A014825 a(n) = 4*a(n-1) + n with n > 1, a(1)=1.
1, 6, 27, 112, 453, 1818, 7279, 29124, 116505, 466030, 1864131, 7456536, 29826157, 119304642, 477218583, 1908874348, 7635497409, 30541989654, 122167958635, 488671834560, 1954687338261, 7818749353066
Offset: 1
Examples
G.f. = x + 6*x^2 + 27*x^3 + 112*x^4 + 453*x^5 + 1818*x^6 + 7279*x^7 + ...
Links
- Vincenzo Librandi, Table of n, a(n) for n = 1..1000
- Dillan Agrawal, Selena Ge, Jate Greene, Tanya Khovanova, Dohun Kim, Rajarshi Mandal, Tanish Parida, Anirudh Pulugurtha, Gordon Redwine, Soham Samanta, and Albert Xu, Chip-Firing on Infinite k-ary Trees, arXiv:2501.06675 [math.CO], 2025. See pp. 9, 18.
- Hacène Belbachir and El-Mehdi Mehiri, Enumerating moves in the optimal solution of the Tower of Hanoi, arXiv:2210.08657 [math.CO], 2022.
- László Tóth, On Schizophrenic Patterns in b-ary Expansions of Some Irrational Numbers, arXiv:2002.06584 [math.NT], 2020. See also Proc. Amer. Math. Soc. 148 (2020), 461-469.
- Index entries for linear recurrences with constant coefficients, signature (6,-9,4).
Crossrefs
Programs
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Magma
[(4^(n+1)-3*n-4)/9: n in [1..30]]; // Vincenzo Librandi, Aug 23 2011
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Mathematica
RecurrenceTable[{a[1]==1,a[n]==4a[n-1]+n},a[n],{n,30}] (* Harvey P. Dale, Oct 12 2011 *) a[ n_]:= SeriesCoefficient[x/((1-4x)(1-x)^2), {x, 0, n}] (* Michael Somos, Jun 20 2012 *) -
PARI
{a(n) = polcoeff( x / ((1 - x)^2 * (1 - 4*x)) + x * O(x^n), n)} /* Michael Somos, Jun 20 2012 */ -
Python
def A014825(n): return (((1<<(n+1<<1))-4)//3-n)//3 # Chai Wah Wu, Nov 12 2024
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Sage
[(4^(n+1) -3*n -4)/9 for n in (1..30)] # G. C. Greubel, Feb 18 2020
Formula
a(n) = (4^(n+1) - 3*n - 4)/9.
G.f.: x/((1-4*x)*(1-x)^2).
a(n) = Sum_{k=0..n} (n-k)*4^k = Sum_{k=0..n} k*4^(n-k). - Paul Barry, Jul 30 2004
a(n) = Sum_{k=0..n} binomial(n+2, k+2)*3^k [Offset 0]. - Paul Barry, Jul 30 2004
a(n) = Sum_{k=0..n} Sum_{j=0..2k} (-1)^(j+1)*J(j)*J(2k-j), J(n) = A001045(n). - Paul Barry, Oct 23 2009
Convolution square of A006314. - Michael Somos, Jun 20 2012
E.g.f.: (4*exp(4*x) - (4+3*x)*exp(x))/9. - G. C. Greubel, Feb 18 2020
a(n) = Sum_{k=0..n} A002450(k). - Joseph Brown, May 11 2021
Last digits give A171654. - Paul Curtz, Oct 10 2021