A010877 -id:A010877 - cp's OEIS frontend

A130486 a(n) = Sum_{k=0..n} (k mod 8) (Partial sums of A010877).

0, 1, 3, 6, 10, 15, 21, 28, 28, 29, 31, 34, 38, 43, 49, 56, 56, 57, 59, 62, 66, 71, 77, 84, 84, 85, 87, 90, 94, 99, 105, 112, 112, 113, 115, 118, 122, 127, 133, 140, 140, 141, 143, 146, 150, 155, 161, 168, 168, 169, 171, 174, 178, 183, 189, 196, 196, 197, 199, 202, 206

Offset: 0

Hieronymus Fischer, May 31 2007

Let A be the Hessenberg n X n matrix defined by A[1,j] = j mod 8, A[i,i]:=1, A[i,i-1]=-1. Then, for n >= 1, a(n)=det(A). - Milan Janjic, Jan 24 2010

Shawn A. Broyles, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,0,0,1,-1).

Cf. A010872, A010873, A010874, A010875, A010876, A010878. A130481, A130482, A130483, A130484, A130485, A130487.

a:=[0,1,3,6,10,15,21,28,28];; for n in [10..71] do a[n]:=a[n-1]+a[n-8]-a[n-9]; od; a; # G. C. Greubel, Aug 31 2019

I:=[0,1,3,6,10,15,21,28,28]; [n le 9 select I[n] else Self(n-1) + Self(n-8) - Self(n-9): n in [1..71]]; // G. C. Greubel, Aug 31 2019

seq(coeff(series(x*(1-8*x^7+7*x^8)/((1-x^8)*(1-x)^3), x, n+1), x, n), n = 0 .. 40); # G. C. Greubel, Aug 31 2019

Array[28 Floor[#1/8] + #2 (#2 + 1)/2 & @@ {#, Mod[#, 8]} &, 61, 0] (* Michael De Vlieger, Apr 28 2018 *)
Accumulate[PadRight[{},100,Range[0,7]]] (* Harvey P. Dale, Dec 21 2018 *)

a(n) = sum(k=0, n, k % 8); \\ Michel Marcus, Apr 28 2018

def A130486_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P(x*(1-8*x^7+7*x^8)/((1-x^8)*(1-x)^3)).list()
A130486_list(70) # G. C. Greubel, Aug 31 2019

a(n) = 28*floor(n/8) + A010877(n)*(A010877(n) + 1)/2.
G.f.: (Sum_{k=1..7} k*x^k)/((1-x^8)*(1-x)).
G.f.: x*(1 - 8*x^7 + 7*x^8)/((1-x^8)*(1-x)^3).

A010888 Digital root of n (repeatedly add the digits of n until a single digit is reached).

Original entry on oeis.org

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

Offset: 0

N. J. A. Sloane

This is sometimes also called the additive digital root of n.
n mod 9 (A010878) is a very similar sequence.
Partial sums are given by A130487(n-1) + n (for n > 0). - Hieronymus Fischer, Jun 08 2007
Decimal expansion of 13717421/111111111 is 0.123456789123456789123456789... with period 9. - Eric Desbiaux, May 19 2008
Decimal expansion of 13717421 / 1111111110 = 0.0[123456789] (periodic) - Daniel Forgues, Feb 27 2017
a(A005117(n)) < 9. - Reinhard Zumkeller, Mar 30 2010
My friend Jahangeer Kholdi has found that 19 is the smallest prime p such that for each number n, a(p*n) = a(n). In fact we have: a(m*n) = a(a(m)*a(n)) so all numbers with digital root 1 (numbers of the form 9k + 1) have this property. See comment lines of A017173. Also we have a(m+n) = a(a(m) + a(n)). - Farideh Firoozbakht, Jul 23 2010

			The digits of 37 are 3 and 7, and 3 + 7 = 10. And the digits of 10 are 1 and 0, and 1 + 0 = 1, so a(37) = 1.

Martin Gardner, Mathematics, Magic and Mystery, 1956.

N. J. A. Sloane, Table of n, a(n) for n = 0..10000
Eric Weisstein's World of Mathematics, Digitaddition
Eric Weisstein's World of Mathematics, Digital Root
Wikipedia, Vedic square
Index entries for Colombian or self numbers and related sequences
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,0,1).

Cf. A007953, A007954, A031347, A113217, A113218, A010878 (n mod 9), A010872, A010873, A010874, A010875, A010876, A010877, A010879, A004526, A002264, A002265, A002266, A017173, A031286 (additive persistence of n), (multiplicative digital root of n), A031346 (multiplicative persistence of n).

a010888 = until (< 10) a007953
-- Reinhard Zumkeller, Oct 17 2011, May 12 2011

[n eq 0 select 0 else 1+(n-1) mod 9: n in [0..110]]; // Bruno Berselli, Mar 18 2016

A010888 := n->if n=0 then 0 else ((n-1) mod 9) + 1; fi; # N. J. A. Sloane, Feb 20 2013

Join[{0}, Array[Mod[ # - 1, 9] + 1 &, 104]] (* Robert G. Wilson v, Jan 04 2006 *)
Join[Range[0, 1], Table[n - 9 Floor[(n - 1) / 9], {n, 2, 100}]] (* José de Jesús Camacho Medina, Nov 10 2014 *) (* Corrected by Vincenzo Librandi, Nov 11 2014 *)
Join[{0},LinearRecurrence[{0, 0, 0, 0, 0, 0, 0, 0, 1},{1, 2, 3, 4, 5, 6, 7, 8, 9},104]] (* Ray Chandler, Aug 26 2015 *)
Table[FixedPoint[Total[IntegerDigits[#, 10]] &, n], {n, 0, 104}] (* IWABUCHI Yu(u)ki, Jun 03 2016 *)

A010888(n)=if(n,(n-1)%9+1) \\ M. F. Hasler, Jan 04 2011

def A010888(n):
    return 1 + (n - 1) % 9 if n else 0 # Chai Wah Wu, Aug 23 2014, Apr 23 2023

0 :: List.fill(10)(1 to 9).flatten // Alonso del Arte, Feb 01 2020

If n = 0 then a(n) = 0; otherwise a(n) = (n reduced mod 9), but if the answer is 0 change it to 9.
Equivalently, if n = 0 then a(n) = 0, otherwise a(n) = (n - 1 reduced mod 9) + 1.
If the initial 0 term is ignored, the sequence is periodic with period 9.
From Hieronymus Fischer, Jun 08 2007: (Start)
a(n) = A010878(n-1) + 1 (for n > 0).
G.f.: g(x) = x*(Sum_{k = 0..8}(k+1)*x^k)/(1 - x^9). Also: g(x) = x(9x^10 - 10x^9 + 1)/((1 - x^9)(1 - x)^2). (End)
a(n) = n - 9*floor((n-1)/9), for n > 0. - José de Jesús Camacho Medina, Nov 10 2014

A010873 a(n) = n mod 4.

Original entry on oeis.org

0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0

Offset: 0

N. J. A. Sloane

Complement of A002265, since 4*A002265(n)+a(n) = n. - Hieronymus Fischer, Jun 01 2007
The rightmost digit in the base-4 representation of n. Also, the equivalent value of the two rightmost digits in the base-2 representation of n. - Hieronymus Fischer, Jun 11 2007
Periodic sequences of this type can be also calculated by a(n) = floor(q/(p^m-1)*p^n) mod p, where q is the number representing the periodic digit pattern and m is the period length. p and q can be calculated as follows: Let D be the array representing the number pattern to be repeated, m = size of D, max = maximum value of elements in D. Than p := max + 1 and q := p^m*sum_{i=1..m} D(i)/p^i. Example: D = (0, 1, 2, 3), p = 4 and q = 57 for this sequence. - Hieronymus Fischer, Jan 04 2013

Antti Karttunen, Table of n, a(n) for n = 0..65536
Index entries for linear recurrences with constant coefficients, signature (0,0,0,1).

Partial sums: A130482. Other related sequences A130481, A130483, A130484, A130485.
Cf. A004526, A002264, A002265, A002266.
Cf. A000035, A010877.

a010873 n = (`mod` 4)
a010873_list = cycle [0..3]  -- Reinhard Zumkeller, Jun 05 2012

seq(chrem( [n,n], [1,4] ), n=0..80); # Zerinvary Lajos, Mar 25 2009

nn=40; CoefficientList[Series[(x+2x^2+3x^3)/(1-x^4), {x,0,nn}], x] (* Geoffrey Critzer, Jul 26 2013 *)
Table[Mod[n,4], {n, 0, 100}] (* T. D. Noe, Jul 26 2013 *)
PadRight[{},120,{0,1,2,3}] (* Harvey P. Dale, Mar 29 2018 *)

a(n)=n%4 \\ Charles R Greathouse IV, Dec 05 2011

(define (A010873 n) (modulo n 4)) ;; Antti Karttunen, Nov 07 2017

a(n) = (1/2)*(3-(-1)^n-2*(-1)^floor(n/2));
also a(n) = (1/2)*(3-(-1)^n-2*(-1)^((2*n-1+(-1)^n)/4));
also a(n) = (1/2)*(3-(-1)^n-2*sin(Pi/4*(2n+1+(-1)^n))).
G.f.: (3x^3+2x^2+x)/(1-x^4). - Hieronymus Fischer, May 29 2007
From Hieronymus Fischer, Jun 11 2007: (Start)
Trigonometric representation: a(n)=2^2*(sin(n*Pi/4))^2*sum{1<=k<4, k*product{1<=m<4,m<>k, (sin((n-m)*Pi/4))^2}}. Clearly, the squared terms may be replaced by their absolute values '|.|'.
Complex representation: a(n)=1/4*(1-r^n)*sum{1<=k<4, k*product{1<=m<4,m<>k, (1-r^(n-m))}} where r=exp(Pi/2*i)=i=sqrt(-1). All these formulas can be easily adapted to represent any periodic sequence.
a(n) = n mod 2+2*(floor(n/2)mod 2) = A000035(n)+2*A000035(A004526(n)). (End)
a(n) = 6 - a(n-1) - a(n-2) - a(n-3) for n > 2. - Reinhard Zumkeller, Apr 13 2008
a(n) = 3/2 + cos((n+1)*Pi)/2 + sqrt(2)*cos((2*n+3)*Pi/4). - Jaume Oliver Lafont, Dec 05 2008
From Hieronymus Fischer, Jan 04 2013: (Start)
a(n) = floor(41/3333*10^(n+1)) mod 10.
a(n) = floor(19/85*4^(n+1)) mod 4. (End)
E.g.f.: 2*sinh(x) - sin(x) + cosh(x) - cos(x). - Stefano Spezia, Apr 20 2021
From Nicolas Bělohoubek, May 30 2024: (Start)
a(n) = (2*a(n-1)-1)*(2-a(n-2)) for n > 1.
a(n) = (2*a(n-1)^2+1)*(3-a(n-1))/3 for n > 0. (End)

First to third formulas re-edited for better readability by Hieronymus Fischer, Dec 05 2011

Incorrect g.f. removed by Georg Fischer, May 18 2019

A046034 Numbers whose digits are primes.

Original entry on oeis.org

2, 3, 5, 7, 22, 23, 25, 27, 32, 33, 35, 37, 52, 53, 55, 57, 72, 73, 75, 77, 222, 223, 225, 227, 232, 233, 235, 237, 252, 253, 255, 257, 272, 273, 275, 277, 322, 323, 325, 327, 332, 333, 335, 337, 352, 353, 355, 357, 372, 373, 375, 377, 522, 523, 525, 527, 532

Offset: 1

Eric W. Weisstein

If n is represented as a zerofree base-4 number (see A084544) according to n=d(m)d(m-1)...d(3)d(2)d(1)d(0) then a(n) = Sum_{j=0..m} c(d(j))*10^j, where c(k)=2,3,5,7 for k=1..4. - Hieronymus Fischer, May 30 2012

According to A153025, it seems that 5, 235 and 72335 are the only terms whose square is also a term, i.e., which are also in the sequence A275971 of square roots of the terms which are squares, listed in A191486. - M. F. Hasler, Sep 16 2016

			a(100)   = 2277,
a(10^3)  = 55327,
a(9881)  = 3233232,
a(10^4)  = 3235757,
a(10922) = 3333333,
a(10^5)  = 227233257.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Robert Baillie and Thomas Schmelzer, Summing Kempner's Curious (Slowly-Convergent) Series, Mathematica Notebook kempnerSums.nb, Wolfram Library Archive, 2008.
Eric Weisstein's World of Mathematics, Smarandache Sequences.
Index entries for 10-automatic sequences.

Cf. A046035, A118950, A019546 (primes), A203263, A035232, A039996, A085823, A052382, A084544, A084984, A017042, A001743, A001744, A014261, A014263, A153025, A191486, A193238, A202267, A202268, A211681, A365471 (complement).

a046034 n = a046034_list !! (n-1)
a046034_list = filter (all (`elem` "2357") . show ) [0..]
-- Reinhard Zumkeller, Jul 19 2011

[n: n in [2..532] | Set(Intseq(n)) subset [2, 3, 5, 7]];  // Bruno Berselli, Jul 19 2011

Table[FromDigits /@ Tuples[{2, 3, 5, 7}, n], {n, 3}] // Flatten (* Michael De Vlieger, Sep 19 2016 *)

is_A046034(n)=Set(isprime(digits(n)))==[1] \\ M. F. Hasler, Oct 12 2013

def A046034(n):
    m = (3*n+1).bit_length()-1>>1
    return int(''.join(('2357'[(3*n+1-(1<<(m<<1)))//(3<<((m-1-j)<<1))&3] for j in range(m)))) # Chai Wah Wu, Feb 08 2023

A055642(a(n)) = A193238(a(n)). - Reinhard Zumkeller, Jul 19 2011
From Hieronymus Fischer, Apr 20, May 30 and Jun 25 2012: (Start)
a(n) = Sum_{j=0..m-1} ((2*b(j)+1) mod 8 + floor(b(j)/4) - floor((b(j)-1)/4))*10^j, where m = floor(log_4(3*n+1)), b(j) = floor((3*n+1-4^m)/(3*4^j)).
a(n) = Sum_{j=0..m-1} A010877(A005408(b(j)) + A002265(b(j)) - A002265(b(j)-1))*10^j.
Special values:
a(1*(4^n-1)/3) = 2*(10^n-1)/9.
a(2*(4^n-1)/3) = 1*(10^n-1)/3.
a(3*(4^n-1)/3) = 5*(10^n-1)/9.
a(4*(4^n-1)/3) = 7*(10^n-1)/9.
Inequalities:
a(n) <= 2*(10^log_4(3*n+1)-1)/9, equality holds for n = (4^k-1)/3, k>0.
a(n) <= 2*A084544(n), equality holds iff all digits of A084544(n) are 1.
a(n) > A084544(n).
Lower and upper limits:
lim inf a(n)/10^log_4(n) = (7/90)*10^log_4(3) = 0.48232167706987..., for n -> oo.
lim sup a(n)/10^log_4(n) = (2/9)*10^log_4(3) = 1.378061934485343..., for n -> oo.
where 10^log_4(n) = n^1.66096404744...
G.f.: g(x) = (x^(1/3)*(1-x))^(-1) Sum_{j>=0} 10^j*z(j)^(4/3)*(2 + z(j) + 2*z(j)^2 + 2*z(j)^3 - 7*z(j)^4)/(1-z(j)^4), where z(j) = x^4^j.
Also g(x) = (x^(1/3)*(1-x))^(-1) Sum_{j>=0} 10^j*z(j)^(4/3)*(1-z(j))*(2 + 3*z(j) + 5*z(j)^2 + 7*z(j)^3)/(1-z(j)^4), where z(j)=x^4^j.
Also: g(x) = (1/(1-x))*(2*h_(4,0)(x) + h_(4,1)(x) + 2*h_(4,2)(x) + 2*h_(4,3)(x) - 7*h_(4,4)(x)), where h_(4,k)(x) = Sum_{j>=0} 10^j*x^((4^(j+1)-1)/3)*x^(k*4^j)/(1-x^4^(j+1)). (End)
Sum_{n>=1} 1/a(n) = 1.857333779940977502574887651449435985318556794733869779170825138954093657197... (calculated using Baillie and Schmelzer's kempnerSums.nb, see Links). - Amiram Eldar, Feb 15 2024

More terms from Cino Hilliard, Aug 06 2006
Typo in second formula corrected by Hieronymus Fischer, May 12 2012
Two typos in example section corrected by Hieronymus Fischer, May 30 2012

A276573 The infinite trunk of least squares beanstalk: The only infinite sequence such that a(0) = 0 and a(n-1) = a(n) - least number of squares (A002828) that sum to a(n).

Original entry on oeis.org

0, 3, 6, 8, 11, 15, 16, 18, 21, 24, 27, 30, 32, 35, 38, 40, 43, 45, 48, 51, 53, 56, 59, 63, 64, 67, 70, 72, 75, 78, 80, 83, 85, 88, 90, 93, 96, 99, 102, 105, 108, 112, 115, 117, 120, 123, 126, 128, 131, 134, 136, 139, 143, 144, 147, 149, 152, 155, 158, 160, 162, 165, 168, 171, 173, 176, 179, 183, 186, 189, 192, 195

Offset: 0

Antti Karttunen, Sep 07 2016

nonn

Antti Karttunen, Table of n, a(n) for n = 0..10028

Cf. A002828, A005563, A255131, A260731, A260733, A262689, A276572, A276574, A276575 (first differences), A277016 (squares present), A277015 (their square roots), A277888 (primes), A278486 (numbers one more than a prime), A278265, A278487, A278488, A278491 (another subsequence), A278497, A278498, A278499, A278513, A278516, A278517, A278518, A278519, A278521, A278522.
Cf. A277890 & A277891 (number of even and odd terms in each range. The latter seem to be slightly more numerous), A277889.
Positions of nonzero terms in A278515.
Subsequence of A278489, no common terms with A278490.
Cf. also A179016, A259934, A276583, A276613, A276623 for similar constructions.

(define (A276573 n) (A276574 (A276572 n)))

a(n) = A276574(A276572(n)).
Other identities and observations. For all n >= 0:
A260731(a(n)) = n.
a(A260733(n+1)) = A005563(n).
A278517(n) <= a(n) <= A278519(n).
A010873(a(n)) = A278499(n). [Terms reduced modulo 4.]
A010877(a(n)) = A278488(n). [modulo 8.]
A046523(a(n)) = A278497(n). [Least number with the same prime signature.]
A008683(a(n)) = A278513(n).
A065338(a(n)) = A278498(n).
A278509(a(n)) = A278265(n).
A278216(a(n)) = A278516(n). [Number of children the n-th node of the trunk has.]

Definition clarified and more identities added to the Formula section by Antti Karttunen, Nov 28 2016

A130481 a(n) = Sum_{k=0..n} (k mod 3) (i.e., partial sums of A010872).

Original entry on oeis.org

0, 1, 3, 3, 4, 6, 6, 7, 9, 9, 10, 12, 12, 13, 15, 15, 16, 18, 18, 19, 21, 21, 22, 24, 24, 25, 27, 27, 28, 30, 30, 31, 33, 33, 34, 36, 36, 37, 39, 39, 40, 42, 42, 43, 45, 45, 46, 48, 48, 49, 51, 51, 52, 54, 54, 55, 57, 57, 58, 60, 60, 61, 63, 63, 64, 66, 66, 67, 69, 69, 70, 72, 72

Offset: 0

Hieronymus Fischer, May 29 2007

Essentially the same as A092200. - R. J. Mathar, Jun 13 2008
Let A be the Hessenberg n X n matrix defined by: A[1,j]=j mod 3, A[i,i]:=1, A[i,i-1]=-1. Then, for n>=1, a(n)=det(A). - Milan Janjic, Jan 24 2010
2-adic valuation of A104537(n+1). - Gerry Martens, Jul 14 2015
Conjecture: a(n) is the exponent of the largest power of 2 that divides all the entries of the matrix {{3,1},{1,-1}}^n. - Greg Dresden, Sep 09 2018

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,1,-1).

Cf. A010872, A010873, A010874, A010875, A010876, A010877, A130482, A130483, A130484.

Cf. A092200, A104537.

List([0..80], n-> Int((n+1)/3) + Int(2*(n+1)/3)); # G. C. Greubel, Aug 31 2019

[Floor((n+1)/3) + Floor(2*(n+1)/3): n in [0..80]]; // G. C. Greubel, Aug 31 2019

seq(coeff(series(x*(1+2*x)/((1-x^3)*(1-x)), x, n+1), x, n), n = 0..80); # G. C. Greubel, Aug 31 2019

a[n_]:= Floor[(n+1)/3] + Floor[2(n+1)/3]; Table[a[n], {n, 0, 80}] (* Clark Kimberling, May 28 2012 *)
a[n_]:= IntegerExponent[A104537[n + 1], 2];
Table[a[n], {n, 0, 80}]  (* Gerry Martens, Jul 14 2015 *)
CoefficientList[Series[x(1+2x)/((1-x^3)(1-x)), {x, 0, 80}], x] (* Stefano Spezia, Sep 09 2018 *)
LinearRecurrence[{1,0,1,-1},{0,1,3,3},100] (* Harvey P. Dale, Jun 14 2021 *)

main(size)=my(n,k);vector(size,n,sum(k=0,n,k%3)) \\ Anders Hellström, Jul 14 2015

first(n)=my(s); concat(0, vector(n,k,s+=k%3)) \\ Charles R Greathouse IV, Jul 14 2015

a(n)=n\3*3+[0,1,3][n%3+1] \\ Charles R Greathouse IV, Jul 14 2015

def A130481_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P(x*(1+2*x)/((1-x^3)*(1-x))).list()
A130481_list(80) # G. C. Greubel, Aug 31 2019

a(n) = 3*floor(n/3) + A010872(n)*(A010872(n) + 1)/2.
G.f.: x*(1 + 2*x)/((1-x^3)*(1-x)).
a(n) = n + 1 - (Fibonacci(n+1) mod 2). - Gary Detlefs, Mar 13 2011
a(n) = floor((n+1)/3) + floor(2*(n+1)/3). - Clark Kimberling, May 28 2010
a(n) = n when n+1 is not a multiple of 3, and a(n) = n+1 when n+1 is a multiple of 3. - Dennis P. Walsh, Aug 06 2012
a(n) = n + 1 - sign((n+1) mod 3). - Wesley Ivan Hurt, Sep 25 2017
a(n) = n + (1-cos(2*(n+2)*Pi/3))/3 + sin(2*(n+2)*Pi/3)/sqrt(3). - Wesley Ivan Hurt, Sep 27 2017
a(n) = n + 1 - (n+1)^2 mod 3. - Ammar Khatab, Aug 14 2020
E.g.f.: ((1 + 3*x)*cosh(x) - (cos(sqrt(3)*x/2) + sqrt(3)*sin(sqrt(3)*x/2))*(cosh(x/2) - sinh(x/2)) + (1 + 3*x)*sinh(x))/3. - Stefano Spezia, May 28 2021
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/(3*sqrt(3)) + log(2)/3. - Amiram Eldar, Sep 17 2022

A130482 a(n) = Sum_{k=0..n} (k mod 4) (Partial sums of A010873).

Original entry on oeis.org

0, 1, 3, 6, 6, 7, 9, 12, 12, 13, 15, 18, 18, 19, 21, 24, 24, 25, 27, 30, 30, 31, 33, 36, 36, 37, 39, 42, 42, 43, 45, 48, 48, 49, 51, 54, 54, 55, 57, 60, 60, 61, 63, 66, 66, 67, 69, 72, 72, 73, 75, 78, 78, 79, 81, 84, 84, 85, 87, 90, 90, 91, 93, 96, 96, 97, 99, 102, 102, 103, 105

Offset: 0

Hieronymus Fischer, May 29 2007

Let A be the Hessenberg n X n matrix defined by: A[1,j]=j mod 4, A[i,i]:=1, A[i,i-1]=-1. Then, for n>=1, a(n)=det(A). - Milan Janjic, Jan 24 2010

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,1,-1).

Cf. A010872, A010874, A010875, A010876, A010877. A130481, A130483, A130484, A130485.

a:=[0,1,3,6,6];; for n in [6..71] do a[n]:=a[n-1]+a[n-4]-a[n-5]; od; a; # G. C. Greubel, Aug 31 2019

I:=[0,1,3,6,6]; [n le 5 select I[n] else Self(n-1) + Self(n-4) - Self(n-5): n in [1..71]]; // G. C. Greubel, Aug 31 2019

a:=n->add(chrem( [n,j], [1,4] ),j=1..n):seq(a(n), n=0..70); # Zerinvary Lajos, Apr 07 2009

Table[(6*n +(1-(-1)^n)*(1+2*I^(n+1)))/4, {n,0,70}] (* G. C. Greubel, Aug 31 2019 *)
LinearRecurrence[{1,0,0,1,-1},{0,1,3,6,6},80] (* Harvey P. Dale, Feb 16 2024 *)

a(n) = (1 - (-1)^n - (2*I)*(-I)^n + (2*I)*I^n + 6*n) / 4 \\ Colin Barker, Oct 15 2015

def A130482_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P(x*(1+2*x+3*x^2)/((1-x^4)*(1-x))).list()
A130482_list(70) # G. C. Greubel, Aug 31 2019

a(n) = 6*floor(n/4) + A010873(n)*(A010873(n)+1)/2.
G.f.: x*(1 + 2*x + 3*x^2)/((1-x^4)*(1-x)).
a(n) = (1 - (-1)^n - (2*i)*(-i)^n + (2*i)*i^n + 6*n) / 4 where i = sqrt(-1). - Colin Barker, Oct 15 2015
a(n) = 3*n/2 + (n mod 2)* ( (n-1) mod 4 ) - (n mod 2)/2. - Ammar Khatab, Aug 27 2020
E.g.f.: (3*x*exp(x) - 2*sin(x) + sinh(x))/2. - Stefano Spezia, Apr 22 2021
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/(4*sqrt(3)) + log(3)/4. - Amiram Eldar, Sep 17 2022

A130483 a(n) = Sum_{k=0..n} (k mod 5) (Partial sums of A010874).

Original entry on oeis.org

0, 1, 3, 6, 10, 10, 11, 13, 16, 20, 20, 21, 23, 26, 30, 30, 31, 33, 36, 40, 40, 41, 43, 46, 50, 50, 51, 53, 56, 60, 60, 61, 63, 66, 70, 70, 71, 73, 76, 80, 80, 81, 83, 86, 90, 90, 91, 93, 96, 100, 100, 101, 103, 106, 110, 110, 111, 113, 116, 120, 120, 121, 123, 126, 130, 130

Offset: 0

Hieronymus Fischer, May 29 2007

Let A be the Hessenberg n X n matrix defined by: A[1,j]=j mod 5, A[i,i]=1, A[i,i-1]=-1. Then, for n>=1, a(n)=det(A). - Milan Janjic, Jan 24 2010

Shawn A. Broyles, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,1,-1).

Cf. A010872, A010873, A010875, A010876, A010877, A130481, A130482, A130484, A130485.

a:=[0,1,3,6,10,10];; for n in [7..71] do a[n]:=a[n-1]+a[n-5]-a[n-6]; od; a; # G. C. Greubel, Aug 31 2019

I:=[0,1,3,6,10,10]; [n le 6 select I[n] else Self(n-1) + Self(n-5) - Self(n-6): n in [1..71]]; // G. C. Greubel, Aug 31 2019

seq(coeff(series(x*(1+2*x+3*x^2+4*x^3)/((1-x^5)*(1-x)), x, n+1), x, n), n = 0..70); # G. C. Greubel, Aug 31 2019

Accumulate[Mod[Range[0,70],5]] (* or *) Accumulate[PadRight[{},70,{0,1,2,3,4}]] (* Harvey P. Dale, Nov 11 2016 *)

a(n) = sum(k=0, n, k % 5); \\ Michel Marcus, Apr 28 2018

def A130483_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P(x*(1+2*x+3*x^2+4*x^3)/((1-x^5)*(1-x))).list()
A130483_list(70) # G. C. Greubel, Aug 31 2019

a(n) = 10*floor(n/5) + A010874(n)*(A010874(n)+1)/2.
G.f.: x*(1 + 2*x + 3*x^2 + 4*x^3)/((1-x^5)*(1-x)).
From Wesley Ivan Hurt, Jul 23 2016: (Start)
a(n) = a(n-5) - a(n-6) for n>5; a(n) = a(n-5) + 10 for n>4.
a(n) = 10 + Sum_{k=1..4} k*floor((n-k)/5). (End)
a(n) = ((n mod 5)^2 - 3*(n mod 5) + 4*n)/2. - Ammar Khatab, Aug 13 2020

A130484 a(n) = Sum_{k=0..n} (k mod 6) (Partial sums of A010875).

Original entry on oeis.org

0, 1, 3, 6, 10, 15, 15, 16, 18, 21, 25, 30, 30, 31, 33, 36, 40, 45, 45, 46, 48, 51, 55, 60, 60, 61, 63, 66, 70, 75, 75, 76, 78, 81, 85, 90, 90, 91, 93, 96, 100, 105, 105, 106, 108, 111, 115, 120, 120, 121, 123, 126, 130, 135, 135, 136, 138, 141, 145, 150, 150, 151, 153

Offset: 0

Hieronymus Fischer, May 31 2007

Let A be the Hessenberg n X n matrix defined by A[1,j] = j mod 6, A[i,i]=1, A[i,i-1]=-1. Then, for n >= 1, a(n)=det(A). - Milan Janjic, Jan 24 2010

Shawn A. Broyles, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,1,-1).

Cf. A010872, A010873, A010874, A010876, A010877, A130481, A130482, A130483, A130485.

a:=[0,1,3,6,10,15,15];; for n in [8..71] do a[n]:=a[n-1]+a[n-6]-a[n-7]; od; a; # G. C. Greubel, Aug 31 2019

I:=[0,1,3,6,10,15,15]; [n le 7 select I[n] else Self(n-1) + Self(n-6) - Self(n-7): n in [1..71]]; // G. C. Greubel, Aug 31 2019

seq(coeff(series(x*(1-6*x^5+5*x^6)/((1-x^6)*(1-x)^3), x, n+1), x, n), n = 0 .. 70); # G. C. Greubel, Aug 31 2019

Accumulate[Mod[Range[0,70],6]] (* or *) Accumulate[PadRight[ {},70, Range[0,5]]] (* Harvey P. Dale, Jul 12 2016 *)

a(n) = sum(k=0, n, k % 6); \\ Michel Marcus, Apr 28 2018

a(n)=n\6*15 + binomial(n%6+1,2) \\ Charles R Greathouse IV, Jan 24 2022

def A130484_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P(x*(1-6*x^5+5*x^6)/((1-x^6)*(1-x)^3)).list()
A130484_list(70) # G. C. Greubel, Aug 31 2019

a(n) = 15*floor(n/6) + A010875(n)*(A010875(n) + 1)/2.

G.f.: (Sum_{k=1..5} k*x^k)/((1-x^6)*(1-x)) = x*(1 - 6*x^5 + 5*x^6)/((1-x^6)*(1-x)^3).

A130485 a(n) = Sum_{k=0..n} (k mod 7) (Partial sums of A010876).

Original entry on oeis.org

0, 1, 3, 6, 10, 15, 21, 21, 22, 24, 27, 31, 36, 42, 42, 43, 45, 48, 52, 57, 63, 63, 64, 66, 69, 73, 78, 84, 84, 85, 87, 90, 94, 99, 105, 105, 106, 108, 111, 115, 120, 126, 126, 127, 129, 132, 136, 141, 147, 147, 148, 150, 153, 157, 162, 168, 168, 169, 171, 174, 178, 183

Offset: 0

Hieronymus Fischer, May 31 2007

Let A be the Hessenberg n X n matrix defined by A[1,j] = j mod 7, A[i,i]:=1, A[i,i-1]=-1. Then, for n >= 1, a(n)=det(A). - Milan Janjic, Jan 24 2010

Shawn A. Broyles, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,0,1,-1).

Cf. A010872, A010873, A010874, A010875, A010877, A130481, A130482, A130483, A130484.

a:=[0,1,3,6,10,15,21,21];; for n in [9..71] do a[n]:=a[n-1]+a[n-7]-a[n-8]; od; a; # G. C. Greubel, Aug 31 2019

I:=[0,1,3,6,10,15,21,21]; [n le 8 select I[n] else Self(n-1) + Self(n-7) - Self(n-8): n in [1..71]]; // G. C. Greubel, Aug 31 2019

a:=n->add(chrem( [n,j], [1,7] ),j=1..n):seq(a(n), n=1..70); # Zerinvary Lajos, Apr 07 2009

LinearRecurrence[{1,0,0,0,0,0,1,-1},{0,1,3,6,10,15,21,21},70] (* Harvey P. Dale, Jul 30 2017 *)

concat(0,Vec((1-7*x^6+6*x^7)/(1-x^7)/(1-x)^3+O(x^70))) \\ Charles R Greathouse IV, Dec 22 2011

def A130485_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P(x*(1-7*x^6+6*x^7)/((1-x^7)*(1-x)^3)).list()
A130485_list(70) # G. C. Greubel, Aug 31 2019

a(n) = 21*floor(n/7) + A010876(n)*(A010876(n) + 1)/2.
G.f.: (Sum_{k=1..6} k*x^k)/((1-x^7)*(1-x)).
G.f.: x*(1 - 7*x^6 + 6*x^7)/((1-x^7)*(1-x)^3).

cp's OEIS Frontend

A130486 a(n) = Sum_{k=0..n} (k mod 8) (Partial sums of A010877).

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A010888 Digital root of n (repeatedly add the digits of n until a single digit is reached).

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A010873 a(n) = n mod 4.

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A046034 Numbers whose digits are primes.

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A276573 The infinite trunk of least squares beanstalk: The only infinite sequence such that a(0) = 0 and a(n-1) = a(n) - least number of squares (A002828) that sum to a(n).

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A130481 a(n) = Sum_{k=0..n} (k mod 3) (i.e., partial sums of A010872).

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