login
a(n) = 4*a(n-1) + n with n > 1, a(1)=1.
21

%I #99 Oct 18 2022 03:25:11

%S 1,6,27,112,453,1818,7279,29124,116505,466030,1864131,7456536,

%T 29826157,119304642,477218583,1908874348,7635497409,30541989654,

%U 122167958635,488671834560,1954687338261,7818749353066

%N a(n) = 4*a(n-1) + n with n > 1, a(1)=1.

%H Vincenzo Librandi, <a href="/A014825/b014825.txt">Table of n, a(n) for n = 1..1000</a>

%H Hacène Belbachir and El-Mehdi Mehiri, <a href="https://arxiv.org/abs/2210.08657">Enumerating moves in the optimal solution of the Tower of Hanoi</a>, arXiv:2210.08657 [math.CO], 2022.

%H László Tóth, <a href="https://arxiv.org/abs/2002.06584">On Schizophrenic Patterns in b-ary Expansions of Some Irrational Numbers</a>, arXiv:2002.06584 [math.NT], 2020. See also <a href="https://doi.org/10.1090/proc/14863">Proc. Amer. Math. Soc.</a> 148 (2020), 461-469.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (6,-9,4).

%F a(n) = (4^(n+1) - 3*n - 4)/9.

%F G.f.: x/((1-4*x)*(1-x)^2).

%F a(n) = Sum_{k=0..n} (n-k)*4^k = Sum_{k=0..n} k*4^(n-k). - _Paul Barry_, Jul 30 2004

%F a(n) = Sum_{k=0..n} binomial(n+2, k+2)*3^k [Offset 0]. - _Paul Barry_, Jul 30 2004

%F 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

%F Convolution square of A006314. - _Michael Somos_, Jun 20 2012

%F E.g.f.: (4*exp(4*x) - (4+3*x)*exp(x))/9. - _G. C. Greubel_, Feb 18 2020

%F a(n) = A014916(-n-1)*4^(n+1) = A091919(2*n-2) for all n in Z. - _Michael Somos_, Oct 02 2020

%F a(n) = Sum_{k=0..n} A002450(k). - _Joseph Brown_, May 11 2021

%F Last digits give A171654. - _Paul Curtz_, Oct 10 2021

%e G.f. = x + 6*x^2 + 27*x^3 + 112*x^4 + 453*x^5 + 1818*x^6 + 7279*x^7 + ...

%t RecurrenceTable[{a[1]==1,a[n]==4a[n-1]+n},a[n],{n,30}] (* _Harvey P. Dale_, Oct 12 2011 *)

%t a[ n_]:= SeriesCoefficient[x/((1-4x)(1-x)^2), {x, 0, n}] (* _Michael Somos_, Jun 20 2012 *)

%o (Magma) [(4^(n+1)-3*n-4)/9: n in [1..30]]; // _Vincenzo Librandi_, Aug 23 2011

%o (PARI) {a(n) = polcoeff( x / ((1 - x)^2 * (1 - 4*x)) + x * O(x^n), n)} /* _Michael Somos_, Jun 20 2012 */

%o (Sage) [(4^(n+1) -3*n -4)/9 for n in (1..30)] # _G. C. Greubel_, Feb 18 2020

%Y Cf. A002450 (first differences), A052161 (partial sums).

%Y Cf. A001045, A006314, A014916, A053142, A091919.

%Y Cf. A171654 (mod 10).

%K nonn,easy

%O 1,2

%A _N. J. A. Sloane_