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Smallest available integer which fits into the repeating pattern 02468.
0

%I #8 Oct 09 2013 14:20:19

%S 0,2,4,6,80,24,680,246,802,46,8024,6802,4680,24680,246802,46802,

%T 468024,68024,680246,80246,8024680,2468024,68024680,24680246,80246802,

%U 4680246,802468024,680246802,468024680,2468024680,24680246802,4680246802

%N Smallest available integer which fits into the repeating pattern 02468.

%C a(n) must be chosen so its rightmost digit is not 8 (so that the next term won't start with 0). - _Sam Alexander_, Jan 04 2005

%C If n>=20, then a(n) is a(n-16) with one period 24680 (or a suitable cyclic permutation thereof) appended (or prepended, or inserted, whatever one prefers). [From _Hagen von Eitzen_, Jun 18 2009]

%F Let (c[0], c[1], ..., c[15]) = (8024600, 2468000, 68024000, 24680000, 80246000, 4680200, 802460000, 680240000, 468020000, 2468000000, 24680000000, 4680200000, 46802000000, 6802400000,68024000000,8024600000), i.e. c[r] = a[r+20] - a[r+4] for 0 <= r < 16. If n>=4, then writing n = 16*k + r + 4 with 0<=r<16 we have a(n) = floor( c[r]*100000^k/99999 ). [From _Hagen von Eitzen_, Jun 18 2009]

%F G.f.: -4 + 2 x - 2 x^2 + 6 x^3 + (4 + 6 x^2 + 80 x^4 + 24 x^5 + 680 x^6 + 246 x^7 + 802 x^8 + 46 x^9 + 8024 x^10 + 6802 x^11 + 4680 x^12 + 24680 x^13 + 246802 x^14 + 46802 x^15 + 68020 x^16 + 68024 x^17 + 80240 x^18 + 80246 x^19 + 24600 x^20 + 68000 x^21 + 24000 x^22 + 80000 x^23 + 46000 x^24 + 80200 x^25 + 60000 x^26 + 40000 x^27 + 20000 x^28 - 200000 x^30)/(1 - 100001 x^16 + 100000 x^32) [From _Hagen von Eitzen_, Jul 19 2009]

%K base,easy,nonn

%O 0,2

%A _Eric Angelini_, Oct 01 2004

%E More terms from _Sam Alexander_, Jan 04 2005

%E More terms from _Hagen von Eitzen_, Jun 18 2009