A111093 - OEIS

%I #24 Apr 19 2024 11:14:21

%S 0,1,6,10,10,15,16,16,20,25,30,36,36,45,50,50,56,61,70,70,70,71,76,80,

%T 80,85,86,86,90,95,100,106,106,115,120,120,126,131,140,140,140,141,

%U 146,150,150,155,156,156,160,165,170,176,176,185,190,190,196,201,210,210

%N Like sequence A111072 but moving right by the squares of the sequence of positive integers.

%C 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

%D G. Balzarotti and P. P. Lava, Le sequenze di numeri interi, Hoepli, 2008, p. 62.

%F a(n+1) = a(n) + ( a(n) - a(n-1) + (n+1)^2 mod 10 ) mod 10, with a(0)=0, a(1)=1.

%F Conjectures from _Chai Wah Wu_, Jan 28 2024: (Start)

%F 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.

%F 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)

%e 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.

%e 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.

%p 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);

%Y Cf. A111072.

%K easy,nonn

%O 0,3

%A _Giorgio Balzarotti_ & _Paolo P. Lava_, Oct 13 2005