A053464 -id:A053464 - cp's OEIS frontend

A230539 a(n) = 3n2^(3*n-1).

0, 12, 192, 2304, 24576, 245760, 2359296, 22020096, 201326592, 1811939328, 16106127360, 141733920768, 1236950581248, 10720238370816, 92358976733184, 791648371998720, 6755399441055744, 57420895248973824, 486388759756013568, 4107282860161892352

Offset: 0

Bruno Berselli, Oct 23 2013

Arithmetic derivative of 8^n: a(n) = A003415(8^n).

Sum of reciprocals of a(n), for n>0: (2/3)*log(8/7).

Bruno Berselli, Table of n, a(n) for n = 0..100
Index entries for linear recurrences with constant coefficients, signature (16,-64).

Cf. A001018, A003415.
Cf. arithmetic derivative of k^n: A001787 (k=2), A027471 (k=3), A018215 (k=4), A053464 (k=5), A212700 (k=6), A027473 (k=7), this sequence, A230540 (k=9), A085708 (k=10), A081127 (k=11).
Row n=8 of A258997.

Magma
```
[3*n*2^(3*n-1): n in [0..20]];
```

A230539:=n->3*n*2^(3*n-1): seq(A230539(n), n=0..30); # Wesley Ivan Hurt, May 03 2017

Table[3 n 2^(3 n - 1), {n,0,20}]
LinearRecurrence[{16,-64},{0,12},20] (* Harvey P. Dale, Dec 25 2022 *)

a(n) = 3*n*2^(3*n-1); \\ Michel Marcus, Oct 23 2013

G.f.: 12*x/(1-8*x)^2.

a(n) = 12*A053539(n).

A230540 a(n) = 2n3^(2*n-1).

Original entry on oeis.org

0, 6, 108, 1458, 17496, 196830, 2125764, 22320522, 229582512, 2324522934, 23245229340, 230127770466, 2259436291848, 22029503845518, 213516729579636, 2058911320946490, 19765548681086304, 189008059262887782, 1801135623563989452, 17110788423857899794

Offset: 0

Bruno Berselli, Oct 23 2013

Arithmetic derivative of 9^n: a(n) = A003415(9^n).

Sum of reciprocals of a(n), for n>0: (3/2)*log(9/8).

Bruno Berselli, Table of n, a(n) for n = 0..100
Index entries for linear recurrences with constant coefficients, signature (18,-81).

Cf. A001019, A003415.

Cf. arithmetic derivative of k^n: A001787 (k=2), A027471 (k=3), A018215 (k=4), A053464 (k=5), A212700 (k=6), A027473 (k=7), A230539 (k=8), this sequence, A085708 (k=10), A081127 (k=11).

Magma
```
[2*n*3^(2*n-1): n in [0..20]];
```
Mathematica
```
Table[2 n 3^(2 n - 1), {n, 0, 20}]
```

a(n) = 2*n*3^(2*n-1); \\ Michel Marcus, Oct 23 2013

G.f.: 6*x/(1-9*x)^2.

a(n) = 6*A053540(n), with A053540(0)=0.

A305837 Triangle read by rows: T(0,0) = 1; T(n,k) = 5*T(n-1,k) + T(n-2,k-1) for k = 0..floor(n/2); T(n,k)=0 for n or k < 0.

Original entry on oeis.org

1, 5, 25, 1, 125, 10, 625, 75, 1, 3125, 500, 15, 15625, 3125, 150, 1, 78125, 18750, 1250, 20, 390625, 109375, 9375, 250, 1, 1953125, 625000, 65625, 2500, 25, 9765625, 3515625, 437500, 21875, 375, 1, 48828125, 19531250, 2812500, 175000, 4375, 30, 244140625, 107421875, 17578125, 1312500, 43750, 525, 1

Offset: 0

Shara Lalo, Jun 11 2018

The numbers in rows of the triangle are along skew diagonals pointing top-left in center-justified triangle given in A013612 ((1+5*x)^n).
The coefficients in the expansion of 1/(1-5x-x^2) are given by the sequence generated by the row sums.
If s(n) is the row sum at n, then the ratio s(n)/s(n-1) is approximately 5.1925824035..., a metallic mean (see A098318), when n approaches infinity.

			Triangle begins:
            1;
            5;
           25,           1;
          125,          10;
          625,          75,          1;
         3125,         500,         15;
        15625,        3125,        150,         1;
        78125,       18750,       1250,        20;
       390625,      109375,       9375,       250,        1;
      1953125,      625000,      65625,      2500,       25;
      9765625,     3515625,     437500,     21875,      375,      1;
     48828125,    19531250,    2812500,    175000,     4375,     30;
    244140625,   107421875,   17578125,   1312500,    43750,    525,     1;
   1220703125,   585937500,  107421875,   9375000,   393750,   7000,    35;
   6103515625,  3173828125,  644531250,  64453125,  3281250,  78750,   700,  1;
  30517578125, 17089843750, 3808593750, 429687500, 25781250, 787500, 10500, 40;

Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, pp. 70, 72, 92, 380, 382.

Shara Lalo, Left-justified triangle
Shara Lalo, Skew diagonals in triangle A013612

Row sums give A052918.
Cf. A000351 (column 0), A053464 (column 1), A081135 (column 2), A081143 (column 3), A036071 (column 4).
Cf. A013612.
Cf. A098318.

t[0, 0] = 1; t[n_, k_] := If[n < 0 || k < 0, 0, 5 t[n - 1, k] + t[n - 2, k - 1]]; Table[t[n, k], {n, 0, 12}, {k, 0, Floor[n/2]}] // Flatten

G.f.: 1/(1 - 5*t*x - t^2).

A173113 a(n) = binomial(n + 10, 10) * 5^n.

Original entry on oeis.org

1, 55, 1650, 35750, 625625, 9384375, 125125000, 1519375000, 17092968750, 180425781250, 1804257812500, 17222460937500, 157872558593750, 1396564941406250, 11970556640625000, 99754638671875000

Offset: 0

Zerinvary Lajos, Feb 10 2010

With a different offset, number of n-permutations (n>=10) of 6 objects: t, u, v, z, x, y with repetition allowed, containing exactly ten (10) u's.

Vincenzo Librandi, Table of n, a(n) for n = 0..400
Index entries for linear recurrences with constant coefficients, signature (55,-1375,20625,-206250,1443750,-7218750,25781250,-64453125,107421875,-107421875,48828125).

Cf. A053464, A084902, A081143, A036071, A050982, A140405, A140220, A140346, A140520.

[5^n*Binomial(n+10, 10): n in [0..30]]; // Vincenzo Librandi, Oct 15 2011

Table[Binomial[n + 10, 10]*5^n, {n, 0, 20}]

a(n) = C(n + 10, 10)*5^n, n>=0.
G.f.: 1/(1-5*x)^11. - Vincenzo Librandi, Oct 15 2011
From Amiram Eldar, Sep 01 2022: (Start)
Sum_{n>=0} 1/a(n) = 184261655/63 - 13107200*log(5/4).
Sum_{n>=0} (-1)^n/a(n) = 503884800*log(6/5) - 11575501585/126. (End)

A320531 T(n,k) = n*k^(n - 1), k > 0, with T(n,0) = A063524(n), square array read by antidiagonals upwards.

Original entry on oeis.org

0, 1, 0, 0, 1, 0, 0, 2, 1, 0, 0, 3, 4, 1, 0, 0, 4, 12, 6, 1, 0, 0, 5, 32, 27, 8, 1, 0, 0, 6, 80, 108, 48, 10, 1, 0, 0, 7, 192, 405, 256, 75, 12, 1, 0, 0, 8, 448, 1458, 1280, 500, 108, 14, 1, 0, 0, 9, 1024, 5103, 6144, 3125, 864, 147, 16, 1, 0, 0, 10, 2304

Offset: 0

Franck Maminirina Ramaharo, Oct 14 2018

T(n,k) is the number of length n*k binary words of n consecutive blocks of length k, respectively, one of the blocks having exactly k letters 1, and the other having exactly one letter 0. First column follows from the next definition.
In Kauffman's language, T(n,k) is the total number of Jordan trails that are obtained by placing state markers at the crossings of the Pretzel universe P(k, k, ..., k) having n tangles, of k half-twists respectively. In other words, T(n,k) is the number of ways of splitting the crossings of the Pretzel knot shadow P(k, k, ..., k) such that the final diagram is a single Jordan curve. The aforementionned binary words encode these operations by assigning each tangle a length k binary words with the adequate choice for splitting the crossings.
Columns are linear recurrence sequences with signature (2*k, -k^2).

			Square array begins:
    0, 0,   0,    0,     0,      0,      0,      0, ...
    1, 1,   1,    1,     1,      1,      1,      1, ...
    0, 2,   4,    6,     8,     10,     12,     14, ... A005843
    0, 3,  12,   27,    48,     75,    108,    147, ... A033428
    0, 4,  32,  108,   256,    500,    864,   1372, ... A033430
    0, 5,  80,  405,  1280,   3125,   6480,  12005, ... A269792
    0, 6, 192, 1458,  6144,  18750,  46656, 100842, ...
    0, 7, 448, 5103, 28672, 109375, 326592, 823543, ...
    ...
T(3,2) = 3*2^(3 - 1) = 12. The corresponding binary words are 110101, 110110, 111001, 111010, 011101, 011110, 101101, 101110, 010111, 011011, 100111, 101011.

Louis H. Kauffman, Formal Knot Theory, Princeton University Press, 1983.

Louis H. Kauffman, State models and the Jones polynomial, Topology, Vol. 26 (1987), 395-407.
Franck Ramaharo, A generating polynomial for the pretzel knot, arXiv:1805.10680 [math.CO], 2018.
Alexander Stoimenow, Everywhere Equivalent 2-Component Links, Symmetry Vol. 7 (2015), 365-375.
Wikipedia, Pretzel link

Antidiagonal sums: A101495.
Column 1 is column 2 of A300453.
Column 2 is column 1 of A300184.
Cf. A104002, A320530.

T[n_, k_] = If [k > 0, n*k^(n - 1), If[k == 0 && n == 1, 1, 0]];
Table[Table[T[n - k, k], {k, 0, n}], {n, 0, 12}]//Flatten

T(n, k) := if k > 0 then n*k^(n - 1) else if k = 0 and n = 1 then 1 else 0$
tabl(nn) := for n:0 thru nn do print(makelist(T(n, k), k, 0, nn))$

T(n,k) = (2*k)*T(n-1,k) - (k^2)*T(n-2,k).
G.f. for columns: x/(1 - k*x)^2.
E.g.f. for columns: x*exp(k*x).
T(n,1) = A001477(n).
T(n,2) = A001787(n).
T(n,3) = A027471(n+1).
T(n,4) = A002697(n).
T(n,5) = A053464(n).
T(n,6) = A053469(n), n > 0.
T(n,7) = A027473(n), n > 0.
T(n,8) = A053539(n).
T(n,9) = A053540(n), n > 0.
T(n,10) = A053541(n), n > 0.
T(n,11) = A081127(n).
T(n,12) = A081128(n).

cp's OEIS Frontend

A230539 a(n) = 3*n*2^(3*n-1).

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A230540 a(n) = 2*n*3^(2*n-1).

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A305837 Triangle read by rows: T(0,0) = 1; T(n,k) = 5*T(n-1,k) + T(n-2,k-1) for k = 0..floor(n/2); T(n,k)=0 for n or k < 0.

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A173113 a(n) = binomial(n + 10, 10) * 5^n.

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Magma

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A320531 T(n,k) = n*k^(n - 1), k > 0, with T(n,0) = A063524(n), square array read by antidiagonals upwards.

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A230539 a(n) = 3n2^(3*n-1).

A230540 a(n) = 2n3^(2*n-1).