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Search: id:A076312
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| A076312 |
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floor(n/10) + 2*(n mod 10). |
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+0
7
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| 0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 7, 9, 11, 13, 15, 17, 19, 21
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OFFSET
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0,2
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COMMENT
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(n==0 modulo 19) iff (a(n)==0 modulo 19); applied recursivly, this property provides a divisibility test for numbers given in base 10 notation.
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REFERENCES
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Karl Menninger, Rechenkniffe, Vandenhoeck & Ruprecht in Goettingen (1961), 79A.
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FORMULA
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G.f.: -x(17x^9-2-2x-2x^2-2x^3-2x^4-2x^5-2x^6-2x^7-2x^8)/((1-x)^2(1+x)(x^4+x^3+x^2+x+1)(x^4-x^3+x^2-x+1)). a(n)=A059995(n)+2*A010879(n). [From R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Jan 24 2009]
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EXAMPLE
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26468 is not a multiple of 19, as 26468 -> 2646+2*8=2662 -> 266+2*2=270 -> 27+2*0=27=19*1+8, therefore the answer is NO;
is 12882 divisible by 19? 12882 -> 1288+2*2=1292 -> 129+2*2=133 -> 13+2*3=19, therefore the answer is YES.
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CROSSREFS
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Cf. A008601, A076309, A076310, A076311.
Sequence in context: A076309 A088133 A115299 this_sequence A061762 A136614 A097586
Adjacent sequences: A076309 A076310 A076311 this_sequence A076313 A076314 A076315
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KEYWORD
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nonn
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AUTHOR
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Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Oct 06 2002
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