A045891 First differences of A045623.
1, 1, 3, 7, 16, 36, 80, 176, 384, 832, 1792, 3840, 8192, 17408, 36864, 77824, 163840, 344064, 720896, 1507328, 3145728, 6553600, 13631488, 28311552, 58720256, 121634816, 251658240, 520093696, 1073741824, 2214592512, 4563402752
Offset: 0
Examples
G.f. = 1 + x + 3*x^2 + 7*x^3 + 16*x^4 + 36*x^5 + 80*x^6 + ... - _Michael Somos_, Mar 26 2022
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..1000
- Paul Barry, A Catalan Transform and Related Transformations on Integer Sequences, Journal of Integer Sequences, Vol. 8 (2005), Article 05.4.5.
- Frank Ellermann, Illustration of binomial transforms
- Milan Janjic, Two Enumerative Functions
- Milan Janjic and B. Petkovic, A Counting Function, arXiv 1301.4550 [math.CO], 2013.
- Milan Janjic and B. Petkovic, A Counting Function Generalizing Binomial Coefficients and Some Other Classes of Integers, J. Int. Seq. 17 (2014) # 14.3.5.
- Thomas Selig and Haoyue Zhu, New combinatorial perspectives on MVP parking functions and their outcome map, arXiv:2309.11788 [math.CO], 2023. See p. 29.
- Index entries for linear recurrences with constant coefficients, signature (4,-4).
Programs
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Magma
[1,1] cat [(n+4)*2^(n-3): n in [2..40]]; // G. C. Greubel, Sep 27 2022
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Mathematica
Join[{1,1,a=3,b=7},Table[c=4*b-4*a;a=b;b=c,{n,100}]] (* Vladimir Joseph Stephan Orlovsky, Jan 15 2011 *) Table[ If[n<2, 1, 2^(n-3)*(n+4)], {n, 0, 30}] (* Jean-François Alcover, Sep 12 2012 *) LinearRecurrence[{4,-4},{1,1,3,7},40] (* Harvey P. Dale, May 03 2019 *)
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PARI
v=[1,1,3,7];for(i=1,99,v=concat(v,4*(v[#v]-v[#v-1])));v \\ Charles R Greathouse IV, Jun 01 2011
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SageMath
[1,1]+[(n+4)*2^(n-3) for n in range(2,40)] # G. C. Greubel, Sep 27 2022
Formula
a(n) = Sum_{k=0..n-2} (k+3)*binomial(n-2,k) for n >= 2. - N. J. A. Sloane, Jan 30 2008
a(n) = (n+4)*2^(n-3), n >= 2, with a(0) = a(1) = 1.
G.f.: (1-x)^3/(1-2*x)^2.
Equals binomial transform of A027656.
Starting 1, 3, 7, 16, ... this is ((n+5)*2^n - 0^n)/4, the binomial transform of (1, 2, 2, 3, 3, ...). - Paul Barry, May 20 2003
From Paul Barry, Nov 29 2004: (Start)
a(n) = ((n+4)*2^(n-1) + 3*C(0, n) - C(1, n))/4;
a(n) = Sum_{k=0..floor(n/2)} C(n, 2*k)*(k+1). (End)
G.f.: 1 + Q(0)*x/(1-x)^2, where Q(k)= 1 + (k+1)*x/(1 - x - x*(1-x)/(x + (k+1)*(1-x)/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, Apr 25 2013
a(n) = Sum_{k=0..n} (k+1)*C(n-2,n-k). Peter Luschny, Apr 20 2015
From Amiram Eldar, Jan 13 2021: (Start)
Sum_{n>=0} 1/a(n) = 128*log(2) - 1292/15.
Sum_{n>=0} (-1)^n/a(n) = 782/15 - 128*log(3/2). (End)
E.g.f.: (2 - x + exp(2*x)*(2 + x))/4. - Stefano Spezia, Mar 26 2022
Comments