A111093 Like sequence A111072 but moving right by the squares of the sequence of positive integers.

0, 1, 6, 10, 10, 15, 16, 16, 20, 25, 30, 36, 36, 45, 50, 50, 56, 61, 70, 70, 70, 71, 76, 80, 80, 85, 86, 86, 90, 95, 100, 106, 106, 115, 120, 120, 126, 131, 140, 140, 140, 141, 146, 150, 150, 155, 156, 156, 160, 165, 170, 176, 176, 185, 190, 190, 196, 201, 210, 210

Offset: 0

Author

Giorgio Balzarotti & Paolo P. Lava, Oct 13 2005

Keywords

• easy
• nonn

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

Examples

			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.
		

References

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

Crossrefs

Cf. A111072.

Programs

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);

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)