A053464 a(n) = n*5^(n-1).

0, 1, 10, 75, 500, 3125, 18750, 109375, 625000, 3515625, 19531250, 107421875, 585937500, 3173828125, 17089843750, 91552734375, 488281250000, 2593994140625, 13732910156250, 72479248046875, 381469726562500

Offset: 0

Author

Barry E. Williams, Jan 13 2000

Keywords

• easy
• nonn

Comments

Arithmetic derivative of 5^n: a(n) = A003415(5^n). - Darrell Minor, Jul 21 2025

References

• A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..500
• Frank Ellermann, Illustration of binomial transforms
• INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 756.
• Index entries for linear recurrences with constant coefficients, signature (10,-25).

Crossrefs

Cf. A002697, A027471, A001787.

Programs

Magma

[n*(5^(n-1)): n in [0..30]]; // Vincenzo Librandi, Jun 09 2011

Mathematica

Join[{a=0,b=1},Table[c=10*b-25*a;a=b;b=c,{n,60}]] (* Vladimir Joseph Stephan Orlovsky, Jan 27 2011 *)
Table[n*5^(n-1),{n,0,20}] (* or *) LinearRecurrence[{10,-25},{0,1},30] (* Harvey P. Dale, Jul 22 2014 *)

PARI

{a(n) = n*5^(n-1)}; /* Michael Somos, Sep 12 2005 */

Sage

[lucas_number1(n,10,25) for n in range(0, 21)] # Zerinvary Lajos, Apr 26 2009

Formula

a(n) = Sum_{k=0..n} 5^(n-k)*binomial(n-k+1, k)*binomial(1, (k+1)/2)*(1-(-1)^k)/2. - Paul Barry, Oct 15 2004

a(n) = 10*a(n-1) - 25*a(n-2); n>1; a(0)=0, a(1)=1.

Fourth binomial transform of n (starting 0, 1, 10...) Convolution of powers of 5.

G.f.: x/(1-5*x)^2; E.g.f.: x*exp(5*x). - Paul Barry, Jul 22 2003

a(n) = - 25^n * a(-n) for all n in Z. - Michael Somos, Jun 26 2017

From Amiram Eldar, Oct 28 2020: (Start)

Sum_{n>=1} 1/a(n) = 5*log(5/4).

Sum_{n>=1} (-1)^(n+1)/a(n) = 5*log(6/5). (End)

Extensions

More terms from James Sellers, Feb 02 2000