Sequence name: A001235 mod 13.

0, 9, 0, 0, 7, 0, 6, 0, 12, 0,
0, 9, 11, 6, 0, 0, 9, 6, 0, 4,
0, 9, 7, 0, 0, 0, 4, 7, 5

Examples: a(1) = 0, because 1729 =7*13*19. a(2) = 9, because 4104 = 32*5*7*13 + 9. a(19) = 0, because 216125 = 53*7*13*19….

PS

Computation was performed by using D and Java. And I welcome more terms.

## 5 thoughts on “A sequence not found in the OEIS: part 14”

1. Dear Jun Mizuki

With this line of code in PARI/GP

for(n=1,20,print(n ” : ” ((n-1)!/(2^(n+1))))

we get

1 : 1/4
2 : 1/8
3 : 1/8
4 : 3/16
5 : 3/8
6 : 15/16
7 : 45/16
8 : 315/32
9 : 315/8
10 : 2835/16
11 : 14175/16
12 : 155925/32
13 : 467775/16
14 : 6081075/32
15 : 42567525/32
16 : 638512875/64
17 : 638512875/8
18 : 10854718875/16
19 : 97692469875/16
20 : 1856156927625/32

How can this sequence be submitted to OEIS?

If you are interested we can do it together.

(I posted here

http://problemasteoremas.wordpress.com/2010/03/22/solucao-do-desafio-sobre-sequencias-sucessoes-descobrir-o-termo-geral-solution-to-the-challenge-find-the-general-term-of-a-sequence/

this sequence as a challenge of mine, with a few more explanations).

Americo Tavares

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2. Dear Jun,

The LaTeX code of the rational sequence is

$x_{n}=\frac{(n-1)!}{2^{n+1}}=\frac{((n-1)!)/\gcd ((n-1)!,2^{n+1})}{2^{n+1}/\gcd ((n-1)!,2^{n+1})}$

and the PARI/GP code can be modified to

for(n=1,50,print(n ” : ” ((n-1)!)/gcd((n-1)!,2^(n+1)))/((2^(n+1)/gcd((n-1)!,2^(n+1)))))))

which is perhaps better for converting it into other languages.

I got as 50th term
8644205195683235286768595007647709520704677734375/32

i.e. the 50th term of your Sequence 1 is

8644205195683235286768595007647709520704677734375

while

32

is the 50th term of Sequence 2.

I hope having written above the correct number of parenthesis.

Américo

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3. To a correct display here:

50 :

864420519568323528676859
5007647709520704677734375

/32

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4. Dear Américo,

Thanks for your LaTeX code and the 50th term!

Regards,

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