A152948 -id:A152948 - cp's OEIS frontend

A227847 Number of tuples (x_1, x_2, ..., x_n) with 1 <= x_1 <= x_2 <= ... <= x_n such that Sum_{i=1..n} x_i^3 = (Sum_{i=1..n} x_i)^2 and Sum_{i=1..n-1} x_i^3 + (x_n-1)^3 + (x_n+1)^3 = (Sum_{i=1..n-1} x_i + 2x_n)^2.

0, 1, 1, 1, 2, 2, 2, 6, 10, 31, 77, 206, 568, 1704, 5037, 15554

Offset: 1

Jimmy Zotos, Aug 01 2013

An n-tuple meeting the first condition is called an n-SCESS ("sum of cubes equals square of sum").
In other words, a(n) is the number of tuples (x_1, x_2, ..., x_n) satisfying SCESS such that (x_1, x_2, ..., x_{n-1}, x_n - 1, x_n + 1) also satisfies SCESS. - Max Alekseyev, Mar 04 2025
x_1 + x_2 + ... + x_{n-1} = A152948(x_n). - Balarka Sen, Aug 01 2013

			a(3) = 1 since the only 3-SCESS is (1, 2, 3) for which the corresponding ordered tuple (1, 2, 2, 4) satisfy the SCESS property. (See Mason et al.)
a(5) = 2 since the only 5-SCESS are (1, 2, 2, 3, 5) and (3, 3, 3, 3, 6) for which the corresponding ordered tuples (1, 2, 2, 3, 4, 6) and (3, 3, 3, 3, 5, 7) satisfy the SCESS property.
a(8) = 6 since the only 8-SCESS are (1, 1, 2, 4, 5, 5, 5, 8), (1, 2, 2, 3, 4, 5, 6, 8), (2, 2, 4, 4, 6, 6, 6, 9), (2, 6, 6, 6, 6, 6, 6, 10), (3, 3, 3, 3, 5, 6, 7, 9) and (3, 5, 5, 5, 6, 7, 7, 10) for which the corresponding ordered tuples (1, 1, 2, 4, 5, 5, 5, 7, 9), (1, 2, 2, 3, 4, 5, 6, 7, 9), (2, 2, 4, 4, 6, 6, 6, 8, 10), (2, 6, 6, 6, 6, 6, 6, 9, 11), (3, 3, 3, 3, 5, 6, 7, 8, 10) and (3, 5, 5, 5, 6, 7, 7, 9, 11) satisfy the SCESS property.

Edward Barbeau and Samer Seraj, Sum of cubes is square of sum, arXiv:1306.5257 [math.NT], 2013.
John Mason, Generalising 'sums of cubes equal to squares of sums', The Mathematical Gazette 85:502 (2001), pp. 50-58.

Cf. A001055, A152948, A158649, A225819.

a(n)=my(v=vector(n, i, 1), N=n^(4/3), k); while(v[#v]N, for(i=2, N, if(v[i]Balarka Sen, Aug 01 2013 */

A001055(n) <= a(n) <= A158649(n). - Balarka Sen, Aug 01 2013

a(11)-a(15) from Balarka Sen, Aug 01 2013
a(16) from Balarka Sen, Aug 11 2013
Definition corrected by Max Alekseyev, Mar 04 2025

A236267 a(n) = 8n^2 + 3n + 1.

Original entry on oeis.org

1, 12, 39, 82, 141, 216, 307, 414, 537, 676, 831, 1002, 1189, 1392, 1611, 1846, 2097, 2364, 2647, 2946, 3261, 3592, 3939, 4302, 4681, 5076, 5487, 5914, 6357, 6816, 7291, 7782, 8289, 8812, 9351, 9906, 10477, 11064, 11667, 12286, 12921, 13572, 14239, 14922, 15621, 16336

Offset: 0

Vladimir Shevelev, Jan 21 2014

Positions a(n) of hexagonal numbers such that h(a(n)) = h(a(n)-1) + h(4*n+1), where h = A000384.

First bisection of A057029. The sequence contains infinitely many squares: 1, 676, 779689, 899760016, ... [Bruno Berselli, Jan 24 2014]

			For n=5, A000384(a(5)) = 93096 = A000384(a(5)-1) + A000384(4*5+1) = 92235 + 861.

Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000384, A057029, A064225, A152948, A236257.

[8*n^2+3*n+1: n in [0..50]]; // Bruno Berselli, Jan 24 2014

Table[8 n^2 + 3 n + 1, {n, 0, 50}] (* Bruno Berselli, Jan 24 2014 *)
LinearRecurrence[{3,-3,1},{1,12,39},50] (* Harvey P. Dale, May 26 2019 *)

Vec(-(6*x^2+9*x+1)/(x-1)^3 + O(x^100)) \\ Colin Barker, Jan 21 2014

From Colin Barker, Jan 21 2014: (Start)
G.f.: -(6*x^2 + 9*x + 1)/(x-1)^3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). (End)
E.g.f.: exp(x)*(1 + 11*x + 8*x^2). - Elmo R. Oliveira, Oct 19 2024

More terms from Colin Barker, Jan 21 2014

a(44)-a(45) from Elmo R. Oliveira, Oct 19 2024

A294070 a(n) = (1/4)(n^2 - 2n)^2 + (9/4)(n^2 - 2n) + 6.

Original entry on oeis.org

4, 6, 15, 40, 96, 204, 391, 690, 1140, 1786, 2679, 3876, 5440, 7440, 9951, 13054, 16836, 21390, 26815, 33216, 40704, 49396, 59415, 70890, 83956, 98754, 115431, 134140, 155040, 178296, 204079, 232566, 263940, 298390, 336111, 377304, 422176, 470940, 523815

Offset: 1

Jan Lakota Nono, Aug 14 2018

			2*2, 2*3, 3*5, 5*8, 8*12, 12*17, 17*23, 23*30, 30*38, ...

Colin Barker, Table of n, a(n) for n = 1..1000
SESC NSU Correspondence School, First assignments for 2018/2019 (in Russian), Kvant, 2018, No. 7, p. 42, Mathematics section, 6th grade, exercise no. 2. "Calculate and show in a reduced fraction form the following sum: 1/(2*3) + 2/(3*5) + 3/(5*8) + 4/(8*12) + 5/(12*17)."
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A027689, A152948.

List([1..40],n->(n^2-3*n+6)*(n^2-n+4)/4); # Muniru A Asiru, Aug 16 2018

[(n^2-3*n+6)*(n^2-n+4)/4: n in [1..40]]; // Vincenzo Librandi, Aug 30 2018

b:=n->(n^2-3*n+6)/2: seq(b(n)*b(n+1),n=1..40); # Muniru A Asiru, Aug 16 2018

Times@@@Partition[Array[(#^2 -3# +6)/2 &, 40], 2, 1] (* Michael De Vlieger, Sep 24 2018 *)
LinearRecurrence[{5,-10,10,-5,1}, {4,6,15,40,96}, 40] (* G. C. Greubel, Feb 10 2019 *)

Vec(x*(4 - 14*x + 25*x^2 - 15*x^3 + 6*x^4)/(1-x)^5 + O(x^40)) \\ Colin Barker, Nov 26 2018

[(n^2-3*n+6)*(n^2-n+4)/4 for n in (1..40)] # G. C. Greubel, Feb 10 2019

a(n) = A152948(n) * A152948(n+1).
From Muniru A Asiru, Aug 16 2018: (Start)
a(n) = (n^2 - 3*n + 6)*(n^2 - n + 4)/4.
a(n) = A152948(n)*A027689(n-1)/2. (End)
a(n) = A266883(A061925(n-1)). - Bruno Berselli, Aug 30 2018
From Colin Barker, Nov 26 2018: (Start)
G.f.: x*(4 - 14*x + 25*x^2 - 15*x^3 + 6*x^4)/(1 - x)^5.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n > 5.
a(n) = (24 - 18*n + 13*n^2 - 4*n^3 + n^4)/4. (End)
E.g.f.: (1/4)*exp(x)*(16 + 8*x + 14*x^2 + 6*x^3 + x^4). - Stefano Spezia, Nov 30 2018

A318054 a(n) = n(n + 1)(n^2 + n + 22)/24.

Original entry on oeis.org

0, 2, 7, 17, 35, 65, 112, 182, 282, 420, 605, 847, 1157, 1547, 2030, 2620, 3332, 4182, 5187, 6365, 7735, 9317, 11132, 13202, 15550, 18200, 21177, 24507, 28217, 32335, 36890, 41912, 47432, 53482, 60095, 67305, 75147, 83657, 92872, 102830, 113570, 125132, 137557

Offset: 0

Luce ETIENNE, Aug 14 2018

			a(1) = 2; a(2)= 5+2 = 7; a(3) = 10+5+2 = 17; a(4) = 18+10+5+2 = 35; a(5) = 30+18+10+5+2 = 65; a(6) = 47+30+18+10+5+2 = 112.

Andrew Howroyd, Table of n, a(n) for n = 0..1000
Tenner, Bridget Eileen Reduced word manipulation: patterns and enumeration, J. Algebr. Comb. 46, No. 1, 189-217 (2017), w in S_n(231): l(w)=4.
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Partial sums of A177787.

Cf. A089071, A022856, A152948.

List([0..30],n->n*(n+1)*(n^2+n+22)/24); # Muniru A Asiru, Aug 15 2018

seq(coeff(series(x*(2*x^2-3*x+2)/(1-x)^5, x,n+1),x,n),n=0..30); # Muniru A Asiru, Aug 15 2018

a(n) = n*(n+1)*(n^2+n+22)/24; \\ Michel Marcus, Aug 17 2018

G.f.: x*(2*x^2-3*x+2)/(1-x)^5.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
a(n) = (1/6)*Sum_{i=1..n} (n-i)*((n-i)^2+11), for n >= 1.

A255315 Lower triangular matrix describing the shape of a half hyperbola in the Dirichlet divisor problem.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 0, 2, 1, 1, 0, 2, 1, 1, 1, 0, 1, 2, 1, 1, 1, 0, 1, 2, 1, 1, 1, 1, 0, 1, 1, 2, 1, 1, 1, 1, 0, 0, 2, 2, 1, 1, 1, 1, 1, 0, 0, 2, 1, 2, 1, 1, 1, 1, 1, 0, 0, 2, 1, 2, 1, 1, 1, 1, 1, 1, 0, 0, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1

Offset: 1

Mats Granvik, May 31 2015

The sum of terms of row n is n. Length of row n is n.
From Mats Granvik, Feb 21 2016: (Start)
A006218(n) = (n^2 - ((2*Sum_{kk=1..n} Sum_{k=1..kk} T(n,k)) - n)) + 2*n - round(1 + (1/2)*(-3 + sqrt(n) + sqrt(1 + n))).
A006218(n) = -((n^2 - ((2*Sum_{kk=1..n} Sum_{k=1..kk} T(n,n-k+1)) - n)) - 2*n + round(1 + (1/2)*(-3 + sqrt(n) + sqrt(1 + n)))).
(End)
From Mats Granvik, May 28 2017: (Start)
A006218(n) = (n^2 - (2*(Sum_{k=1..n} T(n, k)*(n - k + 1)) - n)) + 2*n - round(1 + (1/2)*(-3 + sqrt(n) + sqrt(1 + n))).
A006218(n) = -((n^2 - (2*(Sum_{k=1..n} T(n, n - k + 1)*(n - k + 1)) - n)) - 2*n + round(1 + (1/2)*(-3 + sqrt(n) + sqrt(1 + n)))).
(End)
From Mats Granvik, Sep 07 2017: (Start)
It appears that:
The number of 0's in row n is equal to the number of 2's in row n and their number is given by A000196(n) - 1.
The number of 1's in column k is given by A152948(k+2).
The number of 2's in column k is given by A000096(k-1).
The row index of the last nonzero entry in column k is given by A005563(k).
(End)
From Mats Granvik, Oct 06 2018: (Start)
The smallest k such that T(n,k)=2 is given by A079643(n) = floor(n/floor(sqrt(n))).
This gives the lower bound: A006218(n) >= A094761(n) + A079643(n)*2*(A000196(n)-1).
<=> A006218(n) >= 2*n - (floor(sqrt(n)))^2 + floor(n/floor(sqrt(n)))*2*floor(sqrt(n)-1).
The average of k:s such that T(n,k)=2, for n>3 is given by:
b(n) = Sum_{k=1..n} (k*floor(abs(T(n, k)-1/2)))/floor(sqrt(n)-1).
This gives A006218(n) = 2*n - (floor(sqrt(n)))^2 + b(n)*2*floor(sqrt(n)-1) = 2*n - (floor(sqrt(n)))^2 + (Sum_{k=1..n} (k*floor(abs(T(n, k)-1/2))))*2, for n>3.
The largest k such that T(n,k)=2 is given by A004526(n) = floor(n/2).
This gives the upper bound: A006218(n) <= A094761(n) + A004526(n)*2*(A000196(n)-1).
<=> A006218(n) <= 2*n - (floor(sqrt(n)))^2 + floor(n/2)*2*floor(sqrt(n)-1).
The lower bound starts:  1, 3, 5, 8, 10, 14, 16, 20, 21, 23, ...
Sequence A006218 starts: 1, 3, 5, 8, 10, 14, 16, 20, 23, 27, ...
The upper bound starts:  1, 3, 5, 8, 10, 14, 16, 20, 25, 31, ...
(End)

			1;
1, 1;
1, 1, 1;
0, 2, 1, 1;
0, 2, 1, 1, 1;
0, 1, 2, 1, 1, 1;
0, 1, 2, 1, 1, 1, 1;
0, 1, 1, 2, 1, 1, 1, 1;
0, 0, 2, 2, 1, 1, 1, 1, 1;
0, 0, 2, 1, 2, 1, 1, 1, 1, 1;
0, 0, 2, 1, 2, 1, 1, 1, 1, 1, 1;
0, 0, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1;

Eric Weisstein's World of Mathematics, Dirichlet Divisor Problem

Cf. A000096, A000196, A004526, A005563, A006218, A079643, A094761, A094820, A152948.

(* From Mats Granvik, Feb 21 2016: (Start) *)
nn = 12;
T = Table[
   Sum[Table[
     If[And[If[n*k <= r, If[n >= k, 1, 0], 0] == 1,
       If[(n + 1)*(k + 1) <= r, If[n >= k, 1, 0], 0] == 0], 1, 0], {n,
       1, r}], {k, 1, r}], {r, 1, nn}];
Flatten[T]
A006218a = Table[(n^2 - (2*Sum[Sum[T[[n, k]], {k, 1, kk}], {kk, 1, n}] -
        n)) + 2*n - Round[1 + (1/2)*(-3 + Sqrt[n] + Sqrt[1 + n])], {n,
     1, nn}];
A006218b = -Table[(n^2 - (2*
          Sum[Sum[T[[n, n - k + 1]], {k, 1, kk}], {kk, 1, n}] - n)) -
     2*n + Round[1 + (1/2)*(-3 + Sqrt[n] + Sqrt[1 + n])], {n, 1, nn}];
(A006218b - A006218a);
(* (End) *)
(* From Mats Granvik, May 28 2017: (Start) *)
nn = 12;
T = Table[
   Sum[Table[
     If[And[If[n*k <= r, If[n >= k, 1, 0], 0] == 1,
       If[(n + 1)*(k + 1) <= r, If[n >= k, 1, 0], 0] == 0], 1, 0], {n,
       1, r}], {k, 1, r}], {r, 1, nn}];
Flatten[T]
A006218a = Table[(n^2 - (2*Sum[T[[n, k]]*(n - k + 1), {k, 1, n}] - n)) +
    2*n - Round[1 + (1/2)*(-3 + Sqrt[n] + Sqrt[1 + n])], {n, 1, nn}];
A006218b = Table[-((n^2 - (2*Sum[T[[n, n - k + 1]]*(n - k + 1), {k, 1, n}] -
          n)) - 2*n +
      Round[1 + (1/2)*(-3 + Sqrt[n] + Sqrt[1 + n])]), {n, 1, nn}];
(A006218b - A006218a);
(* (End) *)

See Mathematica program.

A317849 Number of states of the Finite State Automaton Gn accepting the language of maximal (or minimal) lexicographic representatives of elements in the positive braid monoid An.

Original entry on oeis.org

1, 5, 18, 56, 161, 443, 1190, 3156, 8315, 21835, 57246, 149970, 392743, 1028351, 2692416, 7049018, 18454775, 48315461, 126491780, 331160070, 866988641, 2269806085, 5942429868, 15557483796, 40730021821, 106632581993, 279167724510, 730870591916, 1913444051645, 5009461563455

Offset: 1

Michel Marcus, Aug 09 2018

nonn

Ramón Flores, Juan González-Meneses, On lexicographic representatives in braid monoids, arXiv:1808.02755 [math.GR], 2018.
Volker Gebhardt, Juan González-Meneses, Generating random braids, J. Comb. Th. A 120 (1), 2013, 111-128.

List([1..30],n->Sum([1..n],i->(Binomial(n+1-i,2)+1)*Fibonacci(2*i))); # Muniru A Asiru, Aug 09 2018

[&+[(Binomial(n+1-k, 2)+1)*Fibonacci(2*k): k in [1..n]]: n in [1..30]]; // Vincenzo Librandi, Aug 09 2018

Table[Sum[(Binomial[n + 1 - k, 2] + 1) Fibonacci[2 k], {k, n}], {n, 30}] (* Vincenzo Librandi, Aug 09 2018 *)

a(n) = sum(i=1, n, (binomial(n+1-i, 2)+1)*fibonacci(2*i));

a(n) = Sum_{i=1..n} (binomial(n+1-i, 2)+1)*Fibonacci(2*i).

Conjecture: g.f. -x*(1-x+x^2) / ( (x^2-3*x+1)*(x-1)^3 ). a(n) = 2*A001519(n+1) -n*(n+1)/2 -2 = 2*A001519(n+1)-A152948(n+2). - R. J. Mathar, Aug 17 2018

cp's OEIS Frontend

A227847 Number of tuples (x_1, x_2, ..., x_n) with 1 <= x_1 <= x_2 <= ... <= x_n such that Sum_{i=1..n} x_i^3 = (Sum_{i=1..n} x_i)^2 and Sum_{i=1..n-1} x_i^3 + (x_n-1)^3 + (x_n+1)^3 = (Sum_{i=1..n-1} x_i + 2x_n)^2.

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A236267 a(n) = 8*n^2 + 3*n + 1.

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A294070 a(n) = (1/4)*(n^2 - 2*n)^2 + (9/4)*(n^2 - 2*n) + 6.

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A318054 a(n) = n*(n + 1)*(n^2 + n + 22)/24.

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A255315 Lower triangular matrix describing the shape of a half hyperbola in the Dirichlet divisor problem.

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Mathematica

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A317849 Number of states of the Finite State Automaton Gn accepting the language of maximal (or minimal) lexicographic representatives of elements in the positive braid monoid An.

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Author

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GAP

Magma

Mathematica

PARI

Formula

A236267 a(n) = 8n^2 + 3n + 1.

A294070 a(n) = (1/4)(n^2 - 2n)^2 + (9/4)(n^2 - 2n) + 6.

A318054 a(n) = n(n + 1)(n^2 + n + 22)/24.