A022521 -id:A022521 - cp's OEIS frontend

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

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

A341050 Cube array read by upward antidiagonals ignoring zero and empty terms: T(n, k, r) is the number of n-ary strings of length k, containing r consecutive 0's.

Original entry on oeis.org

1, 1, 1, 3, 1, 1, 3, 1, 5, 8, 1, 1, 3, 1, 5, 8, 1, 7, 21, 19, 1, 1, 3, 1, 5, 8, 1, 7, 21, 20, 1, 9, 40, 81, 43, 1, 1, 3, 1, 5, 8, 1, 7, 21, 20, 1, 9, 40, 81, 47, 1, 11, 65, 208, 295, 94, 1, 1, 3, 1, 5, 8, 1, 7, 21, 20, 1, 9, 40, 81, 48, 1, 11, 65, 208, 297, 107, 1, 13, 96, 425, 1024, 1037, 201

Offset: 2

Robert P. P. McKone, Feb 04 2021

			For n = 5, k = 6 and r = 4, there are 65 strings: {000000, 000001, 000002, 000003, 000004, 000010, 000011, 000012, 000013, 000014, 000020, 000021, 000022, 000023, 000024, 000030, 000031, 000032, 000033, 000034, 000040, 000041, 000042, 000043, 000044, 010000, 020000, 030000, 040000, 100000, 100001, 100002, 100003, 100004, 110000, 120000, 130000, 140000, 200000, 200001, 200002, 200003, 200004, 210000, 220000, 230000, 240000, 300000, 300001, 300002, 300003, 300004, 310000, 320000, 330000, 340000, 400000, 400001, 400002, 400003, 400004, 410000, 420000, 430000, 440000}
The first seven slices of the tetrahedron (or pyramid) are:
-----------------Slice 1-----------------
  1
-----------------Slice 2-----------------
    1
  1  3
-----------------Slice 3-----------------
      1
    1  3
  1  5  8
-----------------Slice 4-----------------
        1
      1  3
    1  5   8
  1  7  21  19
-----------------Slice 5-----------------
          1
        1  3
      1  5   8
    1  7  21  20
  1  9  40  81  43
-----------------Slice 6-----------------
              1
           1    3
        1    5     8
      1   7    21    20
    1   9   40    81    47
  1  11  65   208   295   94
-----------------Slice 7-----------------
                 1
              1     3
           1     5     8
         1    7     21    20
      1    9    40     81      48
    1   11   65    208     297     107
  1  13   96   425    1024    1037    201

Robert P. P. McKone, Antidiagonals n = 2..50, flattened

Cf. A340156 (r=2), A340242 (r=3).
Cf. A008466 (n=2, r=2), A186244 (n=3, r=2), A050231 (n=2, r=3), A231430 (n=3, r=3).
Cf. A005408, A003215, A005917, A022521, A022522, A022523, A022524, A022525, A022526, A022527, A022528, A022529, A022530, A022531, A022532, A022533, A022534, A022535, A022536, A022537, A022538, A022539, A022540 (k=x, r=1, where x is the x-th Nexus Number).
Cf. A000567 [(k=4, r=2),(k=5, r=3),(k=6, r=4),...,(k=x, r=x-2)].
Cf. A103532 [(k=6, r=3),(k=7, r=4),(k=8, r=5),...,(k=x, r=x-3)].

m[r_, n_] := Normal[With[{p = 1/n}, SparseArray[{Band[{1, 2}] -> p, {i_, 1} /; i <= r -> 1 - p, {r + 1, r + 1} -> 1}]]]; T[n_, k_, r_] := MatrixPower[m[r, n], k][[1, r + 1]]*n^k; DeleteCases[Transpose[PadLeft[Reverse[Table[T[n, k, r], {k, 2, 8}, {r, 2, k}, {n, 2, r}], 2]], 2 <-> 3], 0, 3] // Flatten

A069478 First differences of A069477, successive differences of (n+1)^5 - n^5.

Original entry on oeis.org

360, 480, 600, 720, 840, 960, 1080, 1200, 1320, 1440, 1560, 1680, 1800, 1920, 2040, 2160, 2280, 2400, 2520, 2640, 2760, 2880, 3000, 3120, 3240, 3360, 3480, 3600, 3720, 3840, 3960, 4080, 4200, 4320, 4440, 4560, 4680, 4800, 4920, 5040, 5160, 5280

Offset: 1

Eli McGowan (ejmcgowa(AT)mail.lakeheadu.ca), Apr 11 2002

nonn

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A022521, A068236, A069477.

[120*(n+2): n in [1..50]]; // G. C. Greubel, Nov 11 2018

Differences[#[[2]]-#[[1]]&/@Partition[Range[50]^5,2,1],3] (* Harvey P. Dale, Jan 28 2013 *)
120(Range[50] +2) (* G. C. Greubel, Nov 11 2018 *)

vector(50, n, 120*(n+2)) \\ G. C. Greubel, Nov 11 2018

a(n) = 120*(n+2).
From Chai Wah Wu, Nov 11 2018: (Start)
a(n) = 2*a(n-1) - a(n-2) for n > 2.
G.f.: x*(-240*x + 360)/(x - 1)^2. (End)
E.g.f.: 120*(-2 + (2+x)*exp(x)). - G. C. Greubel, Nov 11 2018

Corrected by T. D. Noe, Mar 14 2008

A353890 a(n) is the period of the binary sequence {b(m)} defined by b(m) = 1 if (m+1)^n - m^n and (m+2)^n - 2*(m+1)^n + m^n are coprime, 0 otherwise.

Original entry on oeis.org

1, 1, 5, 11, 91, 1247, 3485, 263017, 852841, 1241058127, 74966255, 243641132605417, 181556731572385303, 718802057694183783881, 6582662048285, 943422576750791493013356207217, 487331778345355477261, 607088607861933740557075591887834842297

Offset: 2

Samuel Harkness, May 09 2022

nonn

For any n, consecutive n-th powers will never share a divisor > 1, so now consider the second differences. Specifically, each m > 0, define the binary sequence {b(m)} as follows: b(m) = 1 if the first difference (m+1)^n - m^n and the second difference (m+2)^n - 2*(m+1)^n + m^n are coprime, 0 otherwise. I conjecture that {b(m)} is periodic with period a(n).
If m^n mod p == (m+1)^n mod p == (m+2)^n mod p, then p is in the prime factorization of a(n).
All primes p >= 5 belong to a prime factorization for a(n). p will always belong to the prime factorization of n=p-1 due to Fermat's Little Theorem.
I conjecture that the greatest prime factor for any prime n >= 5 is phi(2^n+1)/2 + 1 = Jacobsthal(n). n*A069925 + 1 = A001045(n).
I conjecture that all prime factors "f" are f=n*k+1, unless n is composite, in which case additionally all prime factors for any divisor of n will also be included in the prime factorization for a(n).

			For n=2 and n=3, the first and second differences are coprime for all m. Each of their sequences {b(m)} consist only of 1's, which can be described trivially as [1] with a period of 1, so a(2) = a(3) = 1.
For n > 3, the first and second differences are coprime for some m values, but not for all. Each repeating periodic sequence {b(m)} begins at m=1, and can be used to predict what b(m) will be at any higher m value for that power n.
n=4 has the 5-term repeating sequence, beginning at m=1:
  [0 0 1 1 1], so a(4) = 5.
The sequence is repeating, so for example, f(41)..f(45) is also [0 0 1 1 1].
n=5 has the 11-term repeating sequence
  [1 1 0 1 1 0 1 1 1 1 1]
so a(5) = 11.
n=6 has the 91-term repeating sequence
  [0 0 0 0 0 0 1 0 0 0 0 1 1
   1 0 0 0 0 0 1 1 0 0 0 0 1
   1 1 0 0 0 0 1 1 1 0 0 0 0
   1 1 1 0 0 0 0 1 1 1 0 0 0
   0 1 1 1 0 0 0 0 1 1 1 0 0
   0 0 1 1 0 0 0 0 0 1 1 1 0
   0 0 0 1 0 0 0 0 0 0 1 1 1]
so a(6) = 91.
The period for higher n values has yet to be found. If they exist, it seems they would be quite large given the large expansion from 5, 11, to 91.
Example: the 233rd term in the sequence of values for n=6 is calculated by using m=233 and n=6. Define the first difference for the 233rd term as 234^6 - 233^6 = 4164782373647. The second difference for the 233rd term is 235^6 - 2*234^6 + 233^6 = 89948228762. The terms 4164782373647 and 89948228762 share a common factor, so the 233rd term of the sequence for 6th powered terms is denoted 0 (not coprime). Because the 6th powered terms repeat their tendency of being coprime or not every 91 terms, we could instead look at 233 mod 91 = 51, and from the table for n=6 above, the 51st term is 0.

Samuel Harkness, MATLAB program

Cf. A005408, A007395, A003215, A008588, A005917, A005914, A022521, A068236, A022522, A069473, A069925, A001045, A002587.

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a(7)-a(19) from Jon E. Schoenfield, May 10 2022

A385897 a(n) = 1 - 5(n + 1)^2 + 5(n + 1)^4.

Original entry on oeis.org

1, 61, 361, 1201, 3001, 6301, 11761, 20161, 32401, 49501, 72601, 102961, 141961, 191101, 252001, 326401, 416161, 523261, 649801, 798001, 970201, 1168861, 1396561, 1656001, 1950001, 2281501, 2653561, 3069361, 3532201, 4045501, 4612801, 5237761, 5924161, 6675901

Offset: 0

Peter Luschny, Jul 21 2025

Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A385896 (column 5), A063497, A000290, A061317, A022521, A008512.

gf := (-x^4 + 4*x^3 - 66*x^2 - 56*x - 1)/(x - 1)^5:
ser := series(gf, x, 35): seq(coeff(ser, x, n), n = 0..33);

a[n_] := With[{h = (n + 1)^2}, 5 (h^2 - h) + 1]; Table[a[n], {n, 0, 33}]

a(n) = [x^n] (-x^4 + 4*x^3 - 66*x^2 - 56*x - 1)/(x - 1)^5.
a(n) = 5! * [x^5] (1 - sin(n*x))^(-1/n) for n > 0.
a(n) = A385896(n + 5, 5).
A000290(n) = (a(n) - 2*a(n-1) + a(n-2)) / 60.
A008512(n) = (a(n) + 2*a(n-1) + a(n-2)) / 2.
A022521(n) = (a(n-1) + a(n)) / 2.
A061317(n) = (a(n) - a(n-2)) / 20.
A063497(n) = a(n) - a(n-1).
gcd(a(n), a(n+1)) = 1.

cp's OEIS Frontend

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

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A341050 Cube array read by upward antidiagonals ignoring zero and empty terms: T(n, k, r) is the number of n-ary strings of length k, containing r consecutive 0's.

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A069478 First differences of A069477, successive differences of (n+1)^5 - n^5.

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A353890 a(n) is the period of the binary sequence {b(m)} defined by b(m) = 1 if (m+1)^n - m^n and (m+2)^n - 2*(m+1)^n + m^n are coprime, 0 otherwise.

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A385897 a(n) = 1 - 5*(n + 1)^2 + 5*(n + 1)^4.

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A385897 a(n) = 1 - 5(n + 1)^2 + 5(n + 1)^4.