A255744 -id:A255744 - cp's OEIS frontend

A255765 Partial sums of A255744.

1, 11, 21, 111, 121, 211, 301, 1111, 1121, 1211, 1301, 2111, 2201, 3011, 3821, 11111, 11121, 11211, 11301, 12111, 12201, 13011, 13821, 21111, 21201, 22011, 22821, 30111, 30921, 38211, 45501, 111111, 111121, 111211, 111301, 112111, 112201, 113011, 113821, 121111

Offset: 1

Omar E. Pol, Mar 05 2015

nonn

Also, this is a row of the square array A255741.

Is this sequence related to positive repunits? (see formula section).

Felix Fröhlich, Table of n, a(n) for n = 1..10000
Hsien-Kuei Hwang, Svante Janson, and Tsung-Hsi Tsai, Identities and periodic oscillations of divide-and-conquer recurrences splitting at half, arXiv:2210.10968 [cs.DS], 2022, p. 33.

Cf. A002275, A005408, A151788, A147562, A151790, A151781, A151792, A151793, A255740, A255741, A255744, A255764, A255766.

Accumulate@ MapAt[Floor, Array[10*9^(DigitCount[# - 1, 2, 1] - 1) &, 40], 1] (* Michael De Vlieger, Nov 03 2022 *)

lista(nn) = {s = 1; for (n=2, nn, print1(s, ", "); s += 10*9^(hammingweight(n-1)-1););} \\ Michel Marcus, Mar 15 2015

a(n) = sum(k=1, n, if (k==1, 1, 10*9^(hammingweight(k-1)-1))); \\ Michel Marcus, Mar 15 2015

Question: a(2^k) = A002275(k+1), k >= 0. Is this true?

More terms from Michel Marcus, Mar 15 2015

A255740 Square array read by antidiagonals upwards: T(n,1) = 1; for k > 1, T(n,k) = (n-1)*(n-2)^(A000120(k-1)-1) with n >= 1.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 2, 0, 0, 1, 4, 3, 2, 1, 0, 1, 5, 4, 6, 2, 0, 0, 1, 6, 5, 12, 3, 2, 0, 0, 1, 7, 6, 20, 4, 6, 2, 0, 0, 1, 8, 7, 30, 5, 12, 6, 2, 1, 0, 1, 9, 8, 42, 6, 20, 12, 12, 2, 0, 0, 1, 10, 9, 56, 7, 30, 20, 36, 3, 2, 0, 0, 1, 11, 10, 72, 8, 42, 30, 80, 4, 6, 2, 0, 0, 1, 12, 11, 90, 9, 56, 42, 150, 5, 12, 6, 2, 0, 0

Offset: 1

Omar E. Pol, Mar 05 2015

The partial sums of row n give the n-th row of the square array A255741.

			The corner of the square array with the first 16 terms of the first 12 rows looks like this:
-------------------------------------------------------------------------
A000007: 1, 0, 0,  0, 0,  0,  0,   0, 0,  0,  0,   0,  0,   0,   0,    0
A255738: 1, 1, 1,  0, 1,  0,  0,   0  1,  0,  0,   0,  0,   0,   0,    0
A040000: 1, 2, 2,  2, 2,  2,  2,   2, 2,  2,  2,   2,  2,   2,   2,    2
A151787: 1, 3, 3,  6, 3,  6,  6,  12, 3,  6,  6,  12,  6,  12,  12,   24
A147582: 1, 4, 4, 12, 4, 12, 12,  36, 4, 12, 12,  36, 12,  36,  36,  108
A151789: 1, 5, 5, 20, 5, 20, 20,  80, 5, 20, 20,  80, 20,  80,  80,  320
A151779: 1, 6, 6, 30, 6, 30, 30, 150, 6, 30, 30, 150, 30, 150, 150,  750
A151791: 1, 7, 7, 42, 7, 42, 42, 252, 7, 42, 42, 252, 42, 252, 252, 1512
A151782: 1, 8, 8, 56, 8, 56, 56, 392, 8, 56, 56, 392, 56, 392, 392, 2744
A255743: 1, 9, 9, 72, 9, 72, 72, 576, 9, 72, 72, 576, 72, 576, 576, 4608
A255744: 1,10,10, 90,10, 90, 90, 810,10, 90, 90, 810, 90, 810, 810, 7290
A255745: 1,11,11,110,11,110,110,1100,11,110,110,1100,110,1100,1100,11000
...

Index entries for sequences related to cellular automata

Cf. A000120, A255741.
Rows 1-12: A000007, A255738, A040000, A151787, A147582, A151789, A151779, A151791, A151782, A255743, A255744, A255745.
Column 1 is A000012.
Columns 2^k+1, for k >=0: A011477.
Columns 4, 6, 7, 10, 11, 13...: 0 together with A002378.

tabl(nn) = {for (n=1, nn, for (k=1, nn, if (k==1, x = 1, x= (n-1)*(n-2)^(hammingweight(k-1)-1)); print1(x, ", ");); print(););} \\ Michel Marcus, Mar 15 2015

T(n,1) = 1; for k > 1, T(n,k) = (n-1)*(n-2)^(A000120(k-1)-1) with n >= 1.

A255743 a(1) = 1; for n > 1, a(n) = 9*8^{A000120(n-1)-1}.

Original entry on oeis.org

1, 9, 9, 72, 9, 72, 72, 576, 9, 72, 72, 576, 72, 576, 576, 4608, 9, 72, 72, 576, 72, 576, 576, 4608, 72, 576, 576, 4608, 576, 4608, 4608, 36864, 9, 72, 72, 576, 72, 576, 576, 4608, 72, 576, 576, 4608, 576, 4608, 4608, 36864, 72, 576, 576, 4608, 576, 4608, 4608

Offset: 1

Omar E. Pol, Mar 05 2015

nonn

Also, this is a row of the square array A255740.

Partial sums give A255764.

			Written as an irregular triangle in which the row lengths are the terms of A011782, the sequence begins:
   1;
   9;
   9, 72;
   9, 72, 72, 576;
   9, 72, 72, 576, 72, 576, 576, 4608;
   ...

Felix Fröhlich, Table of n, a(n) for n = 1..10000
Hsien-Kuei Hwang, Svante Janson, and Tsung-Hsi Tsai, Identities and periodic oscillations of divide-and-conquer recurrences splitting at half, arXiv:2210.10968 [cs.DS], 2022, p. 33.

Cf. A000120, A151787, A147582, A151789, A151779, A151791, A151782, A255740, A255741, A255744, A255745, A255764.

MapAt[Floor, Array[9*8^(DigitCount[# - 1, 2, 1] - 1) &, 55], 1] (* Michael De Vlieger, Nov 03 2022 *)

a(n) = if (n==1, 1, 9*8^(hammingweight(n-1)-1)); \\ Michel Marcus, Mar 15 2015

# Python 3.10+
def A255743(n): return 1 if n == 1 else 9*(1<<((n-1).bit_count()-1)*3) # Chai Wah Wu, Nov 15 2022

More terms from Michel Marcus, Mar 15 2015

A255745 a(1) = 1; for n > 1, a(n) = 11*10^{A000120(n-1)-1}.

Original entry on oeis.org

1, 11, 11, 110, 11, 110, 110, 1100, 11, 110, 110, 1100, 110, 1100, 1100, 11000, 11, 110, 110, 1100, 110, 1100, 1100, 11000, 110, 1100, 1100, 11000, 1100, 11000, 11000, 110000, 11, 110, 110, 1100, 110, 1100, 1100, 11000, 110, 1100, 1100, 11000, 1100, 11000, 11000

Offset: 1

Omar E. Pol, Mar 05 2015

nonn

Also, this is a row of the square array A255740.
Partial sums give A255766.
Is this also A151787 written in base 2?

			Also, written as an irregular triangle in which the row lengths are the terms of A011782, the sequence begins:
1;
11;
11, 110;
11, 110, 110, 1100;
11, 110, 110, 1100, 110, 1100, 1100, 11000;
...

Cf. A000120, A151787, A147582, A151789, A151779, A151791, A151782, A255740, A255741, A255743, A255744, A255766.

a(n) = if (n==1, 1, 11*10^(hammingweight(n-1)-1)); \\ Michel Marcus, Mar 13 2015

More terms from Michel Marcus, Mar 13 2015

cp's OEIS Frontend

A255765 Partial sums of A255744.

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A255740 Square array read by antidiagonals upwards: T(n,1) = 1; for k > 1, T(n,k) = (n-1)*(n-2)^(A000120(k-1)-1) with n >= 1.

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Formula

A255743 a(1) = 1; for n > 1, a(n) = 9*8^{A000120(n-1)-1}.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Python

Extensions

A255745 a(1) = 1; for n > 1, a(n) = 11*10^{A000120(n-1)-1}.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

PARI

Extensions