A016218 -id:A016218 - cp's OEIS frontend

A014827 a(1)=1, a(n) = 5*a(n-1) + n.

1, 7, 38, 194, 975, 4881, 24412, 122068, 610349, 3051755, 15258786, 76293942, 381469723, 1907348629, 9536743160, 47683715816, 238418579097, 1192092895503, 5960464477534, 29802322387690, 149011611938471, 745058059692377, 3725290298461908, 18626451492309564, 93132257461547845

Offset: 1

N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Dillan Agrawal, Selena Ge, Jate Greene, Tanya Khovanova, Dohun Kim, Rajarshi Mandal, Tanish Parida, Anirudh Pulugurtha, Gordon Redwine, Soham Samanta, and Albert Xu, Chip-Firing on Infinite k-ary Trees, arXiv:2501.06675 [math.CO], 2025. See pp. 9, 18.
László Tóth, On Schizophrenic Patterns in b-ary Expansions of Some Irrational Numbers, arXiv:2002.06584 [math.NT], 2020. See also Proc. Amer. Math. Soc. 148 (2020), pp. 461-469.
Index entries for linear recurrences with constant coefficients, signature (7,-11,5).

Cf. A000225, A000392, A002275, A002452, A003462, A003463, A003464, A016123, A016125, A016208, A016218, A016256, A023000, A023001.

[(5^(n+1)-4*n-5)/16: n in [1..30]]; // Vincenzo Librandi, Aug 23 2011

a:=n->sum((5^(n-j)-1^(n-j))/4,j=0..n): seq(a(n), n=1..21); # Zerinvary Lajos, Jan 04 2007

Join[{a=1,b=7},Table[c=6*b-5*a+1;a=b;b=c,{n,60}]] (* Vladimir Joseph Stephan Orlovsky, Feb 06 2011 *)

[(gaussian_binomial(n,1,5)-n)/4 for n in range(2,23)] # Zerinvary Lajos, May 29 2009

a(n) = (5^(n+1) - 4*n - 5)/16.
G.f.: x/((1-5*x)*(1-x)^2).
From Paul Barry, Jul 30 2004: (Start)
a(n) = Sum_{k=0..n} (n-k)*5^k = Sum_{k=0..n} k*5^(n-k).
a(n) = Sum_{k=0..n} binomial(n+2,k+2)*4^k [Offset 0]. (End)
From Elmo R. Oliveira, Mar 29 2025: (Start)
E.g.f.: exp(x)*(5*exp(4*x) - 4*x - 5)/16.
a(n) = 7*a(n-1) - 11*a(n-2) + 5*a(n-3) for n > 3. (End)

A016209 Expansion of 1/((1-x)(1-3x)(1-5x)).

Original entry on oeis.org

1, 9, 58, 330, 1771, 9219, 47188, 239220, 1205941, 6059229, 30384718, 152189310, 761743711, 3811110039, 19062724648, 95335146600, 476740303081, 2383895225649, 11920057258978, 59602029687090

Offset: 0

N. J. A. Sloane

For a combinatorial interpretation following from a(n) = A039755(n+2,2) = h^{(3)}A039755.%20-%20_Wolfdieter%20Lang">n, the complete homogeneous symmetric function of degree n in the symbols {1, 3, 5} see A039755. - _Wolfdieter Lang, May 26 2017

			a(2) = h^{(3)}_2 = 1^2 + 3^2 + 5^2 + 1^1*(3^1 + 5^1) + 3^1*5^1 = 58. - _Wolfdieter Lang_, May 26 2017

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (9,-23,15).

Cf. A016218, A016208, A000392, A000225, A003462, A003463, A003464, A023000, A023001, A002452, A002275, A016123, A016125, A016256, A039755, A021424.

[(5^(n+2)-2*3^(n+2)+1)/8: n in [0..20]]; // Vincenzo Librandi, Sep 17 2011

A016209 := proc(n) (5^(n+2)-2*3^(n+2)+1)/8; end proc: # R. J. Mathar, Mar 22 2011

Join[{a=1,b=9},Table[c=8*b-15*a+1;a=b;b=c,{n,60}]] (* Vladimir Joseph Stephan Orlovsky, Feb 07 2011 *)
CoefficientList[Series[1/((1-x)(1-3x)(1-5x)),{x,0,30}],x] (* or *) LinearRecurrence[ {9,-23,15},{1,9,58},30] (* Harvey P. Dale, Feb 20 2020 *)

PARI
```
a(n)=if(n<0,0,n+=2; (5^n-2*3^n+1)/8)
```

a(n) = A039755(n+2, 2).
a(n) = (5^(n+2) - 2*3^(n+2)+1)/8 = a(n-1) + A005059(n+1) = 8*a(n-1) - 15*a(n-2) + 1 = (A003463(n+2) - A003462(n+2))/2. - Henry Bottomley, Jun 06 2000
G.f.: 1/((1-x)(1-3*x)(1-5*x)). See the name.
E.g.f.: (25*exp(5*x) - 18*exp(3*x) + exp(x))/8, from the e.g.f. of the third column (k=2) of A039755. - Wolfdieter Lang, May 26 2017

A016228 Expansion of 1/((1-x)(1-5x)(1-6*x)).

Original entry on oeis.org

1, 12, 103, 774, 5425, 36456, 238267, 1527258, 9651829, 60352380, 374321311, 2306963022, 14146953913, 86407602384, 526075008835, 3194597025666, 19358317017277, 117103576420068, 707389830102439, 4268180838524790, 25728294320699521, 154965812371951032

Offset: 0

N. J. A. Sloane

Index entries for linear recurrences with constant coefficients, signature (12,-41,30).

Cf. A016218.

a:=n->sum(6^(n-j)-5^(n-j),j=0..n): seq(a(n), n=1..19); # Zerinvary Lajos, Jan 15 2007

Table[(2^(n + 3)*3^(n + 1) - 5^(n + 2) + 1)/20, {n, 40}] (* and *) CoefficientList[Series[1/((1 - z) (1 - 5*z) (1 - 6*z)), {z, 0, 40}], z] (* Vladimir Joseph Stephan Orlovsky, Jul 21 2011 *)
LinearRecurrence[{12,-41,30},{1,12,103},30] (* Harvey P. Dale, Aug 24 2017 *)

Vec(1/((1-x)*(1-5*x)(1-6*x))+O(x^99)) \\ Charles R Greathouse IV, Sep 26 2012

a(n) = (1/20)-(25/4)*5^n+(36/5)*6^n. [Antonio Alberto Olivares, Feb 06 2010]

a(0)=1, a(1)=12, a(n)=11*a(n-1)-30*a(n-2)+1. - Vincenzo Librandi, Feb 10 2011

A341091 Triangle read by rows: Coefficients for calculation of the sum of all the finite differences from order zero to order k. Sum_{n=0..k} T(n, k)*b(n) = b(0) + b(1) + ... + b(k) + (b(1) - b(0)) + ... + (b(k) - b(k-1)) + ((b(2) - b(1)) - (b(1) - b(0))) + ... .

Original entry on oeis.org

1, 0, 2, 1, -1, 3, 0, 3, -3, 4, 1, -2, 7, -6, 5, 0, 4, -8, 14, -10, 6, 1, -3, 13, -21, 25, -15, 7, 0, 5, -15, 35, -45, 41, -21, 8, 1, -4, 21, -49, 81, -85, 63, -28, 9, 0, 6, -24, 71, -129, 167, -147, 92, -36, 10, 1, -5, 31, -94, 201, -295, 315, -238, 129, -45, 11

Offset: 0

Thomas Scheuerle, Feb 13 2022

If we want to calculate the sum of finite differences for a sequence b(n):
  b(0)*T(0, n) + ... + b(n)*T(n, n) = b(0) + b(1) + ... + b(n) + (b(1) - b(0)) + ... + (b(n) - b(n-1)) + ((b(2) - b(1)) - (b(1) - b(0))) + ... This sum includes the sequence b(n) itself. This defines an invertible linear sequence transformation with a deep connection to Bernoulli numbers and other interesting sequences of rational numbers.
From Thomas Scheuerle, Apr 29 2024: (Start)
These are the coefficients of the polynomials defined by the recurrence: P(k, x) = P(k - 1, x) + (x^2 - x)*P(k - 2, x) + 1, with P(-1, x) = 0 and P(0, x) = 1. This can also be expressed as P(k, x) = Sum_{m=1..k+1} binomial(k+2 - m, m)*(x^2 - x)^(m - 1) = Sum_{n=0..k} T(n, k)*x^(k-n). If we would evaluate P(k, t) as sequence for some fixed t then we get the expansion of 1/((1 - x)*(1+(t-1)*x)*(1 - t*x)).
We may replace (x^2 - x) by (x^(-2) - x^(-1)) to get the coefficients in reverse order: x^k*Sum_{m=1..k+1} binomial(k+2 - m, m)*(x^(-2) - x^(-1))^(m - 1) = Sum_{n=0..k} T(n, k)*x^n = F(k, x). If we would evaluate F(k, t) as sequence for some fixed t then we get the expansion of 1/((1 - x)*(1 - (t-1)*x)*(1 - t*x)). (End)

			Triangle begins with T(n, k):
   n=   0,  1,   2,   3,   4,   5,   6,   7,   8
  k=0   1
  k=1   0,  2
  k=2   1, -1,   3
  k=3   0,  3,  -3,   4
  k=4   1, -2,   7,  -6,   5
  k=5   0,  4,  -8,  14, -10,   6
  k=6   1, -3,  13, -21,  25, -15,   7
  k=7   0,  5, -15,  35, -45,  41, -21,   8
  k=8   1, -4,  21, -49,  81, -85,  63, -28,   9
  ...

Cf. A000027, A000032, A000045, A000110, A000217.
Cf. A001045, A004006, A032346, A051744, A052875.
Cf. A084416, A130595, A145766.
Cf. A027642, A164555 (Numerators and denominators of Bernoulli numbers).
Cf. A001008, A002805 (Numerators and denominators of harmonic numbers).
Cf. A344037, A372245.
Sequences below will be obtained by evaluation of the associated polynomials:
Cf. A000295, A000392, A000975, A016208.
Cf. A016218, A016228, A016241, A016249.
Cf. A016256, A016261, A249997.

A341091(n, k) = sum(m=n, k,(-1)^(m+n)*binomial(m+1, n))

A341091(n, k) = (1/2)*(-1)^n*(2*(-1)^k*binomial(2+k, n)*hypergeom([1,k+3],k+3-n,-1)+(-1/2)^n*(2^(n+1)-1)) \\ Thomas Scheuerle, Apr 29 2024

b(0)*T(0, m) + b(1)*T(1, m) + ... + b(m)*T(m, m)
  = Sum_{j=0..m} Sum_{n=0..m-j} Sum_{k=0..n} (-1)^k*binomial(n, k)*b(j+n-k)
  = Sum_{n=0..m} b(n)*Sum_{j=n..m}(-1)^(j+n)*binomial(j+1, n).
T(n, k) = Sum_{m=n..k}(-1)^(m+n)*binomial(m+1, n).
T(n, k) = (1/2)*(-1)^n*(2*(-1)^k*binomial(2+k, n)*Hypergeometric2F1(1, k+3, k+3-n, -1)+(-1/2)^n*(2^(n+1) - 1)), where Hypergeometric2F1 is the Gaussian hypergeometric function 2F1 as defined in Mathematica. - Thomas Scheuerle, Apr 29 2024
T(k, k) = A000027(k+1) The positive integers.
|T(k-1, k)| = A000217(k) The triangular numbers.
T(k-2, k) = A004006(k).
|T(k-3, k)| = A051744(k).
T(0, k*2) = 1.
T(0, k*2 + 1) = 0.
T(1, k*2 + 1) = k + 2.
T(1, k*2 + 2) = -(k + 1).
T(n, k) with constant n and variable k, a linear recurrence relation with characteristic polynomial (x-1)*(x+1)^(n+1).
Sum_{n=0..k} T(n, k)*B_n = 1. B_n is the n-th Bernoulli number with B_1 = 1/2. B_n = A164555(n)/A027642(n).
Sum_{n=0..k} T(n, k)*(1 - B_n) = k.
Sum_{n=0..k} T(n, k)*(2*n - 3+3*B_n) = k^2.
Sum_{n=0..k} T(n, k)*A032346(n) = A032346(k+1).
From Thomas Scheuerle, Apr 29 2024: (Start)
Sum_{n=0..k} T(n, k)*A000110(n+1) = A000110(k+2) - 1.
Sum_{n=0..k} T(n, k)*(1/(1+n)) = H(1+floor(k/2)), where H(k) is the harmonic number A001008(k)/A002805(k). (End)
Sum_{n=0..k} T(n, k)*c(n) = c(k). C(k) = {-1, 0, 1/2, 1/2, 1/8, -7/20, ...} this sequence of rational numbers can be defined recursively: c(0) = -1, c(m) = (-c(m-1) + Sum_{k=0..m-1} A130595(m+1, k)*c(k))/m.
  c(m) is an eigensequence of this transformation, all eigensequences are c(m) multiplied by any factor.
Sum_{n=0..k} T(n, k)*A000045(n) = 2*(A000045(2*floor((k+1)/2) - 1) - 1). A000045 are the Fibonacci numbers.
Sum_{n=0..k} T(n, k)*A000032(n) = A000032(2*floor(k/2)+2) - 2. A000032 are the Lucas numbers.
Sum_{n=0..k} T(n, k)*A001045(n) = A145766(floor((k+1)/2)). A001045 is the Jacobsthal sequence.
This sequence acting as an operator onto a monomial n^w:
Sum_{n=0..k} T(n, k)*n^w = (1/(w+1))*k^(w+1) + Sum_{v=1..w} ((v+B_v)*(w)_v/v!)*k^(w+1-v) - A052875(w) + O_k(w) (w)_v is the falling factorial. If k > w-1 then O_k(w) = 0. If k <= w-1 then O_k(w) is A084416(w, 2+k), the sequence with the exponential generating function: (e^x-1)^(2+k)/(2-e^x).
From Thomas Scheuerle, Apr 29 2024: (Start)
This sequence acting by its inverse operator onto a monomial k^w:
Sum_{n=0..k} T(n, k)*( Sum_{m=0..k} ((-1)^(1+m+k)*binomial(k, m)*(2^(k-m) - 1)*n^m + A344037(m)*B_n) ) = k^w - A372245(w, k+3), note that A372245(w, k+3) = 0 if k+3 > w. B_n is the n-th Bernoulli number with B_1 = 1/2.
How this sequence will act as an operator onto a Dirichlet series may be developed by the formulas below:
Sum_{n=0..k} T(n, k)*2^n = A000295(k+2).
Sum_{n=0..k} T(n, k)*3^n = A000392(k+3).
Sum_{n=0..k} T(n, k)*4^n = A016208(k).
Sum_{n=0..k} T(n, k)*5^n = A016218(k).
Sum_{n=0..k} T(n, k)*6^n = A016228(k).
Sum_{n=0..k} T(n, k)*7^n = A016241(k).
Sum_{n=0..k} T(n, k)*8^n = A016249(k).
Sum_{n=0..k} T(n, k)*9^n = A016256(k).
Sum_{n=0..k} T(n, k)*10^n = A016261(k).
Sum_{n=0..k} T(n, k)*m^n = m^2*m^k/(m-1) - (m-1)^2*(m-1)^k/(m-2) + 1/((m-1)*(m-2)), for m > 2.
Sum_{n=0..k} T(n, k)*( m*B_n + (m-1)*Sum_{t=1..m} t^n )*(1/m^2) = m^k, for m > 0. B_n is the n-th Bernoulli number with B_1 = 1/2.
Sum_{n=0..k} T(n, k) zeta(-n) = Sum_{j=0..k} (-1)^(1+j)/(2+j) = (-1)^(k+1)*LerchPhi(-1, 1, k+3) - 1 + log(2).
Sum_{n=0..k} T(k - n, k)*2^n = A000975(k+1)
Sum_{n=0..k} T(k - n, k)*3^n = A091002(k+2)
Sum_{n=0..k} T(k - n, k)*4^n = A249997(k). (End)

A016241 Expansion of 1/((1-x)(1-6x)(1-7x)).

Original entry on oeis.org

1, 14, 141, 1246, 10277, 81270, 624877, 4710062, 34985973, 256995046, 1871524733, 13536029598, 97364345989, 697223254742, 4974599780109, 35386420442254, 251090274984725, 1777943916226758, 12567479361589405, 88703587219138430, 625312500662044581

Offset: 0

N. J. A. Sloane

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (14,-55,42).

Cf. A016218, A016228, A016249.

[(1 -6^(n+2) +5*7^(n+1))/30 : n in [0..40]]; // G. C. Greubel, Jan 30 2022

Table[(1 -6^(n+2) +5*7^(n+1))/30, {n, 40}] (* or *) CoefficientList[Series[1/((1-z)(1-6*z)(1-7*z)), {z, 0, 40}], z] (* Vladimir Joseph Stephan Orlovsky, Jul 21 2011 *)
LinearRecurrence[{14,-55,42},{1,14,141},40] (* Harvey P. Dale, Aug 05 2011 *)

Vec(1/((1-x)*(1-6*x)*(1-7*x))+O(x^99)) \\ Charles R Greathouse IV, Sep 26 2012

[(1 -6^(n+2) +5*7^(n+1))/30 for n in (0..40)] # G. C. Greubel, Jan 30 2022

a(0)=1, a(1)=14, a(n) = 13*a(n-1) - 42*a(n-2) + 1. - Vincenzo Librandi, Feb 10 2011
a(n) = (1 - 6^(n+2) + 5*7^(n+1))/30. - Vladimir Joseph Stephan Orlovsky, Jul 21 2011
a(0)=1, a(1)=14, a(2)=141, a(n) = 14*a(n-1) - 55*a(n-2) + 42*a(n-3). - Harvey P. Dale, Aug 05 2011
E.g.f.: (1/30)*(exp(x) - 36*exp(6*x) + 35*exp(7*x)). - G. C. Greubel, Jan 30 2022

A016249 Expansion of 1/((1-x)(1-7x)(1-8x)).

Original entry on oeis.org

1, 16, 185, 1880, 17841, 162336, 1435945, 12448360, 106312481, 897579056, 7510186905, 62388376440, 515255179921, 4235078618176, 34671889197065, 282913935338120, 2302083235036161, 18688068146607696, 151404361037090425, 1224533599346325400, 9889359772117817201

Offset: 0

N. J. A. Sloane

Index entries for linear recurrences with constant coefficients, signature (16,-71,56).

Cf. A016218, A016228, A016241.

Table[(3*2^(3*n + 4) - 7^(n + 2) + 1)/42, {n, 40}] (* or *) CoefficientList[Series[1/((1 - z) (1 - 7*z) (1 - 8*z)), {z, 0, 40}], z] (* Vladimir Joseph Stephan Orlovsky, Jul 21 2011 *)
LinearRecurrence[{16,-71,56},{1,16,185},30] (* Harvey P. Dale, May 14 2023 *)

Vec(1/((1-x)*(1-7*x)*(1-8*x))+O(x^99)) \\ Charles R Greathouse IV, Sep 26 2012

a(0)=1, a(1)=16, a(n) = 15*a(n-1) - 56*a(n-2) + 1. - Vincenzo Librandi, Feb 10 2011

A016261 Expansion of 1/((1-x)(1-9x)(1-10x)).

Original entry on oeis.org

1, 20, 291, 3730, 44681, 513240, 5730271, 62683550, 675263061, 7188478660, 75807419051, 793377882570, 8251512054241, 85374719599280, 879483587504631, 9026463398652790, 92349281698986221, 942254646401987100, 9591402928728995011, 97433737469672066210

Offset: 0

N. J. A. Sloane

nonn

Index entries for linear recurrences with constant coefficients, signature (20,-109,90).

Cf. A016218, A016228.

Table[(2^(n + 4)*5^(n + 1) - 9^(n + 2) + 1)/72, {n, 40}] (* or *) CoefficientList[Series[1/((1 - z) (1 - 9*z) (1 - 10*z)), {z, 0, 40}], z] (* Vladimir Joseph Stephan Orlovsky, Jul 21 2011 *)

a(n) = (8*10^(n+2) - 9^(n+3) + 1)/72. - Bruno Berselli, Mar 21 2011

a(n) = 19*a(n-1) - 90*a(n-2) + 1, n >= 2. - Vincenzo Librandi, Mar 21 2011

A016198 Expansion of g.f. 1/((1-x)(1-2x)(1-5x)).

Original entry on oeis.org

1, 8, 47, 250, 1281, 6468, 32467, 162590, 813461, 4068328, 20343687, 101722530, 508620841, 2543120588, 12715635707, 63578244070, 317891351421, 1589457019248, 7947285620527, 39736429151210, 198682147853201, 993410743460308, 4967053725690147, 24835268645227950

Offset: 0

N. J. A. Sloane, Dec 11 1999

Index entries for linear recurrences with constant coefficients, signature (8,-17,10).

Cf. A000225, A000392, A002275, A002452, A003462, A003463, A003464, A016123, A016125, A016208, A016209, A016218, A016256, A023000, A023001.

Cf. A016127.

a:=n->sum((5^(n-j)-2^(n-j))/3,j=0..n): seq(a(n), n=1..20); # Zerinvary Lajos, Jan 04 2007

Join[{a=1,b=8},Table[c=7*b-10*a+1;a=b;b=c,{n,60}]] (* Vladimir Joseph Stephan Orlovsky, Feb 06 2011 *)

a(n) = (25*5^n - 16*2^n + 3)/12. - Bruno Berselli, Feb 09 2011
a(n) = [(5^0-2^0) + (5^1-2^1) + ... + (5^n-2^n)]/3. - r22lou(AT)cox.net, Nov 14 2005
a(0)=1, a(n) = 5*a(n-1) + 2^(n+1) - 1. - Vincenzo Librandi, Feb 07 2011
From Elmo R. Oliveira, Mar 26 2025: (Start)
E.g.f.: exp(x)*(25*exp(4*x) - 16*exp(x) + 3)/12.
a(n) = 8*a(n-1) - 17*a(n-2) + 10*a(n-3).
a(n) = A016127(n+1) - A003463(n+2). (End)

More terms from Wesley Ivan Hurt, May 05 2014

cp's OEIS Frontend

A014827 a(1)=1, a(n) = 5*a(n-1) + n.

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A016209 Expansion of 1/((1-x)(1-3x)(1-5x)).

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A016228 Expansion of 1/((1-x)*(1-5*x)(1-6*x)).

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A341091 Triangle read by rows: Coefficients for calculation of the sum of all the finite differences from order zero to order k. Sum_{n=0..k} T(n, k)*b(n) = b(0) + b(1) + ... + b(k) + (b(1) - b(0)) + ... + (b(k) - b(k-1)) + ((b(2) - b(1)) - (b(1) - b(0))) + ... .

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A016241 Expansion of 1/((1-x)*(1-6*x)*(1-7*x)).

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A016249 Expansion of 1/((1-x)*(1-7*x)*(1-8*x)).

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A016261 Expansion of 1/((1-x)*(1-9*x)*(1-10*x)).

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A016228 Expansion of 1/((1-x)(1-5x)(1-6*x)).

A016241 Expansion of 1/((1-x)(1-6x)(1-7x)).

A016249 Expansion of 1/((1-x)(1-7x)(1-8x)).

A016261 Expansion of 1/((1-x)(1-9x)(1-10x)).

A016198 Expansion of g.f. 1/((1-x)(1-2x)(1-5x)).