|
| |
|
|
A037453
|
|
Positive numbers n such that the base 5 representation of n contains no 3 or 4.
|
|
4
| |
|
|
1, 2, 5, 6, 7, 10, 11, 12, 25, 26, 27, 30, 31, 32, 35, 36, 37, 50, 51, 52, 55, 56, 57, 60, 61, 62, 125, 126, 127, 130, 131, 132, 135, 136, 137, 150, 151, 152, 155, 156, 157, 160, 161, 162, 175, 176, 177, 180, 181, 182, 185, 186, 187
(list; graph; refs; listen; history; internal format)
|
|
|
|
OFFSET
| 1,2
|
|
|
COMMENTS
| 5 divides neither C(2s-1,s) = A001700[ s ] (nor C(2s,s) = A000984[ s ], central column of P ascal's Triangle) if and only if s is one of the terms in this sequence.
n such that the last decimal digit of C(2n,n) is not zero. - Benoit Cloitre (benoit7848c(AT)orange.fr), Aug 18 2002
Let us recall the plan of Apery's irrationality proof. Consider the recurrence (n + 1)^3 * u_(n+1) = (34n^3 + 51n^2 + 27n + 5)u_n - n^3 * u_(n-1). The solution with starting values u_0 = 1; u_1 = 5 has the peculiar property that it has integral terms, despite the fact that at every recursion step we divide by (n + 1)^3. The n-th term is given by f(n) = sum(i=0,n,binomial(n+i,i)^2*binomial(n,i)^2) (see Beukers link) and m such that if Mod(f(m),5) <>0 is same as 2*a(n). - Mohammed Bouayoun (bouyao(AT)wanadoo.fr), Mar 08 2004
|
|
|
LINKS
| Beukers, Title?
|
|
|
FORMULA
| a(3n)=5a(n), a(3n+1)=5a(n)+1, a(3n+2)=5a(n)+2, where by definition a(0)=0. - Emeric Deutsch (deutsch(AT)duke.poly.edu), Mar 23 2004
|
|
|
MATHEMATICA
| Table[FromDigits[IntegerDigits[k, 3], 5], {k, 60}] - T. D. Noe (noe(AT)sspectra.com), Apr 18 2007
|
|
|
PROG
| (PARI) f(n)=sum(i=0, n, binomial(n+i, i)^2*binomial(n, i)^2); for (i=1, 1000, if(Mod(f(i), 5)<>0, print1(i/2, ", ")))
|
|
|
CROSSREFS
| Cf. A050607.
Cf. A005836.
Sequence in context: A177987 A047217 A039015 * A014528 A087791 A002157
Adjacent sequences: A037450 A037451 A037452 * A037454 A037455 A037456
|
|
|
KEYWORD
| nonn,base
|
|
|
AUTHOR
| Clark Kimberling (ck6(AT)evansville.edu)
|
|
|
EXTENSIONS
| Better definition from T. D. Noe (noe(AT)sspectra.com), Apr 18 2007
|
| |
|
|
