A084990 a(n) = n*(n^2+3*n-1)/3.
0, 1, 6, 17, 36, 65, 106, 161, 232, 321, 430, 561, 716, 897, 1106, 1345, 1616, 1921, 2262, 2641, 3060, 3521, 4026, 4577, 5176, 5825, 6526, 7281, 8092, 8961, 9890, 10881, 11936, 13057, 14246, 15505, 16836, 18241, 19722, 21281, 22920, 24641, 26446, 28337, 30316
Offset: 0
Examples
Let n=2. Consider nonnegative multiples of 5 up to 16*2^4 - 1 = 255. There are 52 such numbers and from them only 8 (namely, 35, 50, 55, 115, 140, 200, 205, 220) have an odd digit sum in base 4. Therefore, a(4) = (52 - 8) - 8 = 36. - _Vladimir Shevelev_, May 18 2012
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..1000
- S. Chaiken, C. R. H. Hanusa and T. Zaslavsky, A q-queens problem III. Partial queens, arXiv:1402.4886 [math.CO], 2014-2018. See Conjecture 4.4.
- Vladimir Shevelev, On monotonic strengthening of Newman-like phenomenon on (2m+1)-multiples in base 2m, arXiv:0710.3177 [math.NT], 2007.
- Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).
Programs
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Magma
I:=[0,1,6,17]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..50]]; // Vincenzo Librandi, Mar 28 2014
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Maple
A084990:=n->n*(n^2+3*n-1)/3; seq(A084990(k),k=0..100); # Wesley Ivan Hurt, Oct 19 2013
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Mathematica
Table[n*(n^2+3*n-1)/3,{n,0,100}] (* Vladimir Joseph Stephan Orlovsky, Mar 08 2010 *) CoefficientList[Series[(x + 2 x^2 - x^3)/((1 - x)^4), {x, 0, 50}], x] (* Vincenzo Librandi, Mar 28 2014 *) LinearRecurrence[{4,-6,4,-1},{0,1,6,17},50] (* Harvey P. Dale, Aug 18 2015 *)
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PARI
a(n) = n*(n^2+3*n-1)/3 \\ Charles R Greathouse IV, Sep 24 2015
Formula
a(n) = (n-1)*(n+1)*(n+3)/3 + 1. - Reinhard Zumkeller, Aug 20 2007
a(n) = A077415(n+1) + 1 for n > 0; a(n) = A000290(n) + A007290(n); a(n+1) = Sum_{k=0..n} A028387(k). - Reinhard Zumkeller, Aug 20 2007
a(2*n) = Sum_{i=0..16*n^4, i==0 (mod 2*n+1)} (-1)^s_(2*n)(i), where s_k(n) is the digit sum of n in base k. - Vladimir Shevelev, May 18 2012
a(2*n) = (2/(2*n+1))*Sum_{i=1..n} tan^4(Pi*i/(2*n+1)). - Vladimir Shevelev, May 23 2012
a(n) = Sum_{i=1..n} i*(i+1)-1. - Wesley Ivan Hurt, Oct 19 2013
G.f.: x*(1+2*x-x^2)/(1-x)^4. - Vincenzo Librandi, Mar 28 2014
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n > 3. - Vincenzo Librandi, Mar 28 2014
a(n) = A064043(n)/3. - Alois P. Heinz, Jul 21 2017
E.g.f.: x*exp(x)*(x^2 + 6*x + 3)/3. - Stefano Spezia, Mar 06 2024
Comments