Stefano Maruelli - cp's OEIS frontend

A281999 Half of the height of the right trapezoidal gnomon (of the derivative of Y=X^5).

1, 30, 181, 600, 1501, 3150, 5881, 10080, 16201, 24750, 36301, 51480, 70981, 95550, 126001, 163200, 208081, 261630, 324901, 399000, 485101, 584430, 698281, 828000, 975001, 1140750, 1326781, 1534680, 1766101, 2022750, 2306401, 2618880, 2962081, 3337950, 3748501, 4195800

Offset: 1

Stefano Maruelli, Feb 05 2017

The curves Y = X^m are characterized by the fact that the first derivative Y'= m*X^(m-1) (and all the following derivatives) are squarable in the integers by rectangular columns called gnomons with base=1 and height M_m = X^m - (X-1)^m. Calling Y' = X^m - (X-1)^m the first "integer" derivative, considering the case m=5, {a(n)} represents the values of half of the maximum (right) height of the trapezoidal gnomons. The formula is: a(n) = (n^5 - (n-1)^5) - a(n-1). The broken line given by joining the points (n; 2*a(n)); define a series of trapezoidal areas (gnomons) that have the same area below the curve Y'=5*X^4. It means that the recursive sum of the trapezoidal gnomon's area, (a(n) + a(n-1))*1, from 1 to n, gives n^5.

The general formula, changing the exponent for all the Y = X^m curves, gives infinitely many new sequences: b(m,k) = m^k - (m-1)^k - b(m-1,k). The same can be done for all the following derivatives. For the smallest exponents k of Y = X^k the sequences are known: for k=3 the sequence is A032528, for k=4 the sequence is A007588, and k=5 corresponds to this sequence.

			For n=2, a(2) = (2^5 - 1^5) - (1) = 30.

Colin Barker, Table of n, a(n) for n = 1..1000
Anwar Al Ghabra, K. Gopala Krishna, Patrick Labelle, and Vasilisa Shramchenko, Enumeration of multi-rooted plane trees, arXiv:2301.09765 [math.CO], 2023.
Stefano Maruelli, Trapezoidal gnomon roof, case n=5.
Index entries for linear recurrences with constant coefficients, signature (4,-5,0,5,-4,1).

LinearRecurrence[{4,-5,0,5,-4,1},{1,30,181,600,1501,3150},40] (* Harvey P. Dale, May 03 2024 *)

Vec(x*(1 + 26*x + 66*x^2 + 26*x^3 + x^4)/((1 + x)*(1 - x)^5) + O(x^30)) \\ Colin Barker, Feb 27 2017

G.f.: x*(1 + 26*x + 66*x^2 + 26*x^3 + x^4)/((1 + x)*(1 - x)^5).
a(n) = (5*(n^2 - 1)*n^2 - (-1)^n + 1)/2.
a(n) = (n^5-(n-1)^5) - a(n-1).
a(n) = 4*a(n-1) - 5*a(n-2) + 5*a(n-4) - 4*a(n-5) + a(n-6) for n>6. - Colin Barker, Feb 27 2017

A276158 Triangle read by rows: T(n,k) = 6k(n + 1 - k) for 0 < k <= n; for k = 0, T(n,0) = n + 1.

Original entry on oeis.org

1, 2, 6, 3, 12, 12, 4, 18, 24, 18, 5, 24, 36, 36, 24, 6, 30, 48, 54, 48, 30, 7, 36, 60, 72, 72, 60, 36, 8, 42, 72, 90, 96, 90, 72, 42, 9, 48, 84, 108, 120, 120, 108, 84, 48, 10, 54, 96, 126, 144, 150, 144, 126, 96, 54

Offset: 0

Stefano Maruelli, Aug 22 2016

The row sums of the triangle provide the positive terms of A000578.
Similar triangles can be generated by the formula P(n,k,m) = (Q(k+1,m)-Q(k,m))*(n+1-k), where Q(i,r) = i^r-(i-1)^r, 0 < k <= n, and P(n,0,m) = n+1. T(n,k) is the case m=3, that is T(n,k) = P(n,k,3).
T(9,k) for 0 <= k <= 9 provides the indegrees of the 10 non-leaf nodes of the network graph of the Kaprekar Process on 3 digits when the nodes are listed in numerical order.  Namely, nodes 000, 099, 198, 297, 396, 495, 594, 693, 792, and 891 have indegrees 10, 54, 96, 126, 144, 150, 144, 126, 96, 54, respectively. Result derived empirically. See "Kaprekar Network Graph for 3 Digits". - Norman Whitehead, May 16 2022

			Triangle starts:
----------------------------------------------
n \ k |  0   1    2    3    4    5    6    7
----------------------------------------------
0     |  1;
1     |  2,  6;
2     |  3, 12,  12;
3     |  4, 18,  24,  18;
4     |  5, 24,  36,  36,  24;
5     |  6, 30,  48,  54,  48,  30;
6     |  7, 36,  60,  72,  72,  60,  36;
7     |  8, 42,  72,  90,  96,  90,  72,  42;
...

Norman Whitehead, Kaprekar Network Graph for 3 Digits
Norman Whitehead, Kaprekar Network Graph Node Count Verification

Cf. A000578, A008458, A276189.

[IsZero(k) select n+1 else 6*k*(n+1-k): k in [0..n], n in [0..10]]; // Bruno Berselli, Aug 31 2016

/* As triangle (see the second comment): */ m:=3; Q:=func; P:=func; [[P(n, k, m): k in [0..n]]: n in [0..10]]; // Bruno Berselli, Aug 31 2016

T:= (n, k) -> `if`(k=0, n+1, 6*k*(n+1-k)):
seq(seq(T(n, k), k=0..n), n=0..30); # Robert Israel, Aug 31 2016

Table[If[k == 0, n + 1, 6 k (n + 1 - k)], {n, 0, 10}, {k, 0, n}] // Flatten (* Michael De Vlieger, Aug 25 2016 *)

T(n, k) = if (k==0, n+1, 6*k*(n+1-k));
tabl(nn) = for (n=0, nn, for (k=0, n, print1(T(n, k), ", ")); print); \\ Michel Marcus, Aug 25 2016

Sum_{k=0..n} T(n,k) = T(n,0)^3 = A000578(n+1).
G.f. as triangle: (1+4*x*y + x^2*y^2)/((1-x)^2*(1-x*y)^2). - Robert Israel, Aug 31 2016
T(n,n-h) = (h+1)*A008458(n-h) for 0 <= h <= n. Therefore, the main diagonal of the triangle is A008458. - Bruno Berselli, Aug 31 2016

Corrected and rewritten by Bruno Berselli, Sep 01 2016

A276189 Triangle read by rows: T(n,k) = 2(6k^2 + 1)*(n + 1 - k) for 0 < k <= n; for k = 0, T(n,0) = n + 1.

Original entry on oeis.org

1, 2, 14, 3, 28, 50, 4, 42, 100, 110, 5, 56, 150, 220, 194, 6, 70, 200, 330, 388, 302, 7, 84, 250, 440, 582, 604, 434, 8, 98, 300, 550, 776, 906, 868, 590, 9, 112, 350, 660, 970, 1208, 1302, 1180, 770, 10, 126, 400, 770, 1164, 1510, 1736, 1770, 1540, 974, 11, 140, 450, 880, 1358, 1812, 2170, 2360, 2310, 1948, 1202

Offset: 0

Stefano Maruelli, Aug 24 2016

The row sums of the triangle provide the positive terms of A000583.

Similar triangles can be generated by the formula P(n,k,m) = (Q(k+1,m)-Q(k,m))*(n+1-k), where Q(i,r) = i^r-(i-1)^r, 0 < k <= n, and P(n,0,m) = n+1. T(n,k) is the case m=4, that is T(n,k) = P(n,k,4).

			Triangle starts:
----------------------------------------------
n \ k |  0   1    2    3    4    5    6    7
----------------------------------------------
0     |  1;
1     |  2, 14;
2     |  3, 28,  50;
3     |  4, 42, 100, 110;
4     |  5, 56, 150, 220, 194;
5     |  6, 70, 200, 330, 388, 302;
6     |  7, 84, 250, 440, 582, 604, 434;
7     |  8, 98, 300, 550, 776, 906, 868, 590;
...

Michael De Vlieger, Table of n, a(n) for n = 0..11475 (rows 0 <= n <= 150, flattened)

Cf. A000583, A005914, A276158.

[IsZero(k) select n+1 else 2*(6*k^2+1)*(n+1-k): k in [0..n], n in [0..10]];

/* As triangle (see the second comment): */ m:=4; Q:=func; P:=func; [[P(n,k,m): k in [0..n]]: n in [0..10]];

Table[If[k == 0, n + 1, 2 (6 k^2 + 1) (n + 1 - k)], {n, 0, 9}, {k, 0, n}] // Flatten (* Michael De Vlieger, Aug 29 2016 *)

T(n,n-h) = (h+1)*A005914(n-h) for 0 <= h <= n. Therefore, the main diagonal of the triangle is A005914.

Sum_{k=0..n} T(n,k) = T(n,0)^4 = A000583(n+1).

Corrected, rewritten and extended by Bruno Berselli, Aug 31 2016

a(40) ff. corrected by Georg Fischer, Nov 08 2021

A241496 Expansion of (1 + 4*x + x^2) / (1 - x^2)^3.

Original entry on oeis.org

1, 4, 4, 12, 9, 24, 16, 40, 25, 60, 36, 84, 49, 112, 64, 144, 81, 180, 100, 220, 121, 264, 144, 312, 169, 364, 196, 420, 225, 480, 256, 544, 289, 612, 324, 684, 361, 760, 400, 840, 441, 924, 484, 1012, 529, 1104, 576, 1200, 625, 1300, 676, 1404, 729, 1512

Offset: 0

Stefano Maruelli, Apr 24 2014

Column 3 of the table in A229834.

Index entries for linear recurrences with constant coefficients, signature (0,3,0,-3,0,1).

Cf. A001082, A229834.

Cf. A171920: a(2k) = A000290(k+1); A046092: a(2k+1)= A046092(k+1).

Table[(3 n (n + 4) - (-1)^n (n (n + 4) + 2) + 10)/8, {n, 0, 60}] (* Bruno Berselli, Apr 24 2014 *)
LinearRecurrence[{0,3,0,-3,0,1},{1,4,4,12,9,24},60] (* Harvey P. Dale, Nov 25 2016 *)

G.f.: (1 + 4*x + x^2) / (1 - x^2)^3. [Bruno Berselli, Apr 24 2014]

a(n) = a(-n-4) = 1 + ( 3*n*(n + 4) + 2 - (-1)^n*(n*(n + 4) + 2) )/8. [Bruno Berselli, Apr 24 2014]

Edited by Bruno Berselli, Apr 24 2014

A229834 Expansion of (1+4x+x^2) / ((1-x)^3(1+x)^4).

Original entry on oeis.org

1, 3, 1, 11, -2, 26, -10, 50, -25, 85, -49, 133, -84, 196, -132, 276, -195, 375, -275, 495, -374, 638, -494, 806, -637, 1001, -805, 1225, -1000, 1480, -1224, 1768, -1479, 2091, -1767, 2451, -2090, 2850, -2450, 3290, -2849, 3773, -3289, 4301, -3772, 4876, -4300, 5500, -4875, 6175, -5499, 6903, -6174, 7686, -6902

Offset: 0

Stefano Maruelli, Dec 19 2013

The sequence can be generated in the following way:
---------------------------       --------------------------
      [0]  [1] [2] [3]  [4]  ...             [i]
---------------------------       --------------------------
[0]    1,   1,  1,  1,   1,  ...  t(0,i) =    1
[1]    7,   6,  5,  4,   3,  ...  t(1,i) = t(1,i-1) - t(0,i)
[2]   19,  13,  8,  4,   1,  ...  t(2,i) = t(2,i-1) - t(1,i)
[3]   37,  24, 16, 12,  11,  ...  t(3,i) = t(3,i-1) - t(2,i)
[4]   61,  37, 21,  9,  -2,  ...  t(4,i) = t(4,i-1) - t(3,i)
[5]   91,  54, 33, 24,  26,  ...             etc.
[6]  127,  73, 40, 16, -10,  ...
[7]  169,  96, 56, 40,  50,  ...
[8]  217, 121, 65, 25, -25,  ...
[9]  271, 150, 85, 60,  85,  ...
...
Column 0 is A003215;
column 1 is A032528;
column 2 is A001082;
column 3 is A241496;
column 4 is this sequence.
The third differences are 16, -35, 64, -105, 160, ..., a signed variant of A077415. - R. J. Mathar, Apr 18 2014

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

Cf. A077415; A058373: a(2k) = -A058373(k); A051925: a(2k+1) = A051925(k+2).

Columns of the table in Comments section: A001082, A003215, A032528.

Table[1 + n (n + 5) (9 - (2 n + 5) (-1)^n)/48, {n, 0, 60}] (* Bruno Berselli, Apr 22 2014 *)
CoefficientList[Series[(1+4x+x^2)/((1-x)^3(1+x)^4),{x,0,60}],x] (* or *) LinearRecurrence[{-1,3,3,-3,-3,1,1},{1,3,1,11,-2,26,-10},60] (* Harvey P. Dale, Jan 27 2022 *)

G.f.: (1 + 4*x + x^2) / ((1 - x)^3*(1 + x)^4). - R. J. Mathar, Apr 18 2014

a(n) = a(-n-5) = 1 + n*(n + 5)*(9 - (2*n + 5)*(-1)^n)/48. [Bruno Berselli, Apr 22 2014]

A135989 a(n) = 6n + 3 + 90floor((6*n+3)/10).

Original entry on oeis.org

3, 9, 105, 201, 207, 303, 309, 405, 501, 507, 603, 609, 705, 801, 807, 903, 909, 1005, 1101, 1107, 1203, 1209, 1305, 1401, 1407, 1503, 1509, 1605, 1701, 1707, 1803, 1809, 1905, 2001, 2007, 2103, 2109, 2205, 2301, 2307, 2403, 2409, 2505, 2601, 2607, 2703

Offset: 0

Stefano Maruelli and Paola Paoloni (robotec(AT)netsurf.it), Mar 02 2008

All terms are divisible by 3.

a(n) = a(n-1)+a(n-5)-a(n-6). - Colin Barker, Feb 02 2013

G.f.: 3*(31*x^5 +2*x^4 +32*x^3 +32*x^2 +2*x +1) / ((x -1)^2*(x^4 +x^3 +x^2 +x +1)). - Colin Barker, Feb 02 2013

Edited by Charles R Greathouse IV, Nov 03 2009

cp's OEIS Frontend

User: Stefano Maruelli

A281999 Half of the height of the right trapezoidal gnomon (of the derivative of Y=X^5).

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A276158 Triangle read by rows: T(n,k) = 6*k*(n + 1 - k) for 0 < k <= n; for k = 0, T(n,0) = n + 1.

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A276189 Triangle read by rows: T(n,k) = 2*(6*k^2 + 1)*(n + 1 - k) for 0 < k <= n; for k = 0, T(n,0) = n + 1.

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A241496 Expansion of (1 + 4*x + x^2) / (1 - x^2)^3.

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A229834 Expansion of (1+4*x+x^2) / ((1-x)^3*(1+x)^4).

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A135989 a(n) = 6*n + 3 + 90*floor((6*n+3)/10).

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A276158 Triangle read by rows: T(n,k) = 6k(n + 1 - k) for 0 < k <= n; for k = 0, T(n,0) = n + 1.

A276189 Triangle read by rows: T(n,k) = 2(6k^2 + 1)*(n + 1 - k) for 0 < k <= n; for k = 0, T(n,0) = n + 1.

A229834 Expansion of (1+4x+x^2) / ((1-x)^3(1+x)^4).

A135989 a(n) = 6n + 3 + 90floor((6*n+3)/10).