cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

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

Original entry on oeis.org

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

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Author

Giorgio Balzarotti & Paolo P. Lava, Oct 13 2005

Keywords

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)

A008954 Final digit of triangular number n*(n+1)/2.

Original entry on oeis.org

0, 1, 3, 6, 0, 5, 1, 8, 6, 5, 5, 6, 8, 1, 5, 0, 6, 3, 1, 0, 0, 1, 3, 6, 0, 5, 1, 8, 6, 5, 5, 6, 8, 1, 5, 0, 6, 3, 1, 0, 0, 1, 3, 6, 0, 5, 1, 8, 6, 5, 5, 6, 8, 1, 5, 0, 6, 3, 1, 0, 0, 1, 3, 6, 0, 5, 1, 8, 6, 5, 5, 6, 8, 1, 5, 0, 6, 3, 1, 0, 0, 1, 3, 6, 0, 5, 1, 8, 6, 5, 5, 6, 8, 1, 5, 0, 6, 3, 1, 0
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Links

Crossrefs

Cf. A000217, A010879, A061501, A008953.
First differences of A111072.

Programs

  • GAP
    List([0..100], n-> (Binomial(n+1,2) mod 10) ); # G. C. Greubel, Sep 14 2019
  • Magma
    [Binomial(n+1,2) mod 10: n in [0..100]]; // G. C. Greubel, Sep 14 2019
  • Maple
    seq(mod(binomial(n+1, 2),10), n = 0 .. 100); # G. C. Greubel, Sep 14 2019
  • Mathematica
    Table[Mod[n*(n+1)/2, 10], {n, 0, 100}]
LinearRecurrence[{0,0,0,0,1,0,0,0,0,-1,0,0,0,0,1},{0,1,3,6,0,5,1,8,6,5,5, 6,8,1,5},110] (* Harvey P. Dale, Dec 31 2014 *)
  • PARI
    a(n)=n*(n+1)/2%10 \\ Charles R Greathouse IV, Mar 05 2014
  • Sage
    [Mod(binomial(n+1,2), 10) for n in (0..100)] # G. C. Greubel, Sep 14 2019

Formula

a(1) = 1, a(n+1) = (a(n) + n + 1) mod 10.
Periodic with period 20: repeat [0,1,3,6,0,5,1,8,6,5,5,6,8,1,5,0,6,3, 1,0]. - Franklin T. Adams-Watters, Mar 13 2006
It follows that all triangular numbers end with a digit of 0, 1, 3, 5, 6, or 8, and thus none end with a digit of 2, 4, 7, or 9. - Harvey P. Dale, Dec 31 2014
a(n) = n*(n+1)/2 mod 10. - Ant King, Apr 26 2009
From R. J. Mathar, Apr 15 2010: (Start)
a(n) = a(n-5) - a(n-10) + a(n-15).
G.f.: x*(1 +3*x +6*x^2 +5*x^4 +5*x^6 +5*x^8 +6*x^10 +3*x^11 +x^12)/(1 -x^5 +x^10 -x^15). (End)
a(n) = A010879(A000217(n)). - Michel Marcus, May 26 2022

A111132 a(n+1) = a(n) + (a(n) - a(n-1) + a(n) mod 10) mod 10 with a(0)=0 and a(1)=1.

Original entry on oeis.org

1, 3, 8, 11, 15, 24, 27, 27, 34, 35, 41, 48, 53, 61, 70, 79, 87, 92, 99, 105, 106, 113, 113, 116, 125, 129, 132, 137, 139, 140, 141, 143, 148, 151, 155, 164, 167, 167, 174, 175, 181, 188, 193, 201, 210, 219, 227, 232, 239, 245, 246, 253, 253, 256, 265, 269, 272
Offset: 0

Views

Author

Giorgio Balzarotti and Paolo P. Lava, Oct 17 2005, corrected Sep 29 2006

Keywords

Comments

Similar to A111072 but moving right by a(n).

Examples

			Write the digit string 0123456789 and repeat it infinitely many times. Then, starting from the first "1" digit on the left side, move right by one digit (to the digit "2"), then take the sum 1+2=3. Now move right by 3 digits (the result of the previous sum). We are at the digit "5"; then make the sum 3+5=8. Repeating the process we get 1, 3, 8, 11, 15, 24, 27, 27, 34, ....

Crossrefs

Cf. A111072, A111093.

Programs

  • Maple
    ANM:=proc(N) global anplus1,anminus1; local an,i,anpolus; anminus1:=0; an:=1; print (an); for i from 2 by 1 to N do anplus1:=an+((an-anminus1+ an mod 10) mod 10); print(anplus1); anminus1:=an; an:=anplus1; od; end: ANM(100);
  • Mathematica
    nxt[{a_,b_}]:={b,b+Mod[b-a+Mod[b,10],10]}; NestList[nxt,{0,1},60][[All,1]] (* Harvey P. Dale, Nov 22 2018 *)
Showing 1-3 of 3 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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