A059270 -id:A059270 - cp's OEIS frontend

A250304 Four-column array read by rows: T(n,k) = the coefficient of x^k in the expanded polynomial x^3 + (x+1)^3 + ... + (x+n-1)^3, for 0 <= k <= 3.

0, 0, 0, 1, 1, 3, 3, 2, 9, 15, 9, 3, 36, 42, 18, 4, 100, 90, 30, 5, 225, 165, 45, 6, 441, 273, 63, 7, 784, 420, 84, 8, 1296, 612, 108, 9, 2025, 855, 135, 10, 3025, 1155, 165, 11, 4356, 1518, 198, 12, 6084, 1950, 234, 13, 8281, 2457, 273, 14, 11025, 3045, 315, 15, 14400, 3720, 360, 16

Offset: 1

Derek Orr, Jan 15 2015

A240970 solves the Diophantine equation: k^3 + (k+1)^3 + ... + (k+n-1)^3 = y^3. This array gives the coefficients of the left hand side for specified n.

			Array starts:
n = 1:   0,   0,  0, 1;
n = 2:   1,   3,  3, 2;
n = 3:   9,  15,  9, 3;
n = 4:  36,  42, 18, 4;
n = 5: 100,  90, 30, 5;
n = 6: 225, 165, 45, 6;
n = 7: 441, 273, 63, 7;
n = 8: 784, 420, 84, 8;
...

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

Cf. A000537, A059270, A045943, A240970.

for(n=1,50,for(k=0,3,print1(polcoeff(sum(i=1,n,(x+i-1)^3),k),", ")))

concat([0,0,0], Vec(x^4*(x^12-3*x^11+3*x^10-x^9-3*x^8+6*x^7-4*x^5+3*x^4-3*x^3-3*x^2-x-1) / ((x-1)^5*(x+1)^5*(x^2+1)^5) + O(x^100))) \\ Colin Barker, Jun 02 2015

a(4*k+1) = A000537(k), for k >= 0.
a(4*k+2) = A059270(k), for k >= 0.
a(4*k+3) = A045943(k), for k >= 0.
a(4*k) = k, for k >= 1.
a(n) = ((2*n^4+40*n^3+188*n^2-24*n-558-(2*n^4-24*n^3+188*n^2-792*n-558)*(-1)^n+(2*n^4-20*n^3-130*n^2+772*n+377)*(-1)^((2*n-1+(-1)^n)/4)-(2*n^4+40*n^3-196*n^2-280*n+594)*(-1)^((6*n-1+(-1)^n)/4)-(4*n^3+66*n^2-228*n-217)*(-1)^((10*n-1+(-1)^n)/4)))/8192. - Luce ETIENNE, May 22 2015
G.f.: x^4*(x^12-3*x^11+3*x^10-x^9-3*x^8+6*x^7-4*x^5+3*x^4-3*x^3-3*x^2-x-1) / ((x-1)^5*(x+1)^5*(x^2+1)^5). - Colin Barker, Jun 02 2015

A270861 Irregular triangle read by rows: numerators of the coefficients of polynomials J(2n-1,z) = Sum_(k=1,2, .. n) ((n+1)^2 - k + (n+1-k)z^n)z^(k-1)/k.

Original entry on oeis.org

3, 1, 8, 7, 2, 1, 15, 7, 13, 3, 1, 1, 24, 23, 22, 21, 4, 3, 2, 1, 35, 17, 11, 8, 31, 5, 2, 1, 1, 1, 48, 47, 46, 45, 44, 43, 6, 5, 4, 3, 2, 1, 63, 31, 61, 15, 59, 29, 57, 7, 3, 5, 1, 3, 1, 1, 80, 79, 26, 77, 76, 25, 74, 73, 8, 7, 2, 5, 4, 1, 2, 1

Offset: 1

Paul Curtz, Mar 24 2016

Irregular triangle of fractions:
3,     1,
8,   7/2,    2,  1/2,
15,    7, 13/3,    3,    1,  1/3,
24, 23/2, 22/3, 21/4,    4,  3/2, 2/3, 1/4,
35,   17,   11,    8, 31/5,    5,   2,   1, 1/2, 1/5,
48, 47/2, 46/3, 45/4, 44/5, 43/6,   6, 5/2, 4/3, 3/4, 2/5, 1/6.
etc.
First column: A005563; T(n, 1) = A005563(n).
Main diagonal: T(n, n) - n = n^2+1 = A002522(n).
The first upper diagonal is T(n, n+1) = n.
Consider TT(n, k) = k*T(n, k) for k = 1 to n:
3,
8, 7,
15, 14, 13,
24, 23, 22, 21,
etc.
Row sums: 3, 8+7, ... , are the positive terms of A059270; that is A059270(n).

			Irregular triangle:
3,   1,
8,   7,  2,  1,
15,  7, 13,  3,  1,  1,
24, 23, 22, 21,  4,  3, 2, 1,
35, 17, 11,  8, 31,  5, 2, 1, 1, 1
48, 47, 46, 45, 44, 43, 6, 5, 4, 3, 2, 1
etc.
Second half part by row: A112543.

Jean-François Alcover, Roots of J(2n-1,z) lie close to two concentric circles (example)

Cf. A002260, A002378, A002522, A005563, A059270, A112543, A122197, A004736.

row[n_] := CoefficientList[Sum[(((n + 1)^2 - k + (n + 1 - k)*z^n))*z^(k - 1)/k, {k, n}], z]; Table[row[n] // Numerator, {n, 1, 9}] // Flatten (* Jean-François Alcover, Apr 07 2016 *)

A290168 If n is even then a(n) = n^2(n+2)/8, otherwise a(n) = (n-1)n*(n+1)/8.

Original entry on oeis.org

0, 0, 2, 3, 12, 15, 36, 42, 80, 90, 150, 165, 252, 273, 392, 420, 576, 612, 810, 855, 1100, 1155, 1452, 1518, 1872, 1950, 2366, 2457, 2940, 3045, 3600, 3720, 4352, 4488, 5202, 5355, 6156, 6327, 7220, 7410, 8400

Offset: 0

Jean-François Alcover and Paul Curtz, Jul 23 2017

nonn

Bisection of a(n) [0, 2, 12, 36, 80, 150, 252, ...] is A011379.
Bisection [0, 3, 15, 42, 90, 165, 273, ...] is A059270.
Considering s(n) = [0, 0, 0, 0, 1, 1, 3, 3, 6, 6, 10, 10, 15, 15, ...] (triangular numbers repeated - see A008805), a(n) = n*s(n+2) holds.
Considering the first differences of a(n), b(n) = [0, 2, 1 , 9, 3, 21, 6, 38, 10, 60, 15, 87, ...], b(n) shows bisections A000217 and A005476. In addition, b(n) begins like A249264 up to 12th term, and is an alternation of 4 multiples of 3 and 2 not multiples; b(n) is also such that b(2n) + b(2n+1) = A049450(n).
Considering the second differences c(n), c(n) shows bisections A001105(n+1) and -A000384(n+1), c(n) has 3 consecutive terms multiples of 3 alternating with 3 not multiples; in addition, c(2n) + c(2n+1) = A000027(n).
Considering a(n)/c(n) = [0, 0, 1/4, -1/2, 2/3, -1, 9/8, -3/2, 8/5, -2, 25/12, -5/2, ...], it appears that it is A129194(n)/A022998(n+1) and -A026741(n)/A000034(n) alternating.

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

Cf. A000027, A000034, A000217, A000384, A001105, A005476, A008805, A011379, A022998, A026741, A049450, A059270, A129194, A135713, A161680, A249264.

a[n_] := If[EvenQ[n], n^2*(n + 2)/8, (n - 1)*n*(n + 1)/8]; Table[a[n], {n, 0, 40}]

a(n) = if(n%2==0, n^2*(n+2)/8, (n-1)*n*(n+1)/8) \\ Felix Fröhlich, Jul 23 2017

G.f.: x^2*(2 + x + 3*x^2)/((x-1)^4*(x+1)^3).
a(n) = (1/16)*(-1)^n*n*(1 + (-1)^(n+1) + 2*(1 + (-1)^n)*n + 2*(-1)^n*n^2).
Sum_{n>=2} 1/a(n) = 5 + Pi^2/6 - 8*log(2). - Amiram Eldar, Sep 17 2022

A347823 Triangle read by rows: T(n,k) = (n+k+1)*binomial(n,k), 0 <= k <= n.

Original entry on oeis.org

1, 2, 3, 3, 8, 5, 4, 15, 18, 7, 5, 24, 42, 32, 9, 6, 35, 80, 90, 50, 11, 7, 48, 135, 200, 165, 72, 13, 8, 63, 210, 385, 420, 273, 98, 15, 9, 80, 308, 672, 910, 784, 420, 128, 17, 10, 99, 432, 1092, 1764, 1890, 1344, 612, 162, 19, 11, 120, 585, 1680, 3150, 4032, 3570, 2160, 855, 200, 21

Offset: 0

Jules Beauchamp, Jan 23 2022

			Triangle begins:
  1;
  2,  3;
  3,  8,   5;
  4, 15,  18,   7;
  5, 24,  42,  32,   9;
  6, 35,  80,  90,  50,  11;
  7, 48, 135, 200, 165,  72, 13;
  8, 63, 210, 385, 420, 273, 98, 15;
  ...

Andrew Howroyd, Table of n, a(n) for n = 0..1325 (rows 0..50)

Row sums give A053220.
Columns give A000027, A005563, A212343.
Diagonals give A005408, A001105, A059270, A112742.
Cf. A007318, A094727.

T(n,k) = (n+k+1)*binomial(n,k) \\ Andrew Howroyd, Jan 23 2022

T(n,k) = A094727(n+1,k)*A007318(n,k).

Row g.f.: (1 + x)^(n-1)*(1 + n + x + 2*n*x). - Stefano Spezia, Jan 23 2022

A360665 Square array T(n, k) = k((2n-1)*k+1)/2 read by rising antidiagonals.

Original entry on oeis.org

0, 0, 0, 0, 1, -1, 0, 2, 3, -3, 0, 3, 7, 6, -6, 0, 4, 11, 15, 10, -10, 0, 5, 15, 24, 26, 15, -15, 0, 6, 19, 33, 42, 40, 21, -21, 0, 7, 23, 42, 58, 65, 57, 28, -28, 0, 8, 27, 51, 74, 90, 93, 77, 36, -36, 0, 9, 31, 60, 90, 115, 129, 126, 100, 45, -45

Offset: 0

Paul Curtz, Mar 17 2023

			By rows:
   0,   0,  -1,  -3,  -6,  -10,  -15,  -21,  -28, ...   = -A161680
   0,   1,   3,   6,  10,   15,   21,   28,   36, ...   =  A000217
   0,   2,   7,  15,  26,   40,   57,   77,  100, ...   =  A005449
   0,   3,  11,  24,  42,   65,   93,  126,  164, ...   =  A005475
   0,   4,  15,  33,  58,   90,  129,  175,  228, ...   =  A022265
   0,   5,  19,  42,  74,  115,  165,  224,  292, ...   =  A022267
   0,   6,  23,  51,  90,  140,  201,  273,  356, ...   =  A022269
   ... .

Cf. Antidiagonal sums: A034827(n+1).
Cf. A000217, A000290, A005449, A005475, A022265.
Cf. A022267, A022269, A059270, A081436, A101986.
Cf. A161680, A162254, A179297, A360962, A361226.

T[n_, k_] := ((2*n - 1)*k^2 + k)/2; Table[T[n - k, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Amiram Eldar, Mar 31 2023 *)

T(n, k) = ((2*n-1)*k^2+k)/2 \\ Thomas Scheuerle, Mar 17 2023

T(n,k) = T(n,k-1)+k^2.
T(n,n) = A081436(n-1).
T(n,n+1) = A059270(n).
T(n,n+4) = -3*A179297(n+4).
T(n+3,n) = A162254(n).
T(n+5,n) = 3*A101986(n).
From Stefano Spezia, Mar 31 2023: (Start)
O.g.f.: (x*y - y^2 + 2*x*y^2)/((1 - x)^2*(1 - y)^3).
E.g.f.: exp(x+y)*y*(2*x - y + 2*x*y)/2. (End)

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A250304 Four-column array read by rows: T(n,k) = the coefficient of x^k in the expanded polynomial x^3 + (x+1)^3 + ... + (x+n-1)^3, for 0 <= k <= 3.

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A270861 Irregular triangle read by rows: numerators of the coefficients of polynomials J(2n-1,z) = Sum_(k=1,2, .. n) ((n+1)^2 - k + (n+1-k)*z^n)*z^(k-1)/k.

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A290168 If n is even then a(n) = n^2*(n+2)/8, otherwise a(n) = (n-1)*n*(n+1)/8.

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A347823 Triangle read by rows: T(n,k) = (n+k+1)*binomial(n,k), 0 <= k <= n.

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A360665 Square array T(n, k) = k*((2*n-1)*k+1)/2 read by rising antidiagonals.

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A270861 Irregular triangle read by rows: numerators of the coefficients of polynomials J(2n-1,z) = Sum_(k=1,2, .. n) ((n+1)^2 - k + (n+1-k)z^n)z^(k-1)/k.

A290168 If n is even then a(n) = n^2(n+2)/8, otherwise a(n) = (n-1)n*(n+1)/8.

A360665 Square array T(n, k) = k((2n-1)*k+1)/2 read by rising antidiagonals.