A053541 -id:A053541 - cp's OEIS frontend

A063013 Numbers where k-th digit from right is either 0 or k.

0, 1, 20, 21, 300, 301, 320, 321, 4000, 4001, 4020, 4021, 4300, 4301, 4320, 4321, 50000, 50001, 50020, 50021, 50300, 50301, 50320, 50321, 54000, 54001, 54020, 54021, 54300, 54301, 54320, 54321, 600000, 600001, 600020, 600021, 600300, 600301, 600320, 600321

Offset: 0

Henry Bottomley, Jul 04 2001

a(511) = 987654321 is the last term.

			 a(11) = 4021 since 11 is written as 1011 in binary and the 1's can then be replaced by the relevant digits.

Alois P. Heinz, Table of n, a(n) for n = 0..511

Cf. A063012, A053541.

a:= n-> (l-> parse(cat(seq(l[-i]*(1+nops(l)-i),
         i=1..nops(l)))))(convert(n, base, 2)):
seq(a(n), n=0..2^6-1);  # Alois P. Heinz, Oct 29 2018

FromDigits /@ (Range[9,1,-1]*# & /@ IntegerDigits[Range[0, 511], 2, 9]) (* Giovanni Resta, Oct 28 2018 *)

a(n) = fromdigits(Vec(Pol(binary(2*n))')); \\ Alan Michael Gómez Calderón, May 01 2025

a(n) = a(n-2^floor(log_2(n))) + floor(log_2(n)+1)*10^floor(log_2(n)).

a(n) = Sum_{k>=0} A030308(n,k)*A053541(k+1). - Philippe Deléham, Oct 15 2011

A317055 Triangle read by rows: T(0,0) = 1; T(n,k) = 10*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, 10, 100, 1, 1000, 20, 10000, 300, 1, 100000, 4000, 30, 1000000, 50000, 600, 1, 10000000, 600000, 10000, 40, 100000000, 7000000, 150000, 1000, 1, 1000000000, 80000000, 2100000, 20000, 50, 10000000000, 900000000, 28000000, 350000, 1500, 1, 100000000000, 10000000000, 360000000, 5600000, 35000, 60

Offset: 0

Zagros Lalo, Jul 21 2018

The numbers in rows of the triangle are along skew diagonals pointing top-left in center-justified triangle given in A013617 ((1+10*x)^n) and along skew diagonals pointing top-right in center-justified triangle given in A038303 ((10+x)^n).
The coefficients in the expansion of 1/(1-10*x-x^2) are given by the sequence generated by the row sums.
The row sums are Denominators of continued fraction convergents to sqrt(26), see A041041.
If s(n) is the row sum at n, then the ratio s(n)/s(n-1) is approximately 10.09901951359278483002... (a metallic mean) when n approaches infinity (see A176537: (5+sqrt(26))).

			Triangle begins:
  1;
  10;
  100, 1;
  1000, 20;
  10000, 300, 1;
  100000, 4000, 30;
  1000000, 50000, 600, 1;
  10000000, 600000, 10000, 40;
  100000000, 7000000, 150000, 1000, 1;
  1000000000, 80000000, 2100000, 20000, 50;
  10000000000, 900000000, 28000000, 350000, 1500, 1;
  100000000000, 10000000000, 360000000, 5600000, 35000, 60;

Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, pp. 70, 102

Zagros Lalo, Left-justified triangle
Zagros Lalo, Skew diagonals in center-justified triangle of coefficients in expansion of (1 + 10x)^n
Zagros Lalo, Skew diagonals in center-justified triangle of coefficients in expansion of (10 + x)^n

Row sums give A041041.
Cf. A013617, A038303, A176537.
Cf. A011557 (column 0), A053541 (column 1), A081140 (column 2).

t[0, 0] = 1; t[n_, k_] := t[n, k] = If[n < 0 || k < 0, 0, 10 t[n - 1, k] + t[n - 2, k - 1]]; Table[t[n, k], {n, 0, 11}, {k, 0, Floor[n/2]}] // Flatten

A096302 Number of combinations of two natural numbers that together have n digits.

Original entry on oeis.org

81, 1620, 24300, 324000, 4050000, 48600000, 567000000, 6480000000, 72900000000, 810000000000, 8910000000000, 97200000000000, 1053000000000000, 11340000000000000, 121500000000000000, 1296000000000000000

Offset: 2

Hagai Helman (Helman(AT)actcom.net.il), Jun 25 2004

Vincenzo Librandi, Table of n, a(n) for n = 2..100

Cf. A053541.

[81*(n-1)*10^(n-2): n in [2..30]]; // Vincenzo Librandi, Jun 06 2011

f[n_] := Sum[Binomial[n - 1, i]*9^(i + 1)*i, {i, 0, n}]; Table[ a[n], {n, 2, 17}] (* Robert G. Wilson v, Jun 30 2004 *)

a(n) = Sum_{i=0..n-1} binomial(n-1, i)*9^(i+1)*i.

a(n) = 81*(n-1)*10^(n-2). - Vladeta Jovovic, Jun 26 2004

More terms from Robert G. Wilson v, Jun 30 2004

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).

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A063013 Numbers where k-th digit from right is either 0 or k.

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A317055 Triangle read by rows: T(0,0) = 1; T(n,k) = 10*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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A096302 Number of combinations of two natural numbers that together have n digits.

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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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