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 A084990 n*(n^2+3*n-1)/3. 15
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,3 COMMENTS a(n) = A077415(n+1) + 1 for n>0; a(n) = A000290(n) + A007290(n); a(n+1) = Sum(A028387(k): 0<=k<=n). - Reinhard Zumkeller, Aug 20 2007 a(n) is the number of triples (x,y,z) in {1,2,..,n}^3 with x <= y <= z or x >= y >= z. - Jack Kennedy, Mar 14 2009 a(2*n) is the difference between numbers of nonnegative multiples of 2*n+1  with even  and odd digit sum in base 2*n in interval [0, 16*n^4). - Vladimir Shevelev, May 18 2012 LINKS Index to sequences with linear recurrences with constant coefficients, signature (4,-6,4,-1). FORMULA Row sums of triangle A131782 starting (1, 6, 17, 36, 65, 106,...). - Gary W. Adamson, Jul 14 2007 a(n) = (n-1)*(n+1)*(n+3)/3 + 1. - 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 the 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 EXAMPLE 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 odd digit sum in base 4. Therefore, a(4)=(52-8)-8=36. - Vladimir Shevelev, May 18 2012 MATHEMATICA Table[n*(n^2+3*n-1)/3, {n, 0, 100}] [From Vladimir Joseph Stephan Orlovsky, Mar 08 2010] CROSSREFS a(n)=2*A000292(n-1)-1 (notice offset=-1 in A000292!) Cf. A131782. Sequence in context: A038633 A083045 A012277 * A024181 A023663 A048208 Adjacent sequences:  A084987 A084988 A084989 * A084991 A084992 A084993 KEYWORD nonn,easy AUTHOR Gary Adamson, Jul 16 2003 STATUS approved

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