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A116520
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a(0) = 0, a(1) = 1; a(n) = max { 4*a(k)+a(n-k) | 1 <= k <= n/2 }, for n>1.
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5
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0, 1, 5, 9, 25, 29, 45, 61, 125, 129, 145, 161, 225, 241, 305, 369, 625, 629, 645, 661, 725, 741, 805, 869, 1125, 1141, 1205, 1269, 1525, 1589, 1845, 2101, 3125, 3129, 3145, 3161, 3225, 3241, 3305, 3369, 3625, 3641, 3705, 3769, 4025, 4089, 4345, 4601, 5625
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,3
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COMMENTS
| A 5-divide version of A084230.
Zero together with the partial sums of A102376. [From Omar E. Pol (info(AT)polprimos.com), May 05 2010]
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REFERENCES
| Harborth, H. Number of Odd Binomial Coefficients. Proc. Amer. Math. Soc. 62, 19-22, 1977
D. E. Knuth, Problem submitted to Amer. Math. Monthly, Jun 18 2007.
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LINKS
| Eric Weisstein's World of Mathematics, Stolarsky-Harborth Constant
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FORMULA
| a(0) = 1, a(1) = 1; thereafter a(2n)=5a(n) and a(2n+1)=4a(n)+a(n+1).
Contribution from Gary W. Adamson (qntmpkt(AT)yahoo.com), Mar 26 2010: (Start)
Let r(x) = (1 + 5x + 4x^2). Then (1 + 5x + 9x^2 + 25x^3 + ...) =
r(x) * r(x^2) * r(x^4) * r(x^8) * ... (End)
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MAPLE
| a:=proc(n) if n=0 then 0 elif n=1 then 1 elif n mod 2 = 0 then 5*a(n/2) else 4*a((n-1)/2)+a((n+1)/2) fi end: seq(a(n), n=0..52);
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MATHEMATICA
| b[0] := 0 b[1] := 1 b[n_?EvenQ] := b[n] = 5*b[n/2] b[n_?OddQ] := b[n] = 4*b[(n - 1)/2] + b[(n + 1)/2] a = Table[b[n], {n, 1, 25}]
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CROSSREFS
| Cf. A006046, A077465, A130665, A130667.
Cf. A102376. [From Omar E. Pol (info(AT)polprimos.com), May 05 2010]
Sequence in context: A177240 A074741 A166701 * A024825 A147074 A147192
Adjacent sequences: A116517 A116518 A116519 * A116521 A116522 A116523
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KEYWORD
| nonn
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AUTHOR
| Roger Bagula (rlbagulatftn(AT)yahoo.com), Mar 15 2006
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EXTENSIONS
| Edited by N. J. A. Sloane (njas(AT)research.att.com), Apr 16 2006, Jul 02 2008
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