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 A255818 a(n) = 2*B*C*(n mod A) + A*C*(n mod B) + A*B*(n mod C) with A=3, B=5, C=7. 4
 106, 212, 108, 214, 215, 111, 112, 218, 114, 115, 221, 117, 223, 224, 15, 121, 227, 123, 229, 230, 21, 127, 233, 129, 130, 236, 132, 133, 239, 30, 136, 242, 138, 244, 140, 36, 142, 248, 144, 145, 251, 42, 148, 254, 45, 151, 257, 153, 154, 155, 51, 157, 263, 159 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS After 0 it cycles again from 106 (a(105)=0 so there are 105 (A*B*C) terms). This is another variation on A256496, where a(n) = B*C*(n mod A) + A*C*(n mod B) + A*B*(n mod C), modified to take the values A=3, B=5, C=7 and still maintain the equivalence a(n) mod ABC = n mod ABC. Here modification is required (to maintain that equivalence) so that 'BC' + 'AC' + 'AB' = ABC + 1 where 'BC', 'AC' and 'AB' are the coefficients. Therefore, a(n)= 2B*C*(n mod A) + A*C*(n mod B) + A*B*(n mod C) so that 2*5*7 + 3*7 + 3*5 = 3*5*7 = 70 + 21 + 15 = 106. This is an example with 1 modification. a(n) = n for n: 15, 21, 30, 36, 42, 45, 51, 57, 60, ..., 314. - Robert G. Wilson v, Apr 07 2015 LINKS Aaron Kastel, Table of n, a(n) for n = 1..105 (all terms of a full cycle). FORMULA G.f.: -x*(314*x^11 +836*x^10 +1460*x^9 +1976*x^8 +2384*x^7 +2475*x^6 +2355*x^5 +1921*x^4 +1384*x^3 +850*x^2 +424*x +106) / ((x -1)*(x^2 +x +1)*(x^4 +x^3 +x^2 +x +1)*(x^6 +x^5 +x^4 +x^3 +x^2 +x +1)). - Colin Barker, Apr 14 2015 MATHEMATICA f[n_] := 70 Mod[n, 3] + 21 Mod[n, 5] + 15 Mod[n, 7]; Array[f, 105] (* Robert G. Wilson v, Apr 07 2015 *) PROG (MAGMA) A:=3; B:=5; C:=7; [2*B*C*(n mod A)+A*C*(n mod B)+A*B*(n mod C): n in [1..60]]; // Bruno Berselli, Apr 14 2015 CROSSREFS Cf. A256643 for an example with 2 modifications and A256668 for 3 modifications. Sequence in context: A044338 A044719 A260202 * A090775 A213458 A260937 Adjacent sequences:  A255815 A255816 A255817 * A255819 A255820 A255821 KEYWORD nonn,easy AUTHOR Aaron Kastel, Apr 07 2015 STATUS approved

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Last modified January 19 20:48 EST 2022. Contains 350466 sequences. (Running on oeis4.)