A175676 -id:A175676 - cp's OEIS frontend

A289844 p-INVERT of A175676 (starting at n=3), where p(S) = 1 - S - S^2.

1, 2, 3, 7, 16, 31, 64, 134, 274, 567, 1168, 2405, 4967, 10232, 21094, 43505, 89672, 184892, 381203, 785886, 1620327, 3340606, 6887304, 14199737, 29275538, 60357622, 124439898, 256558196, 528948160, 1090536002, 2248364880, 4635470266, 9556979689, 19703689739

Offset: 0

Clark Kimberling, Aug 12 2017

Suppose s = (c(0), c(1), c(2), ...) is a sequence and p(S) is a polynomial. Let S(x) = c(0) + c(1)*x + c(2)*x^2 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x). Taking p(S) = 1 - S gives the INVERT transform of s, so that p-INVERT is a generalization of the INVERT transform (e.g., A033453).

See A289780 for a guide to related sequences.

Index entries for linear recurrences with constant coefficients, signature (1, 1, 4, -2, 0, -6, 1, 0, 4, 0, 0, -1)

Cf. A175686, A289780.

z = 60; s = x/((x - 1)^2*(1 + x + x^2)^2); p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A175676, shifted *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289844 *)

Vec((1 + x - 2*x^3 + x^6) / (1 - x - x^2 - 4*x^3 + 2*x^4 + 6*x^6 - x^7 - 4*x^9 + x^12) + O(x^60)) \\ Colin Barker, Aug 13 2017

a(n) = a(n-1) + a(n-2) + 4*a(n-3) - 2*a(n-4) - 6*a(n-7) + a(n-8) + 4*a(n-10) - a(n-13).

G.f.: (1 + x - 2*x^3 + x^6) / (1 - x - x^2 - 4*x^3 + 2*x^4 + 6*x^6 - x^7 - 4*x^9 + x^12). - Colin Barker, Aug 13 2017

A213042 Convolution of (1,0,2,0,3,0,...) and (1,0,0,2,0,0,3,0,0,...); i.e., (A027656(n)) and (A175676(n+2)).

Original entry on oeis.org

1, 0, 2, 2, 3, 4, 7, 6, 11, 12, 15, 18, 24, 24, 33, 36, 42, 48, 58, 60, 74, 80, 90, 100, 115, 120, 140, 150, 165, 180, 201, 210, 237, 252, 273, 294, 322, 336, 371, 392, 420, 448, 484, 504, 548, 576, 612, 648, 693, 720, 774, 810, 855, 900, 955, 990, 1055

Offset: 0

Clark Kimberling, Jun 10 2012

			a(6) = (1,0,2,0,3,0,4)**(1,0,0,2,0,0,3) = 1*3 + 0*0 + 2*0 + 0*2 + 3*0 + 0*0 + 4*1 = 7.

Index entries for linear recurrences with constant coefficients, signature (0,2,2,-1,-4,-1,2,2,0,-1).

s = Normal[Series[1/((1 - x^2)^2 (1 - x^3)^2),
{x, 0, 80}]]
c = CoefficientList[s, t]  (* A213042 *)

a(n) = 2*a(n-2)+2*a(n-3)-a(n-4)-4*a(n-5)-a(n-6)+2*a(n-7)+2*a(n-8)-a(n-10).

G.f.: 1/(((1 - x^2)^2)*(1 - x^3)^2).

A289780 p-INVERT of the positive integers (A000027), where p(S) = 1 - S - S^2.

Original entry on oeis.org

1, 4, 14, 47, 156, 517, 1714, 5684, 18851, 62520, 207349, 687676, 2280686, 7563923, 25085844, 83197513, 275925586, 915110636, 3034975799, 10065534960, 33382471801, 110713382644, 367182309614, 1217764693607, 4038731742156, 13394504020957, 44423039068114

Offset: 0

Clark Kimberling, Aug 10 2017

Suppose s = (c(0), c(1), c(2), ...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x).
Taking p(S) = 1 - S gives the INVERT transform of s, so that p-INVERT is a generalization of the INVERT transform (e.g., A033453).
Guide to p-INVERT sequences using p(S) = 1 - S - S^2:
t(A000012) = t(1,1,1,1,1,1,1,...)    = A001906
t(A000290) = t(1,4,9,16,25,36,...)   = A289779
t(A000027) = t(1,2,3,4,5,6,7,8,...)  = A289780
t(A000045) = t(1,2,3,5,8,13,21,...)  = A289781
t(A000032) = t(2,1,3,4,7,11,14,...)  = A289782
t(A000244) = t(1,3,9,27,81,243,...)  = A289783
t(A000302) = t(1,4,16,64,256,...)    = A289784
t(A000351) = t(1,5,25,125,625,...)   = A289785
t(A005408) = t(1,3,5,7,9,11,13,...)  = A289786
t(A005843) = t(2,4,6,8,10,12,14,...) = A289787
t(A016777) = t(1,4,7,10,13,16,...)   = A289789
t(A016789) = t(2,5,8,11,14,17,...)   = A289790
t(A008585) = t(3,6,9,12,15,18,...)   = A289795
t(A000217) = t(1,3,6,10,15,21,...)   = A289797
t(A000225) = t(1,3,7,15,31,63,...)   = A289798
t(A000578) = t(1,8,27,64,625,...)    = A289799
t(A000984) = t(1,2,6,20,70,252,...)  = A289800
t(A000292) = t(1,4,10,20,35,56,...)  = A289801
t(A002620) = t(1,2,4,6,9,12,16,...)  = A289802
t(A001906) = t(1,3,8,21,55,144,...)  = A289803
t(A001519) = t(1,1,2,5,13,34,...)    = A289804
t(A103889) = t(2,1,4,3,6,5,8,7,,...) = A289805
t(A008619) = t(1,1,2,2,3,3,4,4,...)  = A289806
t(A080513) = t(1,2,2,3,3,4,4,5,...)  = A289807
t(A133622) = t(1,2,1,3,1,4,1,5,...)  = A289809
t(A000108) = t(1,1,2,5,14,42,...)    = A081696
t(A081696) = t(1,1,3,9,29,97,...)    = A289810
t(A027656) = t(1,0,2,0,3,0,4,0,5...) = A289843
t(A175676) = t(1,0,0,2,0,0,3,0,...)  = A289844
t(A079977) = t(1,0,1,0,2,0,3,...)    = A289845
t(A059841) = t(1,0,1,0,1,0,1,...)    = A289846
t(A000040) = t(2,3,5,7,11,13,...)    = A289847
t(A008578) = t(1,2,3,5,7,11,13,...)  = A289828
t(A000142) = t(1!, 2!, 3!, 4!, ...)  = A289924
t(A000201) = t(1,3,4,6,8,9,11,...)   = A289925
t(A001950) = t(2,5,7,10,13,15,...)   = A289926
t(A014217) = t(1,2,4,6,11,17,29,...) = A289927
t(A000045*) = t(0,1,1,2,3,5,...)     = A289975 (* indicates prepended 0's)
t(A000045*) = t(0,0,1,1,2,3,5,...)   = A289976
t(A000045*) = t(0,0,0,1,1,2,3,5,...) = A289977
t(A290990*) = t(0,1,2,3,4,5,...)     = A290990
t(A290990*) = t(0,0,1,2,3,4,5,...)   = A290991
t(A290990*) = t(0,0,01,2,3,4,5,...)  = A290992

			Example 1:  s = (1,2,3,4,5,6,...) = A000027 and p(S) = 1 - S.
S(x) = x + 2x^2 + 3x^3 + 4x^4 + ...
p(S(x)) = 1 - (x + 2x^2 + 3x^3 + 4x^4 + ... )
- p(0) + 1/p(S(x)) = -1 + 1 + x + 3x^2 + 8x^3 + 21x^4 + ...
T(x) = 1 + 3x + 8x^2 + 21x^3 + ...
t(s) = (1,3,8,21,...) = A001906.
***
Example 2:  s = (1,2,3,4,5,6,...) = A000027 and p(S) = 1 - S - S^2.
S(x) =  x + 2x^2 + 3x^3 + 4x^4 + ...
p(S(x)) = 1 - ( x + 2x^2 + 3x^3 + 4x^4 + ...) - ( x + 2x^2 + 3x^3 + 4x^4 + ...)^2
- p(0) + 1/p(S(x)) = -1 + 1 + x + 4x^2 + 14x^3 + 47x^4 + ...
T(x) = 1 + 4x + 14x^2 + 47x^3 + ...
t(s) = (1,4,14,47,...) = A289780.

Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5, -7, 5, -1)

Cf. A000027.

P:=[1,4,14,47];; for n in [5..10^2] do P[n]:=5*P[n-1]-7*P[n-2]+5*P[n-3]-P[n-4]; od; P; # Muniru A Asiru, Sep 03 2017

z = 60; s = x/(1 - x)^2; p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289780 *)

x='x+O('x^99); Vec((1-x+x^2)/(1-5*x+7*x^2-5*x^3+x^4)) \\ Altug Alkan, Aug 13 2017

G.f.: (1 - x + x^2)/(1 - 5 x + 7 x^2 - 5 x^3 + x^4).

a(n) = 5*a(n-1) - 7*a(n-2) + 5*a(n-3) - a(n-4).

A211343 Triangle read by rows: T(n,k), n >= 1, k >= 1, in which column k lists the positive integers interleaved with k-1 zeros, and the first element of column k is in row k(k+1)/2.

Original entry on oeis.org

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

Offset: 1

Omar E. Pol, Feb 05 2013

The number of positive terms in row n is A001227(n).
If n = 2^j then the only positive integer in row n is T(n,1) = n
If n is an odd prime then the only two positive integers in row n are T(n,1) = n and T(n,2) = (n - 1)/2.
From Omar E. Pol, Apr 30 2017: (Start)
Conjecture 1: T(n,k) is the smallest part of the partition of n into k consecutive parts, if T(n,k) > 0.
Conjecture 2: the last positive integer in the row n is in the column A109814(n). (End)

			Triangle begins:
   1;
   2;
   3,  1;
   4,  0;
   5,  2;
   6,  0,  1;
   7,  3,  0;
   8,  0,  0;
   9,  4,  2;
  10,  0,  0,  1;
  11,  5,  0,  0;
  12,  0,  3,  0;
  13,  6,  0,  0;
  14,  0,  0,  2;
  15,  7,  4,  0,  1;
  16,  0,  0,  0,  0;
  17,  8,  0,  0,  0;
  18,  0,  5,  3,  0;
  19,  9,  0,  0,  0;
  20,  0,  0,  0,  2;
  21, 10,  6,  0,  0,  1;
  22,  0,  0,  4,  0,  0;
  23, 11,  0,  0,  0,  0;
  24,  0,  7,  0,  0,  0;
  25, 12,  0,  0,  3,  0;
  26,  0,  0,  5,  0,  0;
  27, 13,  8,  0,  0,  2;
  28,  0,  0,  0,  0,  0,  1;
  ...
In accordance with the conjectures, for n = 15 there are four partitions of 15 into consecutive parts: [15], [8, 7], [6, 5, 4] and [5, 4, 3, 2, 1]. The smallest parts of these partitions are 15, 7, 4, 1, respectively, so the 15th row of the triangle is [15, 7, 4, 0, 1]. - _Omar E. Pol_, Apr 30 2017

Robert Price, Table of n, a(n) for n = 1..28864 (rows n = 1..1000, flattened)

Columns 1-3: A000027, A027656, A175676.
Column k starts in row A000217(k).
Row n has length A003056(n).
Cf. A000203, A001227, A196020, A212119, A235791, A236104, A237048, A237591, A237593, A286001.

a196020[n_, k_]:=If[Divisible[n - k(k + 1)/2, k], 2n/k - k, 0]; T[n_, k_]:= Floor[(1 + a196020[n, k])/2]; Table[T[n, k], {n, 28}, {k, Floor[(Sqrt[8n+1]-1)/2]}] // Flatten (* Indranil Ghosh, Apr 30 2017 *)

from sympy import sqrt
import math
def a196020(n, k):return 2*n/k - k if (n - k*(k + 1)/2)%k == 0 else 0
def T(n, k): return int((1 + a196020(n, k))/2)
for n in range(1, 29): print([T(n, k) for k in range(1, int((sqrt(8*n + 1) - 1)/2) + 1)]) # Indranil Ghosh, Apr 30 2017

T(n,k) = floor((1 + A196020(n,k))/2).

T(n,k) = A237048(n,k)*A286001(n,k). - Omar E. Pol, Aug 13 2018

A049992 a(n) is the number of arithmetic progressions of 3 or more positive integers, nondecreasing with sum n.

Original entry on oeis.org

0, 0, 1, 1, 1, 3, 1, 2, 4, 3, 1, 7, 1, 3, 8, 4, 1, 10, 1, 6, 10, 4, 1, 14, 4, 4, 12, 7, 1, 19, 1, 6, 14, 5, 7, 22, 1, 5, 16, 12, 1, 24, 1, 8, 25, 6, 1, 27, 4, 12, 21, 9, 1, 29, 9, 12, 23, 7, 1, 40, 1, 7, 30, 11, 10, 35, 1, 10, 27, 21, 1, 42, 1, 8, 39, 11, 7, 40, 1, 22, 35, 9, 1, 49, 12, 9, 34

Offset: 1

Clark Kimberling

nonn

Antti Karttunen, Table of n, a(n) for n = 1..12580
Sadek Bouroubi and Nesrine Benyahia Tani, Integer partitions into arithmetic progressions, Rostok. Math. Kolloq. 64 (2009), 11-16.
Sadek Bouroubi and Nesrine Benyahia Tani, Integer partitions into arithmetic progressions with an odd common difference, Integers 9(1) (2009), 77-81.
Jon Maiga, Computer-generated formulas for A049992, Sequence Machine.
Graeme McRae, Counting arithmetic sequences whose sum is n.
Graeme McRae, Counting arithmetic sequences whose sum is n [Cached copy]
Augustine O. Munagi and Temba Shonhiwa, On the partitions of a number into arithmetic progressions, Journal of Integer Sequences 11 (2008), Article 08.5.4.
A. N. Pacheco Pulido, Extensiones lineales de un poset y composiciones de números multipartitos, Maestría thesis, Universidad Nacional de Colombia, 2012.
Wikipedia, Arithmetic progression.

Cf. A007862, A014405, A023645, A049980, A049981, A049982, A049983, A049986, A049987, A049988, A049989, A049990, A049991, A049992, A049994, A175676, A240026, A240027, A307824, A320466.

A049992(n) = (A014405(n)+A023645(n)); \\ (Uses the programs given in A014405 and A023645) - Antti Karttunen, Feb 20 2023

G.f.: Sum_{k>=3} x^k/(1-x^(k*(k-1)/2))/(1-x^k). [Leroy Quet from A049988] - Petros Hadjicostas, Sep 29 2019

a(n) = A014405(n) + A023645(n) = A049994(n) + A175676(n). [Two of the formulas listed by Sequence Machine, both obviously true] - Antti Karttunen, Feb 20 2023

More terms from Petros Hadjicostas, Sep 29 2019

A049994 a(n) is the number of arithmetic progressions of 4 or more positive integers, nondecreasing with sum n.

Original entry on oeis.org

0, 0, 0, 1, 1, 1, 1, 2, 1, 3, 1, 3, 1, 3, 3, 4, 1, 4, 1, 6, 3, 4, 1, 6, 4, 4, 3, 7, 1, 9, 1, 6, 3, 5, 7, 10, 1, 5, 3, 12, 1, 10, 1, 8, 10, 6, 1, 11, 4, 12, 4, 9, 1, 11, 9, 12, 4, 7, 1, 20, 1, 7, 9, 11, 10, 13, 1, 10, 4, 21, 1, 18, 1, 8, 14, 11, 7, 14, 1, 22, 8, 9, 1, 21, 12, 9, 5, 15, 1, 29, 8

Offset: 1

Clark Kimberling

nonn

Antti Karttunen, Table of n, a(n) for n = 1..12580
Sadek Bouroubi and Nesrine Benyahia Tani, Integer partitions into arithmetic progressions, Rostok. Math. Kolloq. 64 (2009), 11-16.
Sadek Bouroubi and Nesrine Benyahia Tani, Integer partitions into arithmetic progressions with an odd common difference, Integers 9(1) (2009), 77-81.
Jon Maiga, Computer-generated formulas for A049994, Sequence Machine.
Graeme McRae, Counting arithmetic sequences whose sum is n.
Graeme McRae, Counting arithmetic sequences whose sum is n [Cached copy]
Augustine O. Munagi, Combinatorics of integer partitions in arithmetic progression, Integers 10(1) (2010), 73-82.
Augustine O. Munagi and Temba Shonhiwa, On the partitions of a number into arithmetic progressions, Journal of Integer Sequences 11 (2008), Article 08.5.4.
A. N. Pacheco Pulido, Extensiones lineales de un poset y composiciones de números multipartitos, Maestría thesis, Universidad Nacional de Colombia, 2012.
Wikipedia, Arithmetic progression.

Cf. A014405, A014406, A175676, A049980, A049981, A049982, A049983, A049986, A049987, A049988, A049989, A049990, A049991, A049992, A049993, A127938, A321014.

A049994(n) = (A049992(n)-if(n%3, 0, n/3)); \\ Antti Karttunen, Feb 20 2023

G.f.: Sum_{k >= 4} x^k/(1-x^(k*(k-1)/2))/(1-x^k). [Leroy Quet from A049988] - Petros Hadjicostas, Sep 29 2019

a(n) = A049992(n) - A175676(n) = A049986(n) + A321014(n). [Two of the formulas listed by Sequence Machine, both obviously true] - Antti Karttunen, Feb 20 2023

More terms from Petros Hadjicostas, Sep 29 2019

A366125 The number of prime factors of the cube root of the largest unitary divisor of n that is a cube (A366126), counted with multiplicity.

Original entry on oeis.org

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

Offset: 1

Amiram Eldar, Sep 30 2023

First differs from A295662 at n = 32, and from A295883, A318673, A359472 and A366124 at n = 64.

One third of the sum of exponents that are divisible by 3 in the prime factorization of n.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A001222, A175676, A350386, A366124, A366126.

Cf. A295662, A295883, A318673, A359472.

f[p_, e_] := If[Divisible[e, 3], e/3, 0]; a[1] = 0; a[n_] := Plus @@ f @@@ FactorInteger[n]; Array[a, 100]

a(n) = vecsum(apply(x -> if(!(x%3), x/3, 0), factor(n)[, 2]));

a(n) = A001222(A366126(n))/3.
Additive with a(p^e) = A175676(e) = e if e is divisible by 3, and 0 otherwise.
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Sum_{p prime} p^2*(p-1)/(p^3-1)^2 = 0.35687351842962928035... .

A088439 a(3n) = 3n, otherwise a(n) = 1.

Original entry on oeis.org

0, 1, 1, 3, 1, 1, 6, 1, 1, 9, 1, 1, 12, 1, 1, 15, 1, 1, 18, 1, 1, 21, 1, 1, 24, 1, 1, 27, 1, 1, 30, 1, 1, 33, 1, 1, 36, 1, 1, 39, 1, 1, 42, 1, 1, 45, 1, 1, 48, 1, 1, 51, 1, 1, 54, 1, 1, 57, 1, 1, 60, 1, 1, 63, 1, 1, 66, 1, 1, 69, 1, 1, 72, 1, 1, 75, 1, 1, 78, 1, 1, 81, 1, 1, 84, 1, 1, 87, 1, 1, 90, 1, 1

Offset: 0

Roger L. Bagula, Nov 09 2003

G. C. Greubel, Table of n, a(n) for n = 0..5000
Index entries for linear recurrences with constant coefficients, signature (0,0,2,0,0,-1).

Cf. A011655, A079978, A088440, A175676, A124625.

[n mod 3 eq 0 select n else 1: n in [0..100]];  // Bruno Berselli, Mar 11 2011

Table[If[Divisible[n,3],n,1],{n,0,100}] (* or *) LinearRecurrence[ {0,0,2,0,0,-1},{0,1,1,3,1,1},100] (* Harvey P. Dale, Jun 18 2018 *)

def A088439(n): return 1 if (n%3) else n
[A088439(n) for n in range(121)] # G. C. Greubel, Dec 05 2022

From Bruno Berselli, Mar 11 2011: (Start)
G.f.: x*(1+x+3*x^2-x^3-x^4)/(1-x^3)^2.
a(n) = n^A079978(n).
a(n) = 3*A175676(n-1) + A011655(n) for n>0. (End)
E.g.f.: (1/3)*(x+2)*exp(x) - (2/3)*exp(-x/2)*( cos(sqrt(3)*x/2) + x*sin((Pi + 3*sqrt(3)*x)/6) ). - G. C. Greubel, Dec 05 2022

A117910 Expansion of (1 + x + x^2 + x^4)/((1-x^3)*(1-x^6)).

Original entry on oeis.org

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

Offset: 0

Paul Barry, Apr 01 2006

Diagonal sums of A117908.

Appears to be a permutation of floor((n+5)/5).

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

Cf. A002264, A014125, A024164, A144677, A117908, A152467, A175676.

R:=PowerSeriesRing(Integers(), 100); Coefficients(R!( (1+x+x^2+x^4)/((1-x^3)*(1-x^6)) )); // G. C. Greubel, Oct 21 2021

CoefficientList[Series[(1+x+x^2+x^4)/((1-x^3)(1-x^6)),{x,0,100}],x] (* or *) LinearRecurrence[{0,0,1,0,0,1,0,0,-1},{1,1,1,1,2,1,2,3,2},100] (* Harvey P. Dale, Apr 10 2014 *)
Table[If[Mod[n,3]==1, Mod[Binomial[n+2,3], n+2], Floor[(n+6)/6]], {n, 0, 100}] (* G. C. Greubel, Nov 18 2021 *)

def A117910_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( (1+x+x^2+x^4)/((1-x^3)*(1-x^6)) ).list()
A117910_list(100) # G. C. Greubel, Oct 21 2021

a(n) = a(n-3) + a(n-6) - a(n-9).
a(n) = Sum_{k=0..floor(n/2)} 0^abs(L(C(n-k,2)/3) - 2*L(C(k,2)/3)) where L(j/p) is the Legendre symbol of j and p.
From G. C. Greubel, Nov 18 2021: (Start)
a(n) = A152467(n+3) + A152467(n+6) if n == 1 (mod 3), otherwise A152467(n+6).
a(n) = A175676(n+2) if n == 1 (mod 3), otherwise A152467(n+6).
a(n) = A002264(n+3) if n == 1 (mod 3), otherwise A152467(n+6). (End)

cp's OEIS Frontend

A289844 p-INVERT of A175676 (starting at n=3), where p(S) = 1 - S - S^2.

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A213042 Convolution of (1,0,2,0,3,0,...) and (1,0,0,2,0,0,3,0,0,...); i.e., (A027656(n)) and (A175676(n+2)).

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A289780 p-INVERT of the positive integers (A000027), where p(S) = 1 - S - S^2.

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A211343 Triangle read by rows: T(n,k), n >= 1, k >= 1, in which column k lists the positive integers interleaved with k-1 zeros, and the first element of column k is in row k(k+1)/2.

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A049992 a(n) is the number of arithmetic progressions of 3 or more positive integers, nondecreasing with sum n.

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A049994 a(n) is the number of arithmetic progressions of 4 or more positive integers, nondecreasing with sum n.

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A366125 The number of prime factors of the cube root of the largest unitary divisor of n that is a cube (A366126), counted with multiplicity.

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A088439 a(3n) = 3n, otherwise a(n) = 1.

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