A126646 -id:A126646 - cp's OEIS frontend

A293616 Array of generalized Eulerian number triangles read by ascending antidiagonals, with m >= 0, n >= 0 and 0 <= k <= n.

1, 1, 0, 1, 1, 0, 1, 3, 0, 0, 1, 6, 0, 1, 0, 1, 10, 0, 7, 1, 0, 1, 15, 0, 25, 4, 0, 0, 1, 21, 0, 65, 10, 0, 1, 0, 1, 28, 0, 140, 20, 0, 15, 4, 0, 1, 36, 0, 266, 35, 0, 90, 30, 1, 0, 1, 45, 0, 462, 56, 0, 350, 120, 5, 0, 0, 1, 55, 0, 750, 84, 0, 1050, 350, 15, 0, 1, 0

Offset: 0

Peter Luschny, Oct 14 2017

			Array starts:
m\j| 0   1  2     3    4  5       6       7    8  9      10      11      12
---|----------------------------------------------------------------------------
m=0| 1,  0, 0,    0,   0, 0,      0,      0,   0, 0,      0,      0,      0, ...
m=1| 1,  1, 0,    1,   1, 0,      1,      4,   1, 0,      1,     11,     11, ...
m=2| 1,  3, 0,    7,   4, 0,     15,     30,   5, 0,     31,    146,     91, ...
m=3| 1,  6, 0,   25,  10, 0,     90,    120,  15, 0,    301,    896,    406, ...
m=4| 1, 10, 0,   65,  20, 0,    350,    350,  35, 0,   1701,   3696,   1316, ...
m=5| 1, 15, 0,  140,  35, 0,   1050,    840,  70, 0,   6951,  11886,   3486, ...
m=6| 1, 21, 0,  266,  56, 0,   2646,   1764, 126, 0,  22827,  32172,   8022, ...
m=7| 1, 28, 0,  462,  84, 0,   5880,   3360, 210, 0,  63987,  76692,  16632, ...
m=8| 1, 36, 0,  750, 120, 0,  11880,   5940, 330, 0, 159027, 165792,  31812, ...
m=9| 1, 45, 0, 1155, 165, 0,  22275,   9900, 495, 0, 359502, 331617,  57057, ...
   A000217, A001296,A000292,A001297,A027789,A000332,A001298,A293610,A293611, ...
.
m\j| ...    13  14      15       16       17      18      19 20
---|----------------------------------------------------------------
m=0| ...,    0, 0,       0,       0,       0,      0,      0, 0, ...  [A000007]
m=1| ...,    1, 0,       1,      26,      66,     26,      1, 0, ...  [A173018]
m=2| ...,    6, 0,      63,     588,     868,    238,      7, 0, ...  [A062253]
m=3| ...,   21, 0,     966,    5376,    5586,   1176,     28, 0, ...  [A062254]
m=4| ...,   56, 0,    7770,   30660,   24570,   4200,     84, 0, ...  [A062255]
m=5| ...,  126, 0,   42525,  129780,   84630,  12180,    210, 0, ...
m=6| ...,  252, 0,  179487,  446292,  245322,  30492,    462, 0, ...
m=7| ...,  462, 0,  627396, 1315776,  625086,  68376,    924, 0, ...
m=8| ...,  792, 0, 1899612, 3444012, 1440582, 140712,   1716, 0, ...
m=9| ..., 1287, 0, 5135130, 8198190, 3063060, 270270,   3003, 0, ...
          A000389, A112494, A293612, A293613,A293614,A000579.
.
The parameter m runs over the triangles and j indexes the triangles by reading them by rows. Let T(m, n) denote the row [T(m, n, k) for 0 <= k <= n] and T(m) denote the triangle [T(m, n) for n >= 0]. Then for instance T(2) is the triangle A062253, T(4, 2) is row 2 of A062255 (which is [65, 20, 0]) and T(4, 2, 1) = 20.

Eric Weisstein's World of Mathematics, Nielsen Generalized Polylogarithm.

A000217(n) = T(n, 1, 0), A001296(n) = T(n, 2, 0), A000292(n) = T(n, 2, 1),
A001297(n) = T(n, 3, 0), A027789(n) = T(n, 3, 1), A000332(n) = T(n, 3, 2),
A001298(n) = T(n, 4, 0), A293610(n) = T(n, 4, 1), A293611(n) = T(n, 4, 2),
A000389(n) = T(n, 4, 3), A112494(n) = T(n, 5, 0), A293612(n) = T(n, 5, 1),
A293613(n) = T(n, 5, 2), A293614(n) = T(n, 5, 3), A000579(n) = T(n, 5, 4),
A144969(n) = T(n, 6, 0), A000580(n) = T(n, 6, 5), A000295(n) = T(1, n, 1),
A000460(n) = T(1, n, 2), A000498(n) = T(1, n, 3), A000505(n) = T(1, n, 4),
A000514(n) = T(1, n, 5), A001243(n) = T(1, n, 6), A001244(n) = T(1, n, 7),
A126646(n) = T(2, n, 0), A007820(n) = T(n, n, 0).
Cf. A173018, A062253, A062254, A062255, A293609.

A293616 := proc(m, n, k) option remember:
if m = 0 then m^n elif k < 0 or k > n then 0 elif n = 0 then 1 else
(k+m)*A293616(m,n-1,k) + (n-k)*A293616(m,n-1,k-1) + A293616(m-1,n,k) fi end:
for m in [$0..4] do for n in [$0..6] do print(seq(A293616(m, n, k), k=0..n)) od od;
# Sample uses:
A001298 := n -> A293616(n, 4, 0): A293614 := n -> A293616(n, 5, 3):
# Flatten:
a := proc(n) local w; w := proc(k) local t, s; t := 1; s := 1;
while t <= k do s := s + 1; t := t + s od; [s - 1, s - t + k] end:
seq(A293616(n - k, w(k)[1], w(k)[2]), k=0..n) end: seq(a(n), n = 0..11);

GenEulerianRow[0, n_] := Table[If[n==0 && j==0,1,0], {j,0,n}];
GenEulerianRow[m_, n_] := If[n==0,{1},Join[CoefficientList[x^(-m) (1 - x)^(n+m)
    PolyLog[-n-m, m, x] /. Log[1-x] -> 0, x], {0}]];
(* Sample use: *)
A173018Row[n_] := GenEulerianRow[1, n]; Table[A173018Row[n], {n, 0, 6}]

T(m, n, k) = (k + m)*T(m, n-1, k) + (n - k)*T(m, n-1, k-1) + T(m-1, n, k) with boundary conditions T(0, n, k) = 0^n; T(m, n, k) = 0 if k < 0 or k > n; and T(m, 0, k) = 0^k.

Let h(m, n) = x^(-m)*(1 - x)^(n + m)*PolyLog(-n - m, m, x) and p(m, n) the polynomial given by the expansion of h(m, n) after replacing log(1 - x) by 0. Then T(m, n, k) is the k-th coefficient of p(m, n) for 0 <= k < n.

A341694 Square array T(n, k) read by antidiagonals upwards, n, k > 0: T(n, k) = A227736(n, k) for k = 1..A005811(n), and T(n, k) = T(n, k - A005811(n)) + ... + T(n, k-1) for k > A005811(n).

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 2, 2, 2, 1, 1, 1, 2, 3, 1, 1, 1, 3, 2, 5, 1, 3, 2, 1, 4, 2, 8, 1, 3, 3, 3, 3, 7, 2, 13, 1, 1, 1, 3, 5, 5, 11, 2, 21, 1, 1, 2, 4, 3, 8, 9, 18, 2, 34, 1, 2, 1, 1, 5, 3, 13, 17, 29, 2, 55, 1, 2, 1, 1, 4, 9, 3, 21, 31, 47, 2, 89, 1

Offset: 1

Rémy Sigrist, Feb 17 2021

This table contains all Fibonacci sequences of order m > 0 with positive terms:
- order 1 corresponds to constant sequences (n in A126646),
- order 2 corresponds to Fibonacci-like sequences (n in A043569),
- order 3 corresponds to tribonacci-like sequences (n in A043570),
- order 4 corresponds to tetranacci-like sequences (n in A043571).
For any n > 0, the row A341746(n) corresponds to the n-th row from which the first term has been removed.

			Array T(n, k) begins:
  n\k|  1  2  3  4  5   6   7   8   9   10   11   12   13    14
  ---+---------------------------------------------------------
    1|  1  1  1  1  1   1   1   1   1    1    1    1    1     1 --> A000012
    2|  1  1  2  3  5   8  13  21  34   55   89  144  233   377 --> A000045
    3|  2  2  2  2  2   2   2   2   2    2    2    2    2     2 --> A007395
    4|  2  1  3  4  7  11  18  29  47   76  123  199  322   521 --> A000032
    5|  1  1  1  3  5   9  17  31  57  105  193  355  653  1201 --> A000213
    6|  1  2  3  5  8  13  21  34  55   89  144  233  377   610 --> A000045
    7|  3  3  3  3  3   3   3   3   3    3    3    3    3     3 --> A010701
    8|  3  1  4  5  9  14  23  37  60   97  157  254  411   665 --> A104449
    9|  1  2  1  4  7  12  23  42  77  142  261  480  883  1624 --> A275778
   10|  1  1  1  1  4   7  13  25  49   94  181  349  673  1297 --> A000288

Rémy Sigrist, Table of n, a(n) for n = 1..11325
Rémy Sigrist, Colored representation of the table for n, k = 1..1024
Rémy Sigrist, PARI program for A341694
Wikipedia, Fibonacci numbers of higher order
Index entries for sequences related to Fibonacci numbers

Cf. A005811, A043569, A043570, A043571, A126646, A136480, A227736, A341699, A341746.

Cf. A000012, A000032, A000045, A000213, A000288, A007395, A010701, A104449, A275778.

PARI
```
See Links section.
```

T(A341746(n), k) = T(n, k+1).

T(n, 1) = A136480(n).

A354582 Number of distinct contiguous constant subsequences (or partial runs) in the k-th composition in standard order.

Original entry on oeis.org

0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 2, 3, 4, 1, 2, 2, 3, 2, 3, 2, 4, 2, 2, 3, 3, 3, 3, 4, 5, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 2, 3, 5, 2, 2, 3, 3, 3, 3, 2, 4, 3, 3, 4, 3, 4, 4, 5, 6, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 2, 3, 4, 5, 2, 3, 2, 4, 3, 4, 3

Offset: 0

Gus Wiseman, Jun 13 2022

The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.

			Composition number 981 in standard order is (1,1,1,2,2,2,1), with partial runs (1), (2), (1,1), (2,2), (1,1,1), (2,2,2), so a(981) = 6.
As a triangle:
  1
  1 2
  1 2 2 3
  1 2 2 3 2 2 3 4
  1 2 2 3 2 3 2 4 2 2 3 3 3 3 4 5
  1 2 2 3 2 3 3 4 2 3 3 4 3 2 3 5 2 2 3 3 3 3 2 4 3 3 4 3 4 4 5 6

The version for partitions is A001222, full A001221.
If we allow any constant subsequence we get A063787.
If we allow any contiguous subsequence we get A124771.
Positions of first appearances are A126646.
The version for binary indices is A330036, full A005811.
If we allow any subsequence we get A334299.
The full version is A351014, firsts A351015.
The version for run-sums of partitions is A353861, full A353835.
Counting distinct sums of partial runs gives A354907, full A353849.
A066099 lists all compositions in standard order.
A124767 counts runs in standard compositions.
A238279 and A333755 count compositions by number of runs.
A353852 ranks compositions with all distinct run-sums, counted by A353850.
A353853-A353859 pertain to composition run-sum trajectory.
A353932 lists run-sums of standard compositions, rows ranked by A353847.
Cf. A000120, A003242, A029837, A175413, A274174, A333381, A333489, A353832, A353860, A353864.

stc[n_]:=Differences[Prepend[Join@@ Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
pre[y_]:=NestWhileList[Most,y,Length[#]>1&];
Table[Length[Union[Join@@pre/@Split[stc[n]]]],{n,0,100}]

A363398 Triangle read by rows. T(n, k) = [x^k] P(n, x), where P(n, x) = Sum_{k=0..n} 2^(n - k) * Sum_{j=0..k} (x^j * binomial(k, j) * (2*j + 1)^n), (secant case).

Original entry on oeis.org

1, 3, 3, 7, 36, 25, 15, 297, 625, 343, 31, 2106, 10000, 14406, 6561, 63, 13851, 131250, 369754, 413343, 161051, 127, 87480, 1546875, 7529536, 15411789, 14172488, 4826809, 255, 540189, 17109375, 134237509, 444816117, 721025327, 564736653, 170859375

Offset: 0

Peter Luschny, May 31 2023

Here we give an inclusion-exclusion representation of 2^n*Euler(n) (see A122045 and A002436), in A363399 we give such a representation for 2^n*Euler(n, 1) = A155585(n), and in A363400 one for the combined sequences.

			The triangle T(n, k) starts:
  [0]   1;
  [1]   3,      3;
  [2]   7,     36,       25;
  [3]  15,    297,      625,       343;
  [4]  31,   2106,    10000,     14406,      6561;
  [5]  63,  13851,   131250,    369754,    413343,    161051;
  [6] 127,  87480,  1546875,   7529536,  15411789,  14172488,   4826809;
  [7] 255, 540189, 17109375, 134237509, 444816117, 721025327, 564736653, 170859375;

Index entries for sequences related to Euler numbers.

Cf. A122045 (alternating row sums), A363396 (row sums), A126646 (column 0), A085527 (main diagonal), A141475 (central terms).
Cf. A363399 (tangent case), A363400 (combined case).
Cf. A122045, A002436.

P := (n, x) -> add(add(x^j*binomial(k, j)*(2*j + 1)^n, j=0..k)*2^(n-k), k=0..n):
T := (n, k) -> coeff(P(n, x), x, k): seq(seq(T(n, k), k = 0..n), n = 0..7);

(* From Detlef Meya, Oct 04 2023: (Start) *)
T[n_, k_] := (2*k+1)^n*(2^(n+1) - Sum[Binomial[n+1, j], {j,0,k}]);
(* Or: *)
T[n_, k_] := (2*k+1)^n*Binomial[n+1, k+1]*Hypergeometric2F1[1, k-n, k+2, -1];
Flatten[Table[T[n, k], {n, 0, 7}, {k, 0, n}]]  (* End *)

Sum_{k=0..n} (-1)^k*T(n, k) = 2^n*Euler(n) = 4^n*Euler(n, 1/2).
(Sum_{k=0..n} (-1)^k*T(n, k)) / 2^n = Euler(n) = 2^n*Euler(n, 1/2) = A122045(n).
Sum_{k=0..2*n} (-1)^k*T(2*n, k) = 4^n*Euler(2*n) = 16^n*Euler(2*n, 1/2) = (-1)^n*A002436(n).
From Detlef Meya, Oct 04 2023: (Start)
T(n, k) = (2*k + 1)^n * binomial(n+1, k+1) * hypergeom([1, k-n], [k+2], -1).
T(n, k) = (2*k + 1)^n * (2^(n + 1) - Sum_{j=0..k} binomial(n+1, j)). (End)

A363399 Triangle read by rows. T(n, k) = [x^k] P(n, x), where P(n, x) = Sum_{k=0..n} 2^(n - k) * Sum_{j=0..k} (x^j * binomial(k, j) * (j + 1)^n), (tangent case).

Original entry on oeis.org

1, 3, 2, 7, 16, 9, 15, 88, 135, 64, 31, 416, 1296, 1536, 625, 63, 1824, 10206, 22528, 21875, 7776, 127, 7680, 72171, 262144, 453125, 373248, 117649, 255, 31616, 478953, 2670592, 7265625, 10357632, 7411887, 2097152, 511, 128512, 3057426, 25034752, 100000000, 218350080, 265180846, 167772160, 43046721

Offset: 0

Peter Luschny, May 31 2023

Here we give an inclusion-exclusion representation of 2^n*Euler(n, 1) = A155585(n), in A363398 we give such a representation for 2^n*Euler(n), and in A363400 one for the combined sequences.

			The triangle T(n, k) begins:
  [0]   1;
  [1]   3,     2;
  [2]   7,    16,      9;
  [3]  15,    88,    135,      64;
  [4]  31,   416,   1296,    1536,     625;
  [5]  63,  1824,  10206,   22528,   21875,     7776;
  [6] 127,  7680,  72171,  262144,  453125,   373248,  117649;
  [7] 255, 31616, 478953, 2670592, 7265625, 10357632, 7411887, 2097152;

Index entries for sequences related to Euler numbers.

Cf. A155585 (alternating row sums), A363397 (row sums), A126646 (column 0), A000169 (main diagonal), A163395 (central terms), A084623.

Cf. A363398 (secant case), A363400 (combined case).

P := (n, x) -> add(add(x^j*binomial(k, j)*(j + 1)^n, j=0..k)*2^(n - k), k = 0..n):
T := (n, k) -> coeff(P(n, x), x, k): seq(seq(T(n, k), k = 0..n), n = 0..8);

(* From  Detlef Meya, Oct 04 2023: (Start) *)
T[n_, k_] := (k+1)^n*(2^(n+1)-Sum[Binomial[n+1, j], {j, 0, k}]);
(* Or *)
T[n_, k_] := (k+1)^n*Binomial[n+1, k+1]*Hypergeometric2F1[1, k-n, k+2, -1];
Flatten[Table[T[n, k], {n, 0, 7}, {k, 0, n}]]  (* End *)

Sum_{k=0..n} (-1)^k * T(n, k) = 2^n*Euler(n, 1) = (-2)^n*Euler(n, 0) = A155585(n).
From  Detlef Meya, Oct 04 2023: (Start)
T(n, k) = (k + 1)^n*binomial(n + 1, k + 1)*hypergeom([1, k - n], [k + 2], -1).
T(n, k) = (k + 1)^n * (2^(n + 1) - add(binomial(n + 1, j), j=0..k)). (End)

A363400 Triangle read by rows. T(n, k) = [x^k] P(n, x), where P(n, x) = Sum_{k=0..n} 2^(n - k) * Sum_{j=0..k} (x^j * binomial(k, j) * ((2 - (n mod 2)) * j + 1)^n).

Original entry on oeis.org

1, 3, 2, 7, 36, 25, 15, 88, 135, 64, 31, 2106, 10000, 14406, 6561, 63, 1824, 10206, 22528, 21875, 7776, 127, 87480, 1546875, 7529536, 15411789, 14172488, 4826809, 255, 31616, 478953, 2670592, 7265625, 10357632, 7411887, 2097152

Offset: 0

Peter Luschny, May 31 2023

In A363398 we give an inclusion-exclusion representation for 2^n*Euler(n), and in A363399 we give such a representation of 2^n*Euler(n, 1) = A155585(n). Here the two representations are combined into one of A000111.

			Triangle T(n, k) starts:
[0]   1;
[1]   3,     2;
[2]   7,    36,      25;
[3]  15,    88,     135,      64;
[4]  31,  2106,   10000,   14406,     6561;
[5]  63,  1824,   10206,   22528,    21875,     7776;
[6] 127, 87480, 1546875, 7529536, 15411789, 14172488, 4826809;
[7] 255, 31616,  478953, 2670592,  7265625, 10357632, 7411887, 2097152;

Index entries for sequences related to Euler numbers.

Cf. A126646 (column 0), A363401 (row sums), A000111, A059222, A002436.

Cf. A363398 (secant case), A363399 (tangent case).

P := (n, x) -> add(add(x^j * binomial(k, j) * ((2 - irem(n, 2)) * j + 1)^n,
j = 0..k) * 2^(n - k), k = 0..n): T := (n, k) -> coeff(P(n, x), x, k):
seq(seq(T(n, k), k = 0..n), n = 0..8);

From Detlef Meya, Oct 04 2023: (Start)
T[n_, k_] := (2^(n+1)-Binomial[n+1, n-k+1]*Hypergeometric2F1[1, -k, n-k+2, -1])*(2*k+1-k*Mod[n, 2])^n;
(* Or: *)
T[n_, k_] := (2*k+1-k*Mod[n, 2])^(n-1)*Sum[Binomial[n+1, j], {j, 0, n-k}]*(2*k+1-k*Mod[n, 2]);
Flatten[Table[T[n, k], {n, 0, 7}, {k, 0, n}]]  (* End *)

T(n, k) = A363399(n, k) for 0 <= k <= n if n is odd otherwise A363398(n, k).
(Sum_{k=0..n} (-1)^k * T(n, k)) / h(n) = A000111(n), where h(n) = (-1)^binomial(n, 2) * 2^(n * iseven(n)), see A059222.
From Detlef Meya, Oct 04 2023: (Start)
T(n, k) = (2*k + 1 - k*(n mod 2))^(n - 1)*add(binomial(n + 1, j), j = 0..n - k)*(2*k + 1 - k*(n mod 2)).
T(n, k) = (2^(n + 1) - binomial(n + 1, n - k + 1)*hypergeom([1, -k], [n - k + 2], -1))*(2*k + 1 - k*(n mod 2))^n. (End)

A063016 a(n) is the product of Catalan(n) and (2^(n+1) - 1).

Original entry on oeis.org

1, 3, 14, 75, 434, 2646, 16764, 109395, 730730, 4973826, 34381412, 240728670, 1703826292, 12170930700, 87633375480, 635351667075, 4634365164570, 33985474184970, 250419761106900, 1853107999454250, 13765951702923420, 102618937160787060, 767411273728449480

Offset: 0

Olivier Gérard, Jul 04 2001

From Michael Wallner, Nov 03 2021: (Start)
a(n) is also the number of n X 2 Young tableaux with (possibly) vertical walls. The entries in cells that are separated by such a wall do not have to obey any order constraints. See Banderier, Wallner 2021 and Banderier et al. 2018.
a(n) is also the number of binary trees with n vertices and marked leaves, where at least 1 leaf has to be marked. Banderier, Wallner 2021 give a bijection to n X 2 Young tableaux with vertical walls. (End)

Harry J. Smith, Table of n, a(n) for n = 0..200
C. Banderier, P. Marchal, and M. Wallner, Rectangular Young tableaux with local decreases and the density method for uniform random generation, GASCOM 2018. Vol. 2113. CEUR Workshop Proceedings, 2018.
C. Banderier and M. Wallner, Young tableaux with periodic walls: counting with the density method Sém. Lothar. Combin. 85B (2021), Art. 47, 12 pp.

Cf. A000108, A063017, A126646, A007317, A046521.

Table[CatalanNumber[n]*(2^(n+1)-1),{n,0,20}] (* Harvey P. Dale, Oct 20 2014 *)

a(n) = (2^(n + 1) - 1)*binomial(2*n, n)/(n + 1); \\ Harry J. Smith, Aug 16 2009

def A063016(n) :
    return (8^(n+1)-4^(n+1))*factorial(n-1/2)/(4*sqrt(pi)*factorial(n+1))
[A063016(i) for i in (0..20)] # Peter Luschny, Jul 24 2012

a(n) = Catalan(n)*(2^(n+1) - 1) = A000108(n) * A126646(n).
D-finite with recurrence: a(n) = 2*(2*n-1)*(3*n*a(n-1)-4*(2*n-3)*a(n-2))/((n+1)*n). - Georg Fischer, Jun 06 2021
G.f.: A(x) = (sqrt(1-4*x) - sqrt(1-8*x))/(2*x).
G.f.: G(0)/(2*x) where G(k) = 1 - 2^k/(1 - 2*x*(2*k-1)/(2*x*(2*k-1) - 2^k*(k+1)/G(k+1))); (continued fraction, 3rd kind, 3-step). - Sergei N. Gladkovskii, Jul 24 2012
From Peter Bala, Aug 17 2021: (Start)
a(n) = Sum_{k = 0..n} A046521(n,k)*Catalan(k).
G.f.: A(x) = 1/sqrt(1 - 4*x)*c(x/(1 - 4*x)), where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the g.f. of the Catalan numbers A000108. Inversely, c(x) = 1/sqrt(1 + 4*x)*A(x/(1 + 4*x)).
Series reversion of x*A(x) = x*(1 - 3*x + 4*x^2*c(-2*x^2)). (End)

Initial term 0 removed by Harry J. Smith, Aug 16 2009

A126718 a(n) is the number of nonnegative integers k less than 10^n such that the decimal representation of k lacks the digits 1,2,3, at least one of digits 4,5, at least one of digits 6,7 and at least one of digits 8,9.

Original entry on oeis.org

7, 43, 235, 1171, 5467, 24403, 105595, 447091, 1864027, 7686163, 31440955, 127865011, 517788187, 2090186323, 8417944315, 33843570931, 135890057947, 545108340883, 2185079263675, 8754257900851, 35058860433307, 140360940805843, 561820285607035

Offset: 1

Aleksandar M. Janjic and Milan Janjic, Feb 13 2007

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets
Index entries for linear recurrences with constant coefficients, signature (10,-35,50,-24).

Cf. A125910, A125945, A125946, A125947, A125948, A125880, A125630, A125987, A125904, A125858, A125909, A125908, A126646, A126645, A126644, A126643, A126642, A126641, A126640, A126639, A126635, A126634, A126633, A126632, A126631, A126628, A126627.

[8*4^n-12*3^n+6*2^n-1: n in [1..30]]; // Vincenzo Librandi, May 31 2011

Maple
```
a:=n->8*4^n-12*3^n+6*2^n-1;
```

LinearRecurrence[{10,-35,50,-24},{7, 43, 235, 1171},23] (* James C. McMahon, Dec 27 2024 *)

Vec(-x*(24*x^3-50*x^2+27*x-7) / ((x-1)*(2*x-1)*(3*x-1)*(4*x-1)) + O(x^100)) \\ Colin Barker, Feb 22 2015

a(n) = 8*4^n - 12*3^n + 6*2^n - 1.
a(n) = 10*a(n-1) - 35*a(n-2) + 50*a(n-3) - 24*a(n-4). - Colin Barker, Feb 22 2015
G.f.: -x*(24*x^3 - 50*x^2 + 27*x - 7) / ((x-1)*(2*x-1)*(3*x-1)*(4*x-1)). - Colin Barker, Feb 22 2015

A295077 a(n) = 2n(n-1) + 2^n - 1.

Original entry on oeis.org

0, 1, 7, 19, 39, 71, 123, 211, 367, 655, 1203, 2267, 4359, 8503, 16747, 33187, 66015, 131615, 262755, 524971, 1049335, 2097991, 4195227, 8389619, 16778319, 33555631, 67110163, 134219131, 268436967, 536872535, 1073743563, 2147485507

Offset: 0

Franck Maminirina Ramaharo, Nov 13 2017

We have a(0) = 0, and for n > 0, a(n) is a subsequence of A131098 where the indices are given by the partial sums of A288382.
For n > 0, a(n) gives the number of words of length n over the alphabet A = {a,b,c,d} such that: a word containing 'c' does not contain 'b' or 'd'; a word cannot be fully written with 'a'; a word contains letters from {b,d} if and only if it contains exactly a unique couple of letters from {b,d}. Thus a(1) = 1 where the corresponding word is "c" since 'c' is the only letter allowed to be written alone.
Primes in the sequence are 7, 19, 71, 211, 367, 2267, 16747, 524971, ... which are of the form 4*k + 3 (A002145).
The second difference of this sequence is A140504.

			a(4) = 39. The corresponding words are aabb, aabd, aadb, aadd, abab, abad, abba, abda, adab, adad, adba, adda, aaac, aaca, aacc, acaa, acac, acca, accc, baab, baad, baba, bada, bbaa, bdaa, caaa, caac, caca, cacc, ccaa, ccac, ccca, cccc, daab, daad, daba, dada, dbaa, ddaa.

R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics: A Foundation for Computer Science, Addison-Wesley, 1994.

G. C. Greubel, Table of n, a(n) for n = 0..1000 (terms 0..70 from Franck Maminirina Ramaharo)
Franck Ramaharo, Statistics on some classes of knot shadows, arXiv:1802.07701 [math.CO], 2018.
Franck Ramaharo, A generating polynomial for the pretzel knot, arXiv:1805.10680 [math.CO], 2018.
Index entries for linear recurrences with constant coefficients, signature (5,-9,7,-2)

Cf. A000079, A002145, A008586, A046092, A126646, A131098, A131924, A140504, A288382.

[2*n*(n-1)+2^n-1 : n in [0..40]]; // Wesley Ivan Hurt, Nov 26 2017

A295077:=n->2*n*(n-1)+2^n-1; seq(A295077(n), n=0..70);

Table[2 n (n - 1) + 2^n - 1, {n, 0, 70}]

a(n) = 2*n*(n-1) + 2^n - 1; \\ Michel Marcus, Nov 14 2017

G.f.: (x + 2*x^2 - 7*x^3)/((1 - x)^3*(1 - 2*x)).
a(0)=0, a(1)=1, a(2)=7, a(3)=19; for n>3, a(n) = 5*a(n-1) - 9*a(n-2) + 7*a(n-3) - 2*a(n-4).
a(n) = 2*A131924(n-1) - 1 for n>0, a(0)=0.
a(n) = a(n-1) + A000079(n-1) + A008586(n-1) for n>0, a(0)=0.
a(n) = A126646(n-1) + A046092(n-1) for n>0, a(0)=0.
a(n+1) - 2*a(n) + a(n-1) = A140504(n-1) for n>0, a(0)=0.
E.g.f.: exp(2*x) - (1 - 2*x^2)*exp(x). - G. C. Greubel, Oct 17 2018

A319074 a(n) is the sum of the first n nonnegative powers of the n-th prime.

Original entry on oeis.org

1, 4, 31, 400, 16105, 402234, 25646167, 943531280, 81870575521, 15025258332150, 846949229880161, 182859777940000980, 23127577557875340733, 1759175174860440565844, 262246703278703657363377, 74543635579202247026882160, 21930887362370823132822661921, 2279217547342466764922495586798

Offset: 1

Omar E. Pol, Sep 11 2018

nonn

			For n = 4 the 4th prime is 7 and the sum of the first four nonnegative powers of 7 is 7^0 + 7^1 + 7^2 + 7^3 = 1 + 7 + 49 + 343 = 400, so a(4) = 400.

Main diagonal of A319076.
Cf. A000040, A000203, A006093, A062457, A069459, A093360, A319075.
Cf. A000079, A000244, A000351, A000420, A001020, A001022, A001026, A001029, A009967, A009973, A009975, A009981, A009985, A009987, A009991.
Cf. A126646, A003462, A003463, A023000, A016123, A091030, A091045, A218722, A218726, A218732, A218734, A218740, A218744, A218746, A218750.

a(n) = sum(k=0, n-1, prime(n)^k); \\ Michel Marcus, Sep 13 2018

a(n) = Sum_{k=0..n-1} A000040(n)^k.
a(n) = Sum_{k=0..n-1} A319075(k,n).
a(n) = (A000040(n)^n - 1)/(A000040(n) - 1).
a(n) = (A062457(n) - 1)/A006093(n).
a(n) = A069459(n)/A006093(n).
a(n) = A000203(A000040(n)^(n-1)).
a(n) = A000203(A093360(n)).

cp's OEIS Frontend

A293616 Array of generalized Eulerian number triangles read by ascending antidiagonals, with m >= 0, n >= 0 and 0 <= k <= n.

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A341694 Square array T(n, k) read by antidiagonals upwards, n, k > 0: T(n, k) = A227736(n, k) for k = 1..A005811(n), and T(n, k) = T(n, k - A005811(n)) + ... + T(n, k-1) for k > A005811(n).

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A354582 Number of distinct contiguous constant subsequences (or partial runs) in the k-th composition in standard order.

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A363398 Triangle read by rows. T(n, k) = [x^k] P(n, x), where P(n, x) = Sum_{k=0..n} 2^(n - k) * Sum_{j=0..k} (x^j * binomial(k, j) * (2*j + 1)^n), (secant case).

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A363399 Triangle read by rows. T(n, k) = [x^k] P(n, x), where P(n, x) = Sum_{k=0..n} 2^(n - k) * Sum_{j=0..k} (x^j * binomial(k, j) * (j + 1)^n), (tangent case).

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A363400 Triangle read by rows. T(n, k) = [x^k] P(n, x), where P(n, x) = Sum_{k=0..n} 2^(n - k) * Sum_{j=0..k} (x^j * binomial(k, j) * ((2 - (n mod 2)) * j + 1)^n).

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A063016 a(n) is the product of Catalan(n) and (2^(n+1) - 1).

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A295077 a(n) = 2n(n-1) + 2^n - 1.