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

0,3

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.

Links

Table of n, a(n) for n=0..59.

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

Cf. A111072.

Sequence in context: A127019 A175612 A024746 * A144394 A250742 A315122

Adjacent sequences: A111090 A111091 A111092 * A111094 A111095 A111096

Keyword

easy,nonn

Author

Giorgio Balzarotti & Paolo P. Lava, Oct 13 2005

Status

approved