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 A235699 a(n+1) = a(n) + (a(n) mod 10) + 1, a(0) = 0. 1
 0, 1, 3, 7, 15, 21, 23, 27, 35, 41, 43, 47, 55, 61, 63, 67, 75, 81, 83, 87, 95, 101, 103, 107, 115, 121, 123, 127, 135, 141, 143, 147, 155, 161, 163, 167, 175, 181, 183, 187, 195, 201, 203, 207, 215, 221, 223, 227, 235, 241, 243, 247, 255, 261, 263, 267, 275, 281, 283, 287, 295, 301, 303, 307, 315, 321, 323, 327, 335, 341, 343, 347, 355 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,3 COMMENTS Instead of (a(n) mod 10) one might say "the last (decimal) digit of a(n)". Apart from the initial term, the first differences form the periodic sequence (2,4,8,6)[repeated]. Without the final "+ 1" and starting with 1, one gets A102039: Indeed, the last digit cycles through 2,4,8,6 and therefore the sequence never becomes constant. LINKS Harvey P. Dale, Table of n, a(n) for n = 0..1000 Index entries for linear recurrences with constant coefficients, signature (2,-2,2,-1). FORMULA a(n) = 5*n-6+cos(n*Pi/2)+2*sin(n*Pi/2), for n>0. - Giovanni Resta, Jan 15 2014 EXAMPLE a(n) = -6+(1/2+i)*(-i)^n+(1/2-i)*i^n+5*n for n>0 where i=sqrt(-1). a(n) = 2*a(n-1)-2*a(n-2)+2*a(n-3)-a(n-4) for n>4. G.f.: x*(5*x^3+3*x^2+x+1) / ((x-1)^2*(x^2+1)). - Colin Barker, Jan 16 2014 MATHEMATICA NestList[#+Mod[#, 10]+1&, 0, 80] (* or *) Join[{0}, LinearRecurrence[{2, -2, 2, -1}, {1, 3, 7, 15}, 80]] (* Harvey P. Dale, Dec 21 2014 *) PROG (PARI) print1(a=0); for(i=1, 99, print1(", "a+=a%10+1)) CROSSREFS Cf. A102039, A235698, A045844. Sequence in context: A323650 A144751 A138847 * A077777 A153829 A153830 Adjacent sequences:  A235696 A235697 A235698 * A235700 A235701 A235702 KEYWORD nonn,base,easy AUTHOR M. F. Hasler, Jan 14 2014 STATUS approved

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Last modified December 10 14:27 EST 2019. Contains 329896 sequences. (Running on oeis4.)