A109827 -id:A109827 - cp's OEIS frontend

A026549 Ratios of successive terms are 2, 3, 2, 3, 2, 3, 2, 3, ...

1, 2, 6, 12, 36, 72, 216, 432, 1296, 2592, 7776, 15552, 46656, 93312, 279936, 559872, 1679616, 3359232, 10077696, 20155392, 60466176, 120932352, 362797056, 725594112, 2176782336, 4353564672, 13060694016, 26121388032, 78364164096, 156728328192, 470184984576, 940369969152

Offset: 0

Clark Kimberling

Appears to be the number of permutations p of {1,2,...,n} such that p(i)+p(i+1)>=n for every i=1,2,...,n-1 (if offset is 1). - Vladeta Jovovic, Dec 15 2003
Equals eigensequence of a triangle with 1's in even columns and (1,3,3,3,...) in odd columns. a(5) = 72 = (1, 3, 1, 3, 1, 1) dot (1, 1, 2, 6, 12, 36) = (1 + 3 + 2 + 18 + 12 + 36), where (1, 3, 1, 3, 1, 1) = row 5 of the generating triangle. - Gary W. Adamson, Aug 02 2010
Partial products of A010693. - Reinhard Zumkeller, Mar 29 2012
Satisfies Benford's law [Theodore P. Hill, Personal communication, Feb 06, 2017]. - N. J. A. Sloane, Feb 08 2017
For n >= 2, a(n) is the least k > a(n-1) such that both k and a(n-2) + a(n-1) + k have exactly n prime factors, counted with multiplicity. - Robert Israel, Aug 06 2024

			G.f. = 1 + 2*x + 6*x^2 + 12*x^3 + 36*x^4 + 72*x^5 + 216*x^6 + ... - _Michael Somos_, Apr 09 2022

Arno Berger and Theodore P. Hill, An Introduction to Benford's Law, Princeton University Press, 2015.

Vincenzo Librandi, Table of n, a(n) for n = 0..700
Paul Barry, Embedding structures associated with Riordan arrays and moment matrices, International Journal of Combinatorics, Vol. 2014 (2014), Article ID 301394, 7 pages; arXiv preprint, arXiv:1312.0583 [math.CO], 2013.
Sean A. Irvine, Walks on Graphs.
Index entries for linear recurrences with constant coefficients, signature (0,6).
Index entries for sequences related to Benford's law.

Cf. A026532, A026536, A026551, A026567.

Cf. A010693, A208131, A109827.

a026549 n = a026549_list !! n
a026549_list = scanl (*) 1 $ a010693_list
-- Reinhard Zumkeller, Mar 29 2012

[(1/2)*(3-(-1)^n)*6^Floor(n/2): n in [0..30]]; // Vincenzo Librandi, Jun 08 2011

seq(seq(2^i*3^j, i=j..j+1),j=0..30); # Robert Israel, Aug 06 2024

LinearRecurrence[{0,6},{1,2},30] (* Harvey P. Dale, May 29 2016 *)

{a(n) = 6^(n\2) * (n%2+1)}; /* Michael Somos, Apr 09 2022 */

[(1+(n%2))*6^(n//2) for n in (0..30)] # G. C. Greubel, Apr 09 2022

Equals T(n, 0) + T(n, 1) + ... + T(n, 2n), T given by A026536.
a(n) = 2*A026532(n), for n > 0.
G.f.: (1+2*x)/(1-6*x^2) - Paul Barry, Aug 25 2003
a(n+3) = a(n+2)*a(n+1)/a(n). - Reinhard Zumkeller, Mar 04 2011
a(n) = (1/2)*(3 - (-1)^n)*6^floor(n/2), or a(n) = 6*a(n-2). - Vincenzo Librandi, Jun 08 2011
a(n) = 1/a(-n) if n is even and (2/3)/a(-n) if n is odd for all n in Z. - Michael Somos, Apr 09 2022
Sum_{n>=0} 1/a(n) = 9/5. - Amiram Eldar, Feb 13 2023

New definition from Ralf Stephan, Dec 01 2004

A089293 Sum of digits in the mixed-base enumeration system n=...d(4)d(3)d(2)d(1), where the digits satisfy 0<=d(i)<=1 if i is odd, 0<=d(i)<=2 if i is even.

Original entry on oeis.org

0, 1, 1, 2, 2, 3, 1, 2, 2, 3, 3, 4, 1, 2, 2, 3, 3, 4, 2, 3, 3, 4, 4, 5, 2, 3, 3, 4, 4, 5, 3, 4, 4, 5, 5, 6, 1, 2, 2, 3, 3, 4, 2, 3, 3, 4, 4, 5, 2, 3, 3, 4, 4, 5, 3, 4, 4, 5, 5, 6, 3, 4, 4, 5, 5, 6, 4, 5, 5, 6, 6, 7, 1, 2, 2, 3, 3, 4, 2, 3, 3, 4, 4, 5, 2, 3, 3, 4, 4, 5, 3, 4, 4, 5, 5, 6, 3, 4, 4, 5, 5, 6, 4, 5, 5

Offset: 0

John W. Layman, Jan 15 2004

Counting 0,1,2,3,... (base 10) in this mixed-base system proceeds as follows: 0,1,10,11,20,21,100,101,110,111,120,121,1000,.. = A109827.

			11(base 10) = 121(mixed-base), so a(11)=4.

Cf. A109827 (mixed-base).

Sums of digits in other bases: A000120 (binary), A053735 (ternary), A053827 (base 6).

a(n) = vecsum(digits(n,6)\/2); \\ Kevin Ryde, Aug 03 2021

a(n)=a(n-1)+1 if n=1, 3, 5 mod 6; a(n)=a(n-1) if n=2, 4 mod 5; a(n)=a(n/6) if n=0 mod 6.

A109838 Numbers written in a bits-of-Pi/primorial base system.

Original entry on oeis.org

0, 121, 120, 111, 110, 101, 100, 221, 220, 211, 210, 201, 200, 321, 320, 311, 310, 301, 300, 421, 420, 411, 410, 401, 400, 1021, 1020, 1011, 1010, 1001, 1000, 1121, 1120, 1111, 1110, 1101, 1100, 1221, 1220, 1211, 1210, 1201, 1200, 1321, 1320, 1311, 1310

Offset: 0

Rick L. Shepherd, Jul 04 2005

Exercise 15 on page 30 of the Long textbook is "Let m_1, m_2, m_3, ... and M_0, M_1, M_2, ... be as above. [see A109827.] Let s_0, s_1, s_2, ... be an infinite sequence of zeros and ones containing infinitely many of each. Show that *every* integer r (positive, negative, or zero) can be represented uniquely in the form r = (-1)^s_n c_n M_n + (-1)^s_(n-1) c_(n-1) M_(n-1) + ... + (-1)^s_1 c_1 M_1 + (-1)^s_0 c_0 M_0 where c_n <> 0 for r <> 0 and 0 <= c_i < m_(i+1) for all i. If r is positive show that s_n = 0 and if r is negative show that s_n = 1." Take the primes (A000040) for the m_i. Then the M_i are the primorials (A002110). Take the binary expansion of Pi (A004601) for the s_k. This sequence, a(r) = (c_n c_(n-1) ... c_1 c_0 concatenated), gives the representations of the nonnegative integers. See A109839 for the corresponding negative integers.

			a(13) = 321 as 13 = 18 - 4 - 1 = ((-1)^0)*3*6 + ((-1)^1)*2*2 + ((-1)^1)*1*1, where 1,1,0 are the first three terms of A004601 and 1,2,6 are the first three terms of A002110.

Calvin T. Long, Elementary Introduction to Number Theory, 2nd ed., D.C. Heath and Company, 1972, p. 30.

Cf. A109839 (negative numbers represented similarly), A109827, A004601 (Pi in binary), A000040 (primes), A002110 (primorials), A007623 (factorial base).

A235202 Numbers written in an alternating binary-then-quaternary base.

Original entry on oeis.org

1, 10, 11, 20, 21, 30, 31, 100, 101, 110, 111, 120, 121, 130, 131, 1000, 1001, 1010, 1011, 1020, 1021, 1030, 1031, 1100, 1101, 1110, 1111, 1120, 1121, 1130, 1131, 2000, 2001, 2010, 2011, 2020, 2021, 2030, 2031, 2100

Offset: 1

Jeremy Gardiner, Jan 04 2014

Mixed-radix number representation produced by a serial counter with generating sequence (1, 3, 1, 3, ...) = A010684.
Places reading from the right have values (1, 2, 8, 16, 64, 128, ...) = unsigned A094014.
Conjecture: This sequence interpreted as quaternary (base 4) numbers gives A126001 (hence a simplified scheme for computing that sequence).

			a(15) = 131 since 15 = 1*1+3*2+1*8.

Jeremy Gardiner, Table of n, a(n) for n = 1..1024

Cf. A109827 (Numbers written in an alternating binary-then-ternary base).

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A026549 Ratios of successive terms are 2, 3, 2, 3, 2, 3, 2, 3, ...

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A089293 Sum of digits in the mixed-base enumeration system n=...d(4)d(3)d(2)d(1), where the digits satisfy 0<=d(i)<=1 if i is odd, 0<=d(i)<=2 if i is even.

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A109838 Numbers written in a bits-of-Pi/primorial base system.

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A235202 Numbers written in an alternating binary-then-quaternary base.

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