A131865 -id:A131865 - cp's OEIS frontend

A125118 Triangle read by rows: T(n,k) = value of the n-th repunit in base (k+1) representation, 1<=k<=n.

1, 3, 4, 7, 13, 21, 15, 40, 85, 156, 31, 121, 341, 781, 1555, 63, 364, 1365, 3906, 9331, 19608, 127, 1093, 5461, 19531, 55987, 137257, 299593, 255, 3280, 21845, 97656, 335923, 960800, 2396745, 5380840, 511, 9841, 87381, 488281, 2015539, 6725601, 19173961, 48427561, 111111111

Offset: 1

Reinhard Zumkeller, Nov 21 2006

			First 4 rows:
1: [1]_2
2: [11]_2 ........ [11]_3
3: [111]_2 ....... [111]_3 ....... [111]_4
4: [1111]_2 ...... [1111]_3 ...... [1111]_4 ...... [1111]_5
_
1: 1
2: 2+1 ........... 3+1
3: (2+1)*2+1 ..... (3+1)*3+1 ..... (4+1)*4+1
4: ((2+1)*2+1)*2+1 ((3+1)*3+1)*3+1 ((4+1)*4+1)*4+1 ((5+1)*5+1)*5+1.

G. C. Greubel, Rows n = 1..50 of the triangle, flattened
Eric Weisstein's World of Mathematics, Repunit

This triangle shares some features with triangle A104878.
This triangle is a portion of rectangle A055129.
Each term of A110737 comes from the corresponding row of this triangle.
Diagonals (adjusting offset as necessary): A060072, A023037, A031973, A173468.
Columns (adjusting offset as necessary): A000225, A003462, A002450, A003463, A003464, A023000, A023001, A002452, A002275, A016123, A016125, A091030, A135519, A135518, A131865, A091045, A218721, A218722, A064108, A218724, A218725, A218726, A218727, A218728, A218729, A218730, A218731, A218732, A218733, A218734, A132469, A218736, A218737, A218738, A218739, A218740, A218741, A218742, A218743, A218744, A218745, A218746, A218747, A218748, A218749, A218750, A218751, A218753, A218752.
Cf. A023037, A031973, A125119, A125120 (row sums).

[((k+1)^n -1)/k : k in [1..n], n in [1..12]]; // G. C. Greubel, Aug 15 2022

Table[((k+1)^n -1)/k, {n, 12}, {k, n}]//Flatten (* G. C. Greubel, Aug 15 2022 *)

def A125118(n,k): return ((k+1)^n -1)/k
flatten([[A125118(n,k) for k in (1..n)] for n in (1..12)]) # G. C. Greubel, Aug 15 2022

T(n, k) = Sum_{i=0..n-1} (k+1)^i.
T(n+1, k) = (k+1)*T(n, k) + 1.
Sum_{k=1..n} T(n, k) = A125120(n).
T(2*n-1, n) = A125119(n).
T(n, 1) = A000225(n).
T(n, 2) = A003462(n) for n>1.
T(n, 3) = A002450(n) for n>2.
T(n, 4) = A003463(n) for n>3.
T(n, 5) = A003464(n) for n>4.
T(n, 9) = A002275(n) for n>8.
T(n, n) = A060072(n+1).
T(n, n-1) = A023037(n) for n>1.
T(n, n-2) = A031973(n) for n>2.
T(n, k) = A055129(n, k+1) = A104878(n+k, k+1), 1<=k<=n. - Mathew Englander, Dec 19 2020

A104878 A sum-of-powers number triangle.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 7, 4, 1, 1, 5, 15, 13, 5, 1, 1, 6, 31, 40, 21, 6, 1, 1, 7, 63, 121, 85, 31, 7, 1, 1, 8, 127, 364, 341, 156, 43, 8, 1, 1, 9, 255, 1093, 1365, 781, 259, 57, 9, 1, 1, 10, 511, 3280, 5461, 3906, 1555, 400, 73, 10, 1, 1, 11, 1023, 9841, 21845

Offset: 0

Paul Barry, Mar 28 2005

Columns are partial sums of the columns of A004248. Row sums are A104879. Diagonal sums are A104880.

The rows of this triangle (apart from the initial "1" in each row) are the antidiagonals of rectangle A055129. The diagonals of this triangle (apart from the initial "1") are the rows of rectangle A055129. The columns of this triangle (apart from the leftmost column) are the same as the columns of rectangle A055129 but shifted downward. - Mathew Englander, Dec 21 2020

			Triangle starts:
  1;
  1,  1;
  1,  2,  1;
  1,  3,  3,  1;
  1,  4,  7,  4,  1;
  1,  5, 15, 13,  5,  1;
  1,  6, 31, 40, 21,  6,  1;
  ...

Cf. A004248 (first differences by column), A104879 (row sums), A104880 (antidiagonal sums), A125118 (version of this triangle with fewer terms).
This triangle (ignoring the leftmost column) is a rotation of rectangle A055129.
Columns (adjusting offset as necessary): A000012, A000027, A000225, A003462, A002450, A003463, A003464, A023000, A023001, A002452, A002275, A016123, A016125, A091030, A135519, A135518, A131865, A091045, A218721, A218722, A064108, A218724, A218725, A218726, A218727, A218728, A218729, A218730, A218731, A218732, A218733, A218734, A132469, A218736, A218737, A218738, A218739, A218740, A218741, A218742, A218743, A218744, A218745, A218746, A218747, A218748, A218749, A218750, A218751, A218753, A218752.
T(2n,n) gives A031973.

A104878 :=proc(n,k): if k = 0 then 1 elif k=1 then n elif k>=2 then (k^(n-k+1)-1)/(k-1) fi: end: for n from 0 to 7 do seq(A104878(n,k), k=0..n) od; seq(seq(A104878(n,k), k=0..n), n=0..10); # Johannes W. Meijer, Aug 21 2011

T(n, k) = if(k=1, n, if(k<=n, (k^(n-k+1)-1)/(k-1), 0));
G.f. of column k: x^k/((1-x)(1-k*x)). [corrected by Werner Schulte, Jun 05 2019]
T(n, k) = A069777(n+1,k)/A069777(n,k). [Johannes W. Meijer, Aug 21 2011]
T(n, k) = A055129(n+1-k, k) for n >= k > 0. - Mathew Englander, Dec 19 2020

A131851 Real part of the function z(n)=Sum(d(k)i^k: d as in n=Sum(d(k)2^k), i=sqrt(-1)).

Original entry on oeis.org

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

Offset: 0

Reinhard Zumkeller, Jul 22 2007

sign

A131852(n) = Im(z(n));
z(A000079(n))=(A056594(n),A056594(n+3)); a(A000079(n))=A056594(n);
a(A131854(n))=0; a(A131861(n))>0; a(A131859(n))=1; a(A131863(n))<0;
z(A131853(n))=(0,0); z(A131856(n))=(0,1); z(A131858(n))=(1,0); z(A131860(n))=(1,1);
for n>0: a(A131865(n))=n and ABS(a(m))A131865(n).

R. Zumkeller, Table of n, a(n) for n = 0..10000

Cf. A007088.

z[0] = 0; z[n_] := z[n] = z[Floor[n/2]]*I + Mod[n, 2]; Table[z[n] // Re, {n, 0, 110}] (* Jean-François Alcover, Jul 03 2013 *)

z(n) = if n=0 then (0, 0) else z(floor(n/2))*(0, 1) + (n mod 2, 0), complex multiplication.

A299960 a(n) = (4^(2*n+1) + 1) / 5.

Original entry on oeis.org

1, 13, 205, 3277, 52429, 838861, 13421773, 214748365, 3435973837, 54975581389, 879609302221, 14073748835533, 225179981368525, 3602879701896397, 57646075230342349, 922337203685477581, 14757395258967641293, 236118324143482260685, 3777893186295716170957

Offset: 0

M. F. Hasler, Feb 22 2018

nonn

It is easily seen that 4^(2n+1)+1 is divisible by 5 for all n, since 4 = -1 (mod 5). For even powers this does not hold.
The aerated sequence 1, 0, 13, 0, 205, 0, 3277, ... is a linear divisibility sequence of order 4. It is the case P1 = 0, P2 = -5^2, Q = 4 of the 3-parameter family of 4th-order linear divisibility sequences found by Williams and Guy. Cf. A007583, A095372 and A100706. - Peter Bala, Aug 28 2019
Let G be a sequence satisfying G(i) = 2*G(i-1) + G(i-2) - 2*G(i-3) for arbitrary integers i and without regard to the initial values of G. Then a(n) = (G(i)*2^(4*n+2) + G(i+8*n+4))/(5*G(i+4*n+2)) as long as G(i+4*n+2) != 0. - Klaus Purath, Feb 02 2021
Ch. Gerbr asks (personal comm.) whether we can prove that 13 is the only prime in this sequence. We can prove divisibility conditions for many residue classes of the index n (cf. formulas), but have not yet found a complete covering set. - M. F. Hasler, Jan 07 2025

			For n = 0, a(0) = (4^1+1)/5 = 5/5 = 1.
For n = 1, a(1) = (4^3+1)/5 = 65/5 = 13.

H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences Integers, Volume 12A (2012) The John Selfridge Memorial Volume.
Index entries for linear recurrences with constant coefficients, signature (17,-16).

Cf. A299959 for the smallest prime factor.

Cf. A052539, A007583, A095372, A100706.

A299960 := n -> (4^(2*n+1)+1)/5: seq(A299960(n), n=0..20);

LinearRecurrence[{17, -16}, {1, 13}, 20] (* Jean-François Alcover, Feb 22 2018 *)

PARI
```
A299960(n)=4^(2*n+1)\5+1
```

def A299960(n): return ((1<<(n<<2)+2)+1)//5 # Chai Wah Wu, Jul 29 2022

a(n) = A052539(2*n+1)/5 = A015521(2*n+1) = A014985(2*n+1) = A007910(4*n+1) = A007909(4*n+1) = A207262(n+1)/5.
O.g.f.: (1 - 4*x)/(1 - 17*x + 16*x^2). - Peter Bala, Aug 28 2019
a(n) = 17*a(n-1) - 16*a(n-2). - Wesley Ivan Hurt, Oct 02 2020
From Klaus Purath, Feb 02 2021: (Start)
a(n) = (2^(4*n+2)+1)/5.
a(n) = (A061654(n) + A001025(n))/2.
a(n) = A091881(n+1) + 7*A131865(n-1) for n > 0.
(End)
E.g.f.: (exp(x) + 4*exp(16*x))/5. - Stefano Spezia, Feb 02 2021
We have d | a(n) for all n in R, for the following pairs (d, R) of divisors d and residue classes R: (13, 1 + 3Z), (5, 2 + 5Z), (29, 3 + 7Z), (397, 5 + 11Z),
  (53, 6 + 13Z), (137, 8 + 17Z), (229, 9 + 19Z), (277, 11 + 23Z),
  (107367629, 14 + 29Z), (5581, 15 + 31Z), (149, 18 + 27Z), (10169, 20 + 41Z),
  (173, 21 + 43Z), (3761, 23 + 47Z), (15358129, 26 + 53Z), (1181, 29 + 59Z),
  (733, 30 + 61Z), (269, 33 + 67Z), (569, 35 + 71Z),(293, 36 + 73Z), (317, 39 + 79Z),
  (997, 41 + 83Z), (1069, 44 + 89Z), (389, 48 + 97Z), (809, 50 + 101Z),
  (41201, 51 + 103Z), (857, 53 + 107Z), (5669, 54 + 109Z), (58309, 56 + 113Z),
  (509, 63 + 127Z), (269665073, 65 + 131Z), (189061, 68 + 137Z), (557, 69 + 139Z),
  (1789, 74 + 149Z), (653, 81 + 163Z), (9413, 90 + 181Z), (3821, 95 + 191Z),
  (773, 96 + 193Z), (4729, 98 + 197Z), (797, 99 + 199Z), ... - M. F. Hasler, Jan 07 2025

A144864 a(n) = (4*16^(n-1)-1)/3.

Original entry on oeis.org

1, 21, 341, 5461, 87381, 1398101, 22369621, 357913941, 5726623061, 91625968981, 1466015503701, 23456248059221, 375299968947541, 6004799503160661, 96076792050570581, 1537228672809129301, 24595658764946068821, 393530540239137101141, 6296488643826193618261, 100743818301219097892181

Offset: 1

Artur Jasinski, Sep 23 2008

Old name was: A144863, read as binary numbers, converted to base 10.
All numbers in this sequence for n>1 are congruent to 5 mod 16. - Artur Jasinski, Sep 25 2008
From Omar E. Pol, Sep 10 2011: (Start)
It appears that this is a bisection of A002450.
It appears that this is a bisection of A084241.
It appears that this is a bisection of A153497.
It appears that this is a bisection of A088556, if n>=2.
(End)
All of the above is trivially true. - Joerg Arndt, Aug 19 2014
The aerated sequence (b(n))n>=1 = [1, 0, 21, 0, 341, 0, 5461, 0, 87381, ...] is a fourth-order linear divisibility sequence; that is, a(n) divides a(m) whenever n divides m. It is the case P1 = 0, P2 = -9, Q = -4 of the 3-parameter family of 4th-order linear divisibility sequences found by Williams and Guy. - Peter Bala, Aug 26 2022

Vincenzo Librandi, Table of n, a(n) for n = 1..500
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
Index entries for linear recurrences with constant coefficients, signature (17,-16).

Cf. A001025, A002450, A013776, A056830, A084241, A088556, A094028, A098704, A131865, A135576, A144863, A153497.

Third quadrisection of Jacobsthal numbers A001045; the other quadrisections are A195156 (first), A139792 (second), and A141060 (fourth).

[16^n/12-1/3: n in [1..20]]; // Vincenzo Librandi, Aug 03 2011

Table[1/3 (-1 + 16^(n - 1)) + 16^(n - 1), {n, 1, 17}] (* Artur Jasinski, Sep 25 2008 *)
LinearRecurrence[{17,-16},{1,21},20] (* Harvey P. Dale, Jun 29 2022 *)

vector(66,n,(4*16^(n-1)-1)/3) \\ Joerg Arndt, Aug 19 2014

a(n) = 16^n/12 - 1/3; a(n) = 16*a(n-1) + 5, a(1)=1. - Artur Jasinski, Sep 25 2008
G.f.: x*(1+4*x) / ( (16*x-1)*(x-1) ). - R. J. Mathar, Jan 06 2011
a(n)=b such that Integral_{x=-Pi/2..Pi/2} (-1)^(n+1)*2^(2*n-3)*(cos((2*n-1)*x))/(5/4+sin(x)) dx = c+b*log(3). - Francesco Daddi, Aug 02 2011
a(n) = (2^(4*n-2)-1)/3. - Klaus Purath, Jan 31 2021
From Jianing Song, Aug 30 2022: (Start)
a(n) = A001045(4*n-2).
a(n+1) - a(n) = 10*A013776(n-1) = 20*A001025(n-1) for n >= 1.
a(n) = 10*A098704(n) + 1 = 20*A131865(n-2) + 1 for n >= 2. (End)
E.g.f.: (exp(16*x) - 4*exp(x) + 3)/12. - Stefano Spezia, Apr 18 2024

New name from Joerg Arndt, Aug 19 2014

A182512 a(n) = (16^n - 1)/5.

Original entry on oeis.org

0, 3, 51, 819, 13107, 209715, 3355443, 53687091, 858993459, 13743895347, 219902325555, 3518437208883, 56294995342131, 900719925474099, 14411518807585587, 230584300921369395, 3689348814741910323, 59029581035870565171, 944473296573929042739

Offset: 0

Brad Clardy, May 03 2012

Even bisection of A015521 and also A112627. All of the terms are divisible by 3, even terms by 17.

These are binary numbers 11, 110011, 1100110011, ... - Jamie Simpson, Oct 28 2022

Robert Israel, Table of n, a(n) for n = 0..830
E. Estrada and J. A. de la Pena, From Integer Sequences to Block Designs via Counting Walks in Graphs, arXiv preprint arXiv:1302.1176 [math.CO], 2013. - From _N. J. A. Sloane_, Feb 28 2013
E. Estrada and J. A. de la Pena, Integer sequences from walks in graphs, Notes on Number Theory and Discrete Mathematics, Vol. 19, 2013, No. 3, 78-84
Andreas M. Hinz and Paul K. Stockmeyer, Precious Metal Sequences and Sierpinski-Type Graphs, J. Integer Seq., Vol 25 (2022), Article 22.4.8.
Index entries for linear recurrences with constant coefficients, signature (17,-16).

Cf. A015521, A112627.

Magma
```
[(1/5)*2^(4*i) -(1/5): i in [0..30]];
```

seq((16^n-1)/5, n=0..50); # Robert Israel, Jan 22 2016

(16^Range[0,20]-1)/5 (* Harvey P. Dale, Aug 07 2019 *)
LinearRecurrence[{17,-16},{0,3},20] (* Harvey P. Dale, Aug 07 2019 *)

a(n) = (16^n - 1)/5; \\ Michel Marcus, Jan 22 2016

a(n) = 16*a(n-1) + 3 where a(0)=0.
a(n) = A015521(2n).
a(n) = A112627(2n) for n >= 1; a(0)=0.
G.f.: 3*x / ( (16*x-1)*(x-1) ). - R. J. Mathar, Apr 20 2015
a(n) = 3*A131865(n-1). - R. J. Mathar, Apr 20 2015
a(n) = A108020(n)/4. - Jamie Simpson, Oct 28 2022

A218750 a(n) = (47^n - 1)/46.

Original entry on oeis.org

0, 1, 48, 2257, 106080, 4985761, 234330768, 11013546097, 517636666560, 24328923328321, 1143459396431088, 53742591632261137, 2525901806716273440, 118717384915664851681, 5579717091036248029008, 262246703278703657363377, 12325595054099071896078720

Offset: 0

M. F. Hasler, Nov 04 2012

Partial sums of powers of 47 (A009991).

Vincenzo Librandi, Table of n, a(n) for n = 0..600
Index entries related to partial sums
Index entries related to q-numbers
Index entries for linear recurrences with constant coefficients, signature (48,-47)

Cf. similar sequences of the form (k^n-1)/(k-1): A000225, A003462, A002450, A003463, A003464, A023000, A023001, A002452, A002275, A016123, A016125, A091030, A135519, A135518, A131865, A091045, A218721, A218722, A064108, A218724-A218734, A132469, A218736-A218753, A133853, A094028, A218723.

Cf. A009991.

[n le 2 select n-1 else 48*Self(n-1) - 47*Self(n-2): n in [1..20]]; // Vincenzo Librandi, Nov 08 2012

Table[(47^n - 1)/46, {n, 0, 19}] (* Alonso del Arte, Nov 04 2012 *)
LinearRecurrence[{48, -47}, {0, 1}, 30] (* Vincenzo Librandi, Nov 08 2012 *)

A218750(n):=(47^n-1)/46$ makelist(A218750(n),n,0,30); /* Martin Ettl, Nov 07 2012 */

PARI
```
A218750(n)=47^n\46
```

a(n) = floor(47^n/46).
G.f.: x/(47*x^2-48*x+1) = x/((1-x)*(1-47*x)). [Colin Barker, Nov 06 2012]
a(0)=0, a(n) = 47*a(n-1) + 1. - Vincenzo Librandi, Nov 08 2012
a(n) = 48*a(n-1) - 47*a(n-2). - Wesley Ivan Hurt, Jan 25 2022
E.g.f.: exp(24*x)*sinh(23*x)/23. - Elmo R. Oliveira, Aug 27 2024

A195156 a(n) = (16^n-1)/3.

Original entry on oeis.org

0, 5, 85, 1365, 21845, 349525, 5592405, 89478485, 1431655765, 22906492245, 366503875925, 5864062014805, 93824992236885, 1501199875790165, 24019198012642645, 384307168202282325, 6148914691236517205, 98382635059784275285, 1574122160956548404565

Offset: 0

Omar E. Pol, Sep 10 2011

Numbers of A002450 that are multiples of 5. Also sequence found by reading the line from 0, in the direction 0, 5,..., in the square spiral whose edges are the Jacobsthal numbers A001045 and whose vertices are the numbers A000975. This is a semi-diagonal in the spiral.
In binary, these numbers are 101...01 (see A031982). - Alonso del Arte, May 20 2017
0 together with Jacobsthal numbers ending with the decimal digit 5. - Jianing Song, Aug 30 2022

Vincenzo Librandi, Table of n, a(n) for n = 0..800
Index entries for linear recurrences with constant coefficients, signature (17,-16).

Bisection of A002450.
Cf. A000975, A001025, A013777, A031982, A131865.
First quadrisection of Jacobsthal numbers A001045; the other quadrisections are A139792 (second), A144864 (third), and A141060 (fourth).

[(16^n-1)/3:n in [0..20]]; // Vincenzo Librandi, Sep 20 2011

A195156:=n->(16^n-1)/3; seq(A195156(k), k=0..50); # Wesley Ivan Hurt, Oct 24 2013

Table[(16^n - 1)/3, {n, 0, 63}] (* Wesley Ivan Hurt, Oct 24 2013 *)

for(n=0,50, print1((16^n - 1)/3, ", ")) \\ G. C. Greubel, Oct 11 2017

From Bruno Berselli, Sep 19 2011: (Start)
G.f.: 5*x/((1-x)*(1-16*x)).
a(n) = A002450(2n) = (16^n-1)/3.
a(n) = 5*A131865(n-1) = a(n-1) + 5*A001025(n-1) = 16*a(n-1) + 5 for n > 0. (End)
From Jianing Song, Aug 30 2022: (Start)
a(n) = A001045(4*n).
a(n+1) - a(n) = 10*A013777(n-1) = 80*A001025(n-1) for n >= 1. (End)
E.g.f.: exp(x)*(exp(15*x) - 1)/3. - Stefano Spezia, Dec 17 2022

New sequence name suggested by Charles R Greathouse IV using Berselli's formula. - Sep 19 2011

A218726 a(n) = (23^n - 1)/22.

Original entry on oeis.org

0, 1, 24, 553, 12720, 292561, 6728904, 154764793, 3559590240, 81870575521, 1883023236984, 43309534450633, 996119292364560, 22910743724384881, 526947105660852264, 12119783430199602073, 278755018894590847680, 6411365434575589496641, 147461404995238558422744

Offset: 0

M. F. Hasler, Nov 04 2012

Partial sums of powers of 23, q-integers for q=23: diagonal k=1 in triangle A022187.

Partial sums are in A014909. Also, the sequence is related to A014941 by A014941(n) = n*a(n) - Sum{a(i), i=0..n-1} for n > 0. - Bruno Berselli, Nov 07 2012

Vincenzo Librandi, Table of n, a(n) for n = 0..700
Index entries related to partial sums.
Index entries for linear recurrences with constant coefficients, signature (24,-23).

Cf. similar sequences of the form (k^n-1)/(k-1): A000225, A003462, A002450, A003463, A003464, A023000, A023001, A002452, A002275, A016123, A016125, A091030, A135519, A135518, A131865, A091045, A218721, A218722, A064108, A218724-A218734, A132469, A218736-A218753, A133853, A094028, A218723.

Cf. A014909, A014941, A022187.

[n le 2 select n-1 else 24*Self(n-1)-23*Self(n-2): n in [1..20]]; // Vincenzo Librandi, Nov 07 2012

LinearRecurrence[{24, -23}, {0, 1}, 30] (* Vincenzo Librandi, Nov 07 2012 *)
(23^Range[0,20]-1)/22 (* Harvey P. Dale, Nov 09 2012 *)

A218726(n):=(23^n-1)/22$
makelist(A218726(n),n,0,30); /* Martin Ettl, Nov 07 2012 */

PARI
```
A218726(n)=23^n\22
```

From Vincenzo Librandi, Nov 07 2012: (Start)
G.f.: x/((1-x)*(1-23*x)).
a(n) = floor(23^n/22).
a(n) = 24*a(n-1) - 23*a(n-2). (End)
E.g.f.: exp(12*x)*sinh(11*x)/11. - Elmo R. Oliveira, Aug 27 2024

A218732 a(n) = (29^n - 1)/28.

Original entry on oeis.org

0, 1, 30, 871, 25260, 732541, 21243690, 616067011, 17865943320, 518112356281, 15025258332150, 435732491632351, 12636242257338180, 366451025462807221, 10627079738421409410, 308185312414220872891, 8937374060012405313840, 259183847740359754101361

Offset: 0

M. F. Hasler, Nov 04 2012

Partial sums of powers of 29 (A009973).

Vincenzo Librandi, Table of n, a(n) for n = 0..600
Index entries related to partial sums.
Index entries for linear recurrences with constant coefficients, signature (30,-29).

Cf. similar sequences of the form (k^n-1)/(k-1): A000225, A003462, A002450, A003463, A003464, A023000, A023001, A002452, A002275, A016123, A016125, A091030, A135519, A135518, A131865, A091045, A218721, A218722, A064108, A218724-A218734, A132469, A218736-A218753, A133853, A094028, A218723.

Cf. A009973.

[n le 2 select n-1 else 30*Self(n-1)-29*Self(n-2): n in [1..20]]; // Vincenzo Librandi, Nov 07 2012

LinearRecurrence[{30, -29}, {0, 1}, 30] (* Vincenzo Librandi, Nov 07 2012 *)

A218732(n):=(29^n-1)/28$
makelist(A218732(n),n,0,30); /* Martin Ettl, Nov 07 2012 */

PARI
```
a(n)=29^n\28
```

a(n) = floor(29^n/28).
G.f.: x/((1-x)*(1-29*x)). - Vincenzo Librandi, Nov 07 2012
a(n) = 30*a(n-1) - 29*a(n-2). - Vincenzo Librandi, Nov 07 2012
E.g.f.: exp(15*x)*sinh(14*x)/14. - Elmo R. Oliveira, Aug 27 2024

cp's OEIS Frontend

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