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A047209
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Numbers that are congruent to {1, 4} mod 5.
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53
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1, 4, 6, 9, 11, 14, 16, 19, 21, 24, 26, 29, 31, 34, 36, 39, 41, 44, 46, 49, 51, 54, 56, 59, 61, 64, 66, 69, 71, 74, 76, 79, 81, 84, 86, 89, 91, 94, 96, 99, 101, 104, 106, 109, 111, 114, 116, 119, 121, 124, 126, 129, 131, 134, 136, 139, 141, 144, 146, 149, 151, 154
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OFFSET
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1,2
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COMMENTS
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Apart from initial term(s), dimension of the space of weight 2n cuspidal newforms for Gamma_0( 72 ).
Cf. property described by Gary Detlefs in A113801: more generally, these numbers are of the form (2*h*n+(h-4)*(-1)^n-h)/4 (h, n natural numbers), therefore (2*h*n + (h-4)*(-1)^n - h)/4)^2 - 1 == 0 (mod h); in our case, a(n)^2 - 1 == 0 (mod 5). - Bruno Berselli, Nov 17 2010
The sum of the alternating series (-1)^(n+1)/a(n) from n=1 to infinity is (Pi/5)*cot(Pi/5), that is (1/5)*sqrt(1 + 2/sqrt(5))*Pi. - Jean-François Alcover, May 03 2013
These numbers appear in the product of a Rogers-Ramanujan identity. See A003114 also for references. - Wolfdieter Lang, Oct 29 2016
Let m be a product of any number of terms of this sequence. Then m - 1 or m + 1 is divisible by 5. Closed under multiplication. - David A. Corneth, May 11 2018
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LINKS
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FORMULA
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G.f.: (1+3x+x^2)/((1-x)(1-x^2)).
a(n) = (10*n + (-1)^n - 5)/4.
a(n) - a(n-1) - a(n-2) + a(n-3) = 0 for n > 3.
a(n) = a(n-2) + 5 for n > 2.
a(n) = 5*A000217(n-1) + 1 - 2*Sum_{i=1..n-1} a(i) for n > 1.
a(n)^2 = 5*A036666(n) + 1 (cf. also Comments). (End)
a(n) = 5*floor(n/2) + (-1)^(n+1). - Gary Detlefs, Dec 29 2011
E.g.f.: 1 + ((10*x - 5)*exp(x) + exp(-x))/4. - David Lovler, Aug 23 2022
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MAPLE
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MATHEMATICA
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PROG
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(Haskell)
a047209 = (flip div 2) . (subtract 2) . (* 5)
a047209_list = 1 : 4 : (map (+ 5) a047209_list)
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CROSSREFS
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Cf. A000566, A036666, A003114, A203776, A047336, A047522, A056020, A090771, A175885, A091998, A175886, A175887.
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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