OFFSET
0,3
COMMENTS
Sequences of the form a(n+1) = a(n) + (a(n) - a(n-1) + (n+1)^k mod 10) mod 10 with a(0)=0, a(1)=1 and k=1,2,3,4,5, etc. are identical if the exponents "k" differ by 4. Therefore this sequence, where k = 2, is the same as those with exponents 6, 10, 14, 18, etc. - Paolo P. Lava, Sep 29 2006
REFERENCES
G. Balzarotti and P. P. Lava, Le sequenze di numeri interi, Hoepli, 2008, p. 62.
FORMULA
a(n+1) = a(n) + ( a(n) - a(n-1) + (n+1)^2 mod 10 ) mod 10, with a(0)=0, a(1)=1.
Conjectures from Chai Wah Wu, Jan 28 2024: (Start)
a(n) = a(n-1) + a(n-5) - a(n-6) - a(n-10) + a(n-11) + a(n-15) - a(n-16) for n > 15.
G.f.: x*(9*x^12 + 5*x^11 + 6*x^10 + 5*x^8 - 5*x^6 + 5*x^4 + 4*x^2 + 5*x + 1)/(x^16 - x^15 - x^11 + x^10 + x^6 - x^5 - x + 1). (End)
EXAMPLE
a(8) = 20 because a(7) - a(6) + (8^2 mod 10) = 16 - 16 + 4 = 4 and a(7) + (4 mod 10) = 16 + 4 = 20.
Jumping by the squares of the sequence of positive integers we move to the numbers 0, 1, 5, 4, 0, 5, 1, 0, 4, 5, etc. Summing the numbers we obtain 0, 0+1=1, 1+5=6, 6+4=10, 10+0=10, 10+5=15, etc.
MAPLE
ANM:=proc(N) global anplus1, anminus1; local an, i; anminus1:=0; an:=1; print (anminus1, an); for i from 2 by 1 to N do anplus1:=an+((an-anminus1+ i^2 mod 10) mod 10); print(anplus1); anminus1:=an; an:=anplus1; od; end: ANM(100);
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Giorgio Balzarotti & Paolo P. Lava, Oct 13 2005
STATUS
approved