A002266 -id:A002266 - cp's OEIS frontend

A130519 a(n) = Sum_{k=0..n} floor(k/4). (Partial sums of A002265.)

0, 0, 0, 0, 1, 2, 3, 4, 6, 8, 10, 12, 15, 18, 21, 24, 28, 32, 36, 40, 45, 50, 55, 60, 66, 72, 78, 84, 91, 98, 105, 112, 120, 128, 136, 144, 153, 162, 171, 180, 190, 200, 210, 220, 231, 242, 253, 264, 276, 288, 300, 312, 325, 338, 351, 364, 378, 392, 406, 420, 435, 450

Offset: 0

Hieronymus Fischer, Jun 01 2007

Complementary to A130482 with respect to triangular numbers, in that A130482(n) + 4*a(n) = n(n+1)/2 = A000217(n).
Disregarding the first three 0's the resulting sequence a'(n) is the sum of the positive integers <= n that have the same residue modulo 4 as n. This is the additive counterpart of the quadruple factorial numbers. - Peter Luschny, Jul 06 2011
From Heinrich Ludwig, Dec 23 2017: (Start)
Column sums of (shift of rows = 4):
  1 2 3 4 5 6  7  8  9 10 11 12 13 14 ...
          1 2  3  4  5  6  7  8  9 10 ...
                     1  2  3  4  5  6 ...
                                 1  2 ...
  .......................................
  ---------------------------------------
  1 2 3 4 6 8 10 12 15 18 21 24 28 32 ...
  shift of rows = 1 see A000217
  shift of rows = 2 see A002620
  shift of rows = 3 see A001840
  shift of rows = 5 see A130520
(End)
Conjecture: a(n+2) is the maximum effective weight of a numerical semigroup S of genus n (see Nathan Pflueger). - Stefano Spezia, Jan 04 2019

			G.f. = x^4 + 2*x^5 + 3*x^6 + 4*x^7 + 6*x^8 + 8*x^9 + 10*x^10 + 12*x^11 + ...
[ n] a(n)
---------
[ 4] 1
[ 5] 2
[ 6] 3
[ 7] 4
[ 8] 1 + 5
[ 9] 2 + 6
[10] 3 + 7
[11] 4 + 8

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Bakir Farhi, On the Representation of the Natural Numbers as the Sum of Three Terms of the Sequence floor(n^2/a), Journal of Integer Sequences, Vol. 16 (2013), Article 13.6.4.
Darren Glass and Joshua Wagner, Arithmetical Structures on Paths With a Doubled Edge, arXiv:1903.01398 [math.CO], 2019.
Nathan Pflueger, On non-primitive Weierstrass points, Alg. Number Th. 12 (2018) 1923-1947 and arXiv:1608.0566 [math.AG], 2016.
Index entries for linear recurrences with constant coefficients, signature (2,-1,0,1,-2,1).

Cf. A000217, A001840, A002264, A002265, A002266, A002620, A004526, A010872, A010873, A010874, A130481, A130483, A130520.

Cf. A000290, A007590, A000212, A118015, A056827, A056834, A056838, A056865.

a:=List([0..65],n->Sum([0..n],k->Int(k/4)));; Print(a); # Muniru A Asiru, Jan 04 2019

[Round(n*(n-2)/8): n in [0..70]]; // Vincenzo Librandi, Jun 25 2011

quadsum := n -> add(k, k = select(k -> k mod 4 = n mod 4, [$1 .. n])):
A130519 := n ->`if`(n<3,0,quadsum(n-3)); seq(A130519(n),n=0..58); # Peter Luschny, Jul 06 2011

a[ n_] := Quotient[ (n - 1)^2, 8]; (* Michael Somos, Oct 14 2011 *)

makelist(floor((n-1)^2/8), n, 0, 70); /* Stefano Spezia, Jan 04 2019 */

{a(n) = (n - 1)^2 \ 8}; /* Michael Somos, Oct 14 2011 */

def A130519(n): return (n-1)**2>>3  # Chai Wah Wu, Jul 30 2022

G.f.: x^4/((1-x^4)*(1-x)^2) = x^4/((1+x)*(1+x^2)*(1-x)^3).
a(n) = +2*a(n-1) -1*a(n-2) +1*a(n-4) -2*a(n-5) +1*a(n-6).
a(n) = floor(n/4)*(n - 1 - 2*floor(n/4)) = A002265(n)*(n - 1 - 2*A002265(n)).
a(n) = (1/2)*A002265(n)*(n - 2 + A010873(n)).
a(n) = floor((n-1)^2/8). - Mitch Harris, Sep 08 2008
a(n) = round(n*(n-2)/8) = round((n^2-2*n-1)/8) = ceiling((n+1)*(n-3)/8). - Mircea Merca, Nov 28 2010
a(n) = A001972(n-4), n>3. - Franklin T. Adams-Watters, Jul 10 2009
a(n) = a(n-4)+n-3, n>3. - Mircea Merca, Nov 28 2010
Euler transform of length 4 sequence [ 2, 0, 0, 1]. - Michael Somos, Oct 14 2011
a(n) = a(2-n) for all n in Z. - Michael Somos, Oct 14 2011
a(n) = A214734(n, 1, 4). - Renzo Benedetti, Aug 27 2012
a(4n) = A000384(n), a(4n+1) = A001105(n), a(4n+2) = A014105(n), a(4n+3) = A046092(n). - Philippe Deléham, Mar 26 2013
a(n) = Sum_{i=1..ceiling(n/2)-1} (i mod 2) * (n - 2*i - 1). - Wesley Ivan Hurt, Jan 23 2014
a(n) = ( 2*n^2-4*n-1+(-1)^n+2*((-1)^((2*n-1+(-1)^n)/4)-(-1)^((6*n-1+(-1)^n)/4)) )/16 = ( 2*n*(n-2) - (1-(-1)^n)*(1-2*i^(n*(n-1))) )/16, where i=sqrt(-1). - Luce ETIENNE, Aug 29 2014
E.g.f.: (1/8)*((- 1 + x)*x*cosh(x) + 2*sin(x) + (- 1 - x + x^2)*sinh(x)). - Stefano Spezia, Jan 15 2019
a(n) = (A002620(n-1) - A011765(n+1)) / 2, for n > 0. - Yuchun Ji, Feb 05 2021
Sum_{n>=4} 1/a(n) = Pi^2/12 + 5/2. - Amiram Eldar, Aug 13 2022

Partially edited by R. J. Mathar, Jul 11 2009

A057354 a(n) = floor(2*n/5).

Original entry on oeis.org

0, 0, 0, 1, 1, 2, 2, 2, 3, 3, 4, 4, 4, 5, 5, 6, 6, 6, 7, 7, 8, 8, 8, 9, 9, 10, 10, 10, 11, 11, 12, 12, 12, 13, 13, 14, 14, 14, 15, 15, 16, 16, 16, 17, 17, 18, 18, 18, 19, 19, 20, 20, 20, 21, 21, 22, 22, 22, 23, 23, 24, 24, 24, 25, 25, 26, 26, 26, 27, 27, 28, 28, 28, 29, 29, 30, 30

Offset: 0

Mitch Harris

The cyclic pattern (and numerator of the gf) is computed using Euclid's algorithm for GCD.
The sequence a(n) can be used in determining confidence intervals for the median of a population. Let Y(i) denote the i-th smallest datum in a random sample of size n from any population of values. When estimating the population median with a symmetric interval [Y(r), Y(n-r+1)], the exact confidence coefficient c for the interval is given by c = Sum_{k=r..n-r} C(n, k)(1/2)^n. If  r = a(n-4), then the confidence coefficient will be (i) at least 0.90 for all n>=7, (ii) at least 0.95 for all n>=35, and (iii) at least 0.99 for all n>=115. To use the sequence, for example, decide on the minimum level of confidence desired, say 95%. Hence use a sample size of 35 or greater, say n=40. We then find a(n-4)=a(36)=14, and thus the 14th smallest and 14th largest values in the sample will form the bounds for the confidence interval. If the exact confidence coefficient c is needed, calculate c = Sum_{k=14..26} C(40,k)(1/2)^40, which is 0.9615226917. - Dennis P. Walsh, Nov 28 2011
a(n+2) is also the domination number of the n-antiprism graph. - Eric W. Weisstein, Apr 09 2016
Equals partial sums of 0 together with 0, 0, 1, 0, 1, ... (repeated). - Bruno Berselli, Dec 06 2016
Euler transform of length 5 sequence [1, 1, 0, -1, 1]. - Michael Somos, Dec 06 2016

			G.f. = x^3 + x^4 + 2*x^5 + 2*x^6 + 2*x^7 + 3*x^8 + 3*x^9 + 4*x^10 + 4*x^11 + ...

N. Dershowitz and E. M. Reingold, Calendrical Calculations, Cambridge University Press, 1997.
R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, Addison-Wesley, NY, 1994.

Colin Barker, Table of n, a(n) for n = 0..1000
N. Dershowitz and E. M. Reingold, Calendrical Calculations Web Site.
Dennis Walsh, Median estimation with the point-four-n-minus-two rule.
Eric Weisstein's World of Mathematics, Antiprism Graph.
Eric Weisstein's World of Mathematics, Domination Number.
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,1,-1).

Floors of other ratios: A004526, A002264, A002265, A004523, A057353, A057355, A057356, A057357, A057358, A057359, A057360, A057361, A057362, A057363, A057364, A057365, A057366, A057367.

Cf. A002266.

[2*n div 5: n in [0..80]]; // Bruno Berselli, Dec 06 2016

Table[Floor[2 n/5], {n, 0, 80}] (* Bruno Berselli, Dec 06 2016 *)
a[ n_] := Quotient[2 n, 5]; (* Michael Somos, Dec 06 2016 *)

a(n)=2*n\5 \\ Charles R Greathouse IV, Nov 28 2011

concat(vector(3), Vec(x^3*(1 + x^2) / ((1 - x)^2*(1 + x + x^2 + x^3 + x^4)) + O(x^80))) \\ Colin Barker, Dec 06 2016.

[int(2*n/5) for n in range(80)] # Bruno Berselli, Dec 06 2016

[floor(2*n/5) for n in range(80)] # Bruno Berselli, Dec 06 2016

G.f.: x^3*(1 + x^2) / ((x^4 + x^3 + x^2 + x + 1)*(x - 1)^2). - Numerator corrected by R. J. Mathar, Feb 20 2011
a(n) = a(n-1) + a(n-5) - a(n-6) for n>5. - Colin Barker, Dec 06 2016
a(n) = -a(2-n) for all n in Z. - Michael Somos, Dec 06 2016
a(n) = A002266(2*n). - R. J. Mathar, Jul 21 2020
Sum_{n>=3} (-1)^(n+1)/a(n) = log(2)/2. - Amiram Eldar, Sep 30 2022

A010882 Period 3: repeat [1, 2, 3].

Original entry on oeis.org

1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3

Offset: 0

N. J. A. Sloane

Partial sums are given by A130481(n)+n+1. - Hieronymus Fischer, Jun 08 2007
41/333 = 0.123123123... - Eric Desbiaux, Nov 03 2008
Terms of the simple continued fraction for 3/(sqrt(37)-4). - Paolo P. Lava, Feb 16 2009
This is the lexicographically earliest sequence with no substring of more than 1 term being a palindrome. - Franklin T. Adams-Watters, Nov 24 2013

Index entries for linear recurrences with constant coefficients, signature (0,0,1).
Index entries for sequences that are fixed points of mappings

Cf. A010872, A010873, A010874, A010875, A010876, A004526, A002264, A002265, A002266, A177036 (decimal expansion of (4+sqrt(37))/7), A214090.

Cf. A130481, A180593.

a010882 = (+ 1) . (`mod` 3)
a010882_list = cycle [1,2,3]
-- Reinhard Zumkeller, Mar 20 2013

&cat[[1..3]^^30]; // Vincenzo Librandi, Feb 04 2016

seq(op([1, 2, 3]), n=0..50); # Wesley Ivan Hurt, Jul 05 2016

Nest[ Flatten[ # /. {1 -> {1, 2}, 2 -> {3, 1}, 3 -> {2, 3}}] &, {1}, 7] (* Robert G. Wilson v, Mar 08 2005 *)
PadRight[{},120,{1,2,3}] (* Harvey P. Dale, Apr 09 2018 *)

a(n) = 1 + n%3; \\ Michel Marcus, Feb 04 2016

G.f.: (1+2x+3x^2)/(1-x^3). - Paul Barry, May 25 2003
a(n) = 1 + (n mod 3). - Paolo P. Lava, Nov 21 2006
a(n) = A010872(n) + 1. - Hieronymus Fischer, Jun 08 2007
a(n) = 6 - a(n-1) - a(n-2) for n > 1. - Reinhard Zumkeller, Apr 13 2008
a(n) = n+1-3*floor(n/3) = floor(41*10^(n+1)/333)-floor(41*10^n/333)*10; a(n)-a(n-3)=0 with n>2. - Bruno Berselli, Jun 28 2010
a(n) = A180593(n+1)/3. - Reinhard Zumkeller, Oct 25 2010
a(n) = floor((4*n+3)/3) mod 4. - Gary Detlefs, May 15 2011
a(n) = -cos(2/3*Pi*n)-1/3*3^(1/2)*sin(2/3*Pi*n)+2. - Leonid Bedratyuk, May 13 2012
E.g.f.: 2*(3*exp(3*x/2) - sqrt(3)*cos(Pi/6-sqrt(3)*x/2))*exp(-x/2)/3. - Ilya Gutkovskiy, Jul 05 2016

A008959 Final digit of squares: a(n) = n^2 mod 10.

Original entry on oeis.org

0, 1, 4, 9, 6, 5, 6, 9, 4, 1, 0, 1, 4, 9, 6, 5, 6, 9, 4, 1, 0, 1, 4, 9, 6, 5, 6, 9, 4, 1, 0, 1, 4, 9, 6, 5, 6, 9, 4, 1, 0, 1, 4, 9, 6, 5, 6, 9, 4, 1, 0, 1, 4, 9, 6, 5, 6, 9, 4, 1, 0, 1, 4, 9, 6, 5, 6, 9, 4, 1, 0, 1, 4, 9, 6, 5, 6, 9, 4, 1, 0, 1, 4, 9, 6, 5, 6, 9, 4, 1, 0, 1, 4, 9, 6, 5, 6, 9, 4, 1, 0

Offset: 0

N. J. A. Sloane, Mar 15 1996

a(m*n) = a(m)*a(n) mod 10; a(5*n+k) = a(5*n-k) for k <= 5*n. - Reinhard Zumkeller, Apr 24 2009
a(n) = n^6 mod 10. - Zerinvary Lajos, Nov 06 2009
a(n) = A002015(n) mod 10 = A174452(n) mod 10. - Reinhard Zumkeller, Mar 21 2010
Decimal expansion of 166285490/1111111111. - Alexander R. Povolotsky, Mar 09 2013

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for sequences related to final digits of numbers
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,0,0,1).

Cf. A000290, A070431, A070435, A070438, A070442, A070452, A159852, A010879, A008960, A070514.

[0] cat [Intseq(n^2)[1]: n in [1..80]]; // Bruno Berselli, Feb 14 2013

[n^2 - 10*Floor(n^2/10): n in [0..80]]; // Vincenzo Librandi, Jun 16 2015

A008959:=n->(n^2 mod 10); seq(A008959(n), n=0..50); # Wesley Ivan Hurt, Jun 06 2014

Table[Mod[n^2,10],{n,0,200}] (* Vladimir Joseph Stephan Orlovsky, Apr 21 2011 *)
PowerMod[Range[0,80],2,10] (* or *) LinearRecurrence[{0,0,0,0,0,0,0,0,0,1},{0,1,4,9,6,5,6,9,4,1},120] (* Harvey P. Dale, Oct 16 2012 *)

a(n)=n^2%10 \\ Charles R Greathouse IV, Sep 24 2015

[power_mod(n,2,10) for n in range(0, 81)] # Zerinvary Lajos, Nov 06 2009

Periodic with period 10. - Franklin T. Adams-Watters, Mar 13 2006
a(n) = 4.5 - (1 + 5^(1/2))*cos(Pi*n/5) + (-1 - 3/5*5^(1/2))*cos(2*Pi*n/5) + (5^(1/2) - 1)*cos(3*Pi*n/5) + (-1 + 3/5*5^(1/2))*cos(4*Pi*n/5) - 0.5*(-1)^n. - Richard Choulet, Dec 12 2008
a(n) = A010879(A000290(n)). - Reinhard Zumkeller, Jan 04 2009
G.f.: (x^9+4*x^8+9*x^7+6*x^6+5*x^5+6*x^4+9*x^3+4*x^2+x)/(-x^10+1). - Colin Barker, Aug 14 2012
a(n) = n^2 - 10*floor(n^2/10). - Wesley Ivan Hurt, Jun 12 2013
a(n) = (n - 5*A002266(n + 2))^2 + 5*(5*A002266(n + 2) mod 2). - Wesley Ivan Hurt, Jun 06 2014
a(n) = A033569(n+3) mod 10. - Wesley Ivan Hurt, Dec 06 2014
a(n) = n^k mod 10; for k > 0 where k mod 4 = 2. - Doug Bell, Jun 15 2015

A130518 a(n) = Sum_{k=0..n} floor(k/3). (Partial sums of A002264.)

Original entry on oeis.org

0, 0, 0, 1, 2, 3, 5, 7, 9, 12, 15, 18, 22, 26, 30, 35, 40, 45, 51, 57, 63, 70, 77, 84, 92, 100, 108, 117, 126, 135, 145, 155, 165, 176, 187, 198, 210, 222, 234, 247, 260, 273, 287, 301, 315, 330, 345, 360, 376, 392, 408, 425, 442, 459, 477, 495, 513, 532, 551, 570

Offset: 0

Hieronymus Fischer, Jun 01 2007

Complementary with A130481 regarding triangular numbers, in that A130481(n) + 3*a(n) = n(n+1)/2 = A000217(n).
Apart from offset, the same as A062781. - R. J. Mathar, Jun 13 2008
Apart from offset, the same as A001840. - Michael Somos, Sep 18 2010
The sum of any three consecutive terms is a triangular number. - J. M. Bergot, Nov 27 2014

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (2,-1,1,-2,1).

Cf. A002265, A002266, A004526, A010872, A010873, A010874, A062781, A130482, A130483.

List([0..60], n-> Int(n*(n-1)/6)); # G. C. Greubel, Aug 31 2019

[Round(n*(n-1)/6): n in [0..60]]; // Vincenzo Librandi, Jun 25 2011

seq(floor(n*(n-1)/6),n=0..60); # Robert Israel, Nov 27 2014

Table[n, {n, 0, 19}, {3}] // Flatten // Accumulate (* Jean-François Alcover, Jun 05 2013 *)

a(n)=n*(n-1)\/6 \\ Charles R Greathouse IV, Jun 05 2013

[floor(binomial(n,2)/3) for n in range(0,60)] # Zerinvary Lajos, Dec 01 2009

G.f.: x^3 / ((1-x^3)*(1-x)^2).
a(n) = 2*a(n-1) - a(n-2) + a(n-3) - 2*a(n-4) + a(n-5).
a(n) = (1/2)*floor(n/3)*(2*n - 1 - 3*floor(n/3)) = A002264(n)*(2n - 1 - 3*A002264(n))/2.
a(n) = (1/2)*A002264(n)*(n - 1 + A010872(n)).
a(n) = round(n*(n-1)/6) = round((n^2-n-1)/6) = floor(n*(n-1)/6) = ceiling((n+1)*(n-2)/6). - Mircea Merca, Nov 28 2010
a(n) = a(n-3) + n - 2, n > 2. - Mircea Merca, Nov 28 2010
a(n) = A214734(n, 1, 3). - Renzo Benedetti, Aug 27 2012
a(3n) = A000326(n), a(3n+1) = A005449(n), a(3n+2) = 3*A000217(n) = A045943(n). - Philippe Deléham, Mar 26 2013
a(n) = (3*n*(n-1) - (-1)^n*((1+i*sqrt(3))^(n-2) + (1-i*sqrt(3))^(n-2))/2^(n-3) - 2)/18, where i=sqrt(-1). - Bruno Berselli, Nov 30 2014
Sum_{n>=3} 1/a(n) = 20/3 - 2*Pi/sqrt(3). - Amiram Eldar, Sep 17 2022

A131242 Partial sums of A059995: a(n) = sum_{k=0..n} floor(k/10).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 100, 105, 110, 115, 120, 125, 130, 135, 140, 145, 150, 156, 162, 168, 174, 180, 186, 192, 198

Offset: 0

Hieronymus Fischer, Jun 21 2007

Complementary with A130488 regarding triangular numbers, in that A130488(n)+10*a(n)=n(n+1)/2=A000217(n).

			As square array :
    0,   0,   0,   0,   0,   0,   0,   0,   0,    0
    1,   2,   3,   4,   5,   6,   7,   8,   9,   10
   12,  14,  16,  18,  20,  22,  24,  26,  28,   30
   33,  36,  39,  42,  45,  48,  51,  54,  57,   60
   64,  68,  72,  76,  80,  84,  88,  92,  96,  100
  105, 110, 115, 120, 125, 130, 135, 140, 145,  150
  156, 162, 168, 174, 180, 186, 192, 198, 204,  210
... - _Philippe Deléham_, Mar 27 2013

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,-1,0,0,0,0,0,0,0,1,-2,1).

Cf. A008728, A059995, A010879, A002266, A130488, A000217, A002620, A130518, A130519, A130520, A174709, A174738, A118729, A218470.

Table[(1/2)*Floor[n/10]*(2*n - 8 - 10*Floor[n/10]), {n,0,50}] (* G. C. Greubel, Dec 13 2016 *)
Accumulate[Table[FromDigits[Most[IntegerDigits[n]]],{n,0,110}]] (* or *) LinearRecurrence[{2,-1,0,0,0,0,0,0,0,1,-2,1},{0,0,0,0,0,0,0,0,0,0,1,2},120] (* Harvey P. Dale, Apr 06 2017 *)

for(n=0,50, print1((1/2)*floor(n/10)*(2n-8-10*floor(n/10)), ", ")) \\ G. C. Greubel, Dec 13 2016

a(n)=my(k=n\10); k*(n-5*k-4) \\ Charles R Greathouse IV, Dec 13 2016

a(n) = (1/2)*floor(n/10)*(2n-8-10*floor(n/10)).
a(n) = A059995(n)*(2n-8-10*A059995(n))/2.
a(n) = (1/2)*A059995(n)*(n-8+A010879(n)).
a(n) = (n-A010879(n))*(n+A010879(n)-8)/20.
G.f.: x^10/((1-x^10)(1-x)^2).
From Philippe Deléham, Mar 27 2013: (Start)
a(10n)   = A051624(n).
a(10n+1) = A135706(n).
a(10n+2) = A147874(n+1).
a(10n+3) = 2*A005476(n).
a(10n+4) = A033429(n).
a(10n+5) = A202803(n).
a(10n+6) = A168668(n).
a(10n+7) = 2*A147875(n).
a(10n+8) = A135705(n).
a(10n+9) = A124080(n). (End)
a(n) = A008728(n-10) for n>= 10. - Georg Fischer, Nov 03 2018

A076314 a(n) = floor(n/10) + (n mod 10).

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 8, 9, 10, 11, 12, 13, 14

Offset: 0

Reinhard Zumkeller, Oct 06 2002

For n<100 this is equal to the digital sum of n (see A007953). - Hieronymus Fischer, Jun 17 2007

			a(15) = floor(15 / 10) + (15 mod 10) = 1 + 5 = 6. - _Indranil Ghosh_, Feb 13 2017

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,0,0,0,0,1,-1).

Cf. A076313, A010879, A076309, A076310, A076311, A076312, A055017, A007953, A003132.

a076314 = uncurry (+) . flip divMod 10 -- Reinhard Zumkeller, Jun 01 2013

[n-9*Floor(n/10):n in [0..100]]; // Vincenzo Librandi, Dec 10 2016

Table[n - 9 Floor[n / 10], {n, 0, 100}] (* Vincenzo Librandi Dec 10 2016 *)

a(n)=n\10+n%10 \\ Charles R Greathouse IV, Sep 24 2015

def A076314(n): return (n/10)+(n%10) # Indranil Ghosh, Feb 13 2017

From Hieronymus Fischer, Jun 17 2007: (Start)
a(n) = n - 9*floor(n/10).
a(n) = (n + 9*(n mod 10))/10.
a(n) = n - 9*A002266(A004526(n)) = n - 9*A004526(A002266(n)).
a(n) = (n + 9*A010879(n))/10.
a(n) = (n + 9*A000035(n) + 18*A010874(A004526(n)))/10.
a(n) = (n + 9*A010874(n) + 45*A000035(A002266(n)))/10.
G.f.: x*(8*x^10 - 9*x^9 + 1)/((1 - x^10)*(1 - x)^2). (End)
a(n) = A033930(n) for 1 <= n < 100. - R. J. Mathar, Sep 21 2008
a(n) = +a(n-1) + a(n-10) - a(n-11). - R. J. Mathar, Feb 20 2011

A131296 a(n) = ds_5(a(n-1))+ds_5(a(n-2)), a(0)=0, a(1)=1; where ds_5=digital sum base 5.

Original entry on oeis.org

0, 1, 1, 2, 3, 5, 4, 5, 5, 2, 3, 5, 4, 5, 5, 2, 3, 5, 4, 5, 5, 2, 3, 5, 4, 5, 5, 2, 3, 5, 4, 5, 5, 2, 3, 5, 4, 5, 5, 2, 3, 5, 4, 5, 5, 2, 3, 5, 4, 5, 5, 2, 3, 5, 4, 5, 5, 2, 3, 5, 4, 5, 5, 2, 3, 5, 4, 5, 5, 2, 3, 5, 4, 5, 5, 2, 3, 5, 4, 5, 5, 2, 3, 5, 4, 5, 5, 2, 3, 5, 4, 5, 5, 2, 3, 5, 4, 5, 5, 2, 3

Offset: 0

Hieronymus Fischer, Jun 27 2007

The digital sum analog (in base 5) of the Fibonacci recurrence.
When starting from index n=3, periodic with Pisano period A001175(4)=6.
a(n) and Fib(n)=A000045(n) are congruent modulo 4 which implies that (a(n) mod 4) is equal to (Fib(n) mod 4)=A079343(n). Thus (a(n) mod 4) is periodic with the Pisano period A001175(4)=6 too.
For general bases p>2, the inequality 2<=a(n)<=2p-3 holds for n>2. Actually, a(n)<=5=A131319(5) for the base p=5.

			a(10)=3, since a(8)=5=10(base 5), ds_5(5)=1,
a(9)=2, ds_5(2)=2 and so a(10)=1+2.

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for Colombian or self numbers and related sequences

Cf. A000045, A010073, A010074, A010075, A010076, A010077, A131294, A131295, A131297, A131318, A131319, A131320.

nxt[{a_,b_}]:={b,Total[IntegerDigits[a,5]]+Total[IntegerDigits[b,5]]}; NestList[nxt,{0,1},100][[;;,1]] (* Harvey P. Dale, Sep 01 2024 *)

a(n) = a(n-1)+a(n-2)-4*(floor(a(n-1)/5)+floor(a(n-2)/5)).
a(n) = floor(a(n-1)/5)+floor(a(n-2)/5)+(a(n-1)mod 5)+(a(n-2)mod 5).
a(n) = A002266(a(n-1))+A002266(a(n-2))+A010874(a(n-1))+A010874(a(n-2)).
a(n) = Fib(n)-4*sum{1A000045(n).

Incorrect comment removed by Michel Marcus, Apr 29 2018

A132270 a(n) = floor((n^7-1)/(7*n^6)), which is the same as integers repeated 7 times.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10

Offset: 1

Mohammad K. Azarian, Nov 06 2007

Wolfgang Hornfeck, Chiral spiral cyclic twins. II. A two-parameter family of cyclic twins composed of discrete circle involute spirals, Acta Cryst. (2023) Vol. 79, Part 6, 570-586.
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,0,1,-1).

Cf. A004526 ([n/2]), A002264 ([n/3]), A002265 ([n/4]), A002266 ([n/5]), A054895.
Cf. A152467 ([n/6]), A132292 ([(n-1)/8]).
Cf. A002162.

A132270:=n->floor((n-1)/7); seq(A132270(n), n=1..100); # Wesley Ivan Hurt, Dec 10 2013

Table[Floor[(n - 1)/7], {n, 100}] (* Wesley Ivan Hurt, Dec 10 2013 *)
Table[PadRight[{},7,n],{n,0,10}]//Flatten (* or *) LinearRecurrence[ {1,0,0,0,0,0,1,-1},{0,0,0,0,0,0,0,1},100] (* Harvey P. Dale, Jun 08 2017 *)

a(n)=(n-1)\7 \\ Charles R Greathouse IV, Dec 10 2013

a(n) = floor((n^7-n^6)/(7*n^6-6*n^5)). - Mohammad K. Azarian, Nov 08 2007
G.f.: x^8/(1-x-x^7+x^8). - Robert Israel, Feb 02 2015
a(n) = a(n-1)+a(n-7)-a(n-8). - Wesley Ivan Hurt, May 03 2021
a(n) = floor((n-1)/7). - M. F. Hasler, May 19 2021
Sum_{n>=8} (-1)^n/a(n) = log(2) (A002162). - Amiram Eldar, Sep 30 2022

Offset corrected by Mohammad K. Azarian, Nov 19 2008

A076313 a(n) = floor(n/10) - (n mod 10).

Original entry on oeis.org

0, -1, -2, -3, -4, -5, -6, -7, -8, -9, 1, 0, -1, -2, -3, -4, -5, -6, -7, -8, 2, 1, 0, -1, -2, -3, -4, -5, -6, -7, 3, 2, 1, 0, -1, -2, -3, -4, -5, -6, 4, 3, 2, 1, 0, -1, -2, -3, -4, -5, 5, 4, 3, 2, 1, 0, -1, -2, -3, -4, 6, 5, 4, 3, 2, 1, 0, -1, -2, -3, 7, 6, 5, 4, 3, 2, 1, 0, -1, -2, 8, 7, 6, 5, 4

Offset: 0

Reinhard Zumkeller, Oct 06 2002

For n<100 equal to the negated alternating digital sum of n (see A055017). - Hieronymus Fischer, Jun 17 2007

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,0,0,0,0,1,-1).

Cf. A076314, A010879, A076309, A076310, A076311, A076312, A055017, A076314, A007953, A003132.

a076313 = uncurry (-) . flip divMod 10 -- Reinhard Zumkeller, Jun 01 2013

Table[Floor[n/10]-Mod[n,10],{n,0,100}] (* or *) LinearRecurrence[{1,0,0,0,0,0,0,0,0,1,-1},{0,-1,-2,-3,-4,-5,-6,-7,-8,-9,1},100] (* Harvey P. Dale, Nov 02 2022 *)

a(n)=n\10-n%10 \\ Charles R Greathouse IV, Jan 30 2012

From Hieronymus Fischer, Jun 17 2007: (Start)
a(n) = 11*floor(n/10)-n.
a(n) = (n-11*(n mod 10))/10.
a(n) = 11*A002266(A004526(n))-n=11*A004526(A002266(n))-n.
a(n) = (n-11*A010879(n))/10.
a(n) = (n-11*A000035(n)-22*A010874(A004526(n)))/10.
a(n) = (n-11*A010874(n)-55*A000035(A002266(n)))/10.
G.f.: x*(-8*x^10+11*x^9-1)/((1-x^10)*(1-x)^2). (End)

cp's OEIS Frontend

A130519 a(n) = Sum_{k=0..n} floor(k/4). (Partial sums of A002265.)

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A057354 a(n) = floor(2*n/5).

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A010882 Period 3: repeat [1, 2, 3].

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A008959 Final digit of squares: a(n) = n^2 mod 10.

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A130518 a(n) = Sum_{k=0..n} floor(k/3). (Partial sums of A002264.)

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