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A069129
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Centered 16-gonal numbers.
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35
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1, 17, 49, 97, 161, 241, 337, 449, 577, 721, 881, 1057, 1249, 1457, 1681, 1921, 2177, 2449, 2737, 3041, 3361, 3697, 4049, 4417, 4801, 5201, 5617, 6049, 6497, 6961, 7441, 7937, 8449, 8977, 9521, 10081, 10657, 11249, 11857, 12481, 13121, 13777, 14449
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OFFSET
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1,2
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COMMENTS
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Also, sequence found by reading the line from 1, in the direction 1, 17, ..., in the square spiral whose vertices are the triangular numbers A000217. Opposite numbers to the members of A139098 in the same spiral. - Omar E. Pol, Apr 26 2008
The subsequence of primes begins: 17, 97, 241, 337, 449, 577, 881, 1249, 3041, 3361, 3697, 4049, 4801, 6961, 7937, 9521, 10657, 13121, 14449. See A184899: n such that the n-th centered 12-gonal number is prime. Indices of prime star numbers. - Jonathan Vos Post, Feb 27 2011
Binomial transform of [1, 16, 16, 0, 0, 0, ...] and Narayana transform (A001263) of [1, 16, 0, 0, 0, ...]. - Gary W. Adamson, Jul 28 2011
Centered hexadecagonal numbers or centered hexakaidecagonal numbers. - Omar E. Pol, Oct 03 2011
a(n) = m(n,n) for an array constructed by using the terms in A016813 as the antidiagonals; the first few antidiagonals are 1; 5,9; 13,17,21; 25,29,33,37. - J. M. Bergot, Jul 05 2013
[The first five rows begin: 1,9,21,37,57; 5,17,33,53,77; 13,29,49,73,101; 25,45,69,97,129; 41,65,93,125,161.]
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LINKS
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FORMULA
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a(n) = 8*n^2 - 8*n + 1.
G.f.: -x*(1+14*x+x^2) / (x-1)^3. - R. J. Mathar, Feb 04 2011
Sum_{n>=1} 1/a(n) = Pi * tan(Pi/(2*sqrt(2))) / (4*sqrt(2)). - Vaclav Kotesovec, Jul 23 2019
Sum_{n>=1} a(n)/n! = 9*e - 1.
Sum_{n>=1} (-1)^n * a(n)/n! = 9/e - 1. (End)
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EXAMPLE
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a(5) = 161 because 8*5^2 - 8*5 + 1 = 200 - 40 + 1 = 161.
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MATHEMATICA
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Rest[CoefficientList[Series[-x(1+14x+x^2)/(x-1)^3, {x, 0, 50}], x]] (* Harvey P. Dale, Apr 22 2011 *)
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PROG
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CROSSREFS
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KEYWORD
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easy,nice,nonn
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AUTHOR
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STATUS
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approved
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