A003463 -id:A003463 - cp's OEIS frontend

A263950 Array read by antidiagonals: T(n,k) is the number of lattices L in Z^k such that the quotient group Z^k / L is C_n.

1, 1, 1, 1, 3, 1, 1, 4, 7, 1, 1, 6, 13, 15, 1, 1, 6, 28, 40, 31, 1, 1, 12, 31, 120, 121, 63, 1, 1, 8, 91, 156, 496, 364, 127, 1, 1, 12, 57, 600, 781, 2016, 1093, 255, 1, 1, 12, 112, 400, 3751, 3906, 8128, 3280, 511, 1, 1, 18, 117, 960, 2801, 22932, 19531, 32640

Offset: 1

Álvar Ibeas, Oct 30 2015

All the enumerated lattices have full rank k, since the quotient group is finite.
For m>=1, T(n,k) is the number of lattices L in Z^k such that the quotient group Z^k / L is C_nm x (C_m)^(k-1); and also, (C_nm)^(k-1) x C_m.
Also, number of subgroups of (C_n)^k isomorphic to C_n (and also, to (C_n)^{k-1}), cf. [Butler, Lemma 1.4.1].
T(n,k) is the sum of the divisors d of n^(k-1) such that n^(k-1)/d is k-free. Namely, the coefficient in n^(-(k-1)*s) of the Dirichlet series zeta(s) * zeta(s-1) / zeta(ks).
Also, number of isomorphism classes of connected (C_n)-fold coverings of a connected graph with circuit rank k.
Columns are multiplicative functions.

			There are 7 = A160870(4,2) lattices of volume 4 in Z^2. Among them, only one (<(2,0), (0,2)>) gives the quotient group C_2 x C_2, whereas the rest give C_4. Hence, T(4,2) = 6 and T(1,2) = 1.
Array begins:
      k=1    k=2    k=3    k=4    k=5    k=6
n=1     1      1      1      1      1      1
n=2     1      3      7     15     31     63
n=3     1      4     13     40    121    364
n=4     1      6     28    120    496   2016
n=5     1      6     31    156    781   3906
n=6     1     12     91    600   3751  22932

Lynne M. Butler, Subgroup lattices and symmetric functions. Mem. Amer. Math. Soc., Vol. 112, No. 539, 1994.

Álvar Ibeas, First 100 antidiagonals, flattened
Jin Ho Kwak, Jang-Ho Chun, and Jaeun Lee, Enumeration of regular graph coverings having finite abelian covering transformation groups, SIAM J. Discrete Math. 11(2), 1998, pp. 273-285. [Page 277]
Jin Ho Kwak and Jaeun Lee, Enumeration of graph coverings, surface branched coverings and related group theory, in Combinatorial and Computational Mathematics (Pohang, 2000), ed. S. Hong et al., World Scientific, Singapore 2001, pp. 97-161. [Examples 2 and 3]

Rows (1-11): A000012, A000225, A003462, A006516, A003463, A160869, A023000, A016152, A016142(n-1), A046915(n-1), A016123(n-1).
Columns (1-17): A000012, A001615, A160889, A160891, A160893, A160895, A160897, A160908, A160953, A160957, A160960, A160972, A161010, A161025, A161139, A161167, A161213.
Main diagonal gives A344210.
Cf. A000203, A008683, A160870.

f[p_, e_, k_] := p^((k - 1)*(e - 1))*(p^k - 1)/(p - 1); T[n_, 1] = T[1, k_] = 1; T[n_, k_] := Times @@ (f[First[#], Last[#], k] & /@ FactorInteger[n]); Table[T[n - k + 1, k], {n, 1, 11}, {k, 1, n}] // Flatten (* Amiram Eldar, Nov 08 2022 *)

T(n,k) = J_k(n) / J_1(n) = (Sum_{d|n} mu(n/d) * d^k) / phi(n).
T(n,k) = n^(k-1) * Product_{p|n, p prime} (p^k - 1) / ((p - 1) * p^(k-1)).
Dirichlet g.f. of k-th column: zeta(s-k+1) * Product_{p prime} (1 + p^(-s) + p^(1-s) + ... + p^(k-2-s)).
If n is squarefree, T(n,k) = A160870(n,k) = A000203(n^(k-1)).
From Amiram Eldar, Nov 08 2022: (Start)
Sum_{i=1..n} T(i, k) ~ c * n^k, where c = (1/k) * Product_{p prime} (1 + (p^(k-1)-1)/((p-1)*p^k)).
Sum_{i>=1} 1/T(i, k) = zeta(k-1)*zeta(k) * Product_{p prime} (1 - 2/p^k + 1/p^(2*k-1)), for k > 2. (End)
T(n,k) = (1/n) * Sum_{d|n} mu(n/d)*sigma(d^k). - Ridouane Oudra, Apr 03 2025

A125118 Triangle read by rows: T(n,k) = value of the n-th repunit in base (k+1) representation, 1<=k<=n.

Original entry on oeis.org

1, 3, 4, 7, 13, 21, 15, 40, 85, 156, 31, 121, 341, 781, 1555, 63, 364, 1365, 3906, 9331, 19608, 127, 1093, 5461, 19531, 55987, 137257, 299593, 255, 3280, 21845, 97656, 335923, 960800, 2396745, 5380840, 511, 9841, 87381, 488281, 2015539, 6725601, 19173961, 48427561, 111111111

Offset: 1

Reinhard Zumkeller, Nov 21 2006

			First 4 rows:
1: [1]_2
2: [11]_2 ........ [11]_3
3: [111]_2 ....... [111]_3 ....... [111]_4
4: [1111]_2 ...... [1111]_3 ...... [1111]_4 ...... [1111]_5
_
1: 1
2: 2+1 ........... 3+1
3: (2+1)*2+1 ..... (3+1)*3+1 ..... (4+1)*4+1
4: ((2+1)*2+1)*2+1 ((3+1)*3+1)*3+1 ((4+1)*4+1)*4+1 ((5+1)*5+1)*5+1.

G. C. Greubel, Rows n = 1..50 of the triangle, flattened
Eric Weisstein's World of Mathematics, Repunit

This triangle shares some features with triangle A104878.
This triangle is a portion of rectangle A055129.
Each term of A110737 comes from the corresponding row of this triangle.
Diagonals (adjusting offset as necessary): A060072, A023037, A031973, A173468.
Columns (adjusting offset as necessary): A000225, A003462, A002450, A003463, A003464, A023000, A023001, A002452, A002275, A016123, A016125, A091030, A135519, A135518, A131865, A091045, A218721, A218722, A064108, A218724, A218725, A218726, A218727, A218728, A218729, A218730, A218731, A218732, A218733, A218734, A132469, A218736, A218737, A218738, A218739, A218740, A218741, A218742, A218743, A218744, A218745, A218746, A218747, A218748, A218749, A218750, A218751, A218753, A218752.
Cf. A023037, A031973, A125119, A125120 (row sums).

[((k+1)^n -1)/k : k in [1..n], n in [1..12]]; // G. C. Greubel, Aug 15 2022

Table[((k+1)^n -1)/k, {n, 12}, {k, n}]//Flatten (* G. C. Greubel, Aug 15 2022 *)

def A125118(n,k): return ((k+1)^n -1)/k
flatten([[A125118(n,k) for k in (1..n)] for n in (1..12)]) # G. C. Greubel, Aug 15 2022

T(n, k) = Sum_{i=0..n-1} (k+1)^i.
T(n+1, k) = (k+1)*T(n, k) + 1.
Sum_{k=1..n} T(n, k) = A125120(n).
T(2*n-1, n) = A125119(n).
T(n, 1) = A000225(n).
T(n, 2) = A003462(n) for n>1.
T(n, 3) = A002450(n) for n>2.
T(n, 4) = A003463(n) for n>3.
T(n, 5) = A003464(n) for n>4.
T(n, 9) = A002275(n) for n>8.
T(n, n) = A060072(n+1).
T(n, n-1) = A023037(n) for n>1.
T(n, n-2) = A031973(n) for n>2.
T(n, k) = A055129(n, k+1) = A104878(n+k, k+1), 1<=k<=n. - Mathew Englander, Dec 19 2020

A014827 a(1)=1, a(n) = 5*a(n-1) + n.

Original entry on oeis.org

1, 7, 38, 194, 975, 4881, 24412, 122068, 610349, 3051755, 15258786, 76293942, 381469723, 1907348629, 9536743160, 47683715816, 238418579097, 1192092895503, 5960464477534, 29802322387690, 149011611938471, 745058059692377, 3725290298461908, 18626451492309564, 93132257461547845

Offset: 1

N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Dillan Agrawal, Selena Ge, Jate Greene, Tanya Khovanova, Dohun Kim, Rajarshi Mandal, Tanish Parida, Anirudh Pulugurtha, Gordon Redwine, Soham Samanta, and Albert Xu, Chip-Firing on Infinite k-ary Trees, arXiv:2501.06675 [math.CO], 2025. See pp. 9, 18.
László Tóth, On Schizophrenic Patterns in b-ary Expansions of Some Irrational Numbers, arXiv:2002.06584 [math.NT], 2020. See also Proc. Amer. Math. Soc. 148 (2020), pp. 461-469.
Index entries for linear recurrences with constant coefficients, signature (7,-11,5).

Cf. A000225, A000392, A002275, A002452, A003462, A003463, A003464, A016123, A016125, A016208, A016218, A016256, A023000, A023001.

[(5^(n+1)-4*n-5)/16: n in [1..30]]; // Vincenzo Librandi, Aug 23 2011

a:=n->sum((5^(n-j)-1^(n-j))/4,j=0..n): seq(a(n), n=1..21); # Zerinvary Lajos, Jan 04 2007

Join[{a=1,b=7},Table[c=6*b-5*a+1;a=b;b=c,{n,60}]] (* Vladimir Joseph Stephan Orlovsky, Feb 06 2011 *)

[(gaussian_binomial(n,1,5)-n)/4 for n in range(2,23)] # Zerinvary Lajos, May 29 2009

a(n) = (5^(n+1) - 4*n - 5)/16.
G.f.: x/((1-5*x)*(1-x)^2).
From Paul Barry, Jul 30 2004: (Start)
a(n) = Sum_{k=0..n} (n-k)*5^k = Sum_{k=0..n} k*5^(n-k).
a(n) = Sum_{k=0..n} binomial(n+2,k+2)*4^k [Offset 0]. (End)
From Elmo R. Oliveira, Mar 29 2025: (Start)
E.g.f.: exp(x)*(5*exp(4*x) - 4*x - 5)/16.
a(n) = 7*a(n-1) - 11*a(n-2) + 5*a(n-3) for n > 3. (End)

A104878 A sum-of-powers number triangle.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 7, 4, 1, 1, 5, 15, 13, 5, 1, 1, 6, 31, 40, 21, 6, 1, 1, 7, 63, 121, 85, 31, 7, 1, 1, 8, 127, 364, 341, 156, 43, 8, 1, 1, 9, 255, 1093, 1365, 781, 259, 57, 9, 1, 1, 10, 511, 3280, 5461, 3906, 1555, 400, 73, 10, 1, 1, 11, 1023, 9841, 21845

Offset: 0

Paul Barry, Mar 28 2005

Columns are partial sums of the columns of A004248. Row sums are A104879. Diagonal sums are A104880.

The rows of this triangle (apart from the initial "1" in each row) are the antidiagonals of rectangle A055129. The diagonals of this triangle (apart from the initial "1") are the rows of rectangle A055129. The columns of this triangle (apart from the leftmost column) are the same as the columns of rectangle A055129 but shifted downward. - Mathew Englander, Dec 21 2020

			Triangle starts:
  1;
  1,  1;
  1,  2,  1;
  1,  3,  3,  1;
  1,  4,  7,  4,  1;
  1,  5, 15, 13,  5,  1;
  1,  6, 31, 40, 21,  6,  1;
  ...

Cf. A004248 (first differences by column), A104879 (row sums), A104880 (antidiagonal sums), A125118 (version of this triangle with fewer terms).
This triangle (ignoring the leftmost column) is a rotation of rectangle A055129.
Columns (adjusting offset as necessary): A000012, A000027, A000225, A003462, A002450, A003463, A003464, A023000, A023001, A002452, A002275, A016123, A016125, A091030, A135519, A135518, A131865, A091045, A218721, A218722, A064108, A218724, A218725, A218726, A218727, A218728, A218729, A218730, A218731, A218732, A218733, A218734, A132469, A218736, A218737, A218738, A218739, A218740, A218741, A218742, A218743, A218744, A218745, A218746, A218747, A218748, A218749, A218750, A218751, A218753, A218752.
T(2n,n) gives A031973.

A104878 :=proc(n,k): if k = 0 then 1 elif k=1 then n elif k>=2 then (k^(n-k+1)-1)/(k-1) fi: end: for n from 0 to 7 do seq(A104878(n,k), k=0..n) od; seq(seq(A104878(n,k), k=0..n), n=0..10); # Johannes W. Meijer, Aug 21 2011

T(n, k) = if(k=1, n, if(k<=n, (k^(n-k+1)-1)/(k-1), 0));
G.f. of column k: x^k/((1-x)(1-k*x)). [corrected by Werner Schulte, Jun 05 2019]
T(n, k) = A069777(n+1,k)/A069777(n,k). [Johannes W. Meijer, Aug 21 2011]
T(n, k) = A055129(n+1-k, k) for n >= k > 0. - Mathew Englander, Dec 19 2020

A015004 q-factorial numbers for q=5.

Original entry on oeis.org

1, 1, 6, 186, 29016, 22661496, 88515803376, 1728802155736656, 168827903320618878336, 82435457461295106532780416, 201258420458750640859769304304896, 2456767777551003294245070550498298923776, 149949204558598784020761783280669552547300269056

Offset: 0

Olivier Gérard

Vincenzo Librandi, Table of n, a(n) for n = 0..50
Index entries for sequences related to factorial numbers.

Column q=5 of A069777.
Cf. A015001, A015002, A015005, A015006, A015007, A015008, A015009, A015011.
Cf. A003463, A100222.

[n le 1 select 1 else (5^n-1)*Self(n-1)/4: n in [1..15]]; // Vincenzo Librandi, Oct 25 2012

RecurrenceTable[{a[1]==1, a[n]==((5^n - 1)* a[n-1])/4}, a, {n, 15}] (* Vincenzo Librandi, Oct 25 2012 *)
Table[QFactorial[n, 5], {n, 15}] (* Bruno Berselli, Aug 14 2013 *)

a(n) = { my(q=5); prod(k=1, n, ((q^k - 1) / (q - 1))) } \\ Andrew Howroyd, Feb 18 2018

a(n) = Product_{k=1..n} ((q^k - 1) / (q - 1)) where q = 5.
a(0) = 1, a(n) = (5^n-1)*a(n-1)/4. - Vincenzo Librandi, Oct 25 2012
From Amiram Eldar, Jul 05 2025: (Start)
a(n) = Product_{k=1..n} A003463(k).
a(n) ~ c * 5^(n*(n+1)/2)/4^n, where c = A100222. (End)

a(0)=1 prepended by Alois P. Heinz, Sep 08 2021

A016208 Expansion of 1/((1-x)(1-3x)(1-4x)).

Original entry on oeis.org

1, 8, 45, 220, 1001, 4368, 18565, 77540, 320001, 1309528, 5326685, 21572460, 87087001, 350739488, 1410132405, 5662052980, 22712782001, 91044838248, 364760483725, 1460785327100, 5848371485001, 23409176469808, 93683777468645, 374876324642820, 1499928942876001

Offset: 0

N. J. A. Sloane

Binomial transform of A085277. - Paul Barry, Jun 25 2003

Number of walks of length 2n+5 between two nodes at distance 5 in the cycle graph C_12. - Herbert Kociemba, Jul 05 2004

Muniru A Asiru, Table of n, a(n) for n = 0..1000
Natalia Agudelo Muñetón, Agustín Moreno Cañadas, Pedro Fernando Fernández Espinosa, and Isaías David Marín Gaviria, Brauer Configuration Algebras and Their Applications in Graph Energy Theory, Mathematics (2021) Vol. 9, 3042.
Index entries for linear recurrences with constant coefficients, signature (8,-19,12)

Cf. A000225, A000295, A000392, A002275, A003462, A003463, A003464, A023000, A023001, A002452, A016123, A016125, A016256.

a:=[1,8,45];; for n in [4..30] do a[n]:=8*a[n-1]-19*a[n-2]+12*a[n-3]; od; Print(a); # Muniru A Asiru, Apr 19 2019

Table[(2^(2*n + 3) - 3^(n + 2) + 1)/6, {n, 40}] (* Vladimir Joseph Stephan Orlovsky, Jan 19 2011 *)
CoefficientList[Series[1/((1-x)(1-3x)(1-4x)),{x,0,30}],x] (* or *) LinearRecurrence[ {8,-19,12},{1,8,45},30] (* Harvey P. Dale, Apr 09 2012 *)

Vec(1/((1-x)*(1-3*x)*(1-4*x))+O(x^99)) \\ Charles R Greathouse IV, Sep 23 2012

a(n) = 16*4^n/3 + 1/6 - 9*3^n/2. - Paul Barry, Jun 25 2003
a(0) = 0, a(1) = 8, a(n) = 7*a(n-1) - 12*a(n-2) + 1. - Vincenzo Librandi, Feb 10 2011
a(0) = 1, a(1) = 8, a(2) = 45, a(n) = 8*a(n-1) - 19*a(n-2) + 12*a(n-3). - Harvey P. Dale, Apr 09 2012

A090018 a(n) = 6a(n-1) + 3a(n-2) for n > 2, a(0)=1, a(1)=6.

Original entry on oeis.org

1, 6, 39, 252, 1629, 10530, 68067, 439992, 2844153, 18384894, 118841823, 768205620, 4965759189, 32099171994, 207492309531, 1341251373168, 8669985167601, 56043665125110, 362271946253463, 2341762672896108, 15137391876137037, 97849639275510546, 632510011281474387

Offset: 0

Paul Barry, Nov 19 2003

From Johannes W. Meijer, Aug 09 2010: (Start)

a(n) represents the number of n-move routes of a fairy chess piece starting in a given corner or side square on a 3 X 3 chessboard. This fairy chess piece behaves like a white queen on the eight side and corner squares but on the central square the queen explodes with fury and turns into a red queen, see A180032. The central square leads to A180028. (End)

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (6,3).

Cf. A141041, A180028, A180032.

Sequences with g.f. of the form 1/(1 - 6*x - k*x^2): A106392 (k=-10), A027471 (k=-9), A006516 (k=-8), A081179 (k=-7), A030192 (k=-6), A003463 (k=-5), A084326 (k=-4), A138395 (k=-3), A154244 (k=-2), A001109 (k=-1), A000400 (k=0), A005668 (k=1), A135030 (k=2), this sequence (k=3), A135032 (k=4), A015551 (k=5), A057089 (k=6), A015552 (k=7), A189800 (k=8), A189801 (k=9), A190005 (k=10), A015553 (k=11).

[n le 2 select 6^(n-1) else 6*Self(n-1)+3*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Nov 15 2011

a:= n-> (<<0|1>, <3|6>>^n. <<1,6>>)[1,1]:
seq(a(n), n=0..30);  # Alois P. Heinz, Jan 17 2011

Join[{a=1,b=6},Table[c=6*b+3*a;a=b;b=c,{n,100}]] (* Vladimir Joseph Stephan Orlovsky, Jan 16 2011 *)
LinearRecurrence[{6,3}, {1,6}, 41] (* G. C. Greubel, Oct 10 2022 *)

my(x='x+O('x^30)); Vec(1/(1-6*x-3*x^2)) \\ G. C. Greubel, Jan 24 2018

[lucas_number1(n,6,-3) for n in range(1, 31)] # Zerinvary Lajos, Apr 24 2009

a(n) = (3+2*sqrt(3))^n*(sqrt(3)/4+1/2) + (1/2-sqrt(3)/4)*(3-2*sqrt(3))^n.
a(n) = (-i*sqrt(3))^n * ChebyshevU(n, isqrt(3)), i^2=-1.
From Johannes W. Meijer, Aug 09 2010: (Start)
G.f.: 1/(1 - 6*x - 3*x^2).
Limit_{k->oo} a(n+k)/a(k) = A141041(n) + A090018(n-1)*sqrt(12) for n >= 1.
Limit_{n->oo} A141041(n)/A090018(n-1) = sqrt(12). (End)
a(n) = Sum_{k=0..n} A099089(n,k)*3^k. - Philippe Deléham, Nov 21 2011
E.g.f.: exp(3*x)*(2*cosh(2*sqrt(3)*x) + sqrt(3)*sinh(2*sqrt(3)*x))/2. - Stefano Spezia, Apr 23 2025

Typo in Mathematica program corrected by Vincenzo Librandi, Nov 15 2011

A285061 Sheffer triangle S2[4,1] = (exp(x), exp(4*x) - 1).

Original entry on oeis.org

1, 1, 4, 1, 24, 16, 1, 124, 240, 64, 1, 624, 2656, 1792, 256, 1, 3124, 26400, 33920, 11520, 1024, 1, 15624, 250096, 546560, 331520, 67584, 4096, 1, 78124, 2313360, 8105664, 7822080, 2745344, 372736, 16384, 1, 390624, 21132736, 114627072, 165398016, 88940544, 20299776, 1966080, 65536, 1, 1953124, 191757120, 1574682880, 3270274560, 2529343488, 863256576, 138215424, 10027008, 262144

Offset: 0

Wolfdieter Lang, Apr 13 2017

For Sheffer triangles (infinite lower triangular exponential convolution matrices) see the W. Lang link under A006232, with references.
This is a generalization of the Sheffer triangle Stirling2(n, m) = A048993(n, m) denoted by (exp(x), exp(x)-1), which could be named S2[1,0].
For the Sheffer triangle (exp(x), (1/4)*(exp(4*x) - 1)) see A111578, also the P. Bala link where this triangle, T(n, m)/4^m, is named S_{(4,0,1)}. The triangle T(n, m)*m! is given in A285066.
The a-sequence for this Sheffer triangle has e.g.f. 4*x/log(1+x) and is 4*A006232(n)/A006233(n) (Cauchy numbers of the first kind).
The z-sequence has e.g.f. (4/(log(1+x)))*(1 - 1/(1+x)^(1/4)) and is A285062(n)/A285063(n).
The main diagonal gives A000302.
The row sums give A285064. The alternating row sums give A285065.
The first column sequences are A000012, 4*A003463, 4^2*A016234. For the e.g.f.s and o.g.f.s see the formula section.
This triangle appears in the o.g.f. G(n, x) of the sequence {(1 + 4*m)^n}{m>=0}, as G(n, x) = Sum{m=0..n} T(n, m)*m!*x^m/(1-x)^(m+1), n >= 0. Hence the corresponding e.g.f. is, by the linear inverse Laplace transform, E(n, t) = Sum_{m >=0}(1 + 4*m)^n t^m/m! = exp(t)*Sum_{m=0..n} T(n, m)*t^m.
The corresponding Euler triangle with reversed rows is rEu(n, k) = Sum_{m=0..k} (-1)^(k-m)*binomial(n-m, k-m)*T(n, k)*k!, 0 <= k <= n. This is A225118 with row reversion.
In general the Sheffer triangle S2[d,a] appears in the reordering of the operator (a*1 + d*E_x) = Sum_{m=0..n} S2(d,a;n,m) x^m*(d_x)^m with the derivative d_x and the Euler operator E_x = x*d_x. For [d,a] = [1,0] this becomes the standard Stirling2 property.

			The triangle T(n,m) begins:
  n\m  0      1        2         3         4        6        7       8     9
  0:   1
  1:   1      4
  2:   1     24       16
  3:   1    124      240        64
  4:   1    624     2656      1792       256
  5:   1   3124    26400     33920     11520     1024
  6:   1  15624   250096    546560    331520    67584     4096
  7:   1  78124  2313360   8105664   7822080  2745344   372736   16384
  8:   1 390624 21132736 114627072 165398016 88940544 20299776 1966080 65536
  ...
Three term recurrence: T(4, 1) = 4*1 + (1 + 4*1)*124 = 624.
Recurrence for row polynomial R(3, x) (Meixner type): ((1 + 4*x) + 4*x*d_x)*(1 + 24*x + 16*x^2) = 1 + 124*x + 240*x^2 + 64*x^3.
Boas-Buck recurrence for column m = 2, and n = 4: T(4, 2) = (1/2)*(2*(2 + 4*2)*T(3, 2) + 2*6*(-4)^2*Bernoulli(2)*T(2, 2)) = (1/2)*(20*240 + 12*16*(1/6)*16) = 2656. - _Wolfdieter Lang_, Aug 11 2017

Michael De Vlieger, Table of n, a(n) for n = 0..11475, rows n = 0..150, flattened.
Paweł Hitczenko, A class of polynomial recurrences resulting in (n/log n, n/log^2 n)-asymptotic normality, arXiv:2403.03422 [math.CO], 2024. See p. 9.
Wolfdieter Lang, On Sums of Powers of Arithmetic Progressions, and Generalized Stirling, Eulerian and Bernoulli numbers, arXiv:1707.04451 [math.NT], 2017.

Cf. A000012, A000302, 4*A003463, A111578, A006232/A006233, 16*A016234, A282629, A285062/A285063, A285064, A285065, A285066.

Table[Sum[Binomial[n, k]*4^k*StirlingS2[k, m], {k, 0, n}], {n, 0, 20}, {m, 0, n}] // Flatten (* Indranil Ghosh, May 06 2017 *)

T(n, m) = sum(k=0, n, binomial(n, k)*4^k*stirling(k, m, 2));
for(n=0, 20, for(m=0, n, print1(T(n, m),", ");); print();) \\ Indranil Ghosh, May 06 2017

Three term recurrence (from the conversion property mentioned above, with [d,a] =[4,1]): T(n, -1) = 0, T(0, 0) = 1, T(n, m) = 0 if n < m. T(n, m) = 4*T(n-1, m-1) + (1 + 4*m)*T(n-1, m) for n >= 1, m = 0..n.
T(n, m) = Sum_{k=0..m} binomial(m,k)*(-1)^(k-m)*(1 + 4*k)^n/m!, 0 <= m <= n, satisfying the recurrence.
E.g.f. of the row polynomials R(n, x) = Sum_{m=0..n} T(n, m)*x^m: exp(z)*exp(x*(exp(4*z) - 1)). This is the e.g.f. of the triangle.
E.g.f. for the sequence of column m: exp(x)*((exp(3*x) - 1)^m)/m! (Sheffer property).
O.g.f. for sequence of column m: (4*x)^m/Product_{j=0..m} (1 - (1 + 4*j)*x) (by Laplace transform of the e.g.f.).
T(n, m) = Sum_{k=0..n} binomial(n, k)* 4^k*Stirling2(k, m), 0 <= m <= n, where Stirling2 is given in A048993.
A nontrivial recurrence for the column m=0 entries T(n, 0) = 4^n from the z-sequence given above: T(n,0) = n*Sum_{j=0..n-1} z(j)*T(n-1,j), n >= 1, T(0, 0) = 1.
Recurrence for column m >= 1 entries from the a-sequence given above: T(n, m) = (n/m)* Sum_{j=0..n-m} binomial(m-1+j, m-1)*a(j)*T(n-1, m-1+j), m >= 1.
Recurrence for row polynomials R(n, x) (Meixner type): R(n, x) = ((1 + 4*x) + 4*x*d_x)*R(n-1, x), with differentiation d_x, for n >= 1, with input R(0, x) = 1.
Boas-Buck recurrence for column sequence m: T(n, m) = (1/(n - m))*((n/2)*(2 + 4*m)*T(n-1, m) + m*Sum_{p=m..n-2} binomial(n, p)*(-4)^(n-p)*Bernoulli(n-p)*T(p, m)), for n > m >= 0, with input T(m,m) = 4^m. See a comment and references in A282629. An example is given below. - Wolfdieter Lang, Aug 11 2017

A016209 Expansion of 1/((1-x)(1-3x)(1-5x)).

Original entry on oeis.org

1, 9, 58, 330, 1771, 9219, 47188, 239220, 1205941, 6059229, 30384718, 152189310, 761743711, 3811110039, 19062724648, 95335146600, 476740303081, 2383895225649, 11920057258978, 59602029687090

Offset: 0

N. J. A. Sloane

For a combinatorial interpretation following from a(n) = A039755(n+2,2) = h^{(3)}A039755.%20-%20_Wolfdieter%20Lang">n, the complete homogeneous symmetric function of degree n in the symbols {1, 3, 5} see A039755. - _Wolfdieter Lang, May 26 2017

			a(2) = h^{(3)}_2 = 1^2 + 3^2 + 5^2 + 1^1*(3^1 + 5^1) + 3^1*5^1 = 58. - _Wolfdieter Lang_, May 26 2017

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (9,-23,15).

Cf. A016218, A016208, A000392, A000225, A003462, A003463, A003464, A023000, A023001, A002452, A002275, A016123, A016125, A016256, A039755, A021424.

[(5^(n+2)-2*3^(n+2)+1)/8: n in [0..20]]; // Vincenzo Librandi, Sep 17 2011

A016209 := proc(n) (5^(n+2)-2*3^(n+2)+1)/8; end proc: # R. J. Mathar, Mar 22 2011

Join[{a=1,b=9},Table[c=8*b-15*a+1;a=b;b=c,{n,60}]] (* Vladimir Joseph Stephan Orlovsky, Feb 07 2011 *)
CoefficientList[Series[1/((1-x)(1-3x)(1-5x)),{x,0,30}],x] (* or *) LinearRecurrence[ {9,-23,15},{1,9,58},30] (* Harvey P. Dale, Feb 20 2020 *)

PARI
```
a(n)=if(n<0,0,n+=2; (5^n-2*3^n+1)/8)
```

a(n) = A039755(n+2, 2).
a(n) = (5^(n+2) - 2*3^(n+2)+1)/8 = a(n-1) + A005059(n+1) = 8*a(n-1) - 15*a(n-2) + 1 = (A003463(n+2) - A003462(n+2))/2. - Henry Bottomley, Jun 06 2000
G.f.: 1/((1-x)(1-3*x)(1-5*x)). See the name.
E.g.f.: (25*exp(5*x) - 18*exp(3*x) + exp(x))/8, from the e.g.f. of the third column (k=2) of A039755. - Wolfdieter Lang, May 26 2017

A239473 Triangle read by rows: signed version of A059260: coefficients for expansion of partial sums of sequences a(n,x) in terms of their binomial transforms (1+a(.,x))^n ; Laguerre polynomial expansion of the truncated exponential.

Original entry on oeis.org

1, 0, 1, 1, -1, 1, 0, 2, -2, 1, 1, -2, 4, -3, 1, 0, 3, -6, 7, -4, 1, 1, -3, 9, -13, 11, -5, 1, 0, 4, -12, 22, -24, 16, -6, 1, 1, -4, 16, -34, 46, -40, 22, -7, 1, 0, 5, -20, 50, -80, 86, -62, 29, -8, 1, 1, -5, 25, -70, 130, -166, 148, -91, 37, -9, 1, 0, 6, -30, 95, -200, 296, -314, 239, -128, 46, -10, 1

Offset: 0

Tom Copeland, Mar 19 2014

With T the lower triangular array above and the Laguerre polynomials L(k,x) = Sum_{j=0..k} (-1)^j binomial(k, j) x^j/j!, the following identities hold:
(A) Sum_{k=0..n} (-1)^k L(k,x) = Sum_{k=0..n} T(n,k) x^k/k!;
(B) Sum_{k=0..n} x^k/k! = Sum_{k=0..n} T(n,k) L(k,-x);
(C) Sum_{k=0..n} x^k = Sum_{k=0..n} T(n,k) (1+x)^k = (1-x^(n+1))/(1-x).
More generally, for polynomial sequences,
(D) Sum_{k=0..n} P(k,x) = Sum_{k=0..n} T(n,k) (1+P(.,x))^k,
where, e.g., for an Appell sequence, such as the Bernoulli polynomials, umbrally, (1+ Ber(.,x))^k = Ber(k,x+1).
Identity B follows from A through umbral substitution of j!L(j,-x) for x^j in A. Identity C, related to the cyclotomic polynomials for prime index, follows from B through the Laplace transform.
Integrating C gives Sum_{k=0..n} T(n,k) (2^(k+1)-1)/(k+1) = H(n+1), the harmonic numbers.
Identity A >= 0 for x >= 0 (see MathOverflow link for evaluation in terms of Hermite polynomials).
From identity C, W(m,n) = (-1)^n Sum_{k=0..n} T(n,k) (2-m)^k = number of walks of length n+1 between any two distinct vertices of the complete graph K_m for m > 2.
Equals A112468 with the first column of ones removed. - Georg Fischer, Jul 26 2023

			Triangle begins:
   1
   0    1
   1   -1    1
   0    2   -2    1
   1   -2    4   -3    1
   0    3   -6    7   -4    1
   1   -3    9  -13   11   -5    1
   0    4  -12   22  -24   16   -6    1
   1   -4   16  -34   46  -40   22   -7    1
   0    5  -20   50  -80   86  -62   29   -8    1
   1   -5   25  -70  130 -166  148  -91   37   -9    1

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
J. Adams, On the groups J(x)-II, Topology, Vol. 3, p. 137-171, Pergamon Press, (1965).
MathOverflow, Cyclotomic Polynomials in Combinatorics
Mathoverflow, Inequality for Laguerre polynomials

For column 2: A001057, A004526, A008619, A140106.
Column 3: A002620, A087811.
Column 4: A002623, A173196.
Column 5: A001752.
Column 6: A001753.
Cf. Bottomley's cross-references in A059260.
Embedded in alternating antidiagonals of T are the reversals of arrays A071921 (A225010) and A210220.
Cf. A049310, A112468, A341091.

[[(&+[(-1)^(j+k)*Binomial(j,k): j in [0..n]]): k in [0..n]]: n in [0..10]]; // G. C. Greubel, Feb 06 2018

A239473 := proc(n,k)
    add(binomial(j,k)*(-1)^(j+k),j=k..n) ;
end proc; # R. J. Mathar, Jul 21 2016

Table[Sum[(-1)^(j+k)*Binomial[j,k], {j,0,n}], {n,0,10}, {k,0,n}]//Flatten (* G. C. Greubel, Feb 06 2018 *)

for(n=0,10, for(k=0,n, print1(sum(j=0,n, (-1)^(j+k)*binomial(j, k)), ", "))) \\ G. C. Greubel, Feb 06 2018

Trow = lambda n: sum((x-1)^j for j in (0..n)).list()
for n in (0..10): print(Trow(n)) # Peter Luschny, Jul 09 2019

T(n, k) = Sum_{j=0..n} (-1)^(j+k) * binomial(j, k).
E.g.f: (exp(t) - (x-1)*exp((x-1)*t))/(2-x).
O.g.f. (n-th row): (1-(x-1)^(n+1))/(2-x).
Associated operator identities:
With D=d/dx, :xD:^n=x^n*D^n, and :Dx:^n=D^n*x^n, then bin(xD,n)= binomial(xD,n)=:xD:^n/n! and L(n,-:xD:)=:Dx:^n/n!=bin(xD+n,n)=(-1)^n bin(-xD-1,n),
A-o) Sum_{k=0..n} (-1)^k L(k,-:xD:) = Sum_{k=0..n} :-Dx:^k/k!
     = Sum_{k=0..n} T(n,k) :-xD:^k/k! = Sum_{k=0..n} (-1)^k T(n,k)bin(xD,k)
B-o) Sum_{k=0..n} :xD:^k/k! = Sum_{k=0..n}, T(n,k) L(k,-:xD:)
     = Sum_{k=0..n} T(n,k) :Dx:^k/k! = Sum_{k=0..n}, bin(xD,k).
Associated binomial identities:
A-b) Sum_{k=0..n} (-1)^k bin(s+k,k) = Sum_{k=0..n} (-1)^k T(n,k) bin(s,k)
     = Sum_{k=0..n} bin(-s-1,k) = Sum{k=0..n} T(n,k) bin(-s-1+k,k)
B-b) Sum_{k=0..n} bin(s,k) = Sum_{k=0..n} T(n,k) bin(s+k,k)
     = Sum_{k=0..n} (-1)^k bin(-s-1+k,k)
     = Sum_{k=0..n} (-1)^k T(n,k) bin(-s-1,k).
In particular, from B-b with s=n, Sum_{k=0..n} T(n,k) bin(n+k,k) = 2^n. From B-b with s=0, row sums are all 1.
From identity C with x=-2, the unsigned row sums are the Jacobsthal sequence, i.e., Sum_{k=0..n} T(n,k) (1+(-2))^k = (-1)^n A001045(n+1); for x=2, the Mersenne numbers A000225; for x=-3, A014983 or signed A015518; for x=3, A003462; for x=-4, A014985 or signed A015521; for x=4, A002450; for x=-5, A014986 or signed A015531; and for x=5, A003463; for x=-6, A014987 or signed A015540; and for x=6, A003464.
With -s-1 = m = 0,1,2,..., B-b gives finite differences (recursions):
Sum_{k=0..n} (-1)^k T(n,k) bin(m,k) = Sum_{k=0..n} (-1)^k bin(m+k,k) = T(n+m,m), i.e., finite differences of the columns of T generate shifted columns of T. The columns of T are signed, shifted versions of sequences listed in the cross-references. Since the finite difference is an involution, T(n,k) = Sum_{j=0..k} (-1)^j T(n+j,j) bin(k,j)}. Gauss-Newton interpolation can be applied to give a generalized T(n,s) for s noninteger.
From identity C, S(n,m) = Sum_{k=0..n} T(n,k) bin(k,m) = 1 for m < n+1 and 0 otherwise, i.e., S = T*P, where S = A000012, as a lower triangular matrix and P = Pascal = A007318, so T = S*P^(-1), where P^(-1) = A130595, the signed Pascal array (see A132440), the inverse of P, and T^(-1) = P*S^(-1) = P*A167374 = A156644.
U(n,cos(x)) = e^(-n*i*x)*Sum_{k=0..n} T(n,k)*(1+e^(2*i*x))^k = sin((n+1)x)/sin(x), where U is the Chebyschev polynomial of the second kind A053117 and i^2 = -1. - Tom Copeland, Oct 18 2014
From Tom Copeland, Dec 26 2015: (Start)
With a(n,x) = e^(nx), the partial sums are 1+e^x+...+e^(nx) = Sum_{k=0..n} T(n,k) (1+e^x)^k = [ x / (e^x-1) ] [ e^((n+1)x) -1 ] / x = [ (x / (e^x-1)) e^((n+1)x) -  (x / (e^x-1)) ] / x =  Sum_{k>=0} [ (Ber(k+1,n+1) - Ber(k+1,0)) / (k+1) ] * x^k/k!, where Ber(n,x) are the Bernoulli polynomials (cf. Adams p. 140). Evaluating (d/dx)^m at x=0 of these expressions gives relations among the partial sums of the m-th powers of the integers, their binomial transforms, and the Bernoulli polynomials.
With a(n,x) = (-1)^n e^(nx), the partial sums are 1-e^x+...+(-1)^n e^(nx) = Sum_{k=0..n} T(n,k) (1-e^x)^k = [ (-1)^n e^((n+1)x) + 1 ] / (e^x+1) = [ (-1)^n (2 / (e^x+1)) e^((n+1)x) + (2 / (e^x+1)) ] / 2 = (1/2) Sum_{k>=0} [ (-1)^n Eul(k,n+1) + Eul(k,0) ] * x^k/k!, where Eul(n,x) are the Euler polynomials. Evaluating (d/dx)^m at x=0 of these expressions gives relations among the partial sums of signed m-th powers of the integers; their binomial transforms, related to the Stirling numbers of the second kind and face numbers of the permutahedra; and the Euler polynomials. (End)
As in A059260, a generator in terms of bivariate polynomials with the coefficients of this entry is given by (1/(1-y))*1/(1 + (y/(1-y))*x - (1/(1-y))*x^2) = 1 + y + (x^2 - x*y + y^2) + (2*x^2*y - 2*x*y^2 + y^3) + (x^4 - 2*x^3*y + 4*x^2*y^2 - 3*x*y^3 + y^4) + ... . This is of the form -h2 * 1 / (1 + h1*x + h2*x^2), related to the bivariate generator of A049310 with h1 = y/(1-y) and h2 = -1/(1-y) = -(1+h1). - Tom Copeland, Feb 16 2016
From Tom Copeland, Sep 05 2016: (Start)
Letting P(k,x) = x in D gives Sum_{k=0..n} T(n,k)*Sum_{j=0..k} binomial(k,j) = Sum_{k=0..n} T(n,k) 2^k = n + 1.
The quantum integers [n+1]q = (q^(n+1) - q^(-n-1)) / (q - q^(-1)) = q^(-n)*(1 - q^(2*(n+1))) / (1 - q^2) = q^(-n)*Sum{k=0..n} q^(2k) = q^(-n)*Sum_{k=0..n} T(n,k)*(1 + q^2)^k. (End)
T(n, k) = [x^k] Sum_{j=0..n} (x-1)^j. - Peter Luschny, Jul 09 2019
a(n) = -n + Sum_{k=0..n} A341091(k). - Thomas Scheuerle, Jun 17 2022

Inverse array added by Tom Copeland, Mar 26 2014

Formula re Euler polynomials corrected by Tom Copeland, Mar 08 2024

cp's OEIS Frontend

A263950 Array read by antidiagonals: T(n,k) is the number of lattices L in Z^k such that the quotient group Z^k / L is C_n.

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A125118 Triangle read by rows: T(n,k) = value of the n-th repunit in base (k+1) representation, 1<=k<=n.

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A014827 a(1)=1, a(n) = 5*a(n-1) + n.

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A104878 A sum-of-powers number triangle.

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A015004 q-factorial numbers for q=5.

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A016208 Expansion of 1/((1-x)*(1-3*x)*(1-4*x)).

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A090018 a(n) = 6*a(n-1) + 3*a(n-2) for n > 2, a(0)=1, a(1)=6.

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A016208 Expansion of 1/((1-x)(1-3x)(1-4x)).

A090018 a(n) = 6a(n-1) + 3a(n-2) for n > 2, a(0)=1, a(1)=6.