A265210 -id:A265210 - cp's OEIS frontend

A037096 Periodic vertical binary vectors computed for powers of 3: a(n) = Sum_{k=0 .. (2^n)-1} (floor((3^k)/(2^n)) mod 2) * 2^k.

1, 2, 0, 204, 30840, 3743473440, 400814250895866480, 192435610587299441243182587501623263200, 2911899996313975217187797869354128351340558818020188112521784134070351919360

Offset: 0

Antti Karttunen, Jan 29 1999

This sequence can be also computed with a recurrence that does not explicitly refer to 3^n. See the C program.

Conjecture: For n >= 3, each term a(n), when considered as a GF(2)[X] polynomial, is divisible by the GF(2)[X] polynomial (x + 1) ^ A055010(n-1). If this holds, then for n >= 3, a(n) = A048720(A136386(n), A048723(3,A055010(n-1))).

			When powers of 3 are written in binary (see A004656), under each other as:
  000000000001 (1)
  000000000011 (3)
  000000001001 (9)
  000000011011 (27)
  000001010001 (81)
  000011110011 (243)
  001011011001 (729)
  100010001011 (2187)
it can be seen that the bits in the n-th column from the right can be arranged in periods of 2^n: 1, 2, 4, 8, ... This sequence is formed from those bits: 1, is binary for 1, thus a(0) = 1. 01, reversed is 10, which is binary for 2, thus a(1) = 2, 0000 is binary for 0, thus a(2)=0, 000110011, reversed is 11001100 = A007088(204), thus a(3) = 204.

S. Wolfram, A New Kind of Science, Wolfram Media Inc., (2002), p. 119.

Antti Karttunen, Table of n, a(n) for n = 0..11
Antti Karttunen, C program for computing this sequence.
S. Wolfram, A New Kind of Science, Wolfram Media Inc., (2002), p. 119.

Cf. A036284, A037095, A037097, A136386 for related sequences.
Cf. A000079, A000215, A000225, A000244, A004656, A007088, A048720, A048723, A055010.
Cf. also A004642, A265209, A265210 (for 2^n written in base 3).

a(n) := sum( 'bit_n(3^i, n)*(2^i)', 'i'=0..(2^(n))-1);
bit_n := (x, n) -> `mod`(floor(x/(2^n)), 2);

a(n) = Sum_{k=0 .. A000225(n)} (floor(A000244(k)/(2^n)) mod 2) * 2^k.
Other identities and observations:
For n >= 2, a(n) = A000215(n-1)*A037097(n) = A048720(A037097(n), A048723(3, A000079(n-1))).

Entry revised by Antti Karttunen, Dec 29 2007

Name changed and the example corrected by Antti Karttunen, Dec 05 2015

A264980 Base-3 reversal of 2^n: a(n) = A030102(A000079(n)).

Original entry on oeis.org

1, 2, 4, 8, 16, 64, 32, 184, 352, 704, 1408, 1880, 2824, 14032, 10328, 56128, 100576, 145784, 189472, 370304, 731752, 4388248, 2924096, 11175712, 15965704, 31930448, 63861880, 383165344, 255439712, 1021772344, 510875648, 2550188248, 5619691648, 9689861048, 17830350904, 79068724264, 34109913224, 192259976368, 133338241880

Offset: 0

Antti Karttunen, Dec 05 2015

			2^5 = 32 in base 3 = "1012" (= A007089(32)) as 1*27 + 1*3 + 2*1 = 32. 2^6 = 64 in base 3 = "2101" as 2*27 + 1*9 + 1*1 = 64. "1012" reversed is "2101" and vice versa, thus a(5) = 64 and a(6) = 32.

Antti Karttunen, Table of n, a(n) for n = 0..512

Leftmost column of A265345.
Cf. A000079, A007089, A030102, A263273, A265210, A265342.
Cf. also A036215.

base(n) = {my(a=[n%3]); while(0Altug Alkan, Dec 29 2015

a(n) = A030102(A000079(n)) = A263273(A000079(n)).

a(0) = 1, for n >= 1, a(n) = A265342(a(n-1)).

A265209 Irregular triangle read by rows in which row n lists the base-3 digits of 2^n, n >= 0.

Original entry on oeis.org

1, 2, 1, 1, 2, 2, 1, 2, 1, 1, 0, 1, 2, 2, 1, 0, 1, 1, 1, 2, 0, 2, 1, 0, 0, 1, 1, 1, 2, 0, 0, 2, 2, 2, 1, 1, 0, 1, 2, 2, 1, 2, 2, 1, 0, 2, 1, 2, 1, 2, 1, 2, 1, 2, 0, 1, 1, 0, 2, 0, 2, 0, 1, 0, 2, 2, 1, 1, 1, 1, 0, 2, 1, 1, 1, 1, 2, 2, 2, 2, 1, 1, 2, 2

Offset: 0

L. Edson Jeffery, Dec 04 2015

The length of row n is A020915(n) = 1 + A136409(n).

			Triangle begins:
  1
  2
  1  1
  2  2
  1  2  1
  1  0  1  2
  2  1  0  1
  1  1  2  0  2
  1  0  0  1  1  1
  2  0  0  2  2  2
  1  1  0  1  2  2  1
  2  2  1  0  2  1  2
  1  2  1  2  1  2  0  1
  1  0  2  0  2  0  1  0  2
  2  1  1  1  1  0  2  1  1
  1  1  2  2  2  2  1  1  2  2

Cf. A000079 (powers of 2), A003137, A004642 (powers of 2 written in base 3).

Cf. A265210 (base 3 digits of 2^n in reverse order).

(* Replace Flatten with Grid to display the triangle: *)
Flatten[Table[IntegerDigits[2^n, 3], {n, 0, 15}]]

for(n=0,15,for(k=1,#digits(2^n,3),print1(digits(2^n,3)[k],", "))) \\ Derek Orr, Dec 24 2015

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A037096 Periodic vertical binary vectors computed for powers of 3: a(n) = Sum_{k=0 .. (2^n)-1} (floor((3^k)/(2^n)) mod 2) * 2^k.

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A264980 Base-3 reversal of 2^n: a(n) = A030102(A000079(n)).

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A265209 Irregular triangle read by rows in which row n lists the base-3 digits of 2^n, n >= 0.

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