A005917 -id:A005917 - cp's OEIS frontend

A254870 Seventh partial sums of fourth powers (A000583).

1, 23, 221, 1355, 6239, 23465, 75803, 217373, 566150, 1361802, 3063502, 6508450, 13159666, 25481470, 47493274, 85567222, 149553199, 254336185, 421956275, 684451365, 1087616985, 1695917535, 2598828765, 3918943275, 5822229660, 8530902276, 12339433068

Offset: 1

Luciano Ancora, Feb 17 2015

			Second differences:   2, 14,  50,  110,  194,   302, ...   A120328(2k+1)
First differences:    1, 15,  65,  175,  369,   671, ...   A005917
--------------------------------------------------------------------------
The fourth powers:    1, 16,  81,  256,  625,  1296, ...   A000583
--------------------------------------------------------------------------
First partial sums:   1, 17,  98,  354,  979,  2275, ...   A000538
Second partial sums:  1, 18, 116,  470, 1449,  3724, ...   A101089
Third partial sums:   1, 19, 135,  605, 2054,  5778, ...   A101090
Fourth partial sums:  1, 20, 155,  760, 2814,  8592, ...   A101091
Fifth partial sums:   1, 21, 176,  936, 3750, 12342, ...   A254681
Sixth partial sums:   1, 22, 198, 1134, 4884, 17226, ...   A254470
Seventh partial sums: 1, 23, 221, 1355, 6239, 23465, ...   (this sequence)

Luciano Ancora, Table of n, a(n) for n = 1..1000
Luciano Ancora, Partial sums of m-th powers with Faulhaber polynomials
Luciano Ancora, Pascal’s triangle and recurrence relations for partial sums of m-th powers
Index entries for linear recurrences with constant coefficients, signature (12,-66,220,-495,792,-924,792,-495,220,-66,12,-1).

Cf. A000538, A000583, A005917, A101089, A101090, A101091, A254681, A254470, A254869, A254871, A254872.

[n*(1+n)*(2+n)*(3+n)*(4+n)*(5+n)*(6+n)*(7+n)*(7+2*n)*(7 +42*n+6*n^2)/19958400: n in [1..30]]; // Vincenzo Librandi, Feb 19 2015

Table[n (1 + n) (2 + n) (3 + n) (4 + n) (5 + n) (6 + n) (7 + n) (7 + 2 n)((7 + 42 n + 6 n^2)/19958400), {n, 24}] (* or *)
CoefficientList[Series[(1 + 11 x + 11 x^2 + x^3)/(- 1 + x)^12, {x, 0, 23}], x]

vector(50,n,n*(1 + n)*(2 + n)*(3 + n)*(4 + n)*(5 + n)*(6 + n)*(7 + n)*(7 + 2*n)*(7 + 42*n + 6*n^2)/19958400) \\ Derek Orr, Feb 19 2015

G.f.: (x + 11*x^2 + 11*x^3 + x^4)/(- 1 + x)^12.
a(n) = n*(1 + n)*(2 + n)*(3 + n)*(4 + n)*(5 + n)*(6 + n)*(7 + n)*(7 + 2*n)*(7 + 42*n + 6*n^2)/19958400.
a(n) = 7*a(n-1) - 21*a(n-2) + 35*a(n-3) - 35*a(n-4) + 21*a(n-5) - 7*a(n-6) + a(n-7) + n^4.

A193218 Number of vertices in truncated tetrahedron with faces that are centered polygons.

Original entry on oeis.org

1, 21, 95, 259, 549, 1001, 1651, 2535, 3689, 5149, 6951, 9131, 11725, 14769, 18299, 22351, 26961, 32165, 37999, 44499, 51701, 59641, 68355, 77879, 88249, 99501, 111671, 124795, 138909, 154049, 170251, 187551, 205985, 225589, 246399, 268451, 291781, 316425

Offset: 1

Craig Ferguson, Jul 18 2011

The sequence starts with a central vertex and expands outward with (n-1) centered polygonal pyramids producing a truncated tetrahedron. Each iteration requires the addition of (n-2) edges and (n-1) vertices to complete the centered polygon in each face. For centered triangles see A005448 and centered hexagons A003215.
This sequence is the 18th in the series (1/12)*t*(2*n^3-3*n^2+n)+2*n-1 for t = 2, 4, 6, ... gives A049480, A005894, A063488, A001845, A063489, A005898, A063490, A057813, A063491, A005902, A063492, A005917, A063493, A063494, A063495, A063496 and t = 36. While adjusting for offsets, the beginning sequence A049480 is generated by adding the square pyramidal numbers A000330 to the odd numbers A005408 and each subsequent sequence is found by adding another set of square pyramidals A000330. (T/2) * A000330(n) + A005408(n). At 30 * A000330 + A005408 = centered dodecahedral numbers, 36 * A000330 + A005408 = A193228 truncated octahedron and 90 * A000330 + A005408 = A193248 = truncated icosahedron and dodecahedron. All five of the "Centered Platonic Solids" numbers sequences are in this series of sequences. Also 4 out of five of the "truncated" platonic solid number sequences are in this series. - Bruce J. Nicholson, Jul 06 2018
It would be good to have a detailed description of how the sequence is constructed. Maybe in the Examples section? - N. J. A. Sloane, Sep 07 2018

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
OEIS, (Centered_polygons) pyramidal numbers
Wikipedia, Tetrahedral number
Wikipedia, Triangular number
Wikipedia, Centered polygonal number
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A260810 (partial sums).

[6*n^3-9*n^2+5*n-1: n in [1..40]]; // Vincenzo Librandi, Aug 30 2011

Table[6 n^3 - 9 n^2 + 5 n - 1, {n, 35}] (* Alonso del Arte, Jul 18 2011 *)
CoefficientList[Series[(1+x)*(x^2+16*x+1)/(1-x)^4, {x, 0, 50}], x] (* Stefano Spezia, Sep 04 2018 *)

a(n) = 6*n^3 - 9*n^2 + 5*n - 1.
G.f.: x*(1+x)*(x^2+16*x+1) / (1-x)^4. - R. J. Mathar, Aug 26 2011
a(n) = 18 * A000330(n-1) + A005408(n-1) = A063496(n) + A006331(n-1). - Bruce J. Nicholson, Jul 06 2018

A256141 Square array read by antidiagonals upwards: T(n,k), n>=0, k>=0, in which row n lists the partial sums of the n-th row of the square array of A256140.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 5, 4, 1, 1, 5, 7, 9, 5, 1, 1, 6, 9, 16, 11, 6, 1, 1, 7, 11, 25, 19, 15, 7, 1, 1, 8, 13, 36, 29, 28, 19, 8, 1, 1, 9, 15, 49, 41, 45, 37, 27, 9, 1, 1, 10, 17, 64, 55, 66, 61, 64, 29, 10, 1, 1, 11, 19, 81, 71, 91, 91, 125, 67, 33, 11, 1, 1, 12, 21, 100, 89, 120, 127, 216, 129, 76, 37, 12, 1

Offset: 0

Omar E. Pol, Mar 16 2015

Questions:
Is also A130667 a row of this square array?
Is also A116522 a row of this square array?
Is also A116526 a row of this square array?
Is also A116525 a row of this square array?
Is also A116524 a row of this square array?

			The corner of the square array with the first 15 terms of the first 12 rows looks like this:
--------------------------------------------------------------------------
A000012: 1, 1, 1,  1,  1,  1,  1,   1,   1,   1,   1,   1,   1,   1,   1
A000027: 1, 2, 3,  4,  5,  6,  7,   8,   9,  10,  11,  12,  13,  14,  15
A006046: 1, 3, 5,  9, 11, 15, 19,  27,  29,  33,  37,  45,  49,  57,  65
A130665: 1, 4, 7, 16, 19, 28, 37,  64,  67,  76,  85, 112, 121, 148, 175
A116520: 1, 5, 9, 25, 29, 45, 61, 125, 129, 145, 161, 225, 241, 305, 369
A130667? 1, 6,11, 36, 41, 66, 91, 216, 221, 246, 271, 396, 421, 546, 671
A116522? 1, 7,13, 49, 55, 91,127, 343, 349, 385, 421, 637, 673, 889,1105
A161342: 1, 8,15, 64, 71,120,169, 512, 519, 568, 617, 960,1009,1352,1695
A116526? 1, 9,17, 81, 89,153,217, 729, 737, 801, 865,1377,1441,1953,2465
.......: 1,10,19,100,109,190,271,1000,1009,1090,1171,1900,1981,2710,3439
A116525? 1,11,21,121,131,231,331,1331,1341,1441,1541,2541,2641,3641,4641
.......: 1,12,23,144,155,276,397,1728,1739,1860,1981,3312,3422,4764,6095

Index entries for sequences related to cellular automata

Cf. A000120, A255740, A255741, A256140.
First five rows are A000012, A000027, A006046, A130665, A116520. Row 7 is A161342.
First eight columns are A000012, A000027, A005408, A000290, A028387, A000384, A003215, A000578. Column 9 is A081437. Column 11 is A015237. Columns 13-15 are A005915, A005917, A000583.

A346264 a(n) is the number of distinct possible tilings of type 2 for squares with side = A344332(n) and that can be tiled with squares of two different sizes so that the numbers of large or small squares are equal.

Original entry on oeis.org

1, 2, 2, 3, 1, 2, 4, 2, 4, 2, 3, 4, 2, 1, 6, 4, 4, 4, 5, 2, 3, 6, 2, 6, 4, 2, 4, 2, 2, 8, 1, 3, 8, 4, 6, 2, 8, 2, 2, 6, 4, 4, 4, 6, 9, 2, 4, 7, 8, 2, 8, 2, 4, 6, 1, 6, 4, 3, 2, 2, 10, 3, 2, 6, 4, 12, 2, 8, 4, 8, 2, 4, 4, 2, 2, 12, 4, 2, 4, 6, 7, 8, 8, 2, 6, 4, 5, 12, 2, 12, 2, 3, 3, 4

Offset: 1

Bernard Schott, Aug 09 2021

nonn

Every side of square of type 2 in A344332 is also the side of an elementary square of type 2. An elementary square of type 2 is the smallest square that can be tiled with squares of two different sides a < b satisfying a^2+b^2 = c^2 and so that the numbers of small and large squares are equal.
Some notation: s = side of the tiled square, a = side of small squares, b = side of large squares, and z = number of small squares = number of large squares.
a(n) = 1 iff A344332(n) is a term of A005917 that is not a multiple of another term of A005917 (15, 65, 175, 369, 671, 2465, ...).
The first side that is a multiple of two primitive sides is 195 = 13*15 = 3*65 (see 3rd example).

			---> For a(1), A344332(1) = 15, then, with the formula, we get a(1) = tau(A344332(1)/A005917(2)) = tau(15/15) = tau(1) = 1, and the corresponding tiling of this smallest square 15 X 15 of type 2 consists of z = 9 squares whose sides (a,b) = (3,4) (see below).
            ________ ________ ________ ______
           |        |        |        |      |
           |        |        |        |      |
           |        |        |        |______|
           |_______ |________|________|      |
           |        |        |        |      |
           |        |        |        |______|
           |        |        |        |      |
           |________|________|________|      |
           |        |        |        |______|
           |        |        |        |      |
           |        |        |        |      |
           |_____ __|___ ____|_ ______|______|
           |     |      |      |      |      |
           |     |      |      |      |      |
           |_____|______|______|______|______|
                       a(1) = 1
---> For a(2), A344332(2) = 30, then, with the formula, we get a(2) = tau(A344332(2)/A005917(2)) = tau(30/15) = tau(2) = 2, and these 2 distinct tilings are:
1) 30 = 2*A344332(1) = 2*15, z(30) = 2^2 * z(15) = 4*9 = 36 and square 30 X 30 can be tiled with z = 36 squares whose sides (a,b) = (3,4), that is 4 copies of the elementary and primitive square 15 X 15 (as above). Also,
2) 30 = 1*A344332(2) = 1*30, z(30) = 1^2 * z(15) = 1*9 = 9 and the elementary square 30 X 30 can be tiled with z = 9 squares whose sides (a,b) = (6,8) (see link with corresponding drawings).
---> For a(16), A344332(16) = 195, then, with the formula, we get a(16) = tau(A344333(16)/A005917(2)) + tau(A344333(16)/A005917(3)) = tau(195/15) + tau(195/65) = tau(13) + tau(3) = 2+2 = 4, and these 4 distinct tilings are:
1) 195 = 13*A344332(1) = 13*15, z_1(195) = 13^2 * z(15) = 169*9 = 1521 and square 195 X 195 can be tiled with z = 1521 squares whose sides (a,b) = (3,4), that is 169 copies of the elementary and primitive square 15 X 15, as above;
2) 195 = 1*A344332(16) = 1*195, z_2(195) = 1^2 * z(195) = 1*9 = 9 and the elementary square 195 X 195 can be tiled with z = 9 squares whose sides (a,b) = (39,52);
3) 195 = 3*A344332(5) = 3*65, z_3(195) = 3^2 * z(65) = 9*25 = 225 [z(65) = A346263(5) = T(5,1) = 25] and square 195 X 195 can be tiled with z = 225 squares whose sides (a,b) = (5,12), that is 9 copies of the elementary and primitive square 65 X 65;
4) 195 = 1*A344332(16) = 1*195, z_4(195) = 1^2 * z(195) = 1*25 = 25 and the elementary square 195 X 195 can be tiled with z = 25 squares whose sides (a,b) = (15,36).

Ivan Yashchenko, Invitation to a Mathematical Festival, pp. 10 and 102, MSRI, Mathematical Circles Library, 2013.

Bernard Schott, Tilings of type 2 for square 30 X 30 with a(2) = 2.

Cf. A005917, A016754, A344332, A346263.

\\ using isok2 from A344332; isok7 is for A005917
isok7(k) = my(kk= sqrtnint(k\4, 3)+2); vecsearch(vector(kk, i, (i+1)^4 - i^4), k);
sd(x) = sumdiv(x, d, if (isok7(d), numdiv(x/d)));
lista(nn) = my(v2 = select(isok2, [1..nn])); apply(sd, v2);

a(n) = Sum_{(k>=2) & (A005917(k)|A344332(n))} tau(A344332(n)/A005917(k)).

A104012 Indices of centered dodecahedral numbers (A005904) which are semiprimes (A001358).

Original entry on oeis.org

1, 2, 3, 5, 6, 11, 14, 15, 21, 26, 30, 35, 36, 44, 54, 63, 69, 74, 81, 114, 128, 131, 135, 138, 153, 165, 168, 191, 194, 209, 216, 224, 228, 231, 239, 261, 270, 303, 315, 321, 323, 326, 330, 336, 345, 363, 380, 384, 398, 404, 410, 411, 414, 429, 440, 443, 455, 468, 470

Offset: 1

Jonathan Vos Post, Feb 24 2005

Because the cubic polynomial for centered dodecahedral numbers factors into n time an irreducible quadratic, the dodecahedral numbers can never be prime, but can be semiprime iff (2*n+1) is prime and (5*n^2+5*n+1) is prime. Centered dodecahedral numbers (A005904) are not to be confused with dodecahedral numbers (A006566) = n(3n-1)(3n-2)/2, nor with rhombic dodecahedral numbers (A005917).

Intersection of A005097 and A090563. - Michel Marcus, Apr 30 2016

			a(1) = 1 because A005904(1) = 33 = 3 * 11, which is semiprime.
a(2) = 2 because A005904(2) = 155 = 5 * 31, which is semiprime.
a(3) = 3 because A005904(3) = 427 = 7 * 61, which is semiprime.
a(4) = 5 because A005904(5) = 1661 = 11 * 151.
194 is in this sequence because A005904(194) = 73579739 = 389 * 189151, which is semiprime.

Zak Seidov, Table of n, a(n) for n = 1..1000
B. K. Teo and N. J. A. Sloane, Magic numbers in polygonal and polyhedral clusters, Inorgan. Chem. 24 (1985),4545-4558.

Cf. A001222, A001358, A005904, A090563, A104011.

isok(n) = isprime(2*n+1) && isprime(5*n^2+5*n+1); \\ Michel Marcus, Apr 30 2016

n such that A001222(A005904(n)) = 2. n such that Bigomega((2*n+1)*(5*n^2 + 5*n + 1)) is 2. n such that A104011(n) = 2.

A126062 Array read by antidiagonals: see A128195 for details.

Original entry on oeis.org

1, 1, 1, 1, 4, 1, 1, 9, 15, 1, 1, 16, 65, 64, 1, 1, 25, 175, 511, 325, 1, 1, 36, 369, 2020, 4743, 1956, 1, 1, 49, 671, 5629, 27313, 52525, 13699, 1, 1, 64, 1105, 12736, 100045, 440896, 683657, 109600, 1, 1, 81, 1695, 25099, 280581, 2122449, 8390875, 10256775

Offset: 0

N. J. A. Sloane, Feb 28 2007

			Array begins:
[0]  1,  1,   1,     1,      1,       1,         1,          1,            1
[1]  1,  4,  15,    64,    325,    1956,     13699,     109600,       986409
[2]  1,  9,  65,   511,   4743,   52525,    683657,   10256775,    174369527
[3]  1, 16, 175,  2020,  27313,  440896,   8390875,  184647364,   4616348125
[4]  1, 25, 369,  5629, 100045, 2122449,  53163625, 1542220261,  50895431301
[5]  1, 36, 671, 12736, 280581, 7376356, 229151411, 8252263296, 338358810761

P. Luschny, Variants of Variations.

The second row counts the variations of n distinct objects A007526.

The second column is sequence A000290. The third column is sequence A005917.

A126062 := proc(k,n) if n = 0 then 1 ; else (n*k+1)*(A126062(k,n-1)+k^n) ; fi ; end: for diag from 0 to 10 do for k from diag to 0 by -1 do n := diag-k ; printf("%d, ",A126062(k,n)) ; od ; od ; # R. J. Mathar, May 18 2007

a[, 0] = 1; a[k, n_] := a[k, n] = (n*k+1)*(a[k, n-1]+k^n); Table[a[k-n, n], {k, 0, 10}, {n, 0, k}] // Flatten (* Jean-François Alcover, Jan 08 2014 *)

T(k, n) = (n*k+1)*(T(k, n-1) + k^n), T(k, 0) = 1. - Peter Luschny, Feb 26 2007

More terms from R. J. Mathar, May 18 2007

A272298 a(n) = n^4 + 324.

Original entry on oeis.org

324, 325, 340, 405, 580, 949, 1620, 2725, 4420, 6885, 10324, 14965, 21060, 28885, 38740, 50949, 65860, 83845, 105300, 130645, 160324, 194805, 234580, 280165, 332100, 390949, 457300, 531765, 614980, 707605, 810324, 923845, 1048900, 1186245, 1336660, 1500949, 1679940, 1874485, 2085460

Offset: 0

Bruno Berselli, Apr 25 2016

This is the case k=3 of Sophie Germain's Identity n^4+(2*k^2)^2 = ((n-k)^2+k^2)*((n+k)^2+k^2).

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

Cf. A005917.
Subsequence of A227855.
Cf. A000583 (k=0), A057781 (k=1), A272297 (k=2).

Magma
```
[n^4+324: n in [0..40]];
```

Table[n^4 + 324, {n, 0, 40}]
LinearRecurrence[{5,-10,10,-5,1},{324,325,340,405,580},40] (* Harvey P. Dale, Jan 20 2021 *)

Maxima
```
makelist(n^4+324, n, 0, 40);
```
PARI
```
vector(40, n, n--; n^4+324)
```
Python
```
[n**4+324 for n in range(40)]
```

for n in range(0, 10**5):print(n**4+324,end=", ") # Soumil Mandal, Apr 30 2016

Sage
```
[n^4+324 for n in (0..40)]
```

O.g.f.: (324 - 1295*x + 1955*x^2 - 1285*x^3 + 325*x^4)/(1 - x)^5. [Corrected by Georg Fischer, May 23 2019]
E.g.f.: (324 + x + 7*x^2 + 6*x^3 + x^4)*exp(x).
a(n) = (n^2 - 18)^2 + (6*n)^2.

A342354 M(n,k) = 2n^2 + 2k + 1 for 0 <= k <= n and M(n,k) = 2k^2 + 4k - 2*n + 1 for 0 <= n <= k; square array M(n,k) read by ascending antidiagonals (n, k >= 0).

Original entry on oeis.org

1, 3, 7, 9, 5, 17, 19, 11, 15, 31, 33, 21, 13, 29, 49, 51, 35, 23, 27, 47, 71, 73, 53, 37, 25, 45, 69, 97, 99, 75, 55, 39, 43, 67, 95, 127, 129, 101, 77, 57, 41, 65, 93, 125, 161, 163, 131, 103, 79, 59, 63, 91, 123, 159, 199, 201, 165, 133, 105, 81, 61, 89, 121, 157, 197, 241, 243, 203, 167, 135, 107, 83, 87, 119, 155, 195, 239, 287

Offset: 0

Petros Hadjicostas, Mar 08 2021

This is a square array defined by J. M. Bergot in A005917 (originally by mistake in A047926). Here is the edited description of the array by this contributor.

Construct an array M with M(0,n) = 2*n^2 + 4*n + 1 = A056220(n+1), M(n,0) = 2*n^2 + 1 = A058331(n) and M(n,n) = 2*n*(n+1) + 1 = A001844(n). Row(n) begins with all the increasing odd numbers from A058331(n) to A001844(n) and column(n) begins with all the decreasing odd numbers from A056220(n+1) to A001844(n). The sum of the terms in row(n) plus those in column(n) minus M(n,n) equals A005917(n+1).

			Square array M(n,k) (n, k >= 0) begins:
   1,  7, 17, 31, 49, 71, 97, 127, ...
   3,  5, 15, 29, 47, 69, 95, 125, ...
   9, 11, 13, 27, 45, 67, 93, 123, ...
  19, 21, 23, 25, 43, 65, 91, 121, ...
  33, 35, 37, 39, 41, 63, 89, 119, ...
  51, 53, 55, 57, 59, 61, 87, 117, ...
  73, 75, 77, 79, 81, 83, 85, 115, ...
  ...
The triangular array T(n,k) = M(n-k,k) (with rows n >= 0 and columns k = 0..n) is obtained by reading array M by ascending antidiagonals:
   1;
   3,  7;
   9,  5, 17;
  19, 11, 15, 31;
  33, 21, 13, 29, 49;
  51, 35, 23, 27, 47, 71;
  73, 53, 37, 25, 45, 69, 97;
  99, 75, 55, 39, 43, 67, 95, 127;
  ...

Cf. A001844, A005917, A047926, A056220, A058331.

Antidiagonal sums are in A342362.

tabl(nn) = {my(M=matrix(nn+1,nn+1)); for(n=1, nn+1, for(k=1, nn+1, M[n,k]=if(k == n, 2*n^2-2*n+1, if(k < n, 2*n^2-4*n+2*k+1, 2*k^2-2*n+1)))); M}

O.g.f. for rectangular M: (x^4*y^4 + 4*x^3*y^4 + 3*x^4*y^2 - 18*x^3*y^3 - x^2*y^4 + 8*x^3*y^2 + 4*x^2*y^3 - 10*x^3*y + 10*x^2*y^2 - 2*x*y^3 + 8*x^2*y + 4*x*y^2 + 3*x^2 - 18*x*y - y^2 + 4*y + 1)/((1 - x)^3*(1 - y)^3*(1 - x*y)^2).

O.g.f. for triangular T: (x^8*y^4 + 4*x^7*y^4 - x^6*y^4 - 18*x^6*y^3 + 3*x^6*y^2 + 4*x^5*y^3 + 8*x^5*y^2 - 2*x^4*y^3 + 10*x^4*y^2 - 10*x^4*y + 4*x^3*y^2 + 8*x^3*y - x^2*y^2 - 18*x^2*y + 3*x^2 + 4*x*y + 1)/((1 - x)^3*(1 - x*y)^3*(1 - x^2*y)^2).

A343237 Triangle T obtained from the array A(n, k) = (k+1)^(n+1) - k^(n+1), n, k >= 0, by reading antidiagonals upwards.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 7, 5, 1, 1, 15, 19, 7, 1, 1, 31, 65, 37, 9, 1, 1, 63, 211, 175, 61, 11, 1, 1, 127, 665, 781, 369, 91, 13, 1, 1, 255, 2059, 3367, 2101, 671, 127, 15, 1, 1, 511, 6305, 14197, 11529, 4651, 1105, 169, 17, 1

Offset: 0

Wolfdieter Lang, May 10 2021

This is the row reversed version of the triangle A047969(n, m). The corresponding array A047969 is a(n, k) = A(k, n) (transposed of array A).

A(n-1, k-1) = k^n - (k-1)^n gives the number of n-digit numbers with digits from K = {1, 2, 3, ..., k} such that any digit from K, say k, appears at least once. Motivated by a comment in A005061 by Enrique Navarrete, the instance k=4 for n >= 1 (the column 3 in array A), and the d = 3 (sub)-diagonal sequence of T for m >= 0.

			The array A begins:
n\k  0  1   2    3     4     5     6      7      8      9 ...
-------------------------------------------------------------
0:   1  1   1    1     1     1     1      1      1      1 ...
1:   1  3   5    7     9    11    13     15     17     19 ...
2:   1  7  19   37    61    91   127    169    217    271 ...
3:   1 15  65  175   369   671  1105   1695   2465   3439 ...
4:   1 31 211  781  2101  4651  9031  15961  26281  40951 ...
5:   1 63 665 3367 11529 31031 70993 144495 269297 468559 ...
...
The triangle T begins:
n\m   0    1     2     3     4     5    6    7   8  9 10 ...
-------------------------------------------------------------
0:    1
1:    1    1
2:    1    3     1
3:    1    7     5     1
4:    1   15    19     7     1
5:    1   31    65    37     9     1
6:    1   63   211   175    61    11    1
7:    1  127   665   781   369    91   13    1
8:    1  255  2059  3367  2101   671  127   15   1
9:    1  511  6305 14197 11529  4651 1105  169  17  1
10:   1 1023 19171 58975 61741 31031 9031 1695 217 19  1
...
Combinatorial interpretation (cf. A005061 by _Enrique Navarrete_)
The three digits numbers with digits from K ={1, 2, 3, 4} having at least one 4 are:
j=1 (one 4): 114, 141, 411; 224, 242, 422; 334, 343, 433; 124, 214, 142, 241, 412, 421; 134, 314, 143, 341, 413, 431; 234, 243, 423. That is,  3*3 + 3!*3 = 27 = binomial(3, 1)*(4-1)^(3-1) = 3*3^2;
j=2 (twice 4):  144, 414, 441;  244, 424, 442;  344, 434, 443; 3*3 = 9 = binomial(3, 2)*(4-1)^(3-2) = 3*3;
j=3 (thrice 4) 444; 1 = binomial(3, 3)*(4-1)^(3-3).
Together: 27 + 9 + 1 = 37 = A(2, 3) = T(5, 3).

Cf. A005061, A008292, A047969 (reversed), A045531 (central diagonal), A047970 (row sums of triangle).
Row sequences of array A (nexus numbers): A000012, A005408, A003215, A005917(k+1), A022521, A022522, A022523, A022524, A022525, A022526, A022527, A022528.
Column sequences of array A: A000012, A000225(n+1), A001047(n+1), A005061(n+1), A005060(n+1), A005062(n+1), A016169(n+1), A016177(n+1), A016185(n+1), A016189(n+1), A016195(n+1), A016197(n+1).

egf := exp(exp(x)*y + x)*(exp(x)*y - y + 1): ser := series(egf, x, 12):
cx := n -> series(n!*coeff(ser, x, n), y, 12):
Arow := n -> seq(k!*coeff(cx(n), y, k), k=0..9):
for n from 0 to 5 do Arow(n) od; # Peter Luschny, May 10 2021

A[n_, k_] := (k + 1)^(n + 1) - k^(n + 1); Table[A[n - k, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Amiram Eldar, May 10 2021 *)

Array A(n, k) = (k+1)^(n+1) - k^(n+1), n, k >= 0.
A(n-1, k-1) = Sum_{j=1} binomial(n, j)*(k-1)^(n-j) = Sum_{j=0} binomial(n, j)*(k-1)^(n-j) - (k-1)^n = (1+(k-1))^n - (k-1)^n = k^n - (k-1)^n (from the combinatorial comment on A(n-1, k-1) above).
O.g.f. row n of array A: RA(n, x) = P(n, x)/(1 - x)^n, with P(n, x) = Sum_{m=0..n} A008292(n+1, m+1)*x^m, (the Eulerian number triangle  A008292 has offset 1) for n >= 0. (See the Oct 26 2008 comment in A047969 by Peter Bala). RA(n, x) = polylog(-(n+1), x)*(1-x)/x. (For P(n, x) see the formula by Vladeta Jovovic, Sep 02 2002.)
E.g.f. of e.g.f.s of the rows of array A: EE(x, y) = exp(x)*(1 + y*(exp(x) - 1))*exp(y*exp(x)), that is A(n, k) = [y^k/k!][x^n/n!] EE(x, y).
Triangle T(n, m) = A(n-m, m) = (m+1)^(n-m+1) - m^(n-m+1), n >= 0, m = 0, 1, ..., n.
E.g.f.: -(exp(x)-1)/(x*exp(x)*y-x). - Vladimir Kruchinin, Nov 02 2022

A008511 Number of points on surface of 4-dimensional cube.

Original entry on oeis.org

0, 16, 80, 240, 544, 1040, 1776, 2800, 4160, 5904, 8080, 10736, 13920, 17680, 22064, 27120, 32896, 39440, 46800, 55024, 64160, 74256, 85360, 97520, 110784, 125200, 140816, 157680, 175840

Offset: 0

N. J. A. Sloane

			G.f. = 16*x + 80*x^2 + 240*x^3 + 544*x^4 + 1040*x^5 + 1776*x^6 + 2800*x^7 + ... - _Michael Somos_, Jun 24 2018

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A000583, A005917, A006003.

List([0..30], n-> 8*n*(1+n^2)); # G. C. Greubel, Nov 09 2019

[(n+1)^4-(n-1)^4: n in [0..30]]; // Vincenzo Librandi, Aug 27 2011

seq(8*n*(1+n^2), n=0..30); # G. C. Greubel, Nov 09 2019

Last[#]-First[#]&/@Partition[Range[-1,30]^4,3,1] (* or *) LinearRecurrence[ {4,-6,4,-1},{0,16,80,240},30] (* Harvey P. Dale, Oct 15 2012 *)

vector(31, n, 8*(n-1)*(1+(n-1)^2)) \\ G. C. Greubel, Nov 09 2019

[8*n*(1+n^2) for n in (0..30)] # G. C. Greubel, Nov 09 2019

a(n) = (n+1)^4 - (n-1)^4 = 8*n + 8*n^3.
G.f.: 16*x*(1+x+x^2)/(1-4*x+6*x^2-4*x^3+x^4). - Colin Barker, Jan 02 2012
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4), a(0)=0, a(1)=16, a(2)=80, a(3)=240. - Harvey P. Dale, Oct 15 2012
a(n) = 16 * A006003(n). - J. M. Bergot, Jul 22 2013
For n > 0, a(n) = A005917(n) + A005917(n+1) = A000583(n+1) - A000583(n-1). - Bruce J. Nicholson, Jun 19 2018
a(n) = -a(-n) for all n in Z. - Michael Somos, Jun 24 2018
E.g.f.:  8*x*(2 +3*x +x^2)*exp(x). - G. C. Greubel, Nov 09 2019

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