

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
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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*n6+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+(1/2i)*i^n+5*n for n>0 where i=sqrt(1). a(n) = 2*a(n1)2*a(n2)+2*a(n3)a(n4) for n>4. G.f.: x*(5*x^3+3*x^2+x+1) / ((x1)^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



