A001024 -id:A001024 - cp's OEIS frontend

A165836 Totally multiplicative sequence with a(p) = 15.

1, 15, 15, 225, 15, 225, 15, 3375, 225, 225, 15, 3375, 15, 225, 225, 50625, 15, 3375, 15, 3375, 225, 225, 15, 50625, 225, 225, 3375, 3375, 15, 3375, 15, 759375, 225, 225, 225, 50625, 15, 225, 225, 50625, 15, 3375, 15, 3375, 3375, 225, 15, 759375, 225, 3375

Offset: 1

Jaroslav Krizek, Sep 28 2009

G. C. Greubel, Table of n, a(n) for n = 1..10000

15^PrimeOmega[Range[100]](* G. C. Greubel, Apr 09 2016 *)

a(n) = A001024(A001222(n)) = 15^bigomega(n) = 15^A001222(n).

A239285 a(n) = (15^n - (-2)^n)/17.

Original entry on oeis.org

0, 1, 13, 199, 2977, 44671, 670033, 10050559, 150758257, 2261374111, 33920611153, 508809168319, 7632137522737, 114482062845151, 1717230942669073, 25758464140052479, 386376962100754417, 5795654431511381791, 86934816472670595793, 1304022247090059199039

Offset: 0

Felix P. Muga II, Mar 14 2014

Let k and t be positive integers and consider a(n) = k*a(n-1)+t*a(n-2) for n>=2, with a(0)=0, a(1)=1.
The roots of its characteristic equation are r1 = (k+sqrt(k^2+4t))/2 and r2 = (k-sqrt(k^2+4t))/2. Hence, the solution to the recurrence relation is the sequence {a(n)} where a(n) = alpha1*r1^n + alpha2*r2^n. It can be shown that alpha1 = 1/sqrt(k^2+4t) and alpha2 = -alpha1. It can be shown also that |r2/r1| < 1. Thus, a(n+1)/a(n) converges to r1 as n approaches infinity.
Note that limit a(n+1)/a(n) = 15 with k=13 and t=30.
If n > 20, then |a(n+1)/a(n) - 15| < 10^(-16).
Let b(n) be the number of strings of length n containing the 13-ary digits: 0,...,9,A,B,C or the 30 two-consecutive digits D0,D1,...,D9,DA,...,DT where A corresponds to 10, ..., T corresponds to 29. Then b(0)=1=a(2) and b(1)=13=a(3). The strings q_1q_2...q_n of length n can be partitioned into 2 groups A and B where A contains the strings where q_1=0,1,...,9,A,B,C and B contains the strings where q_1=D. Thus, |A|=13*b(n-1) and |B|=30*b(n-2). Hence, b(n) = 13*b(n-1) + 30*b(n-2) for n>1. Since b(0)=a(2) and b(1)=a(3), we can show that b(n) = a(n+2).

G. C. Greubel, Table of n, a(n) for n = 0..849
Index entries for linear recurrences with constant coefficients, signature (13,30).

Cf. A001024, A122803.

[(15^n -(-2)^n)/17: n in [0..30]]; // G. C. Greubel, May 26 2018

CoefficientList[Series[x/(1-13*x-30*x^2), {x,0,50}], x] (* or *) Table[
(15^n - (-2)^n)/17, {n,0,30}] (* or *) LinearRecurrence[{13,30}, {0,1}, 30] (* G. C. Greubel, May 26 2018 *)

a(n) = (15^n-(-2)^n)/17; \\ Michel Marcus, Mar 16 2014

x='x+O('x^30); concat([0], Vec()) \\ G. C. Greubel, May 26 2018

G.f.: x/(1 - 13*x - 30*x^2).
a(n) = 13*a(n-1) + 30*a(n-2) for n >= 2, a(0)=0, a(1)=1.
a(n) = (1/17)*(A001024(n) - A122803(n)), n >= 0.
a(0)=0, a(n) = Sum_{k=0..n-1} A001024(k)*A122803(n-1-k) for n > 0.
E.g.f.: (exp(15*x) - exp(-2*x))/17. - G. C. Greubel, May 26 2018

A061252 a(n) = 16^n - 15^n.

Original entry on oeis.org

0, 1, 31, 721, 14911, 289201, 5386591, 97576081, 1732076671, 30276117361, 522861237151, 8942430185041, 151728638820031, 2557404559011121, 42864668012537311, 715027614225987601, 11878335717996660991

Offset: 0

Frank Ellermann, Jun 05 2001

Number of ways to assign truth values to n quaternary conjunctions connected by disjunctions such that the proposition is true. For example, a(2) = 31, since for the proposition '(a & b & c & d) v (e & f & g & h)' there are 31 assignments that make the proposition true. - Ori Milstein, Dec 31 2022

Equivalently, the number of length-n words over the alphabet {0,1,..,15} with at least one letter = 15. - Joerg Arndt, Jan 01 2023

Index entries for linear recurrences with constant coefficients, signature (31,-240).

Base 8: A016177, 4: A005061, 2: A000225, 10: A016189.

Table[16^n-15^n,{n,0,20}] (* or  *) LinearRecurrence[{31,-240},{0,1},20] (* Harvey P. Dale, Jan 23 2021 *)

a(n) = 16^n - 15^n; \\ Michel Marcus, Aug 26 2013

a(0)=0, a(n) = 15*a(n-1) + 16^(n-1). - Vincenzo Librandi, Feb 09 2011
a(0)=0, a(1)=1, a(n) = 31*a(n-1) - 240*a(n-2). - Vincenzo Librandi, Feb 09 2011
a(n) = A001025(n) - A001024(n). - Michel Marcus, Aug 26 2013

A091480 Table of multigraphs (by antidiagonals) with n (>=1) nodes and k (>=0) edges. Each type of object labeled from its own label set.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 1, 3, 1, 0, 1, 9, 6, 1, 0, 1, 27, 36, 10, 1, 0, 1, 81, 216, 100, 15, 1, 0, 1, 243, 1296, 1000, 225, 21, 1, 0, 1, 729, 7776, 10000, 3375, 441, 28, 1, 0, 1, 2187, 46656, 100000, 50625, 9261, 784, 36, 1, 0, 1, 6561, 279936, 1000000, 759375

Offset: 1

Christian G. Bower, Jan 13 2004

			1  0   0    0     0 ...
1  1   1    1     1 ...
1  3   9   27    81 ...
1  6  36  216  1296 ...
1 10 100 1000 10000 ...

F. Bergeron, G. Labelle and P. Leroux, Combinatorial Species and Tree-Like Structures, Cambridge, 1998, p. 114 (2.4.44).

Columns 0-8: A000012, A000217(n-1), A000537(n-1), A059827(n-1), A059977(n-1), A059860(n-1), A059978(n-1), A059979(n-1), A059980(n-1).
Rows 1-10: A000007, A000012, A000244, A000400, A011557, A001024, A009965, A009972, A009980, A009989.
Cf. A091478.

a(n, k) = binomial(n, 2)^k.

A239294 a(n) = (15^n - (-3)^n)/18.

Original entry on oeis.org

0, 1, 12, 189, 2808, 42201, 632772, 9492309, 142382448, 2135743281, 32036129532, 480542002029, 7208129853288, 108121948330761, 1621829223367092, 24327438355289349, 364911575314991328, 5473673629767916641, 82105104446389609452

Offset: 0

Felix P. Muga II, Mar 14 2014

Let k and t be positive integers and consider a(n) = k*a(n-1)+t*a(n-2) for n>=2, with a(0)=0, a(1)=1.
The roots of its characteristic equation are r1 = (k+sqrt(k^2+4t))/2 and r2=(k-sqrt(k^2+4t))/2. Hence, the solution to the recurrence relation is the sequence {a(n)} where a(n) = alpha1*r1^n + alpha2*r2^n. It can be shown that alpha1 = 1/sqrt(k^2+4t) and alpha2 = -alpha1. It can be shown also that |r2/r1|< 1. Thus, a(n+1)/a(n) converges to r1 as n approaches infinity.
Note that limit a(n+1)/a(n) = 15 with k=12 and t=45. Thus, we use this as the name of the sequence.
If n > 25, then |a(n+1)/a(n) - 15| < 10^(-16).
a(n+2) is the number of strings of length n containing the 12-ary digits 0,1,...,9,A,B or any of the 45 two-consecutive digits C0,C1,...,C9,CA,...,CZ,Ca,...,Ci where A corresponds to 10, B to 11, ..., Z to 35, a to 36, ..., i to 44.

Index entries for linear recurrences with constant coefficients, signature (12,45).

Table[(15^n-(-3)^n)/18,{n,0,20}] (* or *) LinearRecurrence[{12,45},{0,1},20] (* Harvey P. Dale, Apr 29 2019 *)

a(n) = (15^n - (-3)^n)/18; \\ Michel Marcus, Mar 16 2014

G.f.: x/((1-15*x)*(1+3*x)).
a(n) = 12*a(n-1)+45*a(n-2), for n>=2, a(0)=0, a(1)=1.
a(n) = (1/18)*(A001024(n) - (-1)^n * A000244(n)).
a(n) = A000244(n-1) * Sum_{k=0...n-1} (A000351(k) * A033999(n-1-k)).

cp's OEIS Frontend

A165836 Totally multiplicative sequence with a(p) = 15.

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A239285 a(n) = (15^n - (-2)^n)/17.

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A061252 a(n) = 16^n - 15^n.

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A091480 Table of multigraphs (by antidiagonals) with n (>=1) nodes and k (>=0) edges. Each type of object labeled from its own label set.

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A239294 a(n) = (15^n - (-3)^n)/18.

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