Findk()])*p (internal program code). Note: many slight variations of the "RokType" already exist, such that it has become difficult to assign them all names.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-4,4,-3,1).

Cf. A001844, A004525, A104564, A098180, A059100, A010000, A047599, A007877.

LinearRecurrence[{3, -4, 4, -3, 1}, {0, 1, 3, 5, 8}, 60] (* Amiram Eldar, Dec 14 2024 *)

concat(0, Vec(x*(1 + x)*(1 - x + x^2) / ((1 - x)^3*(1 + x^2)) + O(x^40))) \\ Colin Barker, Apr 29 2019

G.f.: x*(1 + x^3)/((1 + x^2)*(1 - x)^3).
FAMP result: 2*a(n) + 2*A004525(n+1) = A104564(n) + a(n+1).
Superseeker results:
a(2*n+1) = A001844(n) = 2*n*(n+1) + 1 (Centered square numbers);
a(n+1) - a(n) = A098180(n) (Odd numbers with two times the odd numbers repeated in order between them);
a(n) + a(n+2) = A059100(n+1) = A010000(n+1);
a(n+2) - a(n) = A047599(n+1) (Numbers that are congruent to {0, 3, 4, 5} mod 8);
a(n+2) - 2*a(n+1) + a(n) = A007877(n+3) (Period 4 sequence with initial period (0, 1, 2, 1));
Coefficients of g.f.*(1-x)/(1+x) = convolution of this with A280560 gives A004525;
Coefficients of g.f./(1+x) = convolution of this with A033999 gives A054925.
a(n) = (1/2)*(n^2 + 1 - cos(n*Pi/2)). - Ralf Stephan, May 20 2007
From Colin Barker, Apr 29 2019: (Start)
a(n) = (2 - (-i)^n - i^n + 2*n^2) / 4 where i=sqrt(-1).
a(n) = 3*a(n-1) - 4*a(n-2) + 4*a(n-3) - 3*a(n-4) + a(n-5) for n>4. (End)
a(n) = A011848(n-1)+A011848(n+2). - R. J. Mathar, Sep 11 2019
Sum_{n>=1} 1/a(n) = Pi^2/48 + (Pi/2) * tanh(Pi/2) + (Pi/(4*sqrt(2)) * tanh(Pi/(2*sqrt(2)))). - Amiram Eldar, Dec 14 2024

Stephan's formula corrected by Bruno Berselli, Apr 29 2019

A183199 Least integer k such that Floor(kf(n+1))>kf(n), where f(n)=(n^2)/(1+n^2).

Original entry on oeis.org

3, 6, 11, 18, 27, 38, 51, 66, 83, 102, 123, 146, 171, 198, 227, 258, 291, 326, 363, 402, 443, 486, 531, 578, 627, 678, 731, 786, 843, 902, 963, 1026, 1091, 1158, 1227, 1298, 1371, 1446, 1523, 1602, 1683, 1766, 1851, 1938, 2027, 2118, 2211, 2306, 2403, 2502

Offset: 1

Clark Kimberling, Dec 29 2010

nonn

Appears to be essentially the same as A102305, A059100 and A010000. - R. J. Mathar, Jun 07 2011

Cf. A183162.

Table[k=1; While[Floor[k*((n+1)^2)/(1+(n+1)^2)]<=k*(n^2)/(1+(n^2)), k++]; k, {n,100}]

A055630 Table T(k,m) = k^2 + m read by antidiagonals.

Original entry on oeis.org

0, 1, 1, 4, 2, 2, 9, 5, 3, 3, 16, 10, 6, 4, 4, 25, 17, 11, 7, 5, 5, 36, 26, 18, 12, 8, 6, 6, 49, 37, 27, 19, 13, 9, 7, 7, 64, 50, 38, 28, 20, 14, 10, 8, 8, 81, 65, 51, 39, 29, 21, 15, 11, 9, 9, 100, 82, 66, 52, 40, 30, 22, 16, 12, 10, 10, 121, 101, 83, 67, 53, 41, 31, 23, 17, 13

Offset: 0

Henry Bottomley, Jun 05 2000

			Table begins:
..0...1...4...9..16..25..36..49..64..81.100.121.144...
..1...2...5..10..17..26..37..50..65..82.101.122.145...
..2...3...6..11..18..27..38..51..66..83.102.123.146...
..3...4...7..12..19..28..39..52..67..84.103.124.147...
..4...5...8..13..20..29..40..53..68..85.104.125.148...
..5...6...9..14..21..30..41..54..69..86.105.126.149...
..6...7..10..15..22..31..42..55..70..87.106.127.150...
..7...8..11..16..23..32..43..56..71..88.107.128.151...
..8...9..12..17..24..33..44..57..72..89.108.129.152...
..9..10..13..18..25..34..45..58..73..90.109.130.153...
.10..11..14..19..26..35..46..59..74..91.110.131.154...
... - _Philippe Deléham_, Mar 31 2013

First column is A001477, second column is A000027, first row is A000290, second row is A002522, third row (apart from first term) is A010000, main diagonal is A002378, other diagonals include A028387, A028552, A014209, A002061, A014206, A027688-A027694, each row of A055096 (as upper right triangle) is right hand part of some row of this table

Cf. A185910

A097347 Rectangular array, by antidiagonals: T(n,k) = rank of k-th n in A097291.

Original entry on oeis.org

1, 2, 3, 5, 4, 6, 10, 8, 7, 11, 17, 13, 9, 12, 18, 26, 20, 15, 14, 19, 27, 37, 29, 22, 16, 21, 28, 38, 50, 40, 31, 24, 23, 30, 39, 51, 65, 53, 42, 33, 25, 32, 41, 52, 66, 82, 68, 55, 44, 35, 34, 43, 54, 67, 83, 101, 85, 70, 57, 46, 36, 45, 56, 69, 84, 102, 122, 104, 87, 72, 59

Offset: 1

Clark Kimberling, Aug 06 2004

As a sequence, this is a permutation of the natural numbers. Row 1 is A002522 (n^2 + 1); main diagonal is A000290 (n^2); column 1 is A010000 (n^2 + 2).

			Northwest corner:
1 2 5 10
3 4 8 13
6 7 9 15
11 12 14 16

Cf. A097291, A097292.

A113592 Array of quadratic pseudofibonacci sequences, read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 2, 3, 1, 3, 6, 11, 1, 4, 11, 40, 127, 1, 5, 18, 127, 1612, 16151, 1, 6, 27, 332, 16151, 2598264, 260855055, 1, 7, 38, 739, 110260

Offset: 1

Jonathan Vos Post, Jan 26 2006

Row 1 is A113848. Column 1 is A000012 (the simplest sequence of positive numbers: the all 1's sequence). Column 2 is A000027 (the natural numbers) = n. Column 3 is A010000 = A059100(n+1) = n^2 + 2. Column 4 is 2*n + (n^2 + 2)^2 = n^4 + 4*n^2 + 2*n + 4. Column 5 is 2*(n^2 + 2) + (n^4 + 4*n^2 + 2*n + 4)^2 = n^8 + 8*n^6 + 4*n^5 + 24*n^4 + 16*n^3 + 38*n^2 + 16*n + 20.

			Table (upper left corner):
1...1...3...11...127....16151...260855055...
1...2...6...40...1612...2598624.675284696600...
1...3...11..127..16151..260855055...
1...4...18..332..110260.12157268264...
1...5...27..739..546175...
1...6...38..1456.2120012...
1...7...51..2615.6838327...
1...8...66..4372.19114516...
1...9...83..6907.47706815
1..10..102..10424.108659980...

Cf. A000012, A000027, A000278, A000283, A010000, A014253, A059100, A063827, A072878, A112957, A112958, A112959, A112960, A112961, A112969, A113785.

Antidiagonals of table: T(i, j) = j-th iteration of a(i, 0) = 1, a(i, 1) = i and for j>1: a(i, j) = 2*a(i, j-2) + a(i, j-1)^2.

A164515 Positive numbers not of the form n^2+2.

Original entry on oeis.org

1, 4, 5, 7, 8, 9, 10, 12, 13, 14, 15, 16, 17, 19, 20, 21, 22, 23, 24, 25, 26, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80

Offset: 1

Jaroslav Krizek, Aug 14 2009

Complement of A059100, and essentially also of A010000.

With[{upto=80},Complement[Range[upto],Range[0,Floor[Sqrt[upto-2]]]^2+2]] (* Harvey P. Dale, Jun 22 2013 *)

A000027 \ A059100.

a(n) = n + 1 + floor(sqrt(n-1) + 1/2), for n >= 2.

Redefined for simplicity by removing 2, R. J. Mathar, Aug 31 2009

A374224 Integer part of the total Euclidean length of the shortest minimum-link polygonal chains joining all the nodes of the grid {0,1,...,n-1} X {0,1,...,n-1}.

Original entry on oeis.org

0, 3, 12, 20, 28, 40, 53, 68, 85, 104, 125, 148, 173, 200, 229, 260, 293, 328, 365, 404, 445, 488, 533, 580, 629, 680, 733, 788, 845, 904, 965, 1028, 1093, 1160, 1229, 1300, 1373, 1448, 1525, 1604, 1685, 1768, 1853, 1940, 2029, 2120, 2213, 2308, 2405, 2504

Offset: 1

Marco Ripà, Jun 30 2024

This sequence describes the optimal solution of the 2D generalization of the well-known nine dots problem, published in Loyd’s Cyclopedia of Puzzles (1914), p. 301.
Since Solomon Golomb constructively proved that, for any n >= 3, the minimum-link polygonal chain covering a given {0,1,...,n-1} X {0,1,...,n-1} grid consists of (exactly) 2*(n - 1) line segments, we only need to find the shortest trail satisfying the constraint above.
In detail, if n = 2, the trivial spanning path (0,1)-(0,0)-(1,0)-(1,1) is optimal. If n = 3, we have the classic solution of the nine dots problem (0,1)-(0,3)-(3,0)-(0,0)-(2,2). Now, if n > 3, a valid upper bound is given by n^2 + 5*sqrt(2) - 3, but it is possible to improve this solution for the n = 5 case by providing the trail.
(2,3)-(4,3)-(1,0)-(1,3)-(4,0)-(0,0)-(0,4)-(4,4)-(4,1), whose total Euclidean length is 20 + 6*sqrt(2). In the end, assuming n > 5, we can recycle the mentioned solution, then extend the last line segment to reach (4,-1), and finally apply the square spiral pattern to the (extended) ending segment of the {0,1,...,n-1} X {0,1,...,n-1} grid solution in order to get the solution for the {0,1,...,n} X {0,1,...,n} case, joining 2*n + 1 more points by spending two additional line segments having a combined length of 2*n (and this is an iterative strategy which is optimal for any n > 5).

			a(2) = 3 since we can join the points {0,1}^2 with a spanning path consisting of 3 line segments having a total Euclidean length of 2^2 - 1.

Marco Ripà, Shortest polygonal chains covering each planar square grid, arXiv:2207.08708 [math.CO], 2022. See pp. 11-13.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A002193, A010000, A058992, A373537.

Join[{0, 3, 12, 20, 28} Table[Floor[n^2 + 5*Sqrt[2]] - 3 , {n, 6, 50}]]
LinearRecurrence[{3,-3,1},{0,3,12,20,28,40,53,68},50] (* Harvey P. Dale, Jul 19 2025 *)

a(1) = 1, a(2) = 3, a(3) = floor(5*(1+sqrt(2))) = 12, a(5) = floor(20 + 6*sqrt(2)) = 28, and a(n) = floor(n^2 + 5*sqrt(2)) - 3 iff n = 4 or n >= 6.
For n > 5, a(n) = n^2 + 4.
G.f.: x^2*(3 + 3*x - 7*x^2 + x^3 + 4*x^4 - 3*x^5 + x^6)/(1 - x)^3. - Stefano Spezia, Jul 02 2024

cp's OEIS Frontend

A197985 a(n) = round((n+1/n)^2).

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A089000 Square table, read by antidiagonals, of coefficients T(k,n) (row k; column n) defined by: T(k,n) = k*T(k,n-1)+ n; T(k,0) = 0.

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A104563 A floretion-generated sequence relating to centered square numbers.

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A183199 Least integer k such that Floor(k*f(n+1))>k*f(n), where f(n)=(n^2)/(1+n^2).

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A055630 Table T(k,m) = k^2 + m read by antidiagonals.

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A097347 Rectangular array, by antidiagonals: T(n,k) = rank of k-th n in A097291.

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A113592 Array of quadratic pseudofibonacci sequences, read by antidiagonals.

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A164515 Positive numbers not of the form n^2+2.

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A374224 Integer part of the total Euclidean length of the shortest minimum-link polygonal chains joining all the nodes of the grid {0,1,...,n-1} X {0,1,...,n-1}.

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A183199 Least integer k such that Floor(kf(n+1))>kf(n), where f(n)=(n^2)/(1+n^2).