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A084990
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n*(n^2+3*n-1)/3.
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15
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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
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OFFSET
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0,3
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COMMENTS
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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
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LINKS
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Table of n, a(n) for n=0..44.
V. Shevelev, On monotonic strengthening of Newman-like phenomenon on (2m+1)-multiples in base 2m
Index to sequences with linear recurrences with constant coefficients, signature (4,-6,4,-1).
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FORMULA
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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
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EXAMPLE
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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
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MATHEMATICA
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Table[n*(n^2+3*n-1)/3, {n, 0, 100}] [From Vladimir Joseph Stephan Orlovsky, Mar 08 2010]
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CROSSREFS
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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
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KEYWORD
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nonn,easy
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AUTHOR
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Gary Adamson, Jul 16 2003
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STATUS
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approved
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