A047218 -id:A047218 - cp's OEIS frontend

A191741 Dispersion of A047217, (numbers >1 and congruent to 0 or 1 or 2 mod 5), by antidiagonals.

1, 2, 3, 5, 6, 4, 10, 11, 7, 8, 17, 20, 12, 15, 9, 30, 35, 21, 26, 16, 13, 51, 60, 36, 45, 27, 22, 14, 86, 101, 61, 76, 46, 37, 25, 18, 145, 170, 102, 127, 77, 62, 42, 31, 19, 242, 285, 171, 212, 130, 105, 71, 52, 32, 23, 405, 476, 286, 355, 217, 176, 120

Offset: 1

Clark Kimberling, Jun 14 2011

For a background discussion of dispersions and their fractal sequences, see A191426.  For dispersions of congruence sequences mod 3, mod 4, or mod 5, see A191655, A191663, A191667, A191702.
...
Suppose that {2,3,4,5,6} is partitioned as {x1, x2} and {x3,x4,x5}.  Let S be the increasing sequence of numbers >1 and congruent to x1 or x2 mod 5, and let T be the increasing sequence of numbers >1 and congruent to x3 or x4 or x5 mod 5.  There are 10 sequences in S, each matched by a (nearly) complementary sequence in T.  Each of the 20 sequences generates a dispersion, as listed here:
...
A191722=dispersion of A008851 (0, 1 mod 5 and >1)
A191723=dispersion of A047215 (0, 2 mod 5 and >1)
A191724=dispersion of A047218 (0, 3 mod 5 and >1)
A191725=dispersion of A047208 (0, 4 mod 5 and >1)
A191726=dispersion of A047216 (1, 2 mod 5 and >1)
A191727=dispersion of A047219 (1, 3 mod 5 and >1)
A191728=dispersion of A047209 (1, 4 mod 5 and >1)
A191729=dispersion of A047221 (2, 3 mod 5 and >1)
A191730=dispersion of A047211 (2, 4 mod 5 and >1)
A191731=dispersion of A047204 (3, 4 mod 5 and >1)
...
A191732=dispersion of A047202 (2,3,4 mod 5 and >1)
A191733=dispersion of A047206 (1,3,4 mod 5 and >1)
A191734=dispersion of A032793 (1,2,4 mod 5 and >1)
A191735=dispersion of A047223 (1,2,3 mod 5 and >1)
A191736=dispersion of A047205 (0,3,4 mod 5 and >1)
A191737=dispersion of A047212 (0,2,4 mod 5 and >1)
A191738=dispersion of A047222 (0,2,3 mod 5 and >1)
A191739=dispersion of A008854 (0,1,4 mod 5 and >1)
A191740=dispersion of A047220 (0,1,3 mod 5 and >1)
A191741=dispersion of A047217 (0,1,2 mod 5 and >1)
...
For further information about these 20 dispersions, see A191722.
...
Regarding the dispersions A191722-A191741, there are general formulas for sequences of the type "(a or b mod m)" and "(a or b or c mod m)" used in the relevant Mathematica programs.

			Northwest corner:
1....2....5....10...17
3....6....11...20...35
4....7....12...21...36
8....15...26...45...76
9....16...27...46...77
13...22...37...62...105

Ivan Neretin, Table of n, a(n) for n = 1..5050 (first 100 antidiagonals, flattened)

Cf. A047204, A047217, A191722, A191731, A191426.

(* Program generates the dispersion array t of the increasing sequence f[n] *)
r = 40; r1 = 12;  c = 40; c1 = 12;
a=2; b=5; c2=6; m[n_]:=If[Mod[n,3]==0,1,0];
f[n_]:=a*m[n+2]+b*m[n+1]+c2*m[n]+5*Floor[(n-1)/3]
Table[f[n], {n, 1, 30}]  (* A047217 *)
mex[list_] := NestWhile[#1 + 1 &, 1, Union[list][[#1]] <= #1 &, 1, Length[Union[list]]]
rows = {NestList[f, 1, c]};
Do[rows = Append[rows, NestList[f, mex[Flatten[rows]], r]], {r}];
t[i_, j_] := rows[[i, j]];
TableForm[Table[t[i, j], {i, 1, 10}, {j, 1, 10}]] (* A191741 *)
Flatten[Table[t[k, n - k + 1], {n, 1, c1}, {k, 1, n}]] (* A191741  *)

A047203 Numbers that are congruent to {0, 2, 3, 4} mod 5.

Original entry on oeis.org

0, 2, 3, 4, 5, 7, 8, 9, 10, 12, 13, 14, 15, 17, 18, 19, 20, 22, 23, 24, 25, 27, 28, 29, 30, 32, 33, 34, 35, 37, 38, 39, 40, 42, 43, 44, 45, 47, 48, 49, 50, 52, 53, 54, 55, 57, 58, 59, 60, 62, 63, 64, 65, 67, 68, 69, 70, 72, 73, 74, 75, 77, 78, 79, 80, 82, 83, 84, 85, 87, 88, 89

Offset: 1

N. J. A. Sloane

Complement of A016861. - Reinhard Zumkeller, Oct 23 2006

Melvyn B. Nathanson, On the fractional parts of roots of positive real numbers, Amer. Math. Monthly, Vol. 120, No. 5 (2013), pp. 409-429 [see p. 417].
Index entries for linear recurrences with constant coefficients, signature (1,0,0,1,-1).

Cf. A001068, A001622, A002265, A016861, A027445, A047207, A047211, A047218, A177704.

[Floor((5*n-2)/4) : n in [1..100]]; // Wesley Ivan Hurt, May 14 2016

seq(floor(5*n-2)/4), n=1..72); # Gary Detlefs, Mar 06 2010
seq(floor((15*n-5)/12), n=1..72); # Gary Detlefs, Mar 07 2010

Flatten[Table[5*n + {0, 2, 3, 4}, {n, 0, 20}]] (* T. D. Noe, Nov 12 2013 *)

a(n)=(5*n-2)\4 \\ Charles R Greathouse IV, Jun 11 2015

A027445(a(n)) mod 10 = 0. - Reinhard Zumkeller, Oct 23 2006
a(n) = floor((5n-2)/4). - Gary Detlefs, Mar 06 2010
a(n) = floor((15n-5)/12). - Gary Detlefs, Mar 07 2010
G.f.: x^2*(2+x+x^2+x^3)/((1+x)*(1+x^2)*(x-1)^2). - R. J. Mathar, Oct 08 2011
From Wesley Ivan Hurt, May 14 2016: (Start)
a(n) = a(n-1) + a(n-4) - a(n-5) for n>5.
a(n) = (10*n-7+(-1)^n+2*(-1)^((2*n+3+(-1)^n)/4))/8.
a(2n) = A047211(n), a(2n-1) = A047218(n).
a(n) = A047207(n+1) - 1.
a(n+2) = n + 2 + A002265(n) for n>0.
a(n+3)-a(n+2) = A177704(n) for n>0.
a(1-n) = - A001068(n). (End)
Sum_{n>=2} (-1)^n/a(n) = log(5)/4 + sqrt(5)*log(phi)/10 - sqrt(5-2*sqrt(5))*Pi/10, where phi is the golden ratio (A001622). - Amiram Eldar, Dec 07 2021

More terms from Reinhard Zumkeller, Oct 23 2006

A047207 Numbers that are congruent to {0, 1, 3, 4} mod 5.

Original entry on oeis.org

0, 1, 3, 4, 5, 6, 8, 9, 10, 11, 13, 14, 15, 16, 18, 19, 20, 21, 23, 24, 25, 26, 28, 29, 30, 31, 33, 34, 35, 36, 38, 39, 40, 41, 43, 44, 45, 46, 48, 49, 50, 51, 53, 54, 55, 56, 58, 59, 60, 61, 63, 64, 65, 66, 68, 69, 70, 71, 73, 74, 75, 76, 78, 79, 80, 81, 83, 84

Offset: 1

N. J. A. Sloane

Numbers not ending in 2 or 7. - Bruno Berselli, Oct 30 2017

Vincenzo Librandi, Table of n, a(n) for n = 1..8000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,1,-1).

Cf. A001622, A030308, A047209, A047218.

[n : n in [0..100] | n mod 5 in [0, 1, 3, 4]]; // Wesley Ivan Hurt, May 30 2016

seq(floor((5*n-3)/4), n=1..57); # Gary Detlefs, Mar 06 2010

Flatten[Table[5*n + {0, 1, 3, 4}, {n, 0, 20}]] (* T. D. Noe, Nov 12 2013 *)
LinearRecurrence[{1,0,0,1,-1},{0,1,3,4,5},100] (* Harvey P. Dale, Jan 31 2022 *)

forstep(n=0,99,[1,2,1,1],print1(n", ")) \\ Charles R Greathouse IV, Oct 17 2011

a(n) = floor((5*n-3)/4). - Gary Detlefs, Mar 06 2010
G.f.: x^2*(1 + 2*x + x^2 + x^3) / ( (1 + x)*(x^2 + 1)*(x - 1)^2 ). - R. J. Mathar, Oct 08 2011
a(n+1) = Sum_{k>=0} A030308(n,k)*b(k) with b(0)=1, b(1)=3, b(k)=5*2^(k-2) for k>1. - Philippe Deléham, Oct 17 2011
From Wesley Ivan Hurt, May 30 2016: (Start)
a(n) = a(n-1) + a(n-4) - a(n-5) for n>5.
a(n) = (10*n-9-i^(2*n)+(1-i)*i^(-n)+(1+i)*i^n)/8, where i=sqrt(-1).
a(2*k) = A047209(k), a(2*k-1) = A047218(k). (End)
E.g.f.: (4 - sin(x) + cos(x) + (5*x - 4)*sinh(x) + 5*(x - 1)*cosh(x))/4. - Ilya Gutkovskiy, May 30 2016
Sum_{n>=2} (-1)^n/a(n) = log(5)/4 + 3*sqrt(5)*log(phi)/10 + sqrt(1-2/sqrt(5))*Pi/10, where phi is the golden ratio (A001622). - Amiram Eldar, Dec 07 2021

A297469 Solution (bb(n)) of the system of 3 complementary equations in Comments.

Original entry on oeis.org

2, 7, 11, 17, 22, 27, 31, 37, 41, 47, 51, 57, 62, 67, 71, 77, 82, 87, 91, 97, 102, 107, 111, 117, 121, 127, 131, 137, 142, 147, 151, 157, 161, 167, 171, 177, 182, 187, 191, 197, 201, 207, 211, 217, 222, 227, 231, 237, 242, 247, 251, 257, 262, 267, 271, 277

Offset: 0

Clark Kimberling, May 04 2018

Define sequences aa(n), bb(n), cc(n) recursively, starting with aa(0) = 1, bb(0) = 2, cc(0) = 3:
aa(n) = least new;
bb(n) = aa(n) + cc(n-1);
cc(n) = least new;
where "least new k" means the least positive integer not yet placed.
***
The sequences aa,bb,cc partition the positive integers. It appears that cc = A047218 and that for every n >= 0,
(1) 5*n - 1 - 2*aa(n) is in {0,1,2},
(2) (aa(n) mod 5) is in {1,2,4},
(3) 5*n - 3 - bb(n) is in {0,1} for every n >= 0;
(4) (bb(n) mod 5) is in {1,2}.
From N. J. A. Sloane, Nov 05 2019: (Start)
Conjecture: For t >= 0, bb(2t) = 10t + 1 (+1 if binary expansion of t ends in an odd number of 0's), bb(2t+1) = 10t + 7.
The first part may also be written as bb(2t) = 10t + 1 + A328789(t-1).
(End)

			n:  0 1 2 3 4 5 6 7 8 9 10
aa: 1 4 6 9 12 14 16 19 21 24 26
bb: 2 7 11 17 22 27 31 37 41 47 51
cc: 3 5 8 10 13 15 18 20 23 25 28

Clark Kimberling, Table of n, a(n) for n = 0..10000 [This is the sequence bb]

Cf. A299634, A298468 (aa), A047218 (cc), A328789.

z = 500;
mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]);
a = {1}; b = {2}; c = {3};
Do[AppendTo[a, mex[Flatten[{a, b, c}], Last[a]]];
  AppendTo[b, Last[a] + Last[c]];
  AppendTo[c, mex[Flatten[{a, b, c}], Last[a]]], {z}];
Take[a, 100] (* A298468 *)
Take[b, 100] (* A297469 *)
Take[c, 100] (* A047218 *)
(* Peter J. C. Moses, Apr 23 2018 *)

Changed a,b,c to aa,bb,cc to avoid confusion caused by conflict with standard OEIS terminology. - N. J. A. Sloane, Nov 03 2019

A328190 Lexicographically earliest infinite sequence of distinct positive integers such that the sequence and its first differences have no values in common.

Original entry on oeis.org

1, 3, 7, 5, 11, 8, 17, 10, 22, 13, 27, 15, 31, 18, 37, 20, 41, 23, 47, 25, 51, 28, 57, 30, 62, 33, 67, 35, 71, 38, 77, 40, 82, 43, 87, 45, 91, 48, 97, 50, 102, 53, 107, 55, 111, 58, 117, 60, 121, 63, 127, 65, 131, 68, 137, 70, 142, 73, 147, 75, 151, 78, 157

Offset: 1

Peter Kagey, Oct 06 2019

nonn

The graph appears to consist of two lines whose slopes are approximately equal to 1.25 and 2.5.
Conjecture from N. J. A. Sloane, Nov 04 2019: (Start)
a(2t) = floor((5t+1)/2) for t >= 1 (essentially A047218),
a(4t+1) = 10t+1(+1 if binary expansion of t ends in odd number of 0's) for t >= 0 (essentially A297469),
a(4t+3) = 10t+7 for t >= 0.
These formulas explain all the known terms.
One could also say that a(4t+1) = 10t+1+A328979(t+1) for t >= 0.
There is a similar conjecture for A328196.
Call the three sets of conjectured terms S0, S1, and S3. The terms in S0 are == 0 or 3 mod 5; those are in S1 are == 1 or 2 mod 10; and those in S3 are == 7 mod 10. So the sets are disjoint, as required by the definition.
This conjecture would imply that the points a(2t) lie on a line of slope 5/4 and the points a(2t+1) on a line of slope 5/2, as conjectured by Peter Kagey. (End)
Comment from N. J. A. Sloane, Nov 06 2019: (Start)
Let us DEFINE a sequence S by the conjectured formulas given here, and a sequence T by the conjectured formulas given in A328196. Then it is not difficult to prove that the first differences of S are given by T, and that the terms of S and T are disjoint.
So S is certainly a candidate for the lexicographically earliest infinite sequence of distinct positive integers such that the sequence and its first differences have no values in common.
Furthermore Peter Kagey's b-files for this sequence and A328196 show that the first 10000 terms of S are indeed the first 10000 terms of the lexicographically earliest such sequence.
But this is not yet a proof that S IS the lexicographically earliest such sequence. (End)
To construct the bisection a(2n-1), start with [4]. Apply the substitution rule 4 -> 46, 5 -> 46, 6 -> 55. Prepend [1, 6] to the resulting list, then take partial sums. - John Keith, Dec 31 2020

			a(1)  = 1.
a(2) != 1 because a(1) = 1,
a(2) != 2 because then a(2) - a(1) = a(1), so
a(2)  = 3.
The first eight terms of this sequence and first seven terms of its first differences are
[1, 3,  7, 5, 11, 8, 17, 10] and
[2, 4, -2, 6, -3, 9, -7] respectively, and these sequences have no common terms.

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

Cf. A005228, A047218, A080426, A297469, A327460, A328196 (first differences), A328979.

See A328984 and A328985 for simpler sequences which almost have the properties of A329190 and A328196. - N. J. A. Sloane, Nov 07 2019

A298468 Solution (aa(n)) of the system of 3 complementary equations in Comments.

Original entry on oeis.org

1, 4, 6, 9, 12, 14, 16, 19, 21, 24, 26, 29, 32, 34, 36, 39, 42, 44, 46, 49, 52, 54, 56, 59, 61, 64, 66, 69, 72, 74, 76, 79, 81, 84, 86, 89, 92, 94, 96, 99, 101, 104, 106, 109, 112, 114, 116, 119, 122, 124, 126, 129, 132, 134, 136, 139, 141, 144, 146, 149

Offset: 0

Clark Kimberling, May 04 2018

Define sequences aa(n), bb(n), cc(n) recursively, starting with aa(0) = 1, bb(0) = 2, cc(0) = 3:
aa(n) = least new;
bb(n) = aa(n) + cc(n-1);
cc(n) = least new;
where "least new k" means the least positive integer not yet placed.
***
The sequences aa,bb,cc partition the positive integers. It appears that cc = A047218 and that for every n >=0,
(1) 5*n - 1 - 2*aa(n) is in {0,1,2},
(2) (aa(n) mod 5) is in {1,2,4},
(3) 5*n - 3 - bb(n) is in {0,1} for every n >= 0;
(4) (bb(n) mod 5) is in {1,2}.
From N. J. A. Sloane, Nov 05 2019: (Start)
Conjecture: For t >= 1, aa(2t) = 5t+1(+1 if binary expansion of t ends in an odd number of 0's), and for t >= 0, aa(2t+1) = 5t+4.
The first part may also be written as aa(2t) = 5t+1+A328789(t-1).
(End)

			n:  0 1 2 3 4 5 6 7 8 9 10
aa: 1 4 6 9 12 14 16 19 21 24 26
bb: 2 7 11 17 22 27 31 37 41 47 51
cc: 3 5 8 10 13 15 18 20 23 25 28

Clark Kimberling, Table of n, a(n) for n = 0..10000 [This is aa]

Cf. A299634, A297469 (bb), A047218 (cc), A328789.

z = 500;
mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]);
a = {1}; b = {2}; c = {3};
Do[AppendTo[a, mex[Flatten[{a, b, c}], Last[a]]];
  AppendTo[b, Last[a] + Last[c]];
  AppendTo[c, mex[Flatten[{a, b, c}], Last[a]]], {z}];
Take[a, 100] (* A298468 *)
Take[b, 100] (* A297469 *)
Take[c, 100] (* A047218 *)
(* Peter J. C. Moses, Apr 23 2018 *)

Changed a,b,c to aa,bb,cc to avoid confusion caused by conflict with standard OEIS terminology. - N. J. A. Sloane, Nov 03 2019

A301569 Expansion of Product_{k>=1} (1 + x^(5k))(1 + x^(5*k-2)).

Original entry on oeis.org

1, 0, 0, 1, 0, 1, 0, 0, 2, 0, 1, 1, 0, 3, 0, 2, 2, 0, 5, 0, 2, 4, 0, 7, 1, 3, 7, 0, 10, 2, 4, 11, 0, 14, 4, 5, 17, 0, 19, 8, 6, 25, 1, 25, 13, 8, 36, 2, 33, 21, 10, 50, 4, 43, 33, 12, 69, 8, 55, 49, 15, 93, 14, 70, 71, 19, 124, 23, 88, 102, 24, 163, 37, 110, 142, 31

Offset: 0

Ilya Gutkovskiy, Mar 23 2018

nonn

Number of partitions of n into distinct parts congruent to 0 or 3 mod 5.

			a(13) = 3 because we have [13], [10, 3] and [8, 5].

Index entries for sequences related to partitions

Cf. A035369, A036822, A047218, A203776, A219607, A281271, A301562, A301563, A301564, A301565, A301567, A301568, A301570.

nmax = 75; CoefficientList[Series[Product[(1 + x^(5 k)) (1 + x^(5 k - 2)), {k, 1, nmax}], {x, 0, nmax}], x]
nmax = 75; CoefficientList[Series[x^2 QPochhammer[-1, x^5] QPochhammer[-x^(-2), x^5]/(2 (1 + x^2)), {x, 0, nmax}], x]
nmax = 75; CoefficientList[Series[Product[(1 + Boole[MemberQ[{0, 3}, Mod[k, 5]]] x^k), {k, 1, nmax}], {x, 0, nmax}], x]

G.f.: Product_{k>=2} (1 + x^A047218(k)).

a(n) ~ exp(Pi*sqrt(2*n/15)) / (2^(37/20) * 15^(1/4) * n^(3/4)). - Vaclav Kotesovec, Mar 24 2018

A115006 Row 2 of array in A114999.

Original entry on oeis.org

0, 3, 8, 16, 26, 39, 54, 72, 92, 115, 140, 168, 198, 231, 266, 304, 344, 387, 432, 480, 530, 583, 638, 696, 756, 819, 884, 952, 1022, 1095, 1170, 1248, 1328, 1411, 1496, 1584, 1674, 1767, 1862, 1960, 2060, 2163, 2268, 2376, 2486, 2599, 2714, 2832, 2952, 3075, 3200

Offset: 0

N. J. A. Sloane, Feb 23 2006

Number of lattice points (x,y) in the region of the coordinate plane bounded by y < 3x+1, y > x/2 and x <= n. - Wesley Ivan Hurt, Oct 27 2014

Vincenzo Librandi, Table of n, a(n) for n = 0..2000
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

Cf. A114999, A000217 (triangular numbers), A002620 (quarter-squares), A001859 (triangular numbers plus quarter-squares), A017305 (10n+3), A147874 (zero followed by partial sums of A017305).

Partial Sums of A047218.

[ n*(n+1) + (n+1)^2 div 4: n in [0..50] ];

A115006:=n->(10*n^2 + 12*n + 1 - (-1)^n)/8: seq(A115006(n), n=0..50); # Wesley Ivan Hurt, Oct 27 2014

Table[(10*n^2 + 12*n + 1 - (-1)^n)/8, {n, 0, 50}] (* Wesley Ivan Hurt, Oct 27 2014 *)
LinearRecurrence[{2,0,-2,1},{0,3,8,16},60] (* Harvey P. Dale, Jan 13 2015 *)

{for(n=0, 50, print1(n*(n+1)+floor((n+1)^2/4), ","))}

a(n) = floor((n+1)^2/4)+n*(n+1).
G.f.: x*(2*x+3)/((1-x)^3*(1+x)).
From Wesley Ivan Hurt, Oct 27 2014: (Start)
a(n) = 2*a(n-1)-2*a(n-3)+a(n-4).
a(n) = (10*n^2 + 12*n + 1 - (-1)^n)/8.
a(n) = Sum_{i=1..n+1} (10*i + (-1)^i - 9)/4. (End)
E.g.f.: (x*(11 + 5*x)*cosh(x) + (1 + 11*x + 5*x^2)*sinh(x))/4. - Stefano Spezia, Aug 22 2023

Edited by Klaus Brockhaus, Nov 18 2008

A294259 a(n) = n(n^3 + 2n^2 - 5*n + 10)/8.

Original entry on oeis.org

0, 1, 4, 15, 43, 100, 201, 364, 610, 963, 1450, 2101, 2949, 4030, 5383, 7050, 9076, 11509, 14400, 17803, 21775, 26376, 31669, 37720, 44598, 52375, 61126, 70929, 81865, 94018, 107475, 122326, 138664, 156585, 176188, 197575, 220851, 246124, 273505, 303108, 335050, 369451

Offset: 0

Bruno Berselli, Oct 30 2017

a(n) is even for n in A047481.

Also, a(n) is divisible by 5 if and only if n belongs to A047218.

			After 0:
1   =                     -(0) + (1);
4   =                 -(0 + 1) + (2 + 2*3/2);
15  =             -(0 + 1 + 2) + (3 + 4 + 5 + 3*4/2);
43  =         -(0 + 1 + 2 + 3) + (4 + 5 + 6 + 7 + 8 + 9 + 4*5/2);
100 =     -(0 + 1 + 2 + 3 + 4) + (5 + 6 + 7 + 8 + ... + 14 + 5*6/2);
201 = -(0 + 1 + 2 + 3 + 4 + 5) + (6 + 7 + 8 + 9 + ... + 20 + 6*7/2), etc.

Bruno Berselli, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000217, A002817, A176145.
Cf. A101374: the sums in the Example section end in squares.
Subsequence of A047207.

List([0..50], n -> n*(n^3+2*n^2-5*n+10)/8);

[n*(n^3+2*n^2-5*n+10)/8: n in [0..50]];

a := n -> n*(n*(n*(n+2)-5)+10)/8: seq(a(n),n=0..41); # Peter Luschny, Nov 06 2017

Table[n (n^3 + 2 n^2 - 5 n + 10)/8, {n, 0, 50}]
LinearRecurrence[{5,-10,10,-5,1},{0,1,4,15,43},50] (* Harvey P. Dale, Jan 08 2024 *)

makelist(n*(n^3+2*n^2-5*n+10)/8, n, 0, 50);

vector(50, n, n--; n*(n^3+2*n^2-5*n+10)/8)

[n*(n^3+2*n^2-5*n+10)/8 for n in range(50)]

O.g.f.: x*(1 - x + 5*x^2 - 2*x^3)/(1 - x)^5.
E.g.f.: x*(8 + 8*x + 8*x^2 + x^3)*exp(x)/8.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n>4.
a(n) = 2*n + Sum_{i=0..n} i*(i^2 - 3)/2.

A328987 The sequence C(n) defined in the comments (A and B smallest missing numbers, offset 0).

Original entry on oeis.org

3, 10, 15, 20, 27, 32, 39, 44, 51, 56, 61, 68, 73, 80, 85, 90, 97, 102, 109, 114, 119, 126, 131, 138, 143, 150, 155, 160, 167, 172, 179, 184, 189, 196, 201, 208, 213, 220, 225, 230, 237, 242, 249, 254, 259, 266, 271, 278, 283, 290, 295, 300, 307, 312, 319

Offset: 0

N. J. A. Sloane, Nov 07 2019

nonn

Define a triple of sequences A,B,C by A[0]=1, B[0]=2, C[0]=3; for n>=1, A[n] = smallest missing number from the terms of A,B,C defined so far; B[n] = = smallest missing number from the terms of A,B,C defined so far; C[n] = n+A[n]+B[n].
Then A = A086377, B = A080652, C = the present sequence.
Inspired by the triples [A003144, A003145, A004146] and [A298468, A298469, A047218].

			The initial terms are:
n: 0, 1, 2, 3, 4,  5,  6,  7,  8.  9. 10. ...
a: 1, 4, 6, 8, 11, 13, 16, 18, 21, 23,25, ...
b: 2, 5, 7, 9, 12, 14, 17, 19, 22, 24,26, ...
c: 3, 10, 15, 20, 27, 32, 39, 44, 51, 56, 61, 68, 73, ...

Cf. A086377, A080652; A003144, A003145, A003146; A298468, A298469, A047218.

Conjectures from Colin Barker, Nov 08 2019: (Start)
G.f.: (3 + 7*x + 5*x^2 + 5*x^3 + 7*x^4 + 5*x^5 + 7*x^6 + 5*x^7 + 7*x^8 + 5*x^9 + 5*x^10 + 7*x^11 + 2*x^12 - 2*x^20 + 2*x^21) / (1 - x - x^12 + x^13).
a(n) = a(n-1) + a(n-12) - a(n-13) for n>21.
(End)
Conjecture: a(n) ~ 35*n/6. - Stefano Spezia, Nov 02 2021

cp's OEIS Frontend

A191741 Dispersion of A047217, (numbers >1 and congruent to 0 or 1 or 2 mod 5), by antidiagonals.

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A047203 Numbers that are congruent to {0, 2, 3, 4} mod 5.

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A047207 Numbers that are congruent to {0, 1, 3, 4} mod 5.

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A297469 Solution (bb(n)) of the system of 3 complementary equations in Comments.

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A328190 Lexicographically earliest infinite sequence of distinct positive integers such that the sequence and its first differences have no values in common.

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A298468 Solution (aa(n)) of the system of 3 complementary equations in Comments.

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A301569 Expansion of Product_{k>=1} (1 + x^(5*k))*(1 + x^(5*k-2)).

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A115006 Row 2 of array in A114999.

A301569 Expansion of Product_{k>=1} (1 + x^(5k))(1 + x^(5*k-2)).

A294259 a(n) = n(n^3 + 2n^2 - 5*n + 10)/8.